On gravity and time delay

[grav]

This is more of a model-in-the-making than a working hypothesis at this point. But I think it represents the only real possible model for gravity that is allowed within the context of what I have learned so far in this forum, incorporating both Newtonian mechanics and GR. So of course I need help building it up and/or tearing it down.

Here's the way my thinking goes so far. If gravity were transferred through a "stationary" ether, one could think of it as a circle that spreads out in all directions from the point of emission over time. The time delay from the point of emission to another body would offset the direction of gravity from the current position of a moving "emitting" body to the actual point in space in which it was emitted. This, however, does not occur, and so gravity appears to be instantaneous from the current position of the emitting body, although this would require it to travel at up to billions of times the speed of light. Obviously, this is not the case, either. So what is happening here?

The only solution I see, and it seems so simple now, is that gravity travels with the additional speed of the emitting body. It would be like tossing a ball up in a moving vehicle. If the ball did not travel with the additional velocity of the vehicle, but only in relation to the point at which it was tossed up, it would fly straight back through the back window. It does not.

But in this case, the resultant must still remain c. If two bodies were in orbit around each other, the time delay would offset them from being directly across from each other when the gravity from the point of emission is felt. If the gravity moved in direct relation to an ether, like sound does, the direction of gravity would be offset at an angle from the other's current position but travelling at c. With the additional velocity of v in the forward direction, however, the gravity will appear to come directly from the other's current position (if the emitter's direction doesn't change), but with a speed of (c^2-v^2)^.5 . The ratio between this and the original speed of gravity is c/(c^2-v^2)=1/(1-(v/c)^2)^.5=1/gamma. It would be as if the force of gravity travelled directly from the emitter instantaneously, since it would receive the force from the current direction of the emiiting body at the same time that the emitting body is just getting there. This would also make gravity more particle than wave. Particles of gravity would be given the extra velocity of the emitting body and would not refract or interfere with itself (much) as waves would.

So the speed of gravity would appear to travel at c/gamma, but now we must figure in time dilation, which would be different than the way one might think of it in GR, but I'm not sure yet what the correct way is exactly yet, and the mathematics comes out the same, so I will not worry about this at the moment. So c/gamma*gamma=c for the apparent speed of gravity from the current position of the emitter. But that would make it v*gamma in the perpendicular direction, directly along the line of travel of the absorbing body. But this works out as well. For small v in the direction of travel, gravity affects the absorbing body almost as normal. But let's say the absorbing body is travelling at c, or is light. c*gamma makes for zero gravity in the line of travel. So light is not affected in this direction. It will continue at a constant speed. But it is affected perpendicular to this, since the speed in this direction is zero, and so as normal gravity is felt. This will change the path of light but will not slow it down or speed it up. As the path changes, the orientation of the axes change, and so gravity can then affect light from its original direction, but now not in its new direction. Since the speed of light will remain constant, it will not speed it up or slow it down radially, but will change its path at twice the Newtonian rate for gravitational lensing. For matter travelling directly radially from a gravitating body, the force will be felt as F*gamma. For light, it will be zero. But a time dilation will be experienced in the process as well. For light, this translates into a redshift.

For an absorbing body orbitting a stationary emitting one in a circular orbit (both emit and absorb, by the way. This is purely for visual effect), the force is always directed perpendicular to the absorbing body, so gravity is felt as normal, with no apparent time delay. But highly elliptical orbits would be an entirely different matter indeed.

[EDIT-I noticed that my vector analysis for the apparent speed of gravity is slightly off in this thread but I will leave it for now since it will take time to work it out, especially with time dilation as a factor.]

[peteshimmon]

It would be nice to find a physical model of
which GR is an exquisite mathematical description
but I think the mathematicians are going to win
this one. Its fun to think that light passing
gravtating matter is being refracted and indeed
it does slow down while passing. But a medium
is a no no for the experts. And I have pointed
out that the advance of Mercury is indicating
a flux emmanating at c from the Sun. The effect
is due to abberation I read somewhere. So good
luck! I just hope to think of a good experiment
one day.

[RussT]

It would be nice to find a physical model of
which GR is an exquisite mathematical description

If you mean...'of gravity', yes, this would be the unification of GR and QFT.


I think the mathematicians are going to win this one.

No doubt that this will be the ultimate proof, but NOT until they are working on the correct model with the proper associations and relationships!


Its fun to think that light passing gravtating matter is being refracted and indeed it does slow down while passing.

The speed of light in a gravity well should be considered separately from all other EM interactions!


And I have pointed out that the advance of Mercury is indicating
a flux emmanating at c from the Sun.

Where have you pointed this out? I would be interested in seeing more on this. If though, you mean by 'flux'...light (photons) pressure traveling at 'C', this is not the answer IMHO. (The concept is right, just not through photons)

[peteshimmon]

I cannot remember where unfortunately but
Trinitree seemed to remember something as well
on these lines. It was a popular textbook
giving this indication of why the advance
happened. Matter speeds up in a gravity well,
light slows down. I have fun thinking if
a mass has to increase speed in a rarefied
energy field to keep its energy content
constant whereas light as a waveform finds it
slower going. OK naughty...but fun!

[Flaney]

But that would make it v*gamma in the perpendicular direction, directly along the line of travel of the absorbing body. But this works out as well. For small v in the direction of travel, gravity affects the absorbing body almost as normal.

I like working in barycenter coords. Two masses in orbit are always on opposite sides of the Center of Mass.

Also, using a force model introduces abberation effects. Try using a Potential model instead. Any time delays from a moving potential between the masses always seem to back them up such that they remain normal across the barycenter.

See if that helps.

[publius]

Grav,

I've been meaning to say something here, but was too busy in the Geocentrism thread. :)

Here is a paper by Steve Carlip on the "speed of gravity", something he wrote specifically in repsonse to Van Flandern's (erroneous) arguments, that shows how GR gravity "extrapolates" acceleration as well as velocity, better cancelling the propagation delay.

http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9909/9909087.pdf

The math here is very heavy, using 4-vector/tensor stuff, so it may not make much sense. Just think of it as reason to learn the math! Anyway, he starts by showing how EM force "extrapolate" only the velocity, then goes on to show how gravity adds acceleration terms in it.

Something else very important and very fundamental about how GR works. The source term, the stress-energy tensor is required to conserve momentum and energy. What if you just specify a point mass accelerated at some constant rate? Well, that is not conserving momentum and energy. The GR field equations would result in garbage.

So to properly solve the GR field equation, you have to put in what is causing the acceleration, so momentum and energy are conserved. This is very important. Say you have a rocket accelerating. GR requires the contributions of the exhaust being thrown out as well as the rocket itself. So, to get the metric for an accelerating mass, Carlip had to use the "photon rocket" metric. This is mass shooting off EM radiation to propel itself. The relative mass (by E = mc^2) thrown out the exhaust is very small relative to the rocket mass here, so most of the gravity would "stay with the rocket".

That gives a gravitational field as close in GR as you can get to a uniformly accelerating mass.

You'll note he derives an expression showing the acceleration field points at r + v(r/c) + 1/2 a (r/c)^2 roughly.

r is the retarded position, and r/c is the light travel time, and so the above is an extrapolation of acceleration as well as velocity.

What is "emanating" from the particle (and it's exhaust!) are "waves of space-time curvature". The metric describes this curvature.

-Richard

[grav]

It would be nice to find a physical model of
which GR is an exquisite mathematical description
but I think the mathematicians are going to win
this one. Its fun to think that light passing
gravtating matter is being refracted and indeed
it does slow down while passing. But a medium
is a no no for the experts. And I have pointed
out that the advance of Mercury is indicating
a flux emmanating at c from the Sun. The effect
is due to abberation I read somewhere. So good
luck! I just hope to think of a good experiment
one day.Thanks, peteshimmon. But if you choose not to think of it as a medium, it could be considered the "fabric" of space-time, or just the gravitational field. They are all the same to me, at least until it can be shown that there is only one way to think about it. It is really just the "mechanism" I am after (for now), which can probably be found through the precise mathematics of the propagation delay and so forth. It should at least provide a better understanding of what we are really dealing with.

Also, I suppose the precession would indeed be due to abberation of some sort. Either due to the spin of the sun or to the revolution of the planets, or perhaps both. If the sun were not spinning, nor the planets revolving around it (each are static somehow), I suppose no such precession would be observed. A good test of which way it should be is to observe an object (or planet) revolving in the opposite direction as the spin of the sun (or Earth) and to see which way it precesses.

[grav]

Originally Posted by peteshimmon
I think the mathematicians are going to win this one.

No doubt that this will be the ultimate proof, but NOT until they are working on the correct model with the proper associations and relationships!
Precisely, RussT. Thank you for that.

[grav]

I like working in barycenter coords. Two masses in orbit are always on opposite sides of the Center of Mass.

Also, using a force model introduces abberation effects. Try using a Potential model instead. Any time delays from a moving potential between the masses always seem to back them up such that they remain normal across the barycenter.

