Does light have mass?

[TwAgIssmuDe]

Scientist say that even light can't escape the gravity pull of a black, then light must have mass in order to be affected by the gravity pull of the black hole.
So do light have mass?

[KB3HTS]

Yep! :)
Light has two parts: a wave and a very very small piece of mass known as a photon. Photons are emitted by electrons when they drop from a high energy state to a low energy state.
Random fact: EVERYTHING in the universe has both a wave and particle form. So technically you are a wave, but due to the size of a person the wave part of you is negligible.

[Haglund]

No, light do not have mass, but the equivalence principle says that everything, heavy masses as well as massless objects are affected by gravity equally. Gravity is described as the curvature of the spacetime, which all objects has to pass through.

[Deep_Eye]

Hmm we have a contradiction. Personally I think that light does have mass even if it is extremely extremely little. Otherwise it makes no sense that it would be sucked into black holes.

[Haglund]

Otherwise it makes no sense that it would be sucked into black holes.
Yes actually it does. The equivalence principle.

Anyway, some links that might explain it:

http://math.ucr.edu/home/baez/physics/Rela...light_mass.html (http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html)

http://www.discover.com/ask/main52.html

http://van.hep.uiuc.edu/van/qa/section/Lig...t/949533197.htm (http://van.hep.uiuc.edu/van/qa/section/Light_and_Sound/Properties_of_Light/949533197.htm)

[Faulkner]

Do positrons emit photons too? If so, light could be massless, but still be affected by matter, because it exists beyond the frontiers of mere matter & mass etc...Or are there such things as anti-photons?

[Haglund]

Originally posted by Faulkner@Oct 9 2003, 10:50 AM
Do positrons emit photons too? If so, light could be massless, but still be affected by matter, because it exists beyond the frontiers of mere matter & mass etc...Or are there such things as anti-photons?
Not only could light be massless, it is always massless. As for positrons, yes that can happen but the photon has no charge and so there would be no difference between a photon coming from an electron and one from a positron.

[QJones]

Originally posted by Faulkner@Oct 9 2003, 10:50 AM
Or are there such things as anti-photons?
The dark? Kidding!

What about when light waves combine to make a dark spot? I remember some high-school physics about that.

As well, since there is constant background radiation, are photons (of some wavelength) present almost anywhere?

Oh, as well, light can act like it has momentum. That's the concept behind a solar sail.

[GreekJimbo]

Well, I think I can answer our question without even answering your question. Check this out. . . .

All object travel in a stright line in a universe. Doesnt matter is they are weightless or someone's overweight grandmother falling out of a plane. Every object moves in the direction of least resistance (in the cae of the overweight grandmother, that's straight down).

So, we have light . . . it travels in a "straight" path most of the time (by our observations). But whenever sapce-time warps, so does the path of light. Forget about the idea of gravity "pulling" light. Gravity changes the shape of the road that the light is travelling on. Light is just moving in the path of least resistance (and even if it seems to curve to us. . . it is still travelling straight acording to the light particle. In the case of the blackhole, the light is moving on a "road" that infinitally loops back on itself. So, the light is still travelling straight, but the road is bent. Poor light.

Peace!

[rahuldandekar]

Let me clairify one thing- Photons have no antiparticle.

[jkmccrann]

Here is an interesting take on the old Photon debate. Apparently it has no rest mass, but it acquires a mass when moving.....


http://www.physlink.com/Education/AskExperts/ae180.cfm

[Sam5]

Can’t light waves cause a pressure on the objects the light hits, without the light being a “particle” or having “mass”. Can’t the force of the rapidly moving electric and magnetic fields of the light cause the pressure without the fields having “mass”?

[swansont]

Let me clairify one thing- Photons have no antiparticle.

Photons are actually their own antiparticle.

[swansont]

Can’t light waves cause a pressure on the objects the light hits, without the light being a “particle” or having “mass”. Can’t the force of the rapidly moving electric and magnetic fields of the light cause the pressure without the fields having “mass”?

Yes, basically. Photon scattering or absorption does exert a pressure, because the photons has momentum, given by E/c

[alainprice]

Thanks Swansont, for pointing out that photons ARE their own anti-particles.

If photons have mass, how can they go from not existing(an electron in an elevated orbit) to instantly travelling through space at C? All objects with mass must accelerate, and cannot travel at c. On ther other hand, if light is massless, it cannot accelerate and must travel at c. Therefore, we can safely state the photon has no REST MASS.

It is possible to try and measure the mass of the photon, but all attempts have failed. This allows us to say that if the photon does have mass, it is smaller than anything we have tried to measure.

[phonicboom]

I have nothing to add to this but i realy enjoyed reading all the answers, very interesting.

esp. alanprice and GreekJimbo and Swansont

[Ricimer]

Photons do not have mass. This is one of the fundamental principles of Quantum ElectroDynamics. If photons have mass, a lot of stuff in this start to fall apart, and we don't see that happening, and we're talking any mass. Now, I wish I could go into this more, but that's all the info I've been able to glean from chatting with one of the faculty here (a solid state physicist).

Now gravity isn't an interaction between masses (think of the standard force of gravity equation as a Newtonian approximation). Mass bends spacetime in such a way as to alter the paths of objects around it. Without mass, spacetime is flat, and the objects go in a straight line (they always go in a straight line). However, with mass, spacetime is bent, and those straightline paths, which are "straight" but restricted to following spacetime, are bent and curved.

This is analogous to walking a straight line on earth. However, we are restricted to the surface, and so that straight line will eventually curve around on itself. Light, and matter, is restricted to the "surface" of spacetime. So if spacetime bends, so do their trajetories.

[Frog march]

But what if a photon heads towards a blackhole, swings around it and then heads back in the direction that it came from? Will it not have imparted some momentum to that black hole? There fore there must be some kind of interaction between mass'?

[John_Charles_Webb]

Scientist say that even light can't escape the gravity pull of a black hole, then light must have mass in order to be affected by the gravity pull of the black hole.
So do light have mass?

According to Albert Einstein, "mass" expands exponentially as it approaches light speed and such exponential expansion prohibits any mass from attaining the speed of light because of the (also) expanding resistance (like the force against an auto's windscreen grows greater as speed increases).

Light is warped around a black hole as it follows the curvature (warp and compression) of space in the vicinity of a black hole. Theoretically, a black hole is mass that has, in fact, expanded exponentially and is traveling at a speed approaching the speed of light.

Light, it seems, follows the path of least resistance and its natural "physic" is to travel at the speed of light when un-emcumbered by matter. It is my opinion that a black hole contains no light, nor does it stop light but, rather, repels light because of the dense compression of space surrounding it so it seems that no light is eminating from a black hole.

I have written an essay addressing this issue:
http://www.templeofsolomon.org/Quantum_Thought.htg/thought.htm

It is also my belief that matter does not occupy space but, rather, displaces it. It is this belief that has caused me to formulate my own definition of "gravity" which is a result of the compression (from within) and stretching (from without) of space by matter.

Einstein and Hawking seem to disagree about what happens inside of a black hole. Einstein (interpreted) claims that light cannot be "stopped" and Hawking claims that time stops inside of a black hole. If time stops then it is impossible to have "velocity", therefore light , if any, must stop. It is at this junction that Einstein and Hawking part company.

[Tensor]

snip....



Actually, this particular forum (Questions and Answers) is for mainstream answers. That silly essay and its misunderstandings of General Relativity really belong in the ATM forum.

[devilmech]

Yes, basically. Photon scattering or absorption does exert a pressure, because the photons has momentum, given by E/c

Do you have a link to a white paper or essay on this? Light for me is by far the most interesting thing in the universe, and this bit got me thinking about the subject.

[Tensor]

Try here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4). Click on each of the boxes below the top chart. Hope this helps.

[swansont]

In addition to Tensor's link (love the hyperphysics site) try this (http://scienceworld.wolfram.com/physics/RadiationPressure.html) and that (http://en.wikipedia.org/wiki/Radiation_pressure)

[Einzige]

Photons do not have mass.

If photons do not have mass, then do they also not have energy?

If a photon has energy but not mass then I guess I have a fundamental misunderstanding of the equation E=mc^2, which seems to be saying that energy and mass are equivalent, except for a conversion factor of c^2.

Can someone help clear up my confusion, here?

Thanks!

[AstroSmurf]

E = mc^2 is actually the formula for converting between rest energy and mass.

For a moving object, the equation to use is E = E0 + pc, where E0 is the rest energy. For photons, there is no rest energy, so you get E = pc, i.e momentum times the speed of light.

Edit: I might as well add that the photon's energy according to quantum mechanics is E = hf, i.e Planck's constant times the frequency. Combine the two equations, and you get all sorts of interesting factoids such as the momentum of a photon with a certain frequency and so forth. Interesting stuff.

[swansont]

Actually it's E^2 = p^2c^2 + m^2c^4

m is rest mass. You get E = pc because the rest mass is zero.

[AstroSmurf]

I knew there was something I'd forgotten ... serves me right for trying to do this without doing some fact-checking beforehand.

[alainprice]

The only thing you did wrong was to say that E = E0 + pc.
Those 2 vectors(E0, pc) do not add up the same way as E^2 = p^2c^2 + m^2c^4.

[Wina]

Question, and I may have completely and totally misinterpreted what was said here:

Photons are trapped in solar cells to get energy, to my understanding, so my question is, were you to slow a photon to lesser than the speed of light, utilizing E=mc^2, you're converting energy to mass, so what if black holes, rather than bending or repelling light, actually slow photons to sub-light speed and thus your photon becomes particulate matter and is then affected directly by the black hole's gravity (IE. light does not escape a black hole). Your converted photon would only have to weigh a marginal fraction of anything in order to be affected, and by being marginally affected in any way shape or form, it's slowed down even moreso and thus gripped by your black hole.

Warning, my logic hurts even my own brain, I apologize if I'm a blithering idiot.

[alainprice]

Not quite.

A photon's energy is measured by it's color. The shorter the wavelength, the stronger the photon. Photons ALWAYS travel at C(in vacuo). No faster, no slower, but C.

[Wina]

So you're saying that not even a black hole, essentially the most powerful thing in the universe, can affect a photon's speed.....

(brain...........hurting............ow)

[hhEb09'1]

So you're saying that not even a black hole, essentially the most powerful thing in the universe, can affect a photon's speed.....The speed of light C can depend upon the geometry of the local space, though

[PhantomWolf]

Well actually Photons travel at C regardless, they appear to travel slower in some mediums because they get absorbed and retransmitted, so you can't really slow one down. They either are travelling at C, or they don't exist. Pick one and only one state at any one point in time.

edited to add: Just noticed the second page. Photons always travel at C, even in a black hole. The high gravity inside of the black hole distorts space and time, meaning that time slows down. As far as the photon is concerned it's travelling at the speed of light, but to an observer outside of the BH, it would appear to slow down and stop. It's all relative.

[Wina]

By definition, you can't observe a photon from outside a black hole though....... lol gotcha!

[PhantomWolf]

By definition, you can't observe a photon from outside a black hole though....... lol gotcha!

Well in reality you can't observe a photon in -or- out of a Black Hole, so......

[howard2]

If energy and mass are equivalent. Then the photon does have mass. The energy of a single photon is 6.62x 10(-34) Joules or watt/seconds.

I hope I have posted in the correct forum.
Sorry if I haven't.

howard2

[vkp]

Light doesnot have mass and therefore we say that to achieve the speed of light we have to make mass of an object approach zero.

[swansont]

If energy and mass are equivalent. Then the photon does have mass. The energy of a single photon is 6.62x 10(-34) Joules or watt/seconds.

Energy and mass are equivalent, but they are not the same.

Planck's constant gives you the conversion from frequency to energy, but is not the energy of a single photon.

[Ricimer]

energy doesn't have to equal mass. Use the full equation for e=mc^2.

E^2=sqrt(p^2*c^2+m^2*c^4).

Energy is rest mass and momentum. notice, for a stationary object (p=0) it reduces to e=mc^2. For a massless particle (m=0), all the energy is in the momentum (p), unless p=0...in which case there is no energy, so why are you using the equation at all?

Light falls into the second category, stationary masses into the first, moving masses uses the whole thing.

[Titana]

Light dont have mass. Light is composed of photons right? So we could ask if the photon has mass. Well, to my understanding the answer is definitely no. The photon is a massless particle and according to the theory it has energy and momentum but no mass. And this has been confirmed by experiment to within strict limits.



Titana.

[Tensor]

energy doesn't have to equal mass. Use the full equation for e=mc^2.

E^2=sqrt(p^2*c^2+m^2*c^4).


Uh,.... shouldn't either the ^2 on the left or the sqrt on the right be left off?

[G O R T]

Uh,.... shouldn't either the ^2 on the left or the sqrt on the right be left off?



E=pc+mc^2

E^2=(pc)^2+(mc^2)^2

E^2=sqrt(p^2*c^2+m^2*c^4)

No, that's right.

[Tensor]

E=pc+mc^2

E^2=(pc)^2+(mc^2)^2

E^2=sqrt(p^2*c^2+m^2*c^4)

No, that's right.

Wanna try again, a bit more slowly?

[alainprice]

E = pc + mc^2....Are you kidding me?
sqrt(a^2 + b^2) IS NOT EQUAL TO (a+b)

Let us begin with...

E^2 = p^2*c^2 + m^2*c^4 (eq. 1)

Done, now go get yourselves a coffee.

edit: I can't stop chuckling.

[Tensor]

E = pc + mc^2....Are you kidding me?
sqrt(a^2 + b^2) IS NOT EQUAL TO (a+b)

Let us begin with...

E^2 = p^2*c^2 + m^2*c^4 (eq. 1)

Done, now go get yourselves a coffee.

edit: I can't stop chuckling.

