length contraction near a black hole

tommac · #1 (**permalink**) 19-May-2009, 04:40 PM

Sorry can someone clarify this for me as I am a bit confused and would like to make sure that I am correct about something.

If I am very near the EH of a black hole and say my clock is runing 1/10th as fast as someone that is suffiently away from the BH.

Then the length contraction would be needed to keep light constant at C.

my time is running slow so I should see things as being closer to me right? distant objects should appear much closer to me right?

if I my clock is running 1/10 the speed of the distant observer then the distance I see light travelling in 1 year, is equal to the light the distant observer sees travelling in 10 years. So 1ly for the observer near the EH is equal to 10 light years for the observer that is sufficiently away from the BH right?

grant hutchison · #2 (**permalink**) 19-May-2009, 05:57 PM

If you are a stationary observer hovering close to the event horizon, and you compare your own measurements to the Schwarzschild coordinates, you'll note that your clock is running slowly compared to the Schwarzschild timekeeping, and that your metre rule is short when compared to the Schwarzschild radial coordinates. You will (of course) measure local lightspeed at c.
Local measurements are what matter.

Grant Hutchison

WayneFrancis · #3 (**permalink**) 20-May-2009, 05:32 AM

This is part of the problem with some of your questions and statements you tend to mix distances from one reference frame with times of another.

Given that A and B start at ~150,000,000km from a given point in space C.
next a 1 solar mass black hole appears at location C what happens?

Well in the flat space coordinate system the point at which time would be 10x slower would be ~30m from the EH or 2.97km from point C. I'll call this point D

Point D is no longer 149,999,997km away but ~10x further in this new curved space time. This means it is a shorter trip to fly around a black hole then it would be to to fly a "chord" that comes close to the EH.

tommac · #4 (**permalink**) 20-May-2009, 04:54 PM

Quote:

Originally Posted by WayneFrancis

This is part of the problem with some of your questions and statements you tend to mix distances from one reference frame with times of another.

I think this is where it is the most interesting.

captain swoop · #5 (**permalink**) 20-May-2009, 04:57 PM

But it is also where you fall over and you get into trouble.

tommac · #6 (**permalink**) 20-May-2009, 05:07 PM

Quote:

Originally Posted by captain swoop

But it is also where you fall over and you get into trouble.

Oh I agree ... but I usually learn on my way down.

grant hutchison · #7 (**permalink**) 20-May-2009, 07:09 PM

Quote:

Originally Posted by tommac

Oh I agree ... but I usually learn on my way down.

What you should learn (should have learned by now!) is that you should never, ever mix reference frames, because the result is always nonsense.

Grant Hutchison

robross · #8 (**permalink**) 20-May-2009, 08:55 PM

Quote:

Originally Posted by grant hutchison

What you should learn (should have learned by now!) is that you should never, ever mix reference frames, because the result is always nonsense.

Grant Hutchison

This reminds me of that old joke...

Alice: "What are you doing?"

Bob: "Looking for my keys."

Alice: "Where did you lose them?"

Bob: "Up the street a ways, maybe a mile or so."

Alice: "Then why are you looking for them here?"

Bob: "The light is better over here."

Rob

grant hutchison · #9 (**permalink**) 20-May-2009, 09:02 PM

Or Einstein's (alleged) definition of insanity: doing the same thing over and over again, expecting different results.

Grant Hutchison

tommac · #10 (**permalink**) 20-May-2009, 10:22 PM

Quote:

Originally Posted by grant hutchison

What you should learn (should have learned by now!) is that you should never, ever mix reference frames, because the result is always nonsense.

Grant Hutchison

But really it is not. For example, my ATM ... that shows that there is an illusion of superluminal speeds.

if you take X1-X0/T1-T0 relative to an observer who has travelled from frame-initial( t0 ) to frame-final ( t1 ). Where frame initial is very deep into a gravitational well. Your quasi-average speed of light could be much greater than C over the period of time T1-T0.

This is real. This distance at T1 is real ( it is different than the distance he percieved at T0 but equally as real ) . The time that passed for the observer is real.

grant hutchison · #11 (**permalink**) 20-May-2009, 10:34 PM

Quote:

Originally Posted by tommac

But really it is not. For example, my ATM ... that shows that there is an illusion of superluminal speeds.