See if that helps.Yes. I think you and I are going to get along just fine. This is exactly how I like to work this sort of thing out as well. The way I figure it, the masses on each side of the center of mass must always be exactly opposite of each other regardless of the precise mathematics of Newtonian or Einsteinian physics. I cannot be 100% sure about varying masses, but two identical masses would each observe identical forces in an identical manner, and so there can be no doubt that they lie precisely opposite each other and equidistant from the barycenter at all times.

I am not quite sure what you mean by a potential model, however, but I think I may have an idea. It appears to be the only way this can all work out. If I consider that gravity is travelling from some point on the orbit prior to the current position at the speed of light, so that it reaches the other body at a distance of ct in the same amount of time, then it would be travelling through an ether like sound. This doesn't appear to work out. A vector of c as the resultant gives v in the line of motion and c*gamma apparent from the current position. If I consider that the entire field continues to move in a straight line with the velocity of v from the prior position, the vector from the current position lines up with the other as c (except for a slight miss for an orbit), but v in the line of motion still exists, and the resultant is now c*(1+(v/c)^2)^.5 . The only way to get rid of the extra v vector, as far as I can tell, is to consider the field itself as stationary regardless of any actual relative motion through space, but I am not sure if this should be the case, unless the field lines themselves have some sort of effect on each other that make them appear this way.

[RussT]

it could be considered the "fabric" of space-time, or just the gravitational field.

Yes, this is the correct way to think about it. Now, it just needs to be understood that that is a 'background gravity field' which is DM, and is all of 'space'!

In the paper Publius linked above, it explains how to show the aberation (or cancel it), eliminating Van Flanderns 'instantaneous' gravity.

But way down towards the bottom, he also talks about GR still not being precise enough, and that something else has to come along to show the relationships precisely. In this section he also says this...
(with an extra interaction somehow added on to explain the gravitational radiation reaction) bold mine.

So, if you adhere to Einsteins 'ponderable matter distorts space/time' by causing a gravity well in that space/time, then there is no 'gravitational radiation', and DM is the background gravity field that the ponderable matter is distorting. Now, if that DM background gravity field is traveling at 'C', but does not interact with the baryonic matter, then you have gravity propigating at 'C' and you just substitute DM for the currently used 'gravitational radiation', and it should work precisely, I think:think:

[publius]

Grav,

By "potential model", he means, well, what Carlip was using (although the most elegant 4-vector type of potential) in his paper there.

The way to think of the "stuff" emanating from the source particle (at least for EM) is to think of spherical waves. Think of little circles travelling out from the particle moving at c. Now, they always move at c relative to whatever inertial frame you are using to specify your sources for a Maxwell solution.

The potential approach makes this all easier. You can think of the potential as a function of space and time that depends on the sources. *Derivatives* of this function, both spatial and temporal, give the fields, and then the forces.

You see a source charge moving, and have specified its trajectory in some coordinate system. Solve Maxwell in terms of the potentials. You can see those solutions as little waves of "information" enamating from the source as little spherical waves. That information "updates" the field at any point when it gets there.

In GR, all that gets more complex, but the metric is actually like a potential. The "field" (which is a frame dependent thing very much so) involves (complex tensor) derivatives of that potential. But this "potential", the metric has the meaning of defining the geodesics according to giving the ds^2 interval, and the easiest thing to do is just use that to get equations of motions directly and not worry with the field picture.

Potentials are almost always much, much easier to solve for and work with than the fields themselves.

-Richard

[grav]

publius,

New avatar, eh? Who is that?

Thanks for the link. I am not sure if it verifies what I have been working on or not. I will have to break it down further. I have tried working with orbits on my similation program, but I keep forgetting that the units for time integrals that I must use to make it smooth and continuous enough in order to measure the difference between Newtonian physics and GR would probably take longer to run than it would just to observe some actual orbits in the first place. :( So I am forced into pure mathematics and I have just spent the last couple of days staring at a blank sheet of paper trying to determine where to begin. I decided on what Flaney mentioned, finding the offset for two identical masses travelling in a perfect circle. I also had to use a circle because there is no way to formulate the perimeter of an ellipse so that I can backtrack to find the prior position.

If we first place the current positions of the two masses at (0,r) and (0,-r) on each side of the circle with its center at the origin, then we can backtrack from the point (0,-r) to the prior position at (x1,y1). Then, drawing a line straight out from that point tangent to the circle, in the same time, we come to (x2,y2). From that apparent position, the field would travel to the other mass at c in time t. So now we can make some mathematical observations. First, x1^2+x2^2=r^2. Next, (x1-x1)^2+(y2-y1)^2=(vt)^2. Since the second point is on a straight line from the first, tangent to the circle, (vt)^2+r^2=x2^2+y2^2. Also, the distance between the second point and the other mass is (ct)^2=x3^2+(y2-r)^2. Working through these, I find that ([(vt)^2+r^2-u]^.5-x1)^2+(u-[r^2-x1^2]^.5)^2=(vt)^2, where u=[(ct)^2-(vt)^2-2r^2]/2r. I have no idea yet how to work this out further, but one can see that if we extrapolate for x1, we would end up with exponents of (v/c) from zero to four, or five orders ( (v/c)^0 to (v/c)^4 ). Is this similar to what that link demonstrates?

[publius]

But way down towards the bottom, he also talks about GR still not being precise enough, and that something else has to come along to show the relationships precisely. In this section he also says this...
(with an extra interaction somehow added on to explain the gravitational radiation reaction) bold mine.

So, if you adhere to Einsteins 'ponderable matter distorts space/time' by causing a gravity well in that space/time, then there is no 'gravitational radiation', and DM is the background gravity field that the ponderable matter is distorting. Now, if that DM background gravity field is traveling at 'C', but does not interact with the baryonic matter, then you have gravity propigating at 'C' and you just substitute DM for the currently used 'gravitational radiation', and it should work precisely, I think:think:

Russ,

You're misreading that. This paper was in response to some of Van Flandern's stuff, and that particular paragraph about "something extra" was the way it would have to be for Van Flandern's gravity idea to hold.

Van Flandern says gravity propagates much faster than c, but paradoxically agrees that gravitational radiation itself propagates at only c. And he reconciles that by saying that the "force of gravity" and gravitational radiation "forces" are not quite the same thing someway somehow. This is what Carlip was talking about, and it is not apparent unless you know the larger context of the Van Flandern argument.

So Carlip is saying this essentially: GR gravity has the propagation delay built right in. When there is a time varying quadropole moment ( which can roughly be thought of as "jerk" or changes in acceleration), GR predicts the "forces" will "miss" then, and that is gravitational radiation.

But for Van Flandern, if he wants a model of gravity that is near instantaneous, travelling orders of magnitude faster than c, then he has to explain the observed gravitational radiation "missing" effects by some other means, because a faster gravity would predict much less radiation.


-Richard

[publius]

Grav,

Who is that? That is Oliver Heaviside, one of the founders of Classical EM theory. I put that up to have some fun with our Geocentrists. One of them, in his zeal to attack Relativity, had been visiting the fever swamps of kooky EM theory websites. For reasons too involved (and feverish) to get into here, Heaviside is considered the enemy, one who helped surpress the truth about EM. You can thank him in particular for the modern, beautiful form of Maxwell's Equations today. I would call them Heaviside-Maxwell for this reason, actually.

And as you can see from that old photo, Heaviside does not look like a man one would want to mess with. Put a worn cowboy hat and old trail duster on him, with perhaps a wad of chaw or a crudely rolled cigar in his mouth, and he'd be right at home riding with Clint Eastwood in a Sergio Argonnes flick.

I like to see a little showdown on the streets of Dodge with him and Tom Bearden, one of the biggest anti-Heaviside ranters.

"I hear you been talkin' 'bout me, boy....", he might say as he flipped back his trailer duster, revealing his revolver, strapped low at hand level on his thigh..................he'd spit some chaw juice right before he plugged him between the eyes............. :lol:

-Richard

[grav]

You been watchin' taw many westerns. ;)

[RussT]

Russ,

You're misreading that. This paper was in response to some of Van Flandern's stuff, and that particular paragraph about "something extra" was the way it would have to be for Van Flandern's gravity idea to hold.

Van Flandern says gravity propagates much faster than c, but paradoxically agrees that gravitational radiation itself propagates at only c. And he reconciles that by saying that the "force of gravity" and gravitational radiation "forces" are not quite the same thing someway somehow. This is what Carlip was talking about, and it is not apparent unless you know the larger context of the Van Flandern argument.

So Carlip is saying this essentially: GR gravity has the propagation delay built right in. When there is a time varying quadropole moment ( which can roughly be thought of as "jerk" or changes in acceleration), GR predicts the "forces" will "miss" then, and that is gravitational radiation.

But for Van Flandern, if he wants a model of gravity that is near instantaneous, travelling orders of magnitude faster than c, then he has to explain the observed gravitational radiation "missing" effects by some other means, because a faster gravity would predict much less radiation.


-Richard


Okay, I see what you are saying here.