Dang it Alan, he was supposed to work it out. :) :clap:

[mugaliens]

Thought experiment:

Take two perfectly straight, non-diverging lasers and fire them into the flattest available space in perfectly parllel beams.

If light has no mass, the beams would remain parallel. If light has mass, the beams would be drawn together.

The equivalence principle says that everything is equally affected by gravity. But does light's momentum count for anything? Would it curve space, causing the two beams to converge?

[Ricimer]

oops, my bad. I often use one form or the other, and I guess I use both this time.

it is either E^2 or sqrt.

[Ken G]

If light has no mass, the beams would remain parallel. If light has mass, the beams would be drawn together.

The whole issue here is, does one mean rest mass, or energy-equivalent mass (i.e., E/c^2). Note that gravity comes from energy (and pressure), not just rest mass, so it is the latter meaning that is relevant to gravity. So yes, the light beams converge to the extent that you crank up the energy of the beams. If you treat the light as a test particle with essentially zero energy just to probe the background spacetime, then the convergence would be neglected.

[alainprice]

So you're saying that not even a black hole, essentially the most powerful thing in the universe, can affect a photon's speed.....

(brain...........hurting............ow)

Isn't the most powerful thing a first generation star going supernova? Either that or a supermassive black hole feeding.

[rufflychux]

Helloo... have we forgotten the cosmos that soon? as far as I am concerened the cosmos is made up of two parts. Yep.. just two part. the first part of the cosmos is "MASS", the second part of the cosmos is "ENERGY & MOMENTUM". Energy needs a medium to be propagated. That means that Light is the mass through which "LIGHT ENERGY" is propagated. Energy cant travel in a vacuum; so what it does is that it generates light, and through light energy can be successfully transfered in a vacuum. That is why the suns energy reach us.

[Cougar]

That's not even roughly right, Ruffly. :D

[Kaptain K]

The cosmos does consist of two parts (at least you got that right). The rest isn't even close.

Space-time and mass-energy. Everything else is details.

[Sam5]

So you're saying that not even a black hole, essentially the most powerful thing in the universe, can affect a photon's speed.....

(brain...........hurting............ow)

There is a trick to this “always c” business.

As the Einstein theory goes, an atomic clock slows down its tick rate inside a gravity field and light slows down its speed in the same gravity field. When both happen, the atomic clock in the gravity field still “measures” the speed of light to be “c”, even though the light speed is slower in the gravity field than outside it. It is “measured” to be “c” because the atomic clock inside the gravity field has slowed down it’s tick rate.

Of course light slows down in a gravity field, but you’ll need to judge that slowdown by an atomic clock that is located outside of a gravity field.

[Knowledge_Seeker]

Light is massless.

"We believe so. The special theory of relatvity says that the energy of an object with mass traveling at the speed of light would be infinite." - Fortunate

For something to travel at the speed of light it must be massless, and since light travels at the speed of light it has no mass (as far as we know).

[Ken G]

Of course light slows down in a gravity field, but you’ll need to judge that slowdown by an atomic clock that is located outside of a gravity field.
Note that this is an unusual interpretation of special relativity, because SR is basically a set of perfectly natural measurement conventions and their observed ramifications. One of those conventions is that if you want to measure time, you should be able to use your own clock, not refer back to someone else's clock at some distant location. Thus it is with this interpretation of the meaning of time that we say light does not slow down in a gravity field, but from someone else's perspective who is not in the gravity field, it might be said to, although I think if you also use that distant person's sense of distance, you will again reach the same c. The bottom line is, one must avoid absolute statements of reality, and instead be clear on one's measurement conventions, or just use the agreed-on conventions of SR.

[hhEb09'1]

Note that this is an unusual interpretation of special relativity,General relativity?I think if you also use that distant person's sense of distance, you will again reach the same c. How do you do that, and still get results like gravitational lensing?

[Ken G]

General relativity?

Yeah, that's what I meant, you're right.
How do you do that, and still get results like gravitational lensing?
Gravitational lensing need not have a change in lightspeed if it monkeys with the geometry of space itself. I'm no expert, but I think you can always take two different perspectives, which differ in how you are coordinatizing things (after all, coordinates are arbitrary, and so are the words you use to describe what is happening). One is, you treat space as normal but alter the speed of light. That is not the standard coordinatization, however, because it does not correspond to the way space would be locally measured, and as I said, relativity is the ramifications of natural measurement conventions. I think the standard approach actually bends straight lines, without altering the speed of light at all. Note the difference to what is happening to light in a glass lens-- there the geometry of a straight line is not altered, but the speed of light is. So we see that the phrase "gravitational lensing" is probably a misnomer that may cause more confusion than it solves.

[uniqueuponhim]

Scientist say that even light can't escape the gravity pull of a black, then light must have mass in order to be affected by the gravity pull of the black hole.
So do light have mass?
Alright, I'm going to try to get back on topic and actually give answers to this guy's questions.

The short answer is that photons have a rest mass of zero, but they also carry energy, and, as a consquence, momentum. The force of the black hole is able to change the energy and momentum of the photon, so that a photon trying to escape a black hole, while actually travelling away from it at the speed of light, gets redshifted so much by the time it escapes that it has zero energy, which means there is no photon left. The speed of the photon is never changed, but its energy is, because it takes more energy to escape the black hole's gravitational pull than the photon has.

Here's the long answer: As I just mentioned, the two properties we are most interested in of a photon are its energy and its momentum. Both of these are directly proportional to mass, so you might think that since a photon has zero mass that it has zero momentum and energy too. That isn't the case, because it only has zero rest mass. The mass used to calculate momentum and energy is equal to m=γ*m0 where γ(gamma) is the Lorentz factor, or 1/SQRT(1-v^2/c^2) and m0 is the rest mass. A photon always travels at the velocity of light, c, and has a rest mass of zero so that means we get m = 0 * infinity. 0*infinity could be anything though, so that isn't very useful to us, and it turns out that not only do we not need the mass to find the energy and momentum, but actually that you need the energy or momentum to find the mass.

Lucky for us though, there are other ways to find the energy and momentum of a photon. We know, for example, that the energy of a photon is E = hf = hc/λ (where h is planck's constant, f is the frequency of the photon and λ is its wavelength). We also know, from einsteins famous equation: E=mc^2=Sqrt(m0^2*c^4+p^2*c^2) (which, when v<<c, making the second term negligible, reduces to E=m0c^2) since m0=0, the first term is 0, leaving us with E=pc=hc/λ, or E/c=p=h/λ

So for energy, we have E=hc/λ
And for momentum, we have p=h/λ
And the photon's mass, you could find it from either of those: E=mc^2=hc/λ, m=h/λc; p=mc=h/λ, m=h/λc. From now on, I'll be stating everything in terms of E, the energy of the photon though, so simply:

p=E/c
m=E/c^2

But we also know that c is constant, and that all photons travel at c, and can't be slowed down*. So how does a black hole trap a photon? Well, since the photon has a mass while it is moving, which we have just calculated, the black hole also exhibits a force on that photon: F=GMbMγ/r^2 (Mb is the mass of the black hole, Mγ is the mass of the photon, G is the universal gravitational constant and r is the distance from the center of the black hole to the photon). So if the photon is moving away from the black hole, then the hole does a negative amount of work on it equal to W=-FΔr, but since F changes with r, we have to do this in infinitesimal steps and use calculus, so the real equation is dW=-Fdr=-GMbMγdr/r^2. So, say we want the photon to move from a distance r0 to a point where the black hole has no gravitational influence on it at all (infinity). We just integrate this equation from r=r0 to infinity: W=GMbMγ/(infinity)-GMbMγ/r0=0-GMbMγ/r0=-GMbMγ/r0. Now, if we set that value equal to the energy of the photon, then it will take ALL of the energy of the photon to escape from the black hole from that radius r0. If we solve for that radius r0 where W=E, then that will be the radius of the event horizon of the black hole: E=GMbMγ/r0=GMbE/r0c^2, 1=GMb/r0c^2; r0=GMb/c^2. So at any value of r0 smaller than GMb/c^2, it will take more energy than a photon has to escape that black hole, which means it simply can't. And that radius, r0=GMb/c^2 is the radius of the event horizon.

Also keep in mind that a black hole is simply any object with a density large enough that its event horizon has a larger radius than the object itself. At small distances, the density required for this would be enormous (An event horizon of 1Å would require a mass of 1.35*10^17kg, the equivalent of an asteroid 50km across) But since r0 scales with Mb, which itself scales with Rb^3, doubling the radius of an object while maintaining the same density effectively octuples the radius of its event horizon.

[hhEb09'1]

Gravitational lensing need not have a change in lightspeed if it monkeys with the geometry of space itself. I'm no expert, but I think you can always take two different perspectives, which differ in how you are coordinatizing things (after all, coordinates are arbitrary, and so are the words you use to describe what is happening). One is, you treat space as normal but alter the speed of light. That is not the standard coordinatization, however, because it does not correspond to the way space would be locally measured, and as I said, relativity is the ramifications of natural measurement conventions. I think the standard approach actually bends straight lines, without altering the speed of light at all. Note the difference to what is happening to light in a glass lens-- there the geometry of a straight line is not altered, but the speed of light is. So we see that the phrase "gravitational lensing" is probably a misnomer that may cause more confusion than it solves.Still, I disagree with your comment that Sam5's comment is an unusual interpretation. Notice that you even use the word "normal" in describing his point of view!

Also, as you point out, the term "gravitational lensing" is slightly at odds with what you say is the usual interpretation--and it corresponds to Sam5's notion. Gravitational lensing is a very commonly used term, so I doubt you can justify saying that it is an unusual interpretation. In fact, it is probably the common one outside of the physics community, and not uncommon within.

[uniqueuponhim]

I should also point out, as I meant to in my first post, that the photons of light itself do not actually slow down in media other than a vacuum, they simply encounter so many obstacles and bounce around so much that they appear to be travelling more slowly. Light always travels at c.

What occurs with gravitational lensing is sort of the same thing as what I just described with a photon trying to escape an object, except that this time, the light is actually passing by the object, which exerts not only a force parallel to the light as we discussed before, but also a perpendicular one. This perpendicular force is able to change the direction of the momentum of the photon (Fdt=dp, remembering that F and dp are both vectors). This requires absolutely no modification of the shape of spacetime to accomplish, and is simply light being bent by its gravitational attraction to ab object. It also happens that, for black holes, the radius at which the force is big enough to trap the photon in a circular orbit is the same as the event horizon radius we calculated earlier.

[hhEb09'1]

I should also point out, as I meant to in my first post, that the photons of light itself do not actually slow down in media other than a vacuum, they simply encounter so many obstacles and bounce around so much that they appear to be travelling more slowly.That's not true, except maybe in a virtual sense. Any photon that travels through a transparent medium will not have been absorbed.This requires absolutely no modification of the shape of spacetime to accomplish, and is simply light being bent by its gravitational attraction to ab object. I don't think that's true either. That would mean that the bending of light would be half of what actually occurs--that was Einstein's first prediction, before he came up with the theory of general relativity.

[Ken G]

Still, I disagree with your comment that Sam5's comment is an unusual interpretation. Notice that you even use the word "normal" in describing his point of view!

By normal, I meant Euclidean, i.e., what we normally experience. A strong gravity field is not normal, and not Euclidean in the standard coordinatization.

Also, as you point out, the term "gravitational lensing" is slightly at odds with what you say is the usual interpretation--and it corresponds to Sam5's notion. Gravitational lensing is a very commonly used term, so I doubt you can justify saying that it is an unusual interpretation.
You must not argue the physics of something based on its name. Confusing labels are quite common in physics: "centrifugal force", "blackbody radiation", etc.

The real point in all this is that if you want to say that "light slows down", you need to define what you mean by speed. In relativity, one uses local definitions whenever possible, because that would be the normal measurement convention. So speed means the rate of change of locally measured distance with locally measured time. If you want to use some other definition of speed, you can get literally anything, including values slower (and faster!) than c, but you will not be using the standard nomenclature. That's not to say it's wrong, or it's never done. For example, sometimes we do say that things in the Big Bang are receding faster than the speed of light. However, it is unlikely that we would say the light emitted by those objects, away from us, has sped up-- it would be as confusing a nomenclature as saying that gravity changes the speed of light.

[hhEb09'1]

You must not argue the physics of something based on its name.Mostly, I wasn't arguing about the physics :)In relativity, one uses local definitions whenever possible, because that would be the normal measurement convention.Sure, but Sam5's comment was clearly about nonlocal measurement. I don't see anything unusual about his point of view. it would be as confusing a nomenclature as saying that gravity changes the speed of light.Locally, the reference frame in which the speed of light is c (and there always is one) is convenient for some things, but globally it's not so convenient. It's not confusing if one is careful.

[Ken G]

Sure, but Sam5's comment was clearly about nonlocal measurement.
And just what is that, pray tell?

I don't see anything unusual about his point of view.Locally, the reference frame in which the speed of light is c (and there always is one) is convenient for some things, but globally it's not so convenient. It's not confusing if one is careful.
As I stated, if you are going to say time slows down, you also have to say that something happens to lengths. Even from the nonlocal perspective, the deflection of light is due to curvature of spacetime, not a change in the speed of light. In a glass lens, light takes the shortest time, but not the shortest length. In gravity, it takes the shortest proper length.

[hhEb09'1]

And just what is that, pray tell?Surely, our measurement of the deflection of light around the sun has some sort of nonlocal component :)Even from the nonlocal perspective, the deflection of light is due to curvature of spacetime, not a change in the speed of light.That's one way of looking at it. Sam5's way is another, just as valid probably (as you said), and from my experience more common.

Approached from a spacetime curvature point of view, the curvature of the path above the surface of the earth of a thrown softball, and a supersonic bullet, are about the same. Measured in space, as we usually do, they clearly have very much different curvatures. Sam5 was just taking the latter approach.