No, that's exactly the sort of unreal nonsense we're all talking about. These are coordinate effects: you can get all sorts of odd results by choosing the wrong coordinates.

Grant Hutchison

captain swoop · #12 (**permalink**) 20-May-2009, 10:37 PM

Don't bring your ATM into a Q & A thread. It will make bad things happen

WayneFrancis · #13 (**permalink**) 21-May-2009, 01:47 AM

Quote:

Originally Posted by tommac

Quote:

Originally Posted by WayneFrancis

This is part of the problem with some of your questions and statements you tend to mix distances from one reference frame with times of another.

I think this is where it is the most interesting.

I'm glad you think that not understanding SR & GR thus coming up with totally garbage answers because you can't do the math "is the most interesting".

Honestly I'd be much happier if you would understand the math and logic behind these theories can find that "the most interesting"

I guess some of us look at life differently. I see a magician perform a trick and think "How does he do that?" knowing that it is just a trick and that if i try hard enough I can figure it out. I see that you on the other hand will continue to scream "IT'S MAGIC!!!!" at the top of your lungs. We both find the trick interesting but you'll attribute the magician with abilities he/she really does not have because you don't grasp the actual trick.

How's that for an analogy

WayneFrancis · #14 (**permalink**) 21-May-2009, 01:57 AM

Quote:

Originally Posted by grant hutchison

Or Einstein's (alleged) definition of insanity: doing the same thing over and over again, expecting different results.

Grant Hutchison

tommac is doing the same thing over and over and getting the same results...they are wrong but he's consistent with his errors.

pzkpfw · #15 (**permalink**) 21-May-2009, 02:12 AM

Note to all: please, let's not make this personal.

WayneFrancis · #16 (**permalink**) 21-May-2009, 02:37 AM

Quote:

Originally Posted by pzkpfw

Note to all: please, let's not make this personal.

Sorry tommac

grant hutchison · #17 (**permalink**) 21-May-2009, 11:02 AM

The point here is one that Grey has made clearly on another of tommac's threads on this topic. "Distance" is essentially meaningless unless you make the journey, at which point you will achieve a unique correct answer for the distance you travelled between two events.
No-one can retain a distance measurement they make locally, close to the event horizon of a black hole, while completing a journey to a distant point: so it makes no sense to try to measure that entire distance in one set of local coordinates.
I think some of this confusion arises because it's possible under SR to make inferences about a whole journey from the instantaneous measures of a local observer: so long as the observer continues in the same inertial state, her local measures will apply throughout the journey. But that (as the name suggests) is a special case from which we cannot generalize.

Grant Hutchison

tommac · #18 (**permalink**) 21-May-2009, 06:04 PM

Quote:

Originally Posted by WayneFrancis

I'm glad you think that not understanding SR & GR thus coming up with totally garbage answers because you can't do the math "is the most interesting".

Honestly I'd be much happier if you would understand the math and logic behind these theories can find that "the most interesting"

I guess some of us look at life differently. I see a magician perform a trick and think "How does he do that?" knowing that it is just a trick and that if i try hard enough I can figure it out. I see that you on the other hand will continue to scream "IT'S MAGIC!!!!" at the top of your lungs. We both find the trick interesting but you'll attribute the magician with abilities he/she really does not have because you don't grasp the actual trick.

How's that for an analogy

The better analogy is that I actually figure out how he does the trick ... rather than do the math behind pulling the rabit out the hat.

In my opinion there are times where math helps you and other times where it is a distraction.

Quote:

Imagination is more important than knowledge

This to me would hint that einstein himself probably to some extent agreed with me.

tommac · #19 (**permalink**) 21-May-2009, 06:16 PM

I agree that it is confusing ... but I am not convinced that it is wrong.

I also think my lack of explaining power makes it more confusing than it needs to be.

In SR can this experiment be done? :

Observer A and B are standing next to each other.
both A and B each hold a flashlight that releases some photons to a mirror 2 light years away. The flashlight has a timer attached to them. After 4 ly the photons come back to and are detected by receptors held by A and B. When the receptors receive the photons. The caclulator inside of the flashlight does a simple cacluclation T1-T0/4 light years.