[But way down towards the bottom, he also talks about GR still not being precise enough, and that something else has to come along to show the relationships precisely.]

But with this...I am still suggesting that the below 'could' work out to be possible.

So, if you adhere to Einsteins 'ponderable matter distorts space/time' by causing a gravity well in that space/time, then there is no 'gravitational radiation', and DM is the background gravity field that the ponderable matter is distorting. Now, if that DM background gravity field is traveling at 'C', but does not interact with the baryonic matter, then you have gravity propigating at 'C' and you just substitute DM for the currently used 'gravitational radiation', and it should work precisely, I think

[Flaney]

publius,

...
I decided on what Flaney mentioned, finding the offset for two identical masses traveling in a perfect circle.

If we first place the current positions of the two masses at (0,r) and (0,-r) on each side of the circle with its center at the origin, then we can backtrack from the point (0,-r) to the prior position at (x1,y1). Then, drawing a line straight out from that point tangent to the circle, in the same time, we come to (x2,y2). From that apparent position, the field would travel to the other mass at c in time t.

I like to think that potential "fractions" emanate from each mass. When this fraction encounters the other mass, complete information or total potential is available and will be used by the originating mass for force calculations once it has returned home. This makes the gravitational interaction a two way trip.

Since potentials are being used, the math is scalar based simply on the radial distance and size of the masses. Thus the force computed will remain completely radial. So, the needed velocity components of the masses are their radial ones. Use these to "back up" the initial separation distance from the propagation delay to get the effective radius of action.

An interesting observation I note in low velocity, weak field analysis is that regardless of the individual radial velocities, at the emission from the effective radius, the potential fractions return to both masses at the same time at their current positions (the "reflections?" however do not occur simultaneously).

[grav]

I like to think that potential "fractions" emanate from each mass. When this fraction encounters the other mass, complete information or total potential is available and will be used by the originating mass for force calculations once it has returned home. This makes the gravitational interaction a two way trip.

Since potentials are being used, the math is scalar based simply on the radial distance and size of the masses. Thus the force computed will remain completely radial. So, the needed velocity components of the masses are their radial ones. Use these to "back up" the initial separation distance from the propagation delay to get the effective radius of action.

An interesting observation I note in low velocity, weak field analysis is that regardless of the individual radial velocities, at the emission from the effective radius, the potential fractions return to both masses at the same time at their current positions (the "reflections?" however do not occur simultaneously).Yes. That is what I do, too. I back it up using the orbital velocity and then project it in a straight line for the same amount of time and at the same velocity. But I do that to where the new apparent position of the field will travel the distance to the current position of the other body at c (like an expanding spherical wave, as publius said) and in the same time. But I'm still not sure what you mean about the potential field exactly. It seems like if the field is reflected off of the other body and back again, it should indeed interact with the initial body at its current position, but this would be true regardless of the calculations used. So I'm not sure I follow. Then I thought maybe you meant both fields are reflected back simultaneously, but you say that is not so. What I do still get the impression from what you and publius are saying, though, is that the fields from each somehow interact with each other so as to point toward the current position in passing. The fields coming from each would realign themselves in accordance to how the other is moving while passing through each other, or something like that, so that they are in effect aimed toward the other body's current position. Is that right?

[Flaney]

... I do that to where the new apparent position of the field will travel the distance to the current position of the other body at c (like an expanding spherical wave, as publius said) and in the same time.
If this expanding wave is has the velocity of the mass added to it, then its point of origin (center of the sphere) is always centered on the moving mass.


But I'm still not sure what you mean about the potential field exactly. It seems like if the field is reflected off of the other body and back again, it should indeed interact with the initial body at its current position, but this would be true regardless of the calculations used. So I'm not sure I follow.

Think of all the expanding shells as previously being "reflected" back from the other mass. They've been constantly streaming for some time. Only the first interactions have the incomplete potential fraction (that does not affect the never before encountered mass).


Then I thought maybe you meant both fields are reflected back simultaneously, but you say that is not so. What I do still get the impression from what you and publius are saying, though, is that the fields from each somehow interact with each other so as to point toward the current position in passing. The fields coming from each would realign themselves in accordance to how the other is moving while passing through each other, or something like that, so that they are in effect aimed toward the other body's current position. Is that right?

When the expanding field moves with velocity c, its center remains fixed at the point when the mass "emitted or reflected" it. The radius of curvature when it meets the other mass will define whence the potential began. These are the radii to use in the effective potential function incorporating propagation delay (and direction, too). I just add that the return trip must be included to complete the total potential info. So for masses separating from each other, the complete effective radius is found back in time (two-way trip time) to be closer than the actual distance between.

When I say the fields are not reflected simultaneously, all I mean is that when the masses have different radial velocities, it may take a different time interval for each potential fraction to "catch up" with the other mass. But since it's symmetric, they gain back the time diffence when returning (assuming they are moving apart).

[czeslaw]

So, if you adhere to Einsteins 'ponderable matter distorts space/time' by causing a gravity well in that space/time, then there is no 'gravitational radiation', and DM is the background gravity field that the ponderable matter is distorting. Now, if that DM background gravity field is traveling at 'C', but does not interact with the baryonic matter, then you have gravity propigating at 'C' and you just substitute DM for the currently used 'gravitational radiation', and it should work precisely, I think

I agree
As for me, the gravity is an inwards distortion of the space caused by mass (energy). All energy is oscillating and this inwards distortions are oscillating too. It makes a discrete, oscillating space and we observe a Casimir effect and virtual particles.
All space oscillations are a kind of the energy. The gravitational oscillations are special kind of energy because it is a background for photons and real particles. All space oscillations (energy) create an information and it is propagating at "c".
You wrote - DM is the background gravity held that the ponderable matter is distorting. If I good understand, the whole gravitational energy of the Universe is traveling at "c" through the space and may be lensed close to other mass.
This may show effects according to MOND far away behind galaxy halo and additionally a strange clusters of the lensed gravity alone.

[RussT]

If I good understand, the whole gravitational energy of the Universe is traveling at "c" through the space and may be lensed close to other mass.

This seems to be the point that everyone has some question about!

If DM, regardless if it is HOT, COLD, Blue or Brown, if it is WIMPS, then it does not interact with baryonic matter and is traveling right through 99.999999999% of all of it...........then it has 0.00000000000000000000000001 inert energy...it is just Planck size/mass non-baryonic inert DM travelling at 'C', and it does not interact even with itself, so is going in all directions.

[czeslaw]

This seems to be the point that everyone has some question about!

If DM, regardless if it is HOT, COLD, Blue or Brown, if it is WIMPS, then it does not interact with baryonic matter and is traveling right through 99.999999999% of all of it...........then it has 0.00000000000000000000000001 inert energy...it is just Planck size/mass non-baryonic inert DM travelling at 'C', and it does not interact even with itself, so is going in all directions.

But it interacts gravitationally and may supply an kinetic energy as the objects gains in the gravitational field.

[RussT]

But it interacts gravitationally and may supply an kinetic energy as the objects gains in the gravitational field.

These are all baryonic matter related.

[czeslaw]

These are all baryonic matter related.

The baryonic matter in our galaxy has more kinetic energy than it results of a simple Newton calculation. It has also more potential gravitational energy than we see of the baryon matter.
DM is related to baryon matter by a gravitational field.

[RussT]

The baryonic matter in our galaxy has more kinetic energy than it results of a simple Newton calculation.

This has nothing to do with DM. What does it mean?


It has also more potential gravitational energy than we see of the baryon matter.

What is 'it' and what does this sentence mean?


DM is related to baryon matter by a gravitational field.

Okay, what does it mean?

[czeslaw]

Dark Matter is a matter interacting gravitationally with a baryon matter in galaxy and we observe it because our Sun rotates faster around the galaxy and is held by an unknown additional gravity.
Gravity causes an acceleration and it causes an additional kinetic energy.

[RussT]

This seems to be the point that everyone has some question about!

If DM, regardless if it is HOT, COLD, Blue or Brown, if it is WIMPS, then it does not interact with baryonic matter and is traveling right through 99.999999999% of all of it...........then it has 0.00000000000000000000000001 inert energy...it is just Planck size/mass non-baryonic inert DM travelling at 'C', and it does not interact even with itself, so is going in all directions.

The whole point to all of this is to try to show that Non-Baryonic DM "GRAVITY" has no energy! The tiny little Planck size/mass DM 'base element'...has mass, thats what makes it gravity, but the energy is 'locked in', and the only thing that can unlock it, is a TEV or High GEV event.

[czeslaw]

Here is a problem I can not solve.
If there is a mass, there is an energy E=mc^2.
Even a neutrino and bosons have an energy.
TeV, GeV are units of energy too.
It has to be an energy of the gravity but it is not a visible energy like photons or EM energy. All galaxies in the Universe have a potential gravitational energy to the all other galaxies in the Universe:
E(grav)= GMxM/R where M is mass of the Universe.

[RussT]

[If there is a mass, there is an energy E=mc^2.]

This only applies if the mass is has baryonic matter included in it.