Added "our measurement of"

[Ken G]

Surely, our measurement of the deflection of light around the sun has some sort of nonlocal component

No, the measurement is right in our telescopes. The mathematics we use to retrace its steps is relativity, and can be described in local terms entirely. But I confess to be nitpicking, my point was simply that relativity is at its core a theory about local measurement and how one local observer passes off her results to the next in the chain.

That's one way of looking at it. Sam5's way is another, just as valid probably (as you said), and from my experience more common.

Perhaps in the popular literature, but you find all kinds of hooey there that is designed to give people an illusion of understanding. Can you quote a real relativity text that uses a changing speed of light to explain lensing? I think you'll find they use metrics, and the local definition of speed will come out c.

[hhEb09'1]

No, the measurement is right in our telescopes.:)

That is not what is meant by local, in physicsPerhaps in the popular literature, but you find all kinds of hooey there that is designed to give people an illusion of understanding. Can you quote a real relativity text that uses a changing speed of light to explain lensing? I think you'll find they use metrics, and the local definition of speed will come out c.I thought you had me there, I doubted I'd find any recent books at hand that would even describe gravtational lensing in any detail. Even MTW. However, I happened to have a copy of Gravitation and Inertia by Ciufolini and Wheeler (1995) which discusses "gravitational lensing" starting about page 122. They go into detail in section 3.4.2, entitled Delay of Electromagnetic Waves, and starts with the sentence "The other high-precision test of space curvature is the delay in the propagation of radio waves traveling near the Sun, or Shapiro time delay, porposed in 1964." (bolding theirs)

[swansont]

I thought you had me there, I doubted I'd find any recent books at hand that would even describe gravtational lensing in any detail. Even MTW. However, I happened to have a copy of Gravitation and Inertia by Ciufolini and Wheeler (1995) which discusses "gravitational lensing" starting about page 122. They go into detail in section 3.4.2, entitled Delay of Electromagnetic Waves, and starts with the sentence "The other high-precision test of space curvature is the delay in the propagation of radio waves traveling near the Sun, or Shapiro time delay, porposed in 1964." (bolding theirs)

Time delay is not necessarily the same as c changing, though. I was under the impression that the Shapiro delay is completely consistent with a longer path of travel light must take to reach us when it passes through a region of non-flat space. The local description is the time dilation in that region. Neither description (again, to my understanding) relies on c changing.

[hhEb09'1]

Neither descriptionAs Ken G mentioned earlier, any way of looking at the situation is valid. He was just concerned that Sam5's way was an unusual one, and that the term gravitational lensing is misleading (though itself common). I just posted that quote to show that such nomenclature is not limited to just that term.

Of course, locally, there is always a reference frame in which the velocity of light is c, but that reference frame must be free of gravity. Once your reference frame includes gravity, all bets are off. Which reference frame is actually used in practice depends upon the situation.

[Ken G]

Of course, locally, there is always a reference frame in which the velocity of light is c, but that reference frame must be free of gravity.
No, swansont has the same understanding that I do. The delay is due to a different path length, not a change in c, if you take a global integral of path length and use the standard coordinatization of relativity (which is expressed in terms of locally measurable distances, so is purely local, although not necessarily at rest or in zero gravity, as evidenced by the fact that it uses differentials in the metric). Other descriptions may be possible, but are likely to be wrong, are surely confusing, and are generally only used in popular descriptions that are not taken very seriously by experts (of which I am not one, but this is very clearly my impression). The point is, ds/dt = c for light, locally in any reference frame and with any gravity. You can only get different answers if you keep track of time and distance differently, in some kind of quasi-global way. That's what is often done in cosmology, for example, and speeds often exceed c.

[hhEb09'1]

You can only get different answers if you keep track of time and distance differently, in some kind of quasi-global way. That's what is often done in cosmology, for example, and speeds often exceed c.But that's my point. I'm not arguing against any other way of doing things. It's just that Sam5's way of looking at things is not that unusual.

[Ken G]

I've made my point the best I can, and I'm still waiting for an example from outside of popularized literature that says gravity changes the speed of light.

[hhEb09'1]

Another way of saying it, from that previous reference: "The coordinate speed of light is in general different from 1...Of course, the speed of light measured by local inertial observers, with standard clocks and standard rods, is always 1..." Two different ways of saying the same thing, essentially. But that's relativity.

[Ken G]

Yes, two different ways, one which is careful to say the coordinate speed of light (which I referred to earlier), versus the measured speed, which is the standard nomenclature.

[hhEb09'1]

Yes, two different ways,Sam5 is saying the same thing as Wheeler, in a different way

[Sam5]

Sam5 is saying the same thing as Wheeler, in a different way

I agree.

[Ken G]

OK, it's time we got to the bottom of the lensing phenomenon instead of arguing terminology. I think we can all agree that lensing is an interference phenomenon, and as such depends on the rate of accumulation of phase with distance. The light path is bent by things that alter the rate of accumulation of phase with length. That's either a change in speed at the same frequency (a glass lens) or a change in frequency at the same speed (gravitational lensing). If you want to use a picture where the speed changes, but the frequency stays the same even though it is "falling" into a gravitational well, be my guest. I don't think you'll find too many takers.

[hhEb09'1]

Time delay is not necessarily the same as c changing, though. I was under the impression that the Shapiro delay is completely consistent with a longer path of travel light must take to reach us when it passes through a region of non-flat space. The local description is the time dilation in that region. I just checked the Ciufolini and Wheeler calculations, and it appears that the opposite is true. They say the longer path has so little effect that they first assume that the path is straight in calculating the magnitude of the time delay.

That makes sense, since the path difference over even a couple AU at a deflection of 1.75 arcsec (http://www.astro.ucla.edu/~wright/deflection-delay.html) is only 1.75^2/c, which is about 4 usec whereas the Shapiro time delay (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1977JGR....82.4329S) is a couple hundred usec.

[Sam5]

I just checked the Ciufolini and Wheeler calculations, and it appears that the opposite is true. They say the longer path has so little effect that they first assume that the path is straight in calculating the magnitude of the time delay.

That makes sense, since the path difference over even a couple AU at a deflection of 1.75 arcsec (http://www.astro.ucla.edu/~wright/deflection-delay.html) is only 1.75^2/c, which is about 4 usec whereas the Shapiro time delay (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1977JGR....82.4329S) is a couple hundred usec.

Hey, thanks very much for this information! I’ve never been able to refute the “longer path” argument because I didn’t have any information about the calculations of the longer distance.

I’ve been going through my collection of Einstein papers, finding references to this topic, and later I’ll post some of them in his own words. In 1912 Einstein and Max Abraham exchanged some interesting articles about this that were published in the “Annalen der Physik”. These were basically “letters to the editor”, but they were also written as scientific papers.

Abraham tried to condemn the 1905 theory when Einstein, in 1911, changed his point of view about light speed never changing. Einstein’s response was that the 1905 theory did not consider acceleration or gravity fields and (as you know) it represented a “special case” of his developing GR theory.

[Sam5]

I have a copy of the Peter Ustinov documentary about Einstein, and in it Shapiro says that light speed slows down as the light passes near the sun. The program has Shapiro on camera alone saying this, then it cuts to a group of cosmologists sitting in an observatory somewhere trying to refute Shapiro’s claim.

I still don’t understand why so many in the mainstream still try to refute this, since it was Einstein’s own idea which he talked about in a number of papers and he even mentioned it in his 1916 book. By 1911 he believed that the deflection of light near the sun was caused by a slowdown in the light as it passed through the strong gravity field near the surface of the sun. And of course he also believed that the slower speed of light near the sun would not be noticed by someone on the sun measuring the local speed of light with a local atomic clock, since an atomic clock at the surface of the sun would slow down its tick rate, and that would cause a local speed of light calculation to be “c” at the sun, measured by a clock at the specific place where the light speed was measured. But if measured by an earth-based clock (which was the kind Shapiro used) the slowdown as light passed near the sun would definitely be noticed, since the earth based clock would be ticking faster than the sun based clock.

[Ken G]

That makes sense, since the path difference over even a couple AU at a deflection of 1.75 arcsec (http://www.astro.ucla.edu/~wright/deflection-delay.html) is only 1.75^2/c, which is about 4 usec whereas the Shapiro time delay (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1977JGR....82.4329S) is a couple hundred usec.
The issue is, how are you measuring length. If you use rulers laid end to end, you will get a much longer path length. That's the length contraction effect due to the gravity, and is the natural way to measure length-- with local rulers. It's the same reason that it appears to us to take an infinitely long time for matter to fall into a black hole. The alternative is to maintain that matter falling into a black hole slows down, which all depends on coordinates of course, but is an awkward interpretation-- gravity makes things slow down? Nevertheless, that does not make it wrong, and I think hhEb09'1 and Sam5 have sufficiently defended the point that a changing speed of light is one way to interpret gravitational lensing, just as one could say that objects slow down as they fall into a black hole. My point is that there are possibly better ways to think about this, but with pedagogies, I believe it is always best to know and use all the possibilities. You never know which one will lead most naturally to some insight or other. So the discussion has illuminated the possibilities, and the label "standard approach" is pretty unimportant. I started this off because I personally think that talking about changing speeds of light is very confusing when all of relativity is based on the idea that the locally measured speed of light is always the same, but with enough explanatory additions, the changing speed approach does appear to be a valid explanation. I would still prefer talking about lensing being caused by interference from a blueshift in frequency, it's better in tune with concepts like "cosmological redshift" (to be consistent to your approach, you would have to say that the speed of light is increasing as the universe ages, and that's why the background is now seen in the microwave. Of course all other constants would have to change to keep the fundamental unitless parameters the same. Is that where you want to go?)

[hhEb09'1]

The issue is, how are you measuring length. If you use rulers laid end to end, you will get a much longer path length. That's the length contraction effect due to the gravity, and is the natural way to measure length-- with local rulers. It's the same reason that it appears to us to take an infinitely long time for matter to fall into a black hole. The alternative is to maintain that matter falling into a black hole slows down, which all depends on coordinates of course, but is an awkward interpretation-- gravity makes things slow down? Nevertheless, that does not make it wrong, and I think hhEb09'1 and Sam5 have sufficiently defended the point that a changing speed of light is one way to interpret gravitational lensing, just as one could say that objects slow down as they fall into a black hole. My point is that there are possibly better ways to think about this, but with pedagogies, I believe it is always best to know and use all the possibilities. You never know which one will lead most naturally to some insight or other. So the discussion has illuminated the possibilities, and the label "standard approach" is pretty unimportant. I started this off because I personally think that talking about changing speeds of light is very confusing when all of relativity is based on the idea that the locally measured speed of light is always the same, but with enough explanatory additions, the changing speed approach does appear to be a valid explanation. I would still prefer talking about lensing being caused by interference from a blueshift in frequency, it's better in tune with concepts like "cosmological redshift" (to be consistent to your approach, you would have to say that the speed of light is increasing as the universe ages, and that's why the background is now seen in the microwave. Of course all other constants would have to change to keep the fundamental unitless parameters the same. Is that where you want to go?)Sorry, I can't see this post without replying to it, for some reason (http://www.bautforum.com/showthread.php?p=746085#post746085)

There's another issue, how are you measuring the elapsed time? After you lay your rulers end to end down through the gravity well of the sun. In order for your approach to work, wouldn't your clock (and I mean your real clock) have to follow the light path too, as the light goes from one end to the other? That's probably not how I would want to go. :)

PS: Thanks Ken G for sticking with this (and you too JW). I too believe in trying all the possibilities.

[Tensor]

The issue is, how are you measuring length. If you use rulers laid end to end, you will get a much longer path length. That's the length contraction effect due to the gravity, and is the natural way to measure length-- with local rulers.

He, he, he, just to confuse the issue even more, you can think of gravity as changing the equipment used to make our measurements in a flat spacetime, instead of spacetime changing and our measuring devices staying the same. In the above, the rulers would shrink, showing a longer path length. The math works out either way.

[hhEb09'1]

He, he, he, just to confuse the issue even more, you can think of gravity as changing the equipment used to make our measurements in a flat spacetime, instead of spacetime changing and our measuring devices staying the same. In the above, the rulers would shrink, showing a longer path length. The math works out either way.I thought that was the "length contraction effect" that Ken G was talking about.

[Ken G]

There's another issue, how are you measuring the elapsed time? After you lay your rulers end to end down through the gravity well of the sun. In order for your approach to work, wouldn't your clock (and I mean your real clock) have to follow the light path too, as the light goes from one end to the other?
The clock can't follow the light, what happens is, each ruler has a clock on it, and the clocks are synchronized according to Einstein's procedure (which also involves light and is arbitrary, but there really isn't much of an alternative that would seem to make any sense. This is what I mean that relativity is really the ramifications of a particularly natural measurement convention). Then the photon just notices what the local clock said when it left, and what the local clock says when it arrives, and counts the rulers in between. However you slice that, it comes up c. But yes, all of this is completely arbitrary, and other coordinatizations are possible, but at least this one is actually based on measurements and not conceptualizations.

[Ken G]

He, he, he, just to confuse the issue even more, you can think of gravity as changing the equipment used to make our measurements in a flat spacetime, instead of spacetime changing and our measuring devices staying the same. In the above, the rulers would shrink, showing a longer path length. The math works out either way.
Yes, this is what I meant by length contraction, but I think Tensor is noting that even length contraction has multiple interpretations as something happening to rulers (the standard interpretation when motion is involved), or to space (an interpretation more often used when gravity is involved). It's kind of a complicated mess, pedagogically, which is why the experts generally stick to the mathematics (the "metric") and aren't always real careful when they translate the results into words for popular consumption. The bottom line is: avoid absolute statements that sound like the way "things are", and instead choose terms that indicate an arbitrary pedagogy has been adopted. People don't like that, I know, but it's more honest and accurate.

[hhEb09'1]

I know the clock can't follow the light!Then the photon just notices what the local clock said when it left, and what the local clock says when it arrives, and counts the rulers in between. However you slice that, it comes up c.I don't think that works out that way. How could it? The local clocks in between would tick at a different rate--corresponding to the different length contractions as the bar is moved. How is that information preserved, or utilized, in your setup?

[Ken G]

The local clocks in between would tick at a different rate--corresponding to the different length contractions as the bar is moved. How is that information preserved, or utilized, in your setup?
The rulers are made according to some standard, and then laid end to end, and they just sit there. Length contraction takes care of the rest. The same with the clocks-- they are built and standardized, then put in place and synchronized. Then they just tick along. Then you shoot out the light beam and watch it take the shortest path. It's all about measurements that can actually be done, at least in principle.

[hhEb09'1]

The rulers are made according to some standard, and then laid end to end, and they just sit there. Length contraction takes care of the rest. The same with the clocks-- they are built and standardized, then put in place and synchronized. Then they just tick along. Then you shoot out the light beam and watch it take the shortest path. It's all about measurements that can actually be done, at least in principle.Yahbut, in your description, you used the reading on the first and the last clock, both would be farther out of the gravity well. They'd tick at different rates than the other clocks.

[Jeff Root]

Yahbut, in your description, you used the reading on the first
and the last clock, both would be farther out of the gravity
well. They'd tick at different rates than the other clocks.
Which has absolutely no consequences whatsoever. The clocks
in between are interesting but irrelevant. Only the rulers and
the end clocks are needed for this particular measurement.

-- Jeff, in Minneapolis

[Sam5]

I've made my point the best I can, and I'm still waiting for an example from outside of popularized literature that says gravity changes the speed of light.

Below are some comments written by Einstein in 1912 as he further elaborated on this new theory in his 1912 paper, “The Speed of Light and the Statics of the Gravitational Field." The first quote is from the introduction to the paper. The second quote appears later in the paper. In the equations below, the “o” in co and Po is a subscript:

-----------quote follows----------

In a paper that appeared last year, I drew from the hypothesis that the gravitational
field and the state of acceleration of the coordinate system are physically equivalent
a few conclusions that tie in very well with the results of the theory of relativity
(theory of relativity of uniform motion). But at the same time it turned out that one
of the basic principles of that theory, namely, the principle of the constancy of the
velocity of light, is valid only for space-time regions of constant gravitational
potential. Even though this result rules out the universal applicability of the Lorentz
transformation, it should not frighten us away from the further pursuit of the path we
have taken; at the very least the hypothesis that the "acceleration field" is a special
case of the gravitational field has, in my opinion, such a high degree of probability,
especially in view of the conclusions regarding the gravitational mass of the energy
content, already drawn in the first paper, that a more exact consideration of the
conclusions of the above equivalence hypothesis seems indicated.

-----------

§3. Remarks on the Physical Meaning of the Static Gravitational Potential

If we measure the velocity of light in a space of nearly constant gravitational
potential by measuring with a specific clock the time needed by the light to traverse
a closed path of a specified length, then we always obtain the same number for the
velocity of light totally independently of the magnitude of the gravitational potential
in the space in which we carry out the measurement. We always use the same clock for the time measurements; the clock is always brought to the place for which c is to be determined. This follows directly from the equivalence principle. Thus, if we say that the velocity of light at a point P is c/co times greater than at a point Po, this means that in order to measure the time (that is to say, to measure the time denoted in the equations by “t”) at P, we must use a clock that runs c/co times slower than the clock to be used to measure the time at Po if the rates of the two clocks are compared with one another at the
same location. In other words: a clock runs faster the greater the c of the location to
which we bring it. This dependence of the speed of the passage of time on the
gravitational potential (c) holds for the passages of time in any process whatsoever.
This has already been shown in the previous paper.

--------end of quote---------

[Celestial Mechanic]

Below are some comments written by Einstein in 1912 as he further elaborated on this new theory in his 1912 paper, “The Speed of Light and the Statics of the Gravitational Field."
That was 1912. What about the 1915 papers on general relativity? Do you see a variable speed of light there? I don't.

[Ken G]

That's very interesting of course, but the theory has come a long way since 1912. Even Einstein himself iterated the process many times.

[hhEb09'1]

Which has absolutely no consequences whatsoever. The clocks
in between are interesting but irrelevant. Only the rulers and
the end clocks are needed for this particular measurement.Yeah, I popped off on that too quickly. Surely, that's what swansont meant too, so I apologize there as well. Light is its own clock, in this context.

[Ken G]

Yes, the point of the intermediate clocks is just to show that c is always c, when measured that way. To me, the upshot of all this is that as long as one is being clear about how things are being defined and measured, one can use different pictures of what is happening, and a changing speed of light might very well be a perfectly valid approach. I was just trying to avoid confusion relative to the standard description. I have not said that Sam5 or hhEb09'1 are wrong, my intent was to avoid confusion for others reading the thread and give them what might arguably be a better way to look at the lensing phenomenon. What is most fascinating, however, are all the different ways one can look at things, so I do not disapprove of alternate pedagogies, I merely say they must be followed to the point of complete self-consistency, and that can have surprising ramifications.

[Sam5]

That was 1912. What about the 1915 papers on general relativity? Do you see a variable speed of light there? I don't.

Einstein noted the variable speed of light (as related to gravity fields) in his 1916 book, when he said, “In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position.”

Edit, paragraph added:

He also says on the same page, “We can only conclude that the special theory of relativity cannot claim an unlimited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena of light.”

[Nereid]

Einstein noted the variable speed of light (as related to gravity fields) in his 1916 book, when he said, “In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position.”

Edit, paragraph added:

He also says on the same page, “We can only conclude that the special theory of relativity cannot claim an unlimited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena of light.”Sam5, if you wish to present a non-mainstream case wrt GR or SR, please do so by starting a thread in BAUT's ATM section.

It seems to me that you are well aware of the mainstream (relativity) theories, but are choosing (deliberately?) to argue your own, ATM view of these theories, here in this Q&A section (and, I might add, not only in this particular thread).

Please stop doing this.

[Ken G]

While I think it is certainly of historical interest how Einstein liked to think about his theory, almost a hundred years of application has likely caused an evolution in the pedagogy into what we find the standard view today. I'm sure that Einstein's approach is the same theory, so it's not really an ATM idea. What does one do with ATM pedagogies? I'd claim there really is no such thing, only better and worse pedagogies for each situation.

[publius]

I don't think there's anything ATM about this either. In fact, I remember being taught as such that the "coordinate speed" of light does indeed vary with gravitational potential. I dug around some of my old class notes and found the following formula for this coordinate speed for classic spherically symmetric (non-rotating) mass distribution:

c(r) = c0 * (1 - 2GM/r*c0^2 ) That will equal 0 at r = Schwarzchild radius, or the event horizon of a black hole.

The -GM/r in the second term is just the classic gravitational potential,
"Phi", so you can write it as c(r)/c0 = 1 + 2Phi/c0^2.

c0 is the familiar speed of light, and therefore one can think of c0 being the limiting speed of light in flat space. In curved space, it slows down relative to an observer in flat space. An observer down in the gravity well will however measure the local speed to always be c0 because his clock will slow by the same factor.

-Richard

[Sam5]

I don't think there's anything ATM about this either. In fact, I remember being taught as such that the "coordinate speed" of light does indeed vary with gravitational potential. I dug around some of my old class notes and found the following formula for this coordinate speed for classic spherically symmetric (non-rotating) mass distribution:

c(r) = c0 * (1 - 2GM/r*c0^2 ) That will equal 0 at r = Schwarzchild radius, or the event horizon of a black hole.

The -GM/r in the second term is just the classic gravitational potential,
"Phi", so you can write it as c(r)/c0 = 1 + 2Phi/c0^2.

c0 is the familiar speed of light, and therefore one can think of c0 being the limiting speed of light in flat space. In curved space, it slows down relative to an observer in flat space. An observer down in the gravity well will however measure the local speed to always be c0 because his clock will slow by the same factor.

-Richard

That's exactly what I said and what Einstein said. This is certainly not ATM.

[publius]

That's exactly what I said and what Einstein said. This is certainly not ATM.


Yes, I know. The trouble is the notion invariance of the speed of light has become so ingrained it is almost dogma, and many don't appreciate that it applies only in flat space-time.

Another (classical) way to look at this is to consider Maxwell's equations in non-inertial frames. They are not invariant under non-inertial coordinate system transformations, and the wave equation becomes very different from that in inertial frames.

I got "straightened out" about this several years ago, both about the coordinate speed of light, and the non-invariance of Maxwell in non-inertial frames. It was an EM argument about motors/generators, and the equivalence of the rotating field, fixed armature vs rotating armature in a fixed field. It works that for slow speeds (relative to c) that one can easily say E' = v x B, where v is the instaneous velocity of a spot on the armature rotating in a B field and find the same EMF if the B field were rotating around a fixed armature, but things aren't as simple as I was trying to make them at the time.

Technically to calculate the E field relative to the rotating armature, one would have to transform into a non-inertial Coriolis frame and things would get rather complicated in the general case.

-Richard

[Ken G]

Yes, I know. The trouble is the notion invariance of the speed of light has become so ingrained it is almost dogma, and many don't appreciate that it applies only in flat space-time.

No, it applies only locally, in any spacetime. It was never intended as a global concept. It's just that flat spacetimes don't distinguish between local and global properties, it's the same everywhere. Put differently, the equations of relativity are expressed as differentials, not global expressions. The distinction is unimportant in special relativity.

Technically to calculate the E field relative to the rotating armature, one would have to transform into a non-inertial Coriolis frame and things would get rather complicated in the general case.

That's because the Coriolis frame you refer to is a global frame. Maxwell will still work fine locally. The overarching point here is that a coordinate speed is always completely arbitrary, in the sense that you can get any result by choosing appropriate coordinates, because coordinates are completely arbitrary. Relativity is a theory about measurement conventions and their ramifications, not about limits on coordinate speeds. At its heart is what is locally measurable-- you can then transform into any coordinates you like in the normal way. In other words, you are free to do a Galilean transformation to a given coordinate system and come up with a coordinate speed in another set of coordinates, and it will be perfectly correct, it just won't correspond to what you will locally measure.

[publius]

Ken,

You know, I'm not sure if I've really got a handle on the whole local vs global thing to really appreciate this. I've always heard it said that space-time can always be considered "locally flat", which I think I understand means that in some "sufficiently small" neigborhood about a point, we can use local coordinates (And I presume that means using the local point as the origin) with a flat Minkowski metric.

For a simple example, I imagine an arbitray curve in the x-y plane. At any point (x0, y0), I can model that curve as y = x'(x0) * (x - x0) + y0 (where x'(x0) means the derivative at the point), which comes from dy = x'*dx. Or we parameterize and do it terms of tangent vectors and all that good stuff.

So taking that simple idea and trying to extend to any metric, I can evalute that metric at some point (equivalent to taking the derivative at that point as constant), then define local coordinates in terms of the value of the metric components at that point. Well, if I remember what little I learned about this stuff, we'd diagonalize the metric tensor (to eliminate "cross terms" -- ie dx_i*dx_j; i not = j which I think means the coordinates are orthogonal?), defining some local coordinate directions, then scale those diagonalized coordinates so we had
1's down the diagonal.

Is that basically what local coordinates and "locally flat" means? Of course, in the example of the xy-curve, the higher the second derivative, x'', at the point, the smaller the neighborhood about the point we could use the approximation. And I assume the general curvature is analgous to x", or how fast the metric is changing relative to those local coordinates.

So, if I understood this right, if I transform into such local coordinates in any space-time, then Maxwell will be valid in that local neighborhood as written in the those coordinates?

-Richard

[hhEb09'1]

Sam5, if you wish to present a non-mainstream case wrt GR or SR, please do so by starting a thread in BAUT's ATM section.

It seems to me that you are well aware of the mainstream (relativity) theories, but are choosing (deliberately?) to argue your own, ATM view of these theories, here in this Q&A section (and, I might add, not only in this particular thread).

Please stop doing this.I prefer to think that quoting Einstein is not ATM :)

But I see that Ken G has made the same comment. Sam5 does sometimes treat the words of Einstein with more respect than they deserve. :)

[Kaptain K]

Sam5 does sometimes treat the words of Einstein with more respect than they deserve.
While ignoring the math in the very paper that he is quoting (which often refutes his interpretation of the words).

[Ken G]

I've always heard it said that space-time can always be considered "locally flat", which I think I understand means that in some "sufficiently small" neigborhood about a point, we can use local coordinates (And I presume that means using the local point as the origin) with a flat Minkowski metric.

Yes, that's my understanding-- there is always a tangent space on which special relativity may be applied, with less and less accuracy as you move away from the tangent point. So GR is what you need to find what happens over a global spatial regime and an extended period of time.

Well, if I remember what little I learned about this stuff, we'd diagonalize the metric tensor (to eliminate "cross terms" -- ie dx_i*dx_j; i not = j which I think means the coordinates are orthogonal?), defining some local coordinate directions, then scale those diagonalized coordinates so we had
1's down the diagonal.

That sounds right to me, I don't think your knowledge on this is any less than mine.


Is that basically what local coordinates and "locally flat" means? Of course, in the example of the xy-curve, the higher the second derivative, x'', at the point, the smaller the neighborhood about the point we could use the approximation. And I assume the general curvature is analgous to x", or how fast the metric is changing relative to those local coordinates.