[Even a neutrino and bosons have an energy.]

If neutrinos have a Planck mass, which I believe they do, then the energy of that tiny little mass CANNOT be 'active', other wise it would 'interact' with whatever it was trying to 'go through' everytime it went through anything!

[It has to be an energy of the gravity but it is not a visible energy like photons or EM energy]

The bottom line is, the gravity of 'space', DM, has NO energy.

[czeslaw]

If DM has no energy - what balances an additional kinetic energy of the Sun around a galaxy centre ?
A baryon mass can holds the Sun till 160 km/s but it moves about 220 km/s.
Our Sun has an additional kinetic energy not balanced by a baryon potential gravitational energy. Mainstream claims that potential gravitational energy is in the Dark Matter and they count its density 0,3 GeV/cm^3.

[RussT]

If DM has no energy - what balances an additional kinetic energy of the Sun around a galaxy centre ?

Along with the gravity of the baryonic matter (all the other stars and stuff), all the gravity of the dark matter that permeates the galaxy, helps hold on to the stars so they don't go flying out of the galaxy. Kinetic energy has very little if anything to do with this part.

[czeslaw]

Along with the gravity of the baryonic matter (all the other stars and stuff), all the gravity of the dark matter that permeates the galaxy, helps hold on to the stars so they don't go flying out of the galaxy. Kinetic energy has very little if anything to do with this part.

A citation of - http://www.bestsyndication.com/Articles/2006/dan_wilson/sci_tech/02/021506_dark_matter_energy.htm
Galaxies would fly apart if it were not for this unseen dark matter. It is estimated that dark matter makes up about 80-85% of the matter in the Universe. The gravitational energy of the dark matter holds the galaxies together.

[RussT]

They have already accounted for all of this hot gas, and it is baryonic matter.

[czeslaw]

They have already accounted for all of this hot gas, and it is baryonic matter.

The gravitational energy of the Dark matter is not of the Baryon matter.
Galaxies would fly apart if there would be a gravity of the baryon matter only.
Thare is writen - THE GRAVITATIONAL ENERGY OF THE DARK MATTER - exactly.

[RussT]

say that the dark matters is made of hydrogen atoms with temperatures reaching 10,000 degrees C (18,000 F).


Galaxies would fly apart if it were not for this unseen dark matter. It is estimated that dark matter makes up about 80-85% of the matter in the Universe. The gravitational energy of the dark matter holds the galaxies together.

Actually this whole article is off.

You cannot just exchange DM and DE energy like this even in the maistream view.

[dark matter makes up about 80-85% of the matter in the Universe.]

DM is 21%
DE is 75%
Baryonic matter 4%

Dark Energy has nothing to do with galaxy rotation curves and even more than that DE is ANTI-Gravity.

[czeslaw]

Sometimes they call DM+DE = Dark Matter.
Any way DM (21%) and DE (75%) interact gravitationally and together with a baryon matter give enough mass for a flat geometry at critical density Mass/Radius = c^2/2G.
Every mass (even the relativistic) causes a space curvature because of the gravitational force between masses. If there is a gravitational force there is a potential gravitational energy. It is by a definition.
F=GMm/c^2

[RussT]

Gravity is not a 'force'.

By definition, Dark Energy is Anti Gravity and has nothing to do with galaxy rotation curves.

[czeslaw]

Gravity is not a 'force'.

By definition, Dark Energy is Anti Gravity and has nothing to do with galaxy rotation curves.

You right. Gravity is not a force but there is Gravitational energy. Dark Energy is acting against Gravity like a kinetic energy does. Dark Matter is acting attractively as an ordinary Gravity. That way DE and DM have an energy.

[czeslaw]

Dark Matter has a mass and is acting attractively and that way DM has an potential gravitationally energy.

[grav]

If this expanding wave is has the velocity of the mass added to it, then its point of origin (center of the sphere) is always centered on the moving mass.Right. I just think of it as having the extra velocity of the body at the point of emission and expanding from that point as if it were still travelling in a straight line at v. It would be as if the field itself then travels at v. Once emitted, it shouldn't be affected by the actual path of the emitting body if the path then curves, and therefore, the gravitational field should still appear to be travelling in a straight line regardless of the path of the body. As long as the body travels less than the speed of the "wave", its new position along a curved path will never be corrected.


Think of all the expanding shells as previously being "reflected" back from the other mass. They've been constantly streaming for some time. Only the first interactions have the incomplete potential fraction (that does not affect the never before encountered mass).Although the direction of the emitting body can never be corrected for by the field with any new waves emitted from it, since the body always travels at a lesser speed than the wave, it appears here that you are saying that the reflected wave can somehow adjust the incoming wave. This makes sense in a way. It would be as if the reflected wave interacts with the other body in such a way that it carries information about the motions and directions of both bodies after reflection, and somehow transmits this information to the new wave. Only an initial wave (before reflection), then, would not carry this information.



When the expanding field moves with velocity c, its center remains fixed at the point when the mass "emitted or reflected" it. The radius of curvature when it meets the other mass will define whence the potential began. These are the radii to use in the effective potential function incorporating propagation delay (and direction, too). I just add that the return trip must be included to complete the total potential info. So for masses separating from each other, the complete effective radius is found back in time (two-way trip time) to be closer than the actual distance between.

When I say the fields are not reflected simultaneously, all I mean is that when the masses have different radial velocities, it may take a different time interval for each potential fraction to "catch up" with the other mass. But since it's symmetric, they gain back the time diffence when returning (assuming they are moving apart).I do not get this part. By radius of curvature, do you mean the space-time curvature, by using the time of delay as a dimension? If so, then does that mean you are utilizing GR? I ask because what you state above should only occur within a medium. This is because in order for the center of the field to be fixed at the point of emission or reflection, it must still be fixed relative to something. It is obviously not fixed relative to the bodies in motion, so it would either be fixed to a medium or to their barycenter. So unless the field acts in such a way as to identify the barycenter of the entire system, it would have to be transmitted through a fixed medium. Which way do you interpret this to be?

[grav]

When the expanding field moves with velocity c, its center remains fixed at the point when the mass "emitted or reflected" it. The radius of curvature when it meets the other mass will define whence the potential began. These are the radii to use in the effective potential function incorporating propagation delay (and direction, too). I just add that the return trip must be included to complete the total potential info. So for masses separating from each other, the complete effective radius is found back in time (two-way trip time) to be closer than the actual distance between.Okay. I think I see now, somewhat. If I'm guessing right, the radius of curvature might be the way the field line themselves curve. For a stationary mass, the field lines would travel outward in all directions in straight lines. But with a relative velocity to the mass, the lines are no longer straight. As far as the potential goes, we could plot the direction of the field line at some distance from the mass to its prior position with propagation delay, and then from there, it would begin to curve more and more toward the current position at a smaller distance. This would only be the potential, however, because as a second mass actually begins to follow this curve from some point, the original mass would have travelled even further, and so the field lines will curve even more. If this is the correct way of thinking about it, though, and I'm still not sure if this is what's meant or not, but it seems awfully complicated and probably unnecessary (to me, although publius also seems to agree they're much easier). But then, without a propagation delay, the field lines would point almost directly toward the current position anyway.

When I say the fields are not reflected simultaneously, all I mean is that when the masses have different radial velocities, it may take a different time interval for each potential fraction to "catch up" with the other mass. But since it's symmetric, they gain back the time diffence when returning (assuming they are moving apart).This is another part I'm confused about. I still may just not be thinking about it all correctly, maybe, but I don't see how a gravity field can be reflected in the first place, although I can see how it would be applied. It seems to me that instead of a reflected field, the field coming back from the second mass would probably just be its own.

[grav]

:whistle: I can see clearly now, the rain is gone. :whistle: I can see all obstacles in my way. :whistle:

Thank you, publius, for helping along the way with all of your insights into GR and propagation delay. :) As of yesterday, I think I'm finally starting to catch on. I spent part of last night trying to work out some formulas for propagation delay. I think I may have actually made some headway this time, since I was forced to think them through on paper instead of just plugging them into a simulator to see what happens. It's amazing, though, how many slight variations there are to run on. I can see easily filling a whole book with all of the possibilities. But only one can truly be correct, and that is what we (I) must find.

Okay. So I cannot plug in the formulas and what I can model on paper is limited, so I finally just decided to start with the simplest thing I could think of in this case, which is a mass travelling in a straight line, and find out what the formula would have to be for gravity to appear instantaneous from the current position. In order to travel in a straight line without a second mass affecting it, the first must be a large mass, and the second a particle. If we observe this system in such a way so that we are travelling with the particle, so that it is stationary, and the gravitating mass is travelling along the x-axis, then the prior position of the mass is (x,y) and its new (current position is (x+vt,y). The time {t} must be the same time it took for the gravity to propagate the original distance between them, so t=d/c=(x^2+y^2)^.5/c . The new distance, then, is d'=(x'^2+y'^2)^.5=[(x+vt)^2+y^2]^.5=[(x+(v/c)d)^2+y^2]^.5 .