Yes, I think that's it exactly. I think it's analogous to the concept of velocity. Formally, this is a differential concept that applies only over tiny spatial and temporal regions, but a more general "average velocity" can be defined. In the absence of acceleration there is essentially no difference, but when you do have acceleration, you need calculus.


So, if I understood this right, if I transform into such local coordinates in any space-time, then Maxwell will be valid in that local neighborhood as written in the those coordinates?

Yes, Maxwell should still work locally in a rotating frame, but only over time scales much shorter than the rotation period and distance scales much shorter than the radius of curvature. Then again, that's also the scale where you can ignore rotation altogether. The main point is that just as the concept of instantaneous velocity is still useful when there is acceleration, so are the concepts of SR still useful when applied in GR.

[publius]

That sounds right to me, I don't think your knowledge on this is any less than mine.



Ken,

I wouldn't bet the farm on that. :) But assuming that's a half-way kosher way of how the local vs global thing works, here's what I don't understand. What prompted me to jump in here was the seeming objection to saying that gravity slows down light. To put it more precisely, it is gravitational *potential* differences that slow down light, not the strength of the g field, I believe. For example it blows my mind that, for a supermassive blackhole, the classic value of 'g' near the event horizon can be very low -- it goes inverse to total mass (plug in the Schwarzchild radius the GM/r^2 and you get somethink ~c^2/GM). Yet, clocks would be nearly stopped relative to an observer far, far away where. Even though 'g' is very small, the differnence in potential is enormous -- which shows that the notion of a "weak" g-field does not mean simple low g.

At any rate, getting back on track, sitting somewhere in a gravity well, I have to use my local coordinates. Yet, if I want to measure things far away, with global extent, then I have to transform the whole global picture into my coordinates.

And if I do that, the speed of light will indeed speed up and slow down with distance, and can see objects moving at speeds greater than c all in my natural local coordinates.

Coordinates are arbitrary and there are probably an infinite number of coordinate systems I could use. (But then there are coordinates I can't use. For example, I can't have a frame moving faster than c locally as that just wouldn't work.) I could transform to the local coordinates of an observer higher in the well, but then my local speed of light would be much slower.

So I don't see how I can do anything much if I can't "go global" so to speak.

-Richard

[publius]

Ken,

Continuing my ramblings: Let's consider the (accelerating, non-inertial) Coriolis frame, say of the rotating earth. The stars appear to be moving very fast, orbiting around us enormous speeds, which if far enough away will exceed the speed of light we measure here on earth. Is there any objection to this being a "real" measurement, as it is somehow not local?

Of course we know that motion is "not real" since we are in a rotating frame. However, GR and the equivalence principle say there is an equivalent space-time metric that will produce the exact same psuedo-forces of the rotating frame, which we can call the "Coriolis metric". I forget the details, but it is indeed possible to construct such a metric, and it will have a corresponding enery-momentum distribution. The 2w x v Coriolis force is replace by an enormous gravitomagnetic field, and the regular 'g' field supplies the centrifugal acceleration. I think it involves an enormous, rotating mass distribution of some sort, which strikes me as ironic. If we want the forces of a rotating frame, we always have to rotate something. :) And we can indeed think of that massive rotating mass distribution as "dragging" space-time along with it, meaning space is rotating under our feet so to speak, so who is really rotating? Sort of turns the equivalence principle on its head :)

Anyway, in this Coriolis metric, those distant stars would indeed be moving that fast, in terms of my local coordinates near the origin. The speed of light would be many times greater than the local value at that high radius. But, the potential difference would be so great, that clocks would be running so much faster to make local measurments out there still be c. From the frame of one of the distant stars, my clock at the origin would be ticking very slow, and light would be moving very slow as well.

So I ask, what is wrong with that? :) Is there any objection to this?

One more thing I note, which strikes me as interesting about such a Coriolis metric. Just as a "stationary observer" is not possible beyond the event horizon of a black hole, there would be some value of radius in the Coriolis metric at which a stationary observer wouldn't be possible either. And that would be the same radius in a "real" rotating frame where r*w, the linear speed exceeded the speed of the light for an inertial observer. Any object past that radius would have to be moving in the opposite direction of rotation with a minumum speed to keep the inertial speed less than c.

And so that radius isn't really an event horizon (and that would be the contribution of the gravitomagnetic force which allow moving objects to still communicate with the origin -- but there would be an event horizon for objects not moving at that minimum speed somehow).

Anyway, I see no way in this situation to avoid the (ugly?) notion of having to deal with a speed of light that varies very obviously with radius.

And, again by the equivalence principle, this would the exact view of a "really rotating" Coriolis frame as well -- speed of light increases with radius, no stationary observer possible past r = c/w, and clock rate would increase with r as well (less the SR dilation due to motion, which I have no idea how it effects the total effective clock rates).

-Richard

[Ken G]

Even though 'g' is very small, the differnence in potential is enormous -- which shows that the notion of a "weak" g-field does not mean simple low g.

Right-- it's the potential difference, not the g, that matters.

At any rate, getting back on track, sitting somewhere in a gravity well, I have to use my local coordinates. Yet, if I want to measure things far away, with global extent, then I have to transform the whole global picture into my coordinates.

My point is that this would not be the normal approach. Normally, you would simply not use a single global coordinate system, because the whole spirit of relativity is that the equations of physics are to be expressed locally. This is the rock-- you know the local equations, in the local coordinates. But since you do want a global solution, the trick is to continuously transform coordinates, from local frame to local frame, as you go along. This is what you do when you are doing an integral. So the integral is the global calculation, but the local physics is always the recognizable equations of physics, including c. The hard part is doing the infinitesmal transformations from coordinate to coordinate as you step in space. In the end, what you want to know are local answers, like what frequency will you measure and what direction will it come from. It is not necessary to have a single global coordinate system to get at those answers, and indeed I have argued it is rather awkward to do so (kind of like the awkwardness of the rotating frame global coordinates that you mentioned). But that doesn't mean it's impossible, and maybe there are situations where it is a downright clever thing to do, for all I know. That's why I don't object to using a picture where the speed of light changes in a gravity well, I object to the statement "the speed of light changes in a gravity well" because it lacks the clarifying statements needed to avoid creating more confusion than it resolves.

And if I do that, the speed of light will indeed speed up and slow down with distance, and can see objects moving at speeds greater than c all in my natural local coordinates.

That's true. A classic example of all this is in cosmology. If we take the standard pedagogy, then space is expanding, sometimes much faster than c. Thus the coordinate speed, measured in terms of the rate of change of distance between us and some distant photon, can also be much faster than c-- there are distant photons moving at all kinds of different speeds, if one uses that system of measuring distance. But it is much more elegant to split the problem into two pieces, a local piece where the photons are always moving at c, and a global piece where space is expanding so the photon's location is changing for that reason also. Note there is no "velocity addition formula" for combining those contributions, because the latter is a physical speed but the former is just a coordinate correction and involves no physics in including it. The distinction is irrelevant for the only photons we can actually measure, which are local photons and may unambiguously be treated in our own local coordinates. This is why I feel relativity is at its heart a measurement theory. That shouldn't surprise us, as all of physics is, but my point is, you get ambiguities when you describe processes that go beyond what you can actually measure locally.
(But then there are coordinates I can't use. For example, I can't have a frame moving faster than c locally as that just wouldn't work.)
Why not? It's simple-- just do the physics the normal way, and transform the result arbitrarily in any way you like into any coordinates you like, even some moving faster than c. The answer will still be correct, because you also give the transformation. The only limitation will be that the results you get will not be measurable without first transforming back into the local reference frame where you can actually do measurements. This is the difference between a reference frame, where you can do measurements (and transforms via Lorentz transformations if the observer changes speed), and a coordinate system, which is totally arbitrary. But when you can't do measurements, like the speed of light in a distant gravity well, then you are using coordinates and not reference frames so you can really get away with any description you like that ends up with the right answer by the time you reach the local frame where the measurement is done. If, on the other hand, you wish to restrict yourself to coordinates that are actual reference frames where measurements can be done, then you will not see a change in the speed of light in a gravity well, you will see a change in frequency. That's all you need to get the change in interference that causes the lensing.

[Ken G]

Very interesting stuff Richard, here's my take, although it should not be viewed as authoritative:
Let's consider the (accelerating, non-inertial) Coriolis frame, say of the rotating earth. The stars appear to be moving very fast, orbiting around us enormous speeds, which if far enough away will exceed the speed of light we measure here on earth. Is there any objection to this being a "real" measurement, as it is somehow not local?

Yes, it is not a real measurement, it is a coordinate speed that is inferred from real (local) measurements of photons. By rotating, you are in effect making a coordinate transformation that can yield arbitrary speeds, and if you want to do physics in that frame, you will always in effect be removing that coordinate speed before you apply the true physics. However, general relativity does this for you automatically, so it looks like a theory that applies in any coordinate system. In fact it is just that, but beating in its heart is the transformation back to a local reference frame where you could actually measure the speed of those stars locally. But the same could be said for Newtonian mechanics, it also applies in any coordinate system, it just doesn't come seamlessly equipped with a way to do physics in the arbitrary frame, you will have to do the transformation back to a frame where you can write the equations. Some argue that GR is special because it allows you do to physics in any coordinate system so it means that all coordinates are equally valid reference frames, but I argue that we could always do physics in any coordinate system, and the basic distinction between a global coordinate system and a local reference frame has not gone away in GR. I don't really know that I'm right on this point though, it really stretches my knowledge and I'd rather believe someone who actually does these calculations for a living!

And we can indeed think of that massive rotating mass distribution as "dragging" space-time along with it, meaning space is rotating under our feet so to speak, so who is really rotating? Sort of turns the equivalence principle on its head :)

Again I'd say your understanding of the ramifications of the equivalence principle is at least the equal of mine.
So I ask, what is wrong with that? :) Is there any objection to this?

I don't think there's anything wrong with that, I think it's an inferior pedagogy. A superior pedagogy restricts to stringing together local reference frames in which actual measurements may be made. Then nothing would ever exceed the speed of light. In fact, the standard way would be to eliminate the rotation an infinitesmal distance away from the original observer, that would just unwind the rotation immediately and would be like using the nonrotating frame and just doing the last transformation at the very end of the calculation.

Just as a "stationary observer" is not possible beyond the event horizon of a black hole, there would be some value of radius in the Coriolis metric at which a stationary observer wouldn't be possible either. And that would be the same radius in a "real" rotating frame where r*w, the linear speed exceeded the speed of the light for an inertial observer. Any object past that radius would have to be moving in the opposite direction of rotation with a minumum speed to keep the inertial speed less than c.

Yes, that is a profound realization and I think it is very much in tune with my point about the difference between a coordinate system and a reference frame. It seems to me that distinction must be made, though I've actually not seen it made in so many words in my limited reading on this subject.
Anyway, I see no way in this situation to avoid the (ugly?) notion of having to deal with a speed of light that varies very obviously with radius.

As you can no doubt see by now, I think the way to avoid that is similar to the way it is avoided in the cosmology of an infinite expanding universe-- by distinguishing locally observable speeds from coordinate speeds, where the coordinate system is the clearly elegant "comoving frame". That's a coordinate system that clearly consists of locally achievable reference frames that are continously linked by transformations governed by GR. But I agree that the spirit of relativity does admit all coordinate systems, and all kinds of weird pedagogies. They're not wrong, and may even be useful in the right context, but they are often quite awkward and potentially confusing. It's just like the good old "centrifugal force", which is both bashed and used everywhere.

[publius]

Ken,

Thanks for the comments. You know, my "trouble" as it were, may be that I don't really appreciate the difference between a coordinate system and a true reference frame. And the difference between "inferred" quantities and directly measured quantities. I'll have to think about that, and maybe things will click, the music from 2001: A Space Odyssey will play ("Also Sprach Zarathrutra", I think) and all that. :)

Anyway, speaking of the equivalence principle, and going off on yet another tangent, there's some things that stump my little mind (and I think some greater minds as well). And that's considering charge in free fall vs "really" accelerating.

We all know energy has mass equivalence from the famous E=mc^2. Anything with energy "weighs" something, so to speak, and should have inertia as well. But I didn't appreciate how this applies to EM until I sort of stumbled onto it years ago, and it was via equivalence principle thought experiments.

Some character on a forum, "said with a straight face", that a charged object would not fall at the same rate as an neutral object. That struck me as preposterous that anyone would think that, but his reasoning was that, well, accelerating charge radiates, producing a "reaction force", and that force would act counter to gravity and thus be a type of drag.

Well, that's what got me thinking about the equivalence principle applied to electric charge. :) From the GR view, the free fall path is "really inertial", and (with caveats about a small enough neighborhood in space time and all that), inside a black box in free fall we wouldn't know the difference, would feel no force, and think we were moving inertially. So in that frame, that charge wouldn't be radiating.

A stationary observer in the gravitational field, however, sees that charge radiating. And so where does the energy for that come from? And voila, you realize the charge must have a mass equivalent, and that additional m*g*h energy would account for the radiated energy.

And then, via equivalence principle "transforms", the view of the stationary observer, should be that of some observer riding on an accelerated platform in free space. You can see where my fascination with how Maxwell transforms into acclerated frames comes from. :)

Imagine there are inertial charges floating around. What does the accelerated observer see? Well, from his POV, those charges are accelerating, so he must see radiation ( we would think). Where does the energy for that come from? Well, to be "real" there would have to be some way for that observed radiation energy to interact with him. If it did, then (by handwaving trust in the laws of physics :) ), that energy would have to be supplied by whatever engine was keeping him accelerating.

So apparently we can have "apprarent radiation" vs "real radiation", we consider non-inertial frames. A "pure frame" accelerating, with nothing to interact, would see radiation (again, we would think), but we wouldn't have to worry about the energy because it couldn't interact with anything.

Since all matter contains bound charges, any real acclerated observer could interact with that radiation. For a simple case, consider a charge riding in the accelerated frame. From the POV of an inertial frame, that charge is the one that is really radiating, and giving off real energy supplied by the engine acclerating it. But the accelerated observer wouldn't see his charge radiating, but all the inertial charges.

See how hair-pulling this little thought experiment gets? :) I've rambled on long, and am not done yet (yes, we're going to take charge into the rotating frame, too), so I'll take a breather and continue in another message.

But the above is pretty the accepted classical EM theory view of accelerated frames. Co-accelerating charges don't see each other radiating, but they "think" inertial charges are. The interaction energies and forces between those accelerated observers vs the inertial observers gets tricky, and I have many unanswered questions, really.

By the equivalence principle, we translate the the above accelerated platform picture to a stationary observer standing on the ground in a gravitational field. He sees free-falling charge radiating, and that charge should fall at the same rate as anything else. But if that radiation actually transfers any energy to something, the ground pushing up is what actually does the work! (or apparently so....)

Now, what about charge sitting on the ground with me....... Via the equivalence principle reasoning, it should be the one really radiating.......Where does the energy for that come from......that is the question. Again, the "ground pushing up", the real force would seem to be the one doing the work. But we don't have an inexhaustable supply of energy there. Keeping a mass from falling, the ground is "really accelerating it", keeping it from following the curved geodesic, but that force is like a normal force, always perpendicular to the velocity (as long as the mass remains stationary on the ground), and does no real work. But yet with radiation, real radiation, it would have to be throwing energy off. Is it really radiating, or does GR have some further tricks up its sleeve to prevent it from really radiating. Free falling observers would see that radiation.

So which is the real radiation and which is the apparent?

-Richard

[publius]

To continue with this mess:

We've got the burning question of whether charge sitting on the ground, which by GR/equivalence principle reasoning is "really accelerating", is really radiating. If it is, we seem to have a problem of where the supply of energy would come from. We might argue, that well, there would have to be real energy only if that radiation really interacted with something, but the finite propagation speed of EM "information" rules that out.

[If c were infinite and EM forces propagated instaneously, there would be no radiation as we define it, which means "throwing energy away into space", but we could still have coupled systems transfering energy through space -- they would react to each other instaneously. If I had an antenna with no receivers, the load would appear completely reactive until someone, anywhere, turned on the receiver. The load would instantly develop a resistive component. :) With finite speed, such coupling can still occur, with effects most noticeable in the near field -- by putting a large receiver near a transmitter I can change what the antenna's impedence looks like and therefore make it transfer more (or even less) energy, but because of the finite speed, with no receivers, I'm still throwing energy away -- as the distance between transmitter and receiver grows large, this "cross coupling" goes to zero, and where left with the free space radiation]

So, we can't say that no work would be done unless there was some other charge system for that radiation to interact with because c is finite. Anyway, this is the burning question.

Now let's consider charge in a rotating frame. First consider really rotating charge. Put a moving charge in a static B field and it will "orbit" in a circular path, with a characteristic frequency. But, since it is accelerating, it radiates. That radiation reaction produces a *tangential force* on the charge, sort of a drag so to speak. (And this contraption is how a magnetron tube makes microwaves) If I want to keep it moving in a fixed circular path, I have to supply an additional tangential force, which is analogous to spinning something around on a string with air resistance.

I think you can see where this is going. :) This tangential force is not present in the Coriolis frame in any way. Consider watching a stationary charge from within the rotating, that is another case of an accelerating observer watching inertial charge. It appears to be orbiting around in a circle, just like the case of really spinning the charge around on a string, but yet, since it is not really accelerating, there is no radiation and no tangential force is required.

So this seems to say that the rotating observer watching the inertial charge is not going to see the same thing as an inertial observer watching rotating charge..............

Now, by the equivalence principle, transform that to the "Coriolis metric". The "stationary charge" is equivalent to the charge on the ground. By the above, it looks like we would need to supply an additional tangential force to keep that charge stationary. But the g and B_g fields of the metric have no such tangential forces to balance out (or do they...... :) ). And this tangential force is the same problem as whether the charge on the ground is radiating, requiring more force and real energy from the ground pushing up.

This dilemna can be solved if we "break" the equivalence principle for charge. Anyway I was thinking about all this mess again because I stumbled across a paper (and a serious paper, but it was a pay-per-view and I could only read the abstract), that argued exactly this, and claimed to show that "charge does not follow geodesics in curved space-time", which is a fancier way of making that original preposterous statement that charged objects will not fall at the same rate as neutral objects!

If this were true, then it is acceleration in 3-space alone that determines radiation. The charge on the ground doesn't really radiate, it is the free falling charge that really radiates. And in the Coriolis metric, there would be no radiation and no tangential force for the stationary charge, but there would be a tangential force on the "free falling" charge, and it would spiral and radiate.

So I would love to do an experiment to determine if a charged object falls at the same rate as everything else. But this is far, far, easier said than done. Charge is hard to separate over large distances, and we would need to have no other EM forces present at all to do it, which would be even harder. I mean, get a large charge on something, and its field is going to cause charge redistributions in surrounding objects which are going to produce net forces between them.

EDIT: Here is that paper -- didn't have to pay for it here :)

http://216.239.51.104/search?q=cache:HZH1Q4rJIBgJ:arxiv.org/pdf/gr-qc/0009068+Maxwell%27s+equations+in+non-inertial+frames&hl=en&gl=us&ct=clnk&cd=18

This is way over my little head in the tensor math, but the conclusion is curved space-time causes a "self force", the same as real acceleration, and that self-force pushes the charge off the geodesic. He uses something called "Fermi coordinates", whatever that is.

If this is true, I consider it quite profound. I really want an experiment to confirm or deny this.

-Richard

[Ken G]

This is very interesting stuff indeed. It all comes down to the simple question, does a charge undergoing constant acceleration radiate and therefore receive a reaction force? According to this interesting link:
http://www.mathpages.com/home/kmath528/kmath528.htm
the question is not resolved! My own feeling is that a charge under constant accelertion does not radiate, so that is likely the resolution of the paradox you refer to. (Note the "dipole" radiation formula, involving the square of the acceleration, is only applied to sinusoidal motions-- the more general expressions I've seen all involve the third derivative of position, so are zero for constant acceleration like sitting on the surface of the Earth).
The other interesting question you ask is, does a stationary charge radiate in a reference frame that is moving sinusoidally? Again, I feel the answer to this must be no, despite what one might think from general relativity. I think the twin paradox shows us that even in general relativity, there is no reciprocity principle, so there is no requirement that we make the same predictions regardless of which objects are truly being accelerated. We don't need an EM example, we can just look at a simple rope. If one end of a rope is accelerated, waves appear in the rope. However, if we the observer enter a sinusoidal reference frame, the rope remains straight at all times, and just oscillates in a non wavelike way. Again, there is a breakdown of reciprocity, and although the rope is not governed by the speed of light or the EM equations, I think we still see that there is no requirement for nonaccelerated charges to radiate when seen from an accelerated frame. But I could be wrong, and this is very interesting to think about!

[Ken G]

Now, by the equivalence principle, transform that to the "Coriolis metric". The "stationary charge" is equivalent to the charge on the ground. By the above, it looks like we would need to supply an additional tangential force to keep that charge stationary. But the g and B_g fields of the metric have no such tangential forces to balance out (or do they...... :) ).
I think there's an azimuthal electric force, which provides the energy for the radiation in the rotating frame, rather than being provided by the kinetic energy of the charge as it is in the stationary frame. Of course I'm talking instantaneously. My expectation is that whether or not a charge radiates is not a frame-dependent issue, so if you co-oscillate with a charge, it still radiates. I know general relativity says that the physics of a stationary charge is what it is, so it must just be the role of the weird gravity present that causes the radiation (and also cancels out the radiation from any apparently accelerating but actually stationary charges).
As for a charge leaving a geodesic, it might be that the geodesic itself changes due to the charge, without there being any radiation. For example, a charged black hole has a different gravity, so gravity and charge have a strange relationship that goes beyond electrodynamics. You might argue that this can't rear up when analyzing the motion of a test particle in a field from some other source, but note that a particle with finite charge is not really a test particle, because for a true test particle you should be able to let the mass and charge go to zero proportionally, and still get the exact same motion. But if you let the charge go to zero, I'll bet the effect in the paper you cite also goes away. If I'm right, that shows the particle is not a test particle.

[Jeff Root]

How about if every time Ken G uses the term "pedagogy", he
sends BAUT $1.00 ?

-- Jeff, in Minneapolis

[Ken G]

Agreed, if every time someone makes a statement like "space is expanding" or "space curves", in the sense of "the truth", I get $1. :)

[Tensor]

Agreed, if every time someone makes a statement like "space is expanding" or "space curves", in the sense of "the truth", I get $1. :)

Why? Whether or not a particular person thinks spacetime is curved is just so much philosophical male bovine exhaust. I (along with those who make GR their life's work) have studied it enough to know we can model spacetime as curved. It's the simplest, most elegant model, so why not call it that? Those people are also quite aware of the other different types of models that can be used. I'm not going to preface or append every post I make here with every possible different pedagogy, just to satisfy someone's need to have every possible explanation laid out.

I (along with many others here who are said to proclaim the so called "truth") are well aware of the limits and different possibilities of GR, SR, and QM. But, someone coming to this thread and asking a question such as the "Is space really curved" type question, probably doesn't have the philosophical, physical, or mathematical background to understand all the subtle differences in the different models, or they wouldn't have asked that question in just that way. So why confuse them? That there is curvature (or, currently, more properly warpage) of spacetime happens to be the mainstream position, and this is the question and answer thread, where the mainstream position is supposed to be laid out.

[publius]

Ken,

Yes indeed. I was aware of the open question of whether uniformly accelerating charge actually radiates, and I think I've read that very link you posted before. It looked very familiar. I had forgotten that Feynman was in the no radiation camp. All radiating sources we have involve non-uniform acceleration, from the oscillating currents and voltages of antennas to even electrons changing orbitals. I forget the details exactly, but if you take an electron orbital and calculate the expectation value of the charge distrubution from the wave function, I believe there is *no acceleration* of the charge density (or the contribution of the higher order terms in the radiative expressions kick in and make the sum zero). But, when an electron changes orbitals, there is such an acceleration. And, of course, the "quantum rules" require that acceleration/change occur in the exact same way to maintain the energy/frequency relation.

And more to the point, the classical calculation for the radiation field of an accelerating charge involves *endpoints*, where the charge only acclerates for a given time period in between. I forget the details, but it has to be done this way (the more general way might not be tractable). So this too, is non-uniform.

This seems to scream out for an experiment, but like the free-fall case, it would be very hard to do in practice. To start, you would have to have a non-uniform acceleration, but if you could maintain constant acceleration for a long time, it looks like the radiation would die off if uniform acceleration did not cause radiation. And that long time period is the big problem in actually carrying it out.

Anyway, this and the closely related equivalence principle questions with charge involved are something I would love to see experimentally investigated.

-Richard

[publius]

Ken,

That is a very good point that the "test charge" might change the geodesic in such a way to cancel thigns out so it still follows the new geodesic. Very good point, indeed.

EM fields do make gravity so to speak. They always contain energy and possibly momentum if there is both E and B -- this is another sort of "paradox" or weirdness: Put a capacitor in the magnetic field so that at a region in the center, you have E and B at right angles to each other. The Poynting vector is non-zero, and the momentum content is also non-zero. But the flux of everything sums to zero around any closed volume. This is seems to say energy and momentum are "circulating around" through the space but not going anywhere. :) In the case of the case of the fields of a transmission line, you can find a similiar effect. You can integrate the Poynting vector around certain surfaces to find an enormous energy flux, but that will be cancelled out by other surfaces so the total power out of a surface bounding the source sums to the load's power draw as expected. The standard "shut up and calculate" answer is the Poynting vector's interpretation comes from integrals of closed surfaces, not open surfaces, and so one shouldn't put too much meaning into open surface flux.

But, in the big EFE, Einstein's big bad field equations, one needs to be put the EM energy density and momentum in the stress-energy tensor, and that is going to make gravity. In fact, I think the 4-vector form of the Maxwell Stress Tensor (MST) would just be added right in. The 3-vector MST contains the momentum flux/force terms. The 4-vector form adds the 3 components of Poynting vector to the (i, 0) and (0, j) positions with (0,0) being the energy density, and so is the exact same from as the EFE source tensor -- you just add it right to the "regular" mass-energy terms.

And about the "circulating momentum" case above, I think those terms would act like mass currents and make a gravito-magnetic field as well, sort of agreeing with the view that energy/momentum is circulating around in space. Well, I don't really know for sure, because the momentum picture is more complicated than the energy flow picture, but I think this is the case.

Anyway, I'm rambling again, but the point is you're on track, and EM fields make gravity. And this means, really, when ever we have EM, we have curved space, so the familiar inertial frames we all do Maxwell in *don't really exist*! But forturnately, the effect is very small, and it would take very large EM fields with enormous energies and power to make the effect noticeable.

But it is indeed there. And note it is "one-way" so to speak. EM makes gravity, but gravity doesn't make EM. I think this question was what Einstein was working on at the end (and failed), trying to unify EM and gravity into one common framework.

I can remember the details, but I've read a little about this (I think some call this "Einstein UFT", Unified Field Theory), and here, he tried to complete the coupling circuit so to speak, and have gravity indeed making EM.

If curved space could "make EM" under circumstances, that might well be another way to resolve these questions.

-Richard

[publius]

Ken,

I found this buried on one of my hard drives, which I saved a couple of years ago, from the date. This is some abstract quotes from some papers on this very subject. I hope it doesn't violate the forum rules to post this, but I don't know the source, nor any URLs: Anyway, this shows this equivalance principle business about "who is really radiating" is an open question, indeed. Note here that some indeed hold that radiation is frame dependent, and depends on the coordinate system, which is indeed a little disconcerting.