The first thing we notice here is that, in order for this to relate to any reference frame and for any relative velocities, then x'=x+(vx/c)d, y'=y+(vy/c)d, and z'=z+(vz/c)d, where vx, vy, and vz are the relative differences in velocities along these vectors. Following this reasoning, we get

d'^2=x'^2+y'^2+z'^2=
(x+vxd/c)^2+(y+vyd/c)^2+(z+vzd/c)^2=
(x^2+2vxxd/c+(vxd/c)^2)+(y^2+2vyyd/c+(vyd/c)^2)+(z^2+2vzzd/c+(vzd/c)^2)=
(x^2+y^2+z^2)+(2d/c)(vxx+vyy+vzz)+(d/c)^2*(vx^2+vy^2+vz^2)=
d^2+(2d/c)(v dot r)+v^2*(d/c)^2

In your way of writing it, publius, I guess it would read r'^2=r^2+(2r/c)(v dot r)+(vr/c)^2. The acceleration of gravity from a mass travelling in a straight line, with absolutely no propagation delay, would be inversely proportional to r'^2 instead of r^2. So the acceleration would be a=GM/r'^2=GM/[r^2+(2r/c)(v dot r)+(vr/c)^2]. And since this uses the relative velocities, the acceleration of both masses will differ by the same amount, as would any mass with the same distance and relative velocities and directions to a receiving body, so that the effective gravity (with relative motion accounted for) is different from Newtonian gravity (stationary from source) by a factor of (r/r')^2.

I am now trying to apply it to two identical masses travelling in a perfectly circular orbit around each other. Now, my thinking here is that if anything is going to completely eliminate the propagation delay, it would be a circular orbit. The coming and going of an elliptical orbit might come close, but there will always be a slight miss on either side. I would assume that there would be some orbit that all elliptical orbits are attempting to decay into, and that could only be circular, since it is obviously the most symetrical, and that is what would be required. Now, that does not mean that a circular orbit is perfectly stable, but only the most stable. It might still decay slowly with an inward or outward spiral. But if circular orbits are not stable, then as far as I can tell, there would be no perfectly stable orbits possible. But the solar system and galaxies have been around for billions of years, so my thinking is that a circular orbit, or close to it, at least, should be quite stable indeed, and that a perfectly circular orbit, then, may in fact be perfectly stable, where propagation delay is absolutely cancelled out.

[publius]

Grav,

I've been meaning to jump back in here, but that Geocentrism thread and another or two has been getting the most of my attention.

I think you see appreciate this, but let me state it explicitly. In the Schwarzschild metric and even similiar EM and Newtonian situations, one is assuming the source mass is *fixed*, that is not moving. Its field/metric is established throughout all space. We're consider small test particles that themselves do not effect the field (ie cause the source mass to move) as they move.

So in these, propagation delay has absolutely no effect, since by defintion, we're requiring the field be static.

We can also ignore propagation delay when things are moving very slowly. In EM, this would be the magnetostatic and electrostatic solutions without the coupling between them. Ie, charge makes electric field, current makes magnetic and we use the instantaneous positions and velocities of the sources. With gravity, since it cancels the propagation much better than EM, this approach works over a much wider range.

To consider propagation delay, things will of course be different. We have to look at the full, not static metric/potential/field, using the time-delayed positions of the sources.

That makes the equations get much more difficult to solve. :)

When I get the chance, I'll ramble a bit about how this looks at least for EM. I hope you've been reading some of those articles about Maxwell's equations. :)

-Richard

[grav]

But I thought the whole point of incorporating (cancelling) a propagation delay was because it really doesn't matter whether the source mass is stationary and the particle is moving, or if the particle is stationary and the source mass is moving. It would only matter from the perspective of an observer, perhaps moving along with one or the other, but the system itself remains the same, their relative difference in velocities unchanged, so that the field in any frame of reference is identical. It seems like you are saying something else. Are you? The formulas in my last post are for relative velocities only, as it should be, and as such they can be used in any frame of reference. Any absolute velocities, however, stationary or otherwise, would imply an ether. I do believe such an ether to exist, and with any luck I will eventually incorporate this as well, but relativity alone doesn't do this, as far as I know.

[publius]

Grav,

The fields (EM and metric for GR) are indeed frame dependent. Consider a point charge of such large mass that it doesn't move due to some tiny little test charge. In the rest frame of that source charge, we have a simple inverse square Coulomb field. Since the source doesn't move, that field doesn't change. We have simply a radial electric field. The force on the test particle is simply a radial electric force (this is ignoring radiation reaction from the acceleration of the test particle -- if that is included, even if the source is allowed to remain stationary, there will be spiralling, and not a clean orbit like with Netwonian gravity).

Now, switch to the rest frame of the test particle at any instant. Now it sees a moving source, and a magnetic field as well as electric. But in that frame, it is not moving, so it experiences no magnetic force (at that instant). However, the electric field looks different (for low velocities it is very close to the Coulomb radial field). Time dilation and length contraction between frames accounts for the differences in the field.

Now, go to some other frame where both charges are moving. The test particle is experiencing both a magnetic and electric force, and again the clock and ruler changes between frames account for the difference.

But, in any of those frames, we'll see the net field points toward the instananeous position of the source charge. Acceleration and resulting radiation make it miss.

Now, go to GR gravity and not only does it account for velocity, but acceleration as well. It's only changes in acceleration that make it miss.

-Richard

[grav]

I am unsure about the relations you just posted, but this is my understanding- Newtonian gravity only applies to a stationary source and receiver, or for a stationary ether through which gravity propagates and is centered on the emission point for the elapsed time since emission from that point in space even after the body has travelled on. It is very close for relative velocities that are very small compared to that of light, but vary largely for extreme cases such as black holes and tight binary star systems. Relativity, on the other hand, dismisses the ether and therefore also the idea of stationary sources or receivers. All that matters is the relative velocities between them. For EM, a wire passed between two magnets should create the same current as two magnets passed over a wire.

Now, there may be another possibility. The gravity a body feels is really relative directly to the field, not the source. That is, the actual interaction does not occur between the two masses, but between each mass and the other's gravity field. The field itself may be transferred through an ether, or through absolute space (a medium), from the point of emission, but it must still interact with the other body at the velocity at which that body travels through the medium. The force that acts on each will differ slightly, however, and the difference might be distributed to the medium itself, as gravitational waves, maybe?

Now, go to GR gravity and not only does it account for velocity, but acceleration as well. It's only changes in acceleration that make it miss.
If this is the case, then a straight line propagation delay should indeed be completely accounted for. Could you check the post I did this for and see if my result matches that for relativity. I know you posted it here before, probably a few times. I will look for it (unless you could indulge me in posting it just one more time? :)). If it differs, then that comparison might say something as well.

[publius]

Grav,

I posted a link to Carlip's paper about it, not my own calculations. :) That is well beyond my own meager abilities (which I why I keep encouraging you to learn more math and stuff where you can see all this). What he did was get a metric for a meaningful uniform acceleration (remembering that GR's source requires energy and momentum to be conservered, so what it is causing mass to accelerate must be included as well. He used the photon rocket; mass throwing off EM radiation and propelling itself).

Now, he then showed the gravitational acceleration at a distant point pointed right toward the instantaneous position of that accelerating mass(with some minor variations due to GR's "strange effects" different from our Euclidean intuition). And further he showed that that, like EM, was due to velocity (and acceleration) "information" from the time retarded source position.

Now, about the "field" picture. Yes, that's a very useful way to look at it. The "source" makes a "field" that propagates out. This field is what locally interacts with other particles that "respond" to that field.

In GR, that "field" is space-time curvature, the "metric" as we were discussing in that other thread. That metric determines the inertial paths of all matter.

It's best to work through the EM picture of all this, because it is much simpler to see. Those same themes and general principles apply to GR gravity, but in a much more involved and complex way.

In EM the "fields" are well defined and easily found. In GR, the "field" like we think of the Newtonian g field (or g and B_g of the GEM EM-like approximation to GR) is not a simple thing at all. It is far better to work at the metric level (which is like the potentials of EM).

Things like acceleration are very frame dependent in a complex way when space-time is curved.

Remember in those simple Schwarzschild relation how we, the distant observer saw the free-falling particle begin to slow down when it reached the photon sphere? And remember how local stationary observers saw something different because of their different rulers and yardsticks. And the free-fallers themselves saw yet another picture. They all have things in common, but then things that are very different.

Add time variation to the above, where the metric itself is varying with time, as when the sources are moving, and it makes it all the more complicated. There is a way to define a "field" sort of, but it is a very complex tensor thing, the particulars of which, like how much the gravitational acceleration is and what direction it points is a very frame dependent thing.

-Richard

[grav]

Well, the previous posts have gotten me thinking. How can relativity be based on the idea of no ether and yet be similar to ether-like structure? My thinking is, similar to how it's really been all along, that the "fabric" of space-time is an ether, just not an absolutely stationary one. That is, as matter is affected by the ether, and energy is transmitted through it, so is the ether affected by matter and energy. The M-M experiment failed because it was conducted "on the train", sorta' speak. Bumping the apparatus produces a result while light is "in transit", so it cannot always be constant relative to the apparatus. Einstein's paper begins with the motion of a rod relative to a point of origin. That origin is stationary, yet must be relative to something. An ether? SR seems contradictory in many situations, although there are ways to work out any apparent paradoxes, but which are just circular reasoning to my way of thinking. I think these problems arise because the mathematics is not complete, or not applicable to all situations, as gamma, [1-(v/c)^2]^.5, is apparently applied to much more than it is meant to be.

If the ether was perfectly stationary, all light would be transmitted at c relative to it, not an observer. Gravitational fields would be centered on a point that is stationary to the ether. The Doppler shift would apply in the same way as it does for sound. And Gravity would have to travel at about 20 billion times the speed, at least, as Tom Flandern suggests. Light would be a wave and would always travel at c through the ether. If light were purely particle, there would be no ether, light would travel at c from the point of emission plus the speed of the emitting body, or c+v, and would appear as c-v from behind.

So I'm beginning to think now that GR is an incorporated ether/relativity theory. What Einstein appears to have done is a wonderful thing. He recognized that light and gravity fields must appear the same in any frame of reference. This is the main and most important theme here. For instance, if we say that gravity propagates from the point of emission, what does that mean? If the observer is stationary to the emitting body, it is obvious. If the observer (or receiving body) is moving relative to it, then the gravity field must be observed to be moving as well, with the emitting body. But what if the emitting body is constantly changing direction. The field changes with it, but the observed field over a distance, with elapsed time, does not.

It is obvious that the field should remain the same no matter how it is viewed from any particular frame of reference. The irony is that the strength of gravity, then, does not. For a wave travelling through an ether, the speed (for light) is always c, and the strength of the wave is according to the Doppler shift of fobserved=fsource(c+vo)/(c-vs), where the velocities of the source and observer are positive if it is toward the other. For light as a particle and no ether, it is just c+v toward and c-v away. The actual values are somewhere in between. For v=.1c, an ether would always remain c relative to the ether, but otherwise dependent on the velocities of the source and observer through the ether as well. No ether (particle) would be 1.1 c toward and .9c away. Relativity says gamma and 1/gamma (although I've it as 1/gamma for both as well), for .994987c and 1.005038c.

Okay. Now here's the thing. If there is no gravitational time delay for a body that travels in a straight line, then the formula in post #42 should follow. I have tried to find it on the internet and haven't yet. Nor have I found a post where publius shows it yet, either. But it applies to any and all frames of reference, as it should, and I have gone over it quite a few times and find no error. So if it is true that there is absolutely no gravitational time lapse for a body travelling in a straight line, as relativity seems to imply, then it must follow.

So we have d'^2=d^2+(2d/c)(v dot d)+v^2*(d/c)^2. Now let's say that we want to find the apparent distance of a body that is travelling directly or toward an observer. In this case, v dot d becomes just l vxx l=l vd l. For x and v are the same sign, the bodies are moving away from each other, so vxx=vd. When they are not the same sign, the bodies are moving toward each other, so vxx=-vd. So now we have

d'^2=d^2+(2d/c)vd+v^2*(d/c)^2=
d^2*(1 +/- (2/c)v+v^2(1/c)^2)=
d^2*(1 +/- 2(v/c)+(v/c)^2).

For vd is negative (toward observer), we get d'^2=d^2*(1-2(l v l/c)+(l v l/c)^2), where l v l is the absolute velocity, so d'=d*[1-(l v l/c)]=d*(c-l v l)/c. But for vd is positive (travelling away from the observer), we get d'^2=d^2*(1+2(l v l/c)+(l v l/c)^2), so d'=d*[1+(l v l/c)]=d*(c+l v l)/c. All in all, this becomes d'=d*[1+(v/c)]=d*(c+v)/c. If gravity propagates with absolutely no time delay for a straight line, then this must be the case. Otherwise, there will be a miss even when travelling in a straight line.

[grav]

Well, I thought that the last post might lead somewhere, but it didn't. I thought toward the end that I was arriving at something similar to Einstein's equations, but they aren't. It is still just that for light and gravity moving with the additional speed of the source. It does provide the same result for any frame of reference, but I'm not so sure now that is necessary. An ether wouldn't do this, but any motion is still relative, relative to the ether itself. So the M-M experiment wouldn't work out either way, then. With an ether, we are riding "on the train" and wouldn't observe an interference or Doppler shift. Without an ether, with light moving with the additional speed of the source, the source is also moving with the apparatus, so this tells us nothing as well.

In order to achieve something like Einstein's equations, the formula I came up with would have to be d'^2=d^2*[1+/-2(v/c)^2+(v/c)^4] instead of d'^2=d^2*[1+/-2(v/c)+(v/c)^2], for (v/c)^2 in the place of all (v/c). This would give us d'=d*[1-(v/c)^2]^.5 toward and d'=d*[1+(v/c)^2]^.5 away. The [1-(v/c)^2]^.5 is equal to gamma, of course, and the [1+(v/c)^2]^.5 part would reduce to about 1/gamma for small v. But even as large as something like v=.1c, they would still be very close, since 1/gamma=1/[1-(v/c)^2]^.5=1.005037815 and [1+(v/c)^2]^.5=1.004987562 .

I'm wondering if this might mean that not necessarily motion, but the difference in energy must be the same for all frames of reference. This would give us d'=d*[1+/-2Ekinetic/Erest]^.5, where Ekinetic=mv^2 /2 and Erest=mc^2, so that we achieve d'=d*[1+/-(v/c)^2]^.5 . For small Ekinetic, this would give us approximately d'=d*[1+/-Ekinetic/Erest], so d'*Erest=d*(Erest+/-Ekinetic).

[grav]

I am now wondering if the frames of reference need not be identical for motion, as they wouldn't be with an ether. Perhaps the ether is neither pure wave nor pure particle-like, but something in between. I'm thinking that instead of relative motions (of fields) appearing the same in every frame, momentum and energy are what must appear the same, in order to conserve them in any frame of reference, as it would seem they should.

[RussT]

I am now wondering if the frames of reference need not be identical for motion, as they wouldn't be with an ether. Perhaps the ether is neither pure wave nor pure particle-like, but something in between. I'm thinking that instead of relative motions (of fields) appearing the same in every frame, momentum and energy are what must appear the same, in order to conserve them in any frame of reference, as it would seem they should.

This is very intersting, but using the word 'aether' is like using the term DM.

Unless it is defined it is much too ambigious.

[Perhaps the ether is neither pure wave nor pure particle-like,]

There could be something to this...Planck paricles traveling at 'C'...all of space.

http://en.wikipedia.org/wiki/Aether

http://en.wikipedia.org/wiki/Inertial_frame

[The Lorentz transformation is equivalent to the Galilean transformation in the limit or, equivalently, (low speeds).

Under Lorentz transformations, the time and distance between events may differ among inertial reference frames; however, the Lorentz scalar distance s2 between two events is the same in all inertial reference frames


where c is the speed of light. From this perspective, the speed of light is only accidentally a property of light, and is rather a property of spacetime, a conversion factor between conventional time units (such as seconds) and length units (such as meters).]

[From this perspective, the speed of light is only accidentally a property of light, and is rather a property of spacetime,]

And this???

[grav]

This is very intersting, but using the word 'aether' is like using the term DM.

Unless it is defined it is much too ambigious.

[Perhaps the ether is neither pure wave nor pure particle-like,]

There could be something to this...Planck paricles traveling at 'C'...all of space.

http://en.wikipedia.org/wiki/Aether

http://en.wikipedia.org/wiki/Inertial_frame

[The Lorentz transformation is equivalent to the Galilean transformation in the limit or, equivalently, (low speeds).

Under Lorentz transformations, the time and distance between events may differ among inertial reference frames; however, the Lorentz scalar distance s2 between two events is the same in all inertial reference frames


where c is the speed of light. From this perspective, the speed of light is only accidentally a property of light, and is rather a property of spacetime, a conversion factor between conventional time units (such as seconds) and length units (such as meters).]

[From this perspective, the speed of light is only accidentally a property of light, and is rather a property of spacetime,]

And this???Yes. My thinking is that light is really just energy being transferred through the ether itself. It has no substance of its own, except through the substance of the medium.

[grav]

I am coming to realize, I think, that the first formula would be a conservetion of momentum in any frame of reference, since p=mv, where the mass doesn't change depending on one's perspective, so the relativity of velocities is the same as a relativity of momentum. The second formula is one for energy, which must also be conserved. The first, for momentum (motion), is similar to Newtonian mechanics, and the second, for energy, is similar to Einstein's mechanics. But the two formulas are different, so they can't both be right. The only thing that I can think of in order to conserve both, then, is some transformation for mass itself must take place as viewed from different reference frames. This would make mass a variable as well, which might unite the two formulas in some way.