-----------------------------------------
Classical Radiation from a Uniformly Accelerated Charge, Thomas Fulton, Fritz Rohrlich, Annals of Physics: 9, 499-517 (1960)

5. Does Radiation In Hyperbolic Motion Contradict The Principle of Equivalence?

[...]. The solution to this apparent difficulty is to be found by considering an actual measurement of radiation, using our definition in Section3. Radiation is defined by the behavior of the fields in the limit of large distances from the source. Correspondingly, an observer who wants to detect the radiation cannot do so in the neighborhood of the particle's geodesic. Rather, he must be at a large distance from it, where gravitational fields have different values. The principle of equivalence, however, is a locally valid principle, referring to the geodesic of the particle, whereas the discussion above shows that an observation of radiation is not a local observation.

[...] Whatever gravitational field we introduce for the purpose of comparing it with an inertial field, we must be sure to have a distribution of distant stars which define our inertial systems. This means in particular that any homogeneous gravitational field is necessarily of finite extent, imbedded as it were, in an inertial coordinate system. We remark parenthetically that an infinite homogeneous gravitational field does not exist within the framework of general relativity either. The nonexistence of infinite homogeneous gravitational fields assures us that observation of radiation (observer at large distance from the source) takes place outside the homogeneous part of the gravitational field.

[...] "An electron which falls freely in a uniform gravitational field embedded in an inertial frame will radiate, and one which sits at rest on a table in the same field will not radiate; and these two statements do not contradict each other."

See also Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173


--------------------------------------------------------------------------------

Radiation Damping in a Gravitational Field, Bryce S. DeWitt, Robert W. Brehme, Annals of Physics: 9, 220-259 (1960)

The charged particle tries to do its best to satisfy the equivalence principle, and on a local basis, in fact, does so. In the absence of an externally applied electromagnetic field the motion of the particle deviates from geodetic motion only because of the unavoidable tail in the propagation function of the electromagnetic field, which enters into the picture nonlocally by appearing in an integral over the past history of the particles.


--------------------------------------------------------------------------------

Principle of Equivalence, F. Rohrlich, Annals of Physics: 22, 169-191, (1963), page 173

(C3) An acceleration field is locally equivalent to a gravitational field.13

[...]

If one argues on the basis of (C3) that this situation involves an accelerated charge which should always radiate, the argument is erroneous, because the fact that a charge is accelerated does not necessarily imply that it radiates, unless the acceleration takes place relative to an inertial observer. A noninertial observer uses different clocks and yardsticks. Thus, even though the charge is accelerated, it follows that, because the observer is also accelerated, the co-accelerated observer sees no radiation. Since radiation is not a generally covariant concept the question whether the charge really radiates is meaningless unless it is referred to a particular coordinate system. Finally, since the Schwarzschild metric, locally, for small G, and nonrelativsitically, is identical with the [static homogeneous gravitational field] metric, the above conclusion also holds for a charge at rest as seen by an observer in a Schwarzschild field.

[Footnote page 185: "An inertial frame in which there is a gravitational field present" is meaningless and self-contradictory"]


--------------------------------------------------------------------------------

Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969

Abstract: The connection between an accelerated charge an and one at rest in a (weak) gravitational field is discussed in accordance with the principle of equivalence principle. For that purpose, the fields produced by a freely falling charge and a supported one (i.e. at rest in a gravitational field) are transformed to the rest frame of the observer, who may be similarly supported or freely falling. A nonvanishing energy flux is found only if the charge is freely falling and the observer supported, or vice versa. This agrees with previously established results.

[...]

It may be interesting to discuss the foregoing results from a different view point - that of the photon picture: Is it possible that one of the two observers we have been considering counts a number of photons , while the other, looking at the same charge, does not encounter any of them? In order to answer this question we take the case of a supported charge charge and the supported observer. Projecting the four-potential of the Born field onto the orthonomal tetrad carried by the supported observer, we find that only the fourth component is different than zero. This means that a radiation detector carried by the observer will not record any transitions in which transverse photons are involved. This is the quantum electrodynamical explanation for the absence of radiation from the supported charge. It is not enough that photons are there; to be observable, they must be of the transverse kind, and this property (like the nonvanishing of a magnetic field) is not Lorentz invariant.


--------------------------------------------------------------------------------

Radiation from a Uniformly Accelerated Charge, David G. Boulware, Annals of Physics: 124, 169-188 (1980)

Abstract: The electromagnetic field associated with a uniformly accelerated charge is studied in some detail. The equivalence principle paradox that the co-accelerating observer measures no radiation while a freely falling observer measures the standard radiation of an accelerated charge is resolved by noting that all the radiation goes into the region of space time inaccessible to the co-accelerating observer.


--------------------------------------------------------------------------------

Hawking-Unruh Radiation and Radiation of a Uniformly Accelerated Charge, Kirk T. MacDonald, Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544

Abstract: ...an accelerated observer in a gravity-free environment experiences the same physics (locally) as an observer at rest in a gravitational field. Therefore, an accelerated observer (in zero gravity) should find him(her)self in a thermal bath of radiation characterized by temperature

T = hbar*a/2pck

where a is the acceleration as measured in the observer’s instantaneous rest frame.


--------------------------------------------------------------------------------

[Ken G]

Why? Whether or not a particular person thinks spacetime is curved is just so much philosophical male bovine exhaust.

Please note that I said space, not spacetime. I will grant you that the theory of general relativity invokes the curvature of spacetime, and this is a terrific and successful theory. My statement applies to curvature in space, which is not at all a unique or required element of general relativity but rather of a particular choice of coordinates. Curvature in spacetime is an important theory, curvature of space is an arbitrary pedagogy. This oft-forgotten fact is entirely my point!
But, someone coming to this thread and asking a question such as the "Is space really curved" type question, probably doesn't have the philosophical, physical, or mathematical background to understand all the subtle differences in the different models, or they wouldn't have asked that question in just that way. So why confuse them? .
Because they deserve to know the truth. Many of the people who come here do not come from a materialistic standpoint, they come from a more philosophical stance and they want to know what truths science has to offer. They need, therefore, to be given a clear representation of the difference between a theory and a pedagogy. (There, Jeff, that's $2 for you.)

[Ken G]

Put a capacitor in the magnetic field so that at a region in the center, you have E and B at right angles to each other. The Poynting vector is non-zero, and the momentum content is also non-zero. But the flux of everything sums to zero around any closed volume. This is seems to say energy and momentum are "circulating around" through the space but not going anywhere. :)
That's quite an interesting point, and the role this has in making gravity.
And this means, really, when ever we have EM, we have curved space, so the familiar inertial frames we all do Maxwell in *don't really exist*! But forturnately, the effect is very small, and it would take very large EM fields with enormous energies and power to make the effect noticeable.
You've certainly convinced me that unifying Maxwell and Einstein involves more than meets the eye! I suspect it's well beyond my expertise, unfortunately.
If curved space could "make EM" under circumstances, that might well be another way to resolve these questions.

I agree, although surely Einstein would have been on that track, and it didn't seem to play out. Maybe there is an unturned stone in there somewhere-- for the next Einstein.

[Ken G]

Hi again Richard, this is the one that really strikes my interest:

Radiation from a Uniformly Accelerated Charge, David G. Boulware, Annals of Physics: 124, 169-188 (1980)

Abstract: The electromagnetic field associated with a uniformly accelerated charge is studied in some detail. The equivalence principle paradox that the co-accelerating observer measures no radiation while a freely falling observer measures the standard radiation of an accelerated charge is resolved by noting that all the radiation goes into the region of space time inaccessible to the co-accelerating observer.


That sounds like a significant argument, but I'm not sure I buy it. You are right, it's amazing all the different attempts to get a handle on this. The above one seems to be saying that whether or not you may observe radiation is frame dependent, but whether or not you expect the presence of radiation is not! I would have expected exactly the opposite. Further, he seems to be contradicting Feynman's claim that a uniformly accelerating charge does not radiate, and that makes me doubt his conclusions. I preferred the idea that perfectly uniform acceleration does not radiate, but real acceleration has a start and a finish so does radiate. This question is the most subtle thing I've ever seen on this forum!

[Skyywatcher]

That is a very interestng question. I'd go with mass but very little of it. I'll admit it completely Wina I'm the idiot. Just going on intuition which can be completely wrong.

[publius]

Ken,

Sorry I didn't get back, but I was offline for the past couple of days due to a little EM incident -- lightning (but I didn't notice any gravity that EM made. :lol:) . We had a little cloud come through the other day and a pop came pretty close, knocking out the cable TV. I have a cable modem for internet access. Well, they got the TV back on in a few hours, but the internet still wouldn't work. Signal strength on the downstream channel was so low, the modem couldn't get a lock. They finally fixed it just an hour or so ago -- turned out to be an amplifier, which was still passing the TV freqs, but wasn't doing so hot with the higher frequencies the downstream data signals ride on (upstream is on much lower frequencies, below the TV bands, usually around 30ish MHz -- downstream is up there around 700+ MHz, even as high 1GHz sometimes, which means you need good splitters and cabling to pass that stuff).

I was riding them high about fixing it, and they had told me it would be Tuesday, but I get it was affecting enough people that they got on the ball -- of all the utilities around here, the cable people are the worst.

Now, back to our discussion: Yes, that abstract quote you noticed is interesting. However, lurking there at the last is one that is even more profound when you think about it, but the wording is so bland, you wouldn't notice it. I'll quote it again:


Hawking-Unruh Radiation and Radiation of a Uniformly Accelerated Charge, Kirk T. MacDonald, Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544

Abstract: ...an accelerated observer in a gravity-free environment experiences the same physics (locally) as an observer at rest in a gravitational field. Therefore, an accelerated observer (in zero gravity) should find him(her)self in a thermal bath of radiation characterized by temperature

T = hbar*a/2pck

where a is the acceleration as measured in the observer’s instantaneous rest frame.


An accelerated observer in gravity-free space sees, (drum roll), **HAWKING radiation**! This is where the Equivalence Principle can take you if you follow it through to its logical conclusions. (And consider where the energy for this Hawking radiation comes from --it seems to me it has to come from the engine accelerating the observer, somehow. This would seem to mean there is "additional work" that must be done to accelerate something, if it interacts with that Hawking radiation -- boy, this gets fascinatingly complex)

Now, imagine how that Hawking radiation, which only the accelerated observer will see, will interact with the apparent radiation he would see from inertial charges floating around................... This may be the answer.

And yet another profound question. Is normal Hawking radiation an example of "gravity making EM"? It would seem to be so. If it is, then this effect is quantum, not classical, which seems to suggest the answer to these gravity/EM paradoxes does indeed lie in a quantum theory of gravity.

-Richard

[publius]

Ken,

Einstein was trying to unify EM and gravity, and that (failed) attempt is sometimes called Einstein UFT (unified field theory). The math of GR is well beyond me, but the math of UFT was even more complicated. The geometry of this was to Riemann as Riemann was to Euclid. :) Or I've heard it described that way. Again, the details are beyond me, but I've heard that traditional GR cannot describe intrinsic angular momentum, that is "spin" or angular momentum associated with a point particle (for any lurkers, this is not to be confused with classical "spin", like that of a rotating body such as the earth; GR can handle this fine; it is the idea of "intrinsic" L that is the problem)-- there is an extenstion to GR called "Einstein-Cartan", I believe, which can handle it. As I understand it, the differences between GR and the modifed version are so small they are below the threshold of measurement for anything we can currently observe.

Anyway, UFT was going well beyond this into the "non-Riemannian" geometry, somehow.

There are even conspiracy theories and kook physics associated with this UFT. The conspiracy goes that Einstein did indeed succeed, but the US government covered it up and kept it "black" because it was so powerful, and allowed for all sorts of sci-fi like technology. Of course, the big secret that dropped on Hiroshima became very public soon, so I doubt they could keep anything like UFT under wraps for too long. :)

Note there is nothing unexpected about "EM making gravity". EM has energy and momentum, and goes into the stress-energy tensor source of the field equation. However, it would take some very powerful EM fields to make significant gravity (imagine how powerful a blast it would be if the entire mass of the Earth were converted completely into photons), but the kook-conspiracy belief is that UFT's "additional terms" allowed for making very powerful gravitational fields with much less EM input. And so the military has been using this to make UFO-like stuff, maybe even space and time warps and all that crap.

And finally, there was guy named Kron, Gabriel Kron I believe, who was an electric engineer among other things. He developed some tensor-based description for analyzing the behavior of synchronous machines (ie generators and motors that use a rotating DC field, which are the work-horse of the AC power system). It was pretty elegant, but so complex that it didn't catch on.

But supposedly he went on and came up with something like Einstein EFTs, and claimed there were gravitational effects going on in syncrhonous machines. I don't know enough about it to comment if Kron's stuff is serious or not, but there are some effects, taken to be just random noise, known as "hunting" and other mechanical perturbations. The speed of a synchronous machine is dead-locked to frequency, and "hunting" is slight little deviations in this synchronous speed, sort of like a wobbling tire said to be "hunting tacks" (large deviations in synchronous speed are called "swing", which can be caused by large faults and other failures -- basically, the system looses stability, loosing frequency control).

Anyway, it is claimed Kron tensor framework explained and predicted hunting, and further that it was due to *gravitational effects*. But this is dimissed as just a flight of fancy, of course, but it is something that caught my eye when I first read it.