[grav]

It's amazing how my thoughts seem to be in a constant state of metamorphose, isn't it? If only they would converge on the answer. I am now thinking that the conservation of momentum would only be conserved in respect to an ether, but that the energy is conserved in all frames of reference. If this is the case, then the second formula is correct, and almost agrees with Einstein's. I'm not sure, however, how this would affect the time delay of gravity yet. It appears that all of the pieces of the puzzle are present, though. We must conserve momentum and energy, each in respect to either the bodies or an ether, but in such a way as to make gravity appear (at least almost) instantaneous.

[grav]

It would appear, then, that the formula would reduce to d'=d*[1-(cos L)(v/c)^2]^.5, where L is zero when the lines of travel are directly toward each other.

[RussT]

This would make mass a variable as well, which might unite the two formulas in some way.

I am not quite sure how to wrap my head around this yet, but in one way or another the following has to apply.

If neutrinos or whatever the base Planck mass particle turns out to be, HAVE MASS, then...

Since they are collisionless and going right through baryonic matter (And therfore should be traveling at 'C', as far as I can see), that has to mean that whatever mass a body (earth, sun) has, it also has whatever volume of extra mass from these particles in the volume of the body at any given time of measurement. So the larger the body, the more 'extra' Planck matter/mass it will have in it at any given time.

[grav]

I am not quite sure how to wrap my head around this yet, but in one way or another the following has to apply.

If neutrinos or whatever the base Planck mass particle turns out to be, HAVE MASS, then...

Since they are collisionless and going right through baryonic matter (And therfore should be traveling at 'C', as far as I can see), that has to mean that whatever mass a body (earth, sun) has, it also has whatever volume of extra mass from these particles in the volume of the body at any given time of measurement. So the larger the body, the more 'extra' Planck matter/mass it will have in it at any given time.I thought about that possibility also, like with DM/DE of galaxies. But the thing is, the surrounding medium (intergalactic space) would also include the same density of these neutrinos, and would cancel out its own mass (or gravity of it). If it were actually trapped within a mass, with a greater density than without, it would still be measured as part of the mass of the body. And to make matters worse, the energy density of the neutrinos would actually be repulsive, acting as a pressure that pushes outward in every direction, so it would actually have to be a lack of neutrinos that causes it. It would only become attractive when the neutrinos (or neutrino-like particles) are actually absorbed, creating a negative energy density at the point of absorption of a massive body, which is what would create the positive attraction for mass in the first place. In other words, it is the area of particles that determine how much of the pressure is actually absorbed, but the "mass" of the neutrino, based on its area as well, would be too small to produce gravity on its own, but only with collaboration of a large number of other neutrinos acting together, being the operators of gravity in the first place, but only in tremendous numbers. Neutrinos themselves, then, would not have any gravity of their own, except with the "gravity creates gravity" thing, where the mean free path toward a massive body is increased due to the absence of some of the original energy density, but would otherwise follow QM (at least with individual neutrinos) rather than GR.

[grav]

If light, traveling at 'C' is time dilated in a gravity well, then shouldn't gravity traveling at 'C' in a gravity well, be time dilated too?

That's a good one. Kind of like my question about how gravitons escape from black holes.


Originally Posted by LayMan
Maybe a lay question again, but if (virtual) gravitons are assumed to be the mediators of gravity itself, what should they be escaping from? The gravitational waves? If the waves don't get bothered by it, since they have no mass, and the gravitons are 'riding' the waves like photons 'ride' EM, then I don't really see the problem...? Basically, what I'm trying to say is, whatever is causing gravity, can't be expected to be affected by it, no?

That's true. If these particles (or waves, or whatever) are all travelling outward from a mass at the same rate, then they would spread out away from each other, and wouldn't interact with each other at all (except perhaps for where the gradient of the gravitating body is concerned), and therefore wouldn't be affected by gravity themselves. This seems so obvious that it would seem anyone would have already thought of it. I would have thought I had, but perhaps not in exactly the same way you are thinking about it. It appears to be the only way it could be, in fact! Good going.

[EDIT-Actually, that would explain the gravity escaping black holes thing, but perhaps not RussT's original question, I think. And so the paradox might still exist. Oh, well. I still think you made a good point. And the solution might be related.]


I didn't know where else I could put this, but this seems sufficient, especially since the topics are related.

Now, as Layman said, gravity would not affect gravity, since all of its "components" would be spreading outward in all directions at the same rate. But this still does not answer the original question. My thinking is that "stray" neutrinos (ones that did not pass through the gravitating body), would experience gravity in a manner (through QM, more of a Brownian effect or some such) themselves because the mean free path toward the gravitating body is increased due to the absorption of some of the original neutrinos as they passed through, so gravity creates gravity. The neutrinos that left the body might then experience a greater resistance from those that are moving more directly toward it, but since they are the same mass, momentum and energy will just be transferred from one to the next on average. The original neutrino will take on the momentum and energy of the incoming one and be turned back, but the incoming neutrino would be turned forward with the momentum and energy of the original (on average). On a whole, it would be like they are completely transparent to each other, but this would only be true on average (throughout the entire medium locally), and even out over large numbers. Light, being propagated through space as a pure wave by the neutrino medium, would not experience the original neutrinos, since they also travel outward at the same speed, but the gravity creates gravity thing that affects the "stray" neutrinos coming from space would also affect any energy transmitted through it, so that light would not slow down, it would always travel with the average speed of the neutrinos, but its energy would be redshifted by the same increase in the free mean path that affects the incoming neutrinos, since these are the neutrinos through which it would propagate, and it would lose energy more quickly to a greater number of incoming neutrinos, which would be in direct proportion to the gravity itself.

[RussT]

but the "mass" of the neutrino, based on its area as well, would be too small to produce gravity on its own, but only with collaboration of a large number of other neutrinos acting together, being the operators of gravity in the first place, but only in tremendous numbers.

Sorry grav, I missed this post at first because I only saw the post below it.

But yes, this is it. But they are really not 'acting together'...they are just 'there together'. It is difficult to think of the gravity of baryonic matter interacting, and then including the 'collisionless' DM neutrinos, especially when they are almost certainly travelling at 'C', and just like trinitree88 indicated MUST be subliminal to retain the validity of SR.


My thinking is that "stray" neutrinos (ones that did not pass through the gravitating body), would experience gravity in a manner (through QM, more of a Brownian effect or some such) themselves because the mean free path toward the gravitating body is increased due to the absorption of some of the original neutrinos as they passed through, so gravity creates gravity.

See, this is where the thinking breaks down, because they are collisionless, they have none of these effects, which means they really don't have 'energy' as such...they are 'inert', but the tiny little mass is there.

Which means there is no 'gravitational radiation', it's just that 'all' of 'space', including what is in the 'bodies' at any given point in time, and what they have 'trapped' in the electrons, protons, and neutrons + heavier elements that are 'trapped' during supernova, is 100% gravity.

[grav]

See, this is where the thinking breaks down, because they are collisionless, they have none of these effects, which means they really don't have 'energy' as such...they are 'inert', but the tiny little mass is there.

Which means there is no 'gravitational radiation', it's just that 'all' of 'space', including what is in the 'bodies' at any given point in time, and what they have 'trapped' in the electrons, protons, and neutrons + heavier elements that are 'trapped' during supernova, is 100% gravity.If they have mass, then they are not point particles. Thinking in terms of point particles is just a convenience. It simplifies equations. But if they have mass, then they have area, and they will collide (interact) to some degree. Now, the energy and momentum that will be transferred between them will depend on QM and in what specific way each collision takes place, but the overall transmission of the energy of large numbers of neutrinos would appear smooth. If this transmission is not completely smooth, however, it will have less than 100% efficiency, and lose some of the energy over time to "stray" neutrinos in free space.

[RussT]

But if they have mass, then they have area, and they will collide (interact) to some degree.

Why? They are already defined as 'collisionless'. But, just think of how absolutely tiny they are 10 ^-35.


Now, the energy and momentum that will be transferred between them will depend on QM and in what specific way each collision takes place, but the overall transmission of the energy of large numbers of neutrinos would appear smooth.

Just can't get away from the energy, can you (or anybody else)?

Without the 'energy', it is smooth on any small to medium scales.
How they get here (think expansion from Voids) and the subliminal frame they are in has to cause motion of some kind, but I am not sure that could even be considered a wave motion, as the whole field is travelling at 'c'.

[zeus]

The speed of gravity is almost instantaneous.

Newton, who said it was exactly instantaneous, was closer to it than Einstein who said it acts at the speed of light.

I side with Tom Van Flandern on this one gang.

My book will be out in December and in it is this very topic.

Web Site is http://www.amperefitz.com

You can preview the book right now by clicking the link below:

http://www.amperefitz.com/us_20061020_ck_ds_jm_ds.pdf

Send reviews to zeusrdx@yahoo.com

I'll put the favorable ones in my web Site.


z

[RussT]

The speed of gravity is almost instantaneous.

Newton, who said it was exactly instantaneous, was closer to it than Einstein who said it acts at the speed of light.

I side with Tom Van Flandern on this one gang.

My book will be out in December and in it is this very topic.

Web Site is http://www.amperefitz.com

You can preview the book right now by clicking the link below:

http://www.amperefitz.com/us_20061020_ck_ds_jm_ds.pdf

Send reviews to zeusrdx@yahoo.com

I'll put the favorable ones in my web Site.


z

z, you should start your own thread on this!

[RussT]

Why? They are already defined as 'collisionless'. But, just think of how absolutely tiny they are 10 ^-35.



Just can't get away from the energy, can you (or anybody else)?

Without the 'energy', it is smooth on any small to medium scales.
How they get here (think expansion from Voids) and the subliminal frame they are in has to cause motion of some kind, but I am not sure that could even be considered a wave motion, as the whole field is travelling at 'c'.

grav...do you want to continue?

[grav]

grav...do you want to continue?Sure. Sorry. I thought your last post was rhetorical and I wasn't quite sure what you meant by that last part.


Originally Posted by grav
But if they have mass, then they have area, and they will collide (interact) to some degree.

Why? They are already defined as 'collisionless'. But, just think of how absolutely tiny they are 10 ^-35. Yes. They would be extremely small and they would appear practically inert in comparison to electromagnetic phenomena and such, and could pass through light-years of matter without any interaction at all, but that would be on the scale of individual particles, and a very miniscule chance of interaction still exists. For a large abundance of them, however, the interactions would be more common, although still minute. The ratio of gravitational to electric force for the electron, for example, is only one part in 10^42. Very small indeed, and virtually non-existent for most practical purposes with a mass that small, but still there.


Originally Posted by grav
Now, the energy and momentum that will be transferred between them will depend on QM and in what specific way each collision takes place, but the overall transmission of the energy of large numbers of neutrinos would appear smooth.

Just can't get away from the energy, can you (or anybody else)?

Without the 'energy', it is smooth on any small to medium scales.
How they get here (think expansion from Voids) and the subliminal frame they are in has to cause motion of some kind, but I am not sure that could even be considered a wave motion, as the whole field is travelling at 'c'.What do you mean by this?

[RussT]

Yes. They would be extremely small and they would appear practically inert in comparison to electromagnetic phenomena and such, and could pass through light-years of matter without any interaction at all, but that would be on the scale of individual particles, and a very miniscule chance of interaction still exists.

Aand this is what is observed, as far as I know.


For a large abundance of them, however, the interactions would be more common, although still minute.

As is this, however, I am not exactly clear on when and how they are seeing this 'interaction' and then how exactly this relates to only the weak force interaction. My main problem with this though is...that they interact with the weak force 1 time out of Billions and Billions, so does that really constitute 'interaction'? if they interact, thet should do so regularly!

I don't remember now where I said it, but I 'supposed' that the Higgs Ocean and the Neutrino Sea...were one and the same...'space'.

Everyone seems to think that if any kind of background gravity field is identified, that that would nullify SR, and that is simply not true.

What they do not realize (although Publius seems to be working his way there), is that "IF" the background gravity field (BGF) is 'all' traveling at the velocity of light, then light is just traveling on the BGF, and SR is safe.


The ratio of gravitational to electric force for the electron, for example, is only one part in 10^42.

Yes but remember...gravity is different and not a force.


Very small indeed, and virtually non-existent for most practical purposes with a mass that small, but still there.

See above. And yes, I still have to resolve this somehow, with my view, because, we wouldn't even know the neutrinos were there if we were not seeing some kind of interaction.

[grav]

Aand this is what is observed, as far as I know.



As is this, however, I am not exactly clear on when and how they are seeing this 'interaction' and then how exactly this relates to only the weak force interaction. My main problem with this though is...that they interact with the weak force 1 time out of Billions and Billions, so does that really constitute 'interaction'? if they interact, thet should do so regularly!They would do so regularly with large numbers. It is just that they are neutral and so small that they must "strike" in just such and such a way to actually be absorbed. Otherwise, they would just be reflected, or their path only altered in some way, or pass right through, or miss altogether (depending on how the actual interactions take place).

I don't remember now where I said it, but I 'supposed' that the Higgs Ocean and the Neutrino Sea...were one and the same...'space'.Yes, I agree, except that I still think of neutrinos as existing within space, but that they would give resistance to matter in the form of mass, and they would make up the energy density of space, which would then be its "fabric".

Everyone seems to think that if any kind of background gravity field is identified, that that would nullify SR, and that is simply not true.I think the formulas for QM, SR, and GR must all be brought together in one unifying formula for gravity, and that this would be recognized in the characteristics of the energy density.

What they do not realize (although Publius seems to be working his way there), is that "IF" the background gravity field (BGF) is 'all' traveling at the velocity of light, then light is just traveling on the BGF, and SR is safe.Yes. That is what I think also. Maybe in a different way, though. I'm not sure. Can you expand on this? The model seems to point toward light being purely waves of energy travelling in the neutrino field and following its "curvature", but would probably differ tremendously for extreme cases such as with black holes.



Yes but remember...gravity is different and not a force.Well, that is one way to look at it, but I choose not to. I can't easily measure it this way. I need the force and associated acceleration in order to relate it to the pressure and so forth. All of these are very important and cannot be dismissed. In this case, it is the entire energy field that is warping, the energy density itself. However, I can also see how it can be related that way, and I try to keep an open mind. It think the main reason for this sort of thinking is that gravity would appear to travel "between" space with s^2=d^2-(ct)^2, and this may be useful, but is really only something else that must be incorporated into a full theory of gravity. We make observations and come up with formulas and theories to fit those observations as best we can, but they may not be absolutely precise.

For a simple example, it is the same way way with the formula for gravity itself. a=Gm/d^2 is the general formula for the acceleration of gravity, but it may really be something like a=Gm/(d+r)^2 or a=Gm/(d^2+r^2), where r is the radius of the particles that gravitate. Also, below the surface of a large gravitating mass, the acceleration suddenly changes to a=Gmd/r^3. These two formulas, a=Gm/d^2 and a=Gmd/r^3 for above and below the surface are the same at the surface itself, but it would seem they should be incorporated into one all encompassing formula that includes both cases. It is the same with that gravitational acceleration that kicks in when the "normal" acceleration becomes too low, but I can't remember the name (Pioneer anomaly?). And probably the same with DM/DE, for galaxies. Perhaps the experiments of the borehole anomaly could tell us something about this, but I haven't studied that yet. Also, since gravitating masses are not really point particles, but have a gradient, and may be moving or rotating as well, these must also be incorporated, which is where SR and GR come in as well.



See above. And yes, I still have to resolve this somehow, with my view, because, we wouldn't even know the neutrinos were there if we were not seeing some kind of interaction.Well, we do see interactions all around us every day, in the form of gravity. We just don't recognize it as such. We are attempting to detect "fresh" stellar neutrinos that haven't "oscillated out of phase" yet, by the rare interactions that might release light or some form of radioactivity, but we are ignoring the interactions going on all around us all the time, right under our noses. Mass might cause them to oscillate slightly back "into phase". But then, to be fair, if all mass gravitates, then it would be extremely difficult to compare these interactions to that of a control group anyway, so we are only left with the overall observations of gravity itself.

[RussT]

Yes. That is what I think also. Maybe in a different way, though. I'm not sure. Can you expand on this? The model seems to point toward light being purely waves of energy travelling in the neutrino field and following its "curvature", but would probably differ tremendously for extreme cases such as with black holes.

Yes, this is it!

Look at like this..."IF" we are correct, that the Higgs Ocean and the Neutrino sea are one and the same...that is DM and DE are 'all' "collisionless non-baryonic dark matter"....that is 'all' of 'space'... that would be a BGF travelling at 'C'.


Perhaps the experiments of the borehole anomaly could tell us something about this, but I haven't studied that yet.

Cahill has the BGF correct, and shows how it can apply to the solar system as well as the galaxy rotation curves. BUT, BUT, I am very reluctant to even use this anymore because he has so many other relationships and the reasons for them wrong. The major thing he doesn't realize is...THAT THE BACKGROUND GRAVITY FIELD IS TRAVELLING AT 'C'!

Because he doesn't realize this, in much of his work he is trashing Einstein (although he is still totally willing to use and talk about black holes, but has parts of that wrong too, when he relates elliptical galaxy black holes to globular star cluster black holes).

Also, I am not convinced that his application of the Fine Structure Constant is the correct 'base element' of how protons and electrons get their mass, but...how he uses that does mimic the same effect as DM, especially for the galaxy rotation curves.

Anyone looking at his stuff, is going to have a very hard time only picking out what actually applies correctly, and the 'rest of what he thinks he sees'. But as I have indicated before, trying to work with relationships from the bottom up (quantum to macro) without understanding it from the top down first (Massive Black Holes/Galaxy formation), is vitually impossible. Just like trying to figure out the universe as a whole by starting with an initial premise of how the universe starts (Impossible to see) without understanding MBH/Galaxy formation first.