-Richard

[Ken G]

The plot thickens. By the way, I had a course from Kirk MacDonald-- smart guy. Unfortunately, since I don't understand Hawking radiation, saying that accelerated observers see Hawking radiation doesn't help me too much, but it certainly is stimulating. It sounds like it comes from all directions, whereas I always expected Hawking radiation to emanate from the direction of the black hole, but it sounds like Hawking radiation is only nonisotropic because of the gradient in the magnitude of g, not the direction of g.

[publius]

The plot thickens. By the way, I had a course from Kirk MacDonald-- smart guy. Unfortunately, since I don't understand Hawking radiation, saying that accelerated observers see Hawking radiation doesn't help me too much, but it certainly is stimulating. It sounds like it comes from all directions, whereas I always expected Hawking radiation to emanate from the direction of the black hole, but it sounds like Hawking radiation is only nonisotropic because of the gradient in the magnitude of g, not the direction of g.

Ken,

Well, I'll be darned; "small world" as they say. If you know him, maybe you could call him up and ask him what he thinks about all this business. :)

I don't understand Hawking radiation either. And while I have never really thought about, I would've assumed it was coming from the direction of the black hole too, plus would've thought you needed an actual black hole to make it.

However, this seems to imply that all you need for Hawking radiation is a gravitational field (or accelerate yourself by the Equivalence Principle).

Or maybe rather than saying you need a black hole, maybe what you need is an *event horizon*. In the case of an accelerated frame, there is an event horizon. From an inertial frame watching the whole show, there is a region beyond which light could never reach the accelerated observer (if he continued to accelerate). The light would get asymptotically close, but never quite reach it. The distance to that event horizon is frame dependent -- there's Lorentz contraction and all that.

Now, the Earth below us has no event horizon, but standing on the ground, we are equivalent to an acclerating observer locally. Do we see Hawking radiation as well? I guess the answer is in the question does the gravitational field itself tell us whether there is an event horizon lurking below us somewhere? :) But taking the Equivalance Principle "at its full word" so to speak, the answer should be yes, we should see Hawking radiation (not in free fall) in any gravitational field.

That would mean the "Coriolis metric" would have Hawking radiation, as well as rotating frame........

Another question I have is if EM radiation implies gravitational radiation. That is plug in the stress-energy of a propagating EM wave into the GR field equation. Does this yield gravitational wave solutions, and does it always give such wave solutions? (any such gravitational waves would be vanishingly small, but are they always non-zero?).

Now, if Hawking radiation is present for any gravitational field, there would be such EM radiation associated with any gravitational wave (but again, vanishingly small). This would then complete the coupling between EM and gravity. Gravitational and EM waves would go hand-in-hand.

-Richard

[Kaptain K]

Anybody else notice that it is "Hawking-Unruh Radiation" that is mentioned in the Kirk T. MacDonald paper, not "Hawking Radiation". I don't pretend to know enough to know if there is a difference, but I would assume that there is a difference.

[publius]

KaptainK,

I just Googled on this. The notion that an accelerated observer will see radiation goes all the way back to 1976, and was first proposed by a guy named Unruh from quantum-thermodynamic considerations. The ground states of a true quantum vacuum for an inertial observer will appear to be excited states to an accelerated observer somehow, and so should emit thermal radiation from his POV. Hence, "Unruh radiation" is the name given to this acceleration radiation.

Hawking radiation means the black hole/gravity radiation. By Equivalance Principle reasoning (including the sort of surprising result that an accelerating observer sees an event horizon), these two should be the same, and so the names were combined.

-Richard

[vkp]

According to Heisenberg's uncetaininty principle deltax.deltap>=h/4pi

Since we cannot determine the exactly position and speed of a particle lying in atomic range simultaneously, how can this be applied to light ?

Light is itself composed to this atomic particples. :confused:

[Ken G]

Since we cannot determine the exactly position and speed of a particle lying in atomic range simultaneously, how can this be applied to light ?

You are right that the uncertainty principle also applies to light. It doesn't contradict the Unruh radiation idea, the excited states in electrodynamics are taken as infinite plane waves. It is just a convention, other states can also be formed as superpositions of these.

[daxloves]

So I gather that the answer to the origional question of this post is that photons have no rest mass, yet have mass while in motion. A photon can not slow down or stop so is allways in motion. So wouldnt that mean that they allways have mass?

I read someware that if you could trap light in a container it would increase the mass off the box while not actually having mass itself. Isnt our universe kinda like a container, with all the light within it conributing to the mass of the total?

[Ken G]

So I gather that the answer to the origional question of this post is that photons have no rest mass, yet have mass while in motion. A photon can not slow down or stop so is allways in motion. So wouldnt that mean that they allways have mass?
It means they always have the mass equivalent of energy, but the typical terminology assumes the word "mass" means rest mass.
I read someware that if you could trap light in a container it would increase the mass off the box while not actually having mass itself. Isnt our universe kinda like a container, with all the light within it conributing to the mass of the total?
Yes it is, but again the issue is what do you mean by mass. It would add to the inertia of the box, and the gravity of the box, so in that sense it is mass, but these are both really the mass equivalent of energy. The rest mass is a bit different, and that's what the photon doesn't have.

[daxloves]

Yes it is, but again the issue is what do you mean by mass. It would add to the inertia of the box, and the gravity of the box, so in that sense it is mass, but these are both really the mass equivalent of energy. The rest mass is a bit different, and that's what the photon doesn't have.

What I mean by mass is something that contributes to the mass of the universe as a whole. Is it posible that the inertia and gravity contributed by light/photons is in part the missing dark matter of the universe? Posibly a small part but a part none the less?

It is difficult if not impossible to find a place in the universe with no light present. If all the space between all objects in space is filled with light from all directions from all sources, how much gravitational mass would light have in the universe?

[Ken G]

What I mean by mass is something that contributes to the mass of the universe as a whole. Is it posible that the inertia and gravity contributed by light/photons is in part the missing dark matter of the universe? Posibly a small part but a part none the less?
Yes, it counts in the gravity of the universe, though not as "cold dark matter"-- light is not cold, nor dark. But more to the point, light energy dwindles quite significantly when you expand its "container", so there is very little light energy left in the universe, relatively speaking. It's not a significant contributor.

[daxloves]

So then light was figured in when they measure or weigh or whatever they did to figure out how much of the universe is a form of dark matter?

Also, what do you mean by "light energy dwindles quite significantly when you expand its "container"" I had assumed that light was fairly constant. Expanding its container may make it more diffuse but shouldnt make it any less there. Havent we found traces of the first light from the big bang? Like an echo so to speak. Isnt that light still out there, trveling on the edge of our universe?

Im sorry for all the questions. Im just currious.

[Ken G]

So then light was figured in when they measure or weigh or whatever they did to figure out how much of the universe is a form of dark matter?
Yes, but it's insignificant.

Also, what do you mean by "light energy dwindles quite significantly when you expand its "container"" I had assumed that light was fairly constant. No it isn't. Rest mass energy doesn't care if you expand the box, but light energy (and kinetic energy in general) dwindles rapidly. In a real box, that's because the light transfers energy to the walls, but in the expanding universe, one generally just says the light, including the CMB, redshifts.

[gnickol1]

Just as anything else in history time tells. It may be based off of an equation of mass that is yet to be identified. Photon's obvesily exsist in light, but without value and the constent speed indicates other wise. I personally beleive one of the biggest break throughs in the future will be identifying and using the properties of light mass. Chances are it may be in another lifetime, but future time travel depends on it!!!

Sincerely,

-Who Really Knows

[e129745]

OK, if light has no mass, then how might light energy convert to heat energy as its line of travel is impeeded and redirected by an object? And please speak to the interesting thought of lowest energy state yielding an equalibrium, no added energy equals no motion I.e., a black hole being at its lowest energy state. Is this possible? Another thought is if a particle, oops, forgot...no mass...if a photon takes a longer path from point A to point B by bumping into another?, then it is shedding energy or momentum, adding this force to the other, yet is effected by photons traveling with it on the mean path, and so on. It gives up and then receives. Essentialy remaing constant velocity as an average. It seems that as light energy converts its kinetic to heat and transfers kinetic to what it strikes along the way, it would eventualy convert all its energy. And how many photons are neccassary for the average human eye in complete darkness to register it as light? I would guess that as the quanity of photons traveling in a finit group would raise the % of a few traveling farther in distance, due to statistical reasoning.

I won't bother you folks further as it is quite obvious I am not educated to all of your level.

Thanks,

Patrick

[mugaliens]

OK, if light has no mass, then how might light energy convert to heat energy as its line of travel is impeeded and redirected by an object?

How does your hand impart heat to a cool object you pick up without transferring mass to it?

Or how does your radiator impart heat to you?

Light is energy. Depending on the frequency of the light, and the characteristic properties of the materials it encounters, one of several things happens:

- transparent - (glass) the light passes right through it
- translucent - (wax paper) the light is scattered by it, little to a great deal of it is absorbed
- reflective - (mirror, polished metal) the light is reflected uniformly off it's surface, little to a great deal is absorbed
- scattered - (carpet) light is reflected in a non-uniform manner off it's surface, little to a great deal of it is absorbed

Light, being itself energy, does not require mass to impart itself as energy to objects that do have mass. Upon interacting with that mass, light's energy usually changes, most often to thermal energy, but occasionally to electrical energy (photovoltaics) and chemical energy (chlorophyll).

[AonSao]

Scientists have actually been able to stop light completely in a controlled environment (for very short periods of time). From those experiments, we know for a fact that light has no mass when it is stopped relative to the observer.

I think this is the link: http://www.hno.harvard.edu/gazette/2001/01.24/01-stoplight.html

Edit: that link doesn't actually say anything about mass, but I've read a similar article that did.

[swansont]

Scientists have actually been able to stop light completely in a controlled environment (for very short periods of time). From those experiments, we know for a fact that light has no mass when it is stopped relative to the observer.

I think this is the link: http://www.hno.harvard.edu/gazette/2001/01.24/01-stoplight.html

Edit: that link doesn't actually say anything about mass, but I've read a similar article that did.

That's not really "stopped light" — it's a horrible pop-sci term to use. The light was absorbed, not stopped. The way-cool thing, though, is that it was absorbed in a special way so that the original pulse could be recreated, which makes it different from most examples of absorbed light.

[AonSao]

hmmm... absorbed? Well if you put it that way... wouldn't the mass of the material absorbing the light increase? After all, when you turn a flashlight on, it slowly loses mass.

[antoniseb]

hmmm... absorbed? Well if you put it that way... wouldn't the mass of the material absorbing the light increase?


Yes, A nucleus absorbing a gamma ray gains mass. That does not mean that a photon has rest mass.

[Smoke Ring]

If photons are massless, how do we explain this equation, M = fh/C^2 derived from fh = E = MC^2?

[mugaliens]

If photons are massless, how do we explain this equation, M = fh/C^2 derived from fh = E = MC^2?

The rest mass of a photon is 0. But photons are not at rest. The difference between the two is one of energy. However, through Einstein's E=mc2, the energy of a photon also has mass and mass-effects (gravity). Due to the balance of the equation, where a tiny mass contains a boatload of energy, however, the amounts of mass associated with laboratory uses of light are inconsequential. Since we know how to calculate them (we know an energy's mass-equivalence), if we ever need to do so, that's what we do instead of trying to measure them.

[FP name TBD]

From what I understand, If light were not to have a mass formulas for values such as kinetic energy could not be correct. Whith photoelectricity, a photon colliding with an electron must have both a frequency which exceeds that of the threshold frequency, and Kinetic energy great enough to exceed the work function. If the photon didnt cary kinetic energy it could not cause the electron to absorb it and escape from the metal.
As the formula for kinetic energy is 1/2 x MASS x Velocity^2 , the photon must have some mass for it to have a kinetic energy greater than zero.
The mass of a photon must be very small, and we are most probably unable to detect it at the moment; however, I do not doubt its existance.

[Jeff Root]

TBD,

All measurements so far are consistent with photons having zero mass.

Photons have kinetic energy and momentum. The energy equals Planck's
constant times the photon's frequency. The momentum equals the energy
divided by the speed of light, or Planck's constant divided by the photon's
wavelength.

The full expression for energy is E2 = p2c2 + m2c4 . ("p" is momentum.)

Since photons have zero mass, the second term of the expression is zero.
The Newtonian formula you quoted only applies to objects having mass.

-- Jeff, in Minneapolis

[Fiery Phoenix]

I remember Pamela saying light doesn't have mass in one episode of Astronomy Cast. Fraser then asked her why it doesn't, and she replied with E=MC^2. I was never really sure what she meant; in other words, I still have no idea whether or not light has mass.

[Jeff Root]

Fiery Phoenix,

E=mc^2 gives the energy of any mass. Light does not have mass, but any
quantity of light will have an equivalent mass which can be calculated and
used in Newton's gravity law just as if it were mass. The formula is fairly
simple for a ball of completely incoherent light, and much more complex for
something like a beam. However the equivalent mass is so miniscule that
only in cosmology would one ever need to calculate it, and even then only
for the earliest moments of the Big Bang.

An article in the May 1976 issue of Scientific American titled "The Mass
of the Photon" by Goldhaber and Nieto explains the history of attempts
to measure photon mass, and says:

Limits on the mass of the photon have been improved by a
factor of a billion since Robison's first test of Coulomb's law
more than 200 years ago. The best limits based on direct
measurements are deduced from the failure of experiments
to detect any exponential decay in large magnetic fields.
Measurements of Jupiter's magnetic field, obtained by the
Pioneer 10 spacecraft ... imply that the mass of the photon
cannot exceed 8x10^-49 gram.
It also suggests observations of the Galaxy's gravitational field
which could reduce the maximum possible mass even further.

Why does light not have mass? Why should it? Some particles have
mass, some don't. Some particles have electric charge, some don't.

-- Jeff, in Minneapolis

[Fiery Phoenix]

That explains it. Thanks, Jeff. :)