A + B test it out and sure enough the results for both A + B return C.

Then they try it again release the photon streams ... but this time after releasing the photons B travels off at near the speed of light perpendicular to the direction of the light ... A percieves B to come back after 364 days, B rests a day and then that next day A + B detect the photons and their calculator comes back with results. A says that light travelled at C, but Bs caculator disagrees and computers 20 x C.

Remember B released these photons and is now recieving them.

Quote:

Originally Posted by grant hutchison

The point here is one that Grey has made clearly on another of tommac's threads on this topic. "Distance" is essentially meaningless unless you make the journey, at which point you will achieve a unique correct answer for the distance you travelled between two events.
No-one can retain a distance measurement they make locally, close to the event horizon of a black hole, while completing a journey to a distant point: so it makes no sense to try to measure that entire distance in one set of local coordinates.
I think some of this confusion arises because it's possible under SR to make inferences about a whole journey from the instantaneous measures of a local observer: so long as the observer continues in the same inertial state, her local measures will apply throughout the journey. But that (as the name suggests) is a special case from which we cannot generalize.

Grant Hutchison

tommac · #20 (**permalink**) 21-May-2009, 06:17 PM

Quote:

Originally Posted by WayneFrancis

Sorry tommac

No offence taken ... I can understand your frustration. I am just trying to learn something here. Please read the scenerio I posted. Maybe you can help explain that to me.

tommac · #21 (**permalink**) 21-May-2009, 06:18 PM

One quick follow up ...
Is the light redshifted for either A or B?

Quote:

Originally Posted by tommac

I agree that it is confusing ... but I am not convinced that it is wrong.

I also think my lack of explaining power makes it more confusing than it needs to be.

In SR can this experiment be done? :

Observer A and B are standing next to each other.
both A and B each hold a flashlight that releases some photons to a mirror 2 light years away. The flashlight has a timer attached to them. After 4 ly the photons come back to and are detected by receptors held by A and B. When the receptors receive the photons. The caclulator inside of the flashlight does a simple cacluclation T1-T0/4 light years.

A + B test it out and sure enough the results for both A + B return C.

Then they try it again release the photon streams ... but this time after releasing the photons B travels off at near the speed of light perpendicular to the direction of the light ... A percieves B to come back after 364 days, B rests a day and then that next day A + B detect the photons and their calculator comes back with results. A says that light travelled at C, but Bs caculator disagrees and computers 20 x C.

Remember B released these photons and is now recieving them.

rommel543 · #22 (**permalink**) 21-May-2009, 06:49 PM

Quote:

Originally Posted by tommac

I agree that it is confusing ... but I am not convinced that it is wrong.

I also think my lack of explaining power makes it more confusing than it needs to be.

In SR can this experiment be done? :

Observer A and B are standing next to each other.
both A and B each hold a flashlight that releases some photons to a mirror 2 light years away. The flashlight has a timer attached to them. After 4 ly the photons come back to and are detected by receptors held by A and B. When the receptors receive the photons. The caclulator inside of the flashlight does a simple cacluclation T1-T0/4 light years.

A + B test it out and sure enough the results for both A + B return C.

Then they try it again release the photon streams ... but this time after releasing the photons B travels off at near the speed of light perpendicular to the direction of the light ... A percieves B to come back after 364 days, B rests a day and then that next day A + B detect the photons and their calculator comes back with results. A says that light travelled at C, but Bs caculator disagrees and computers 20 x C.

Remember B released these photons and is now recieving them.

Regardless of B leaving and coming back, both photon streams were released from the same point. Thats what is important.

It's like watching someone play with a basket ball. They throw the ball up 10 ft and it comes down where they catch the ball. The ball traveled a distance of 20 ft, 10 up and 10 down. If the person throws the ball up and runs in a circle it doesn't effect the distance that the ball travels, only the person. The same for your flashlight analogy. The photon stream goes it's 2 light years one way, and 2 light years back from point of origin . It just so happened that the flash light also traveled for a bit in there as well, but it doesn't effect the distance that the photons traveled.

If you and I are driving down the street and I drop you off in front of your house, then drive 10 more blocks to my house. That night you walk 5 blocks to the store and back. Next morning I drive 10 blocks to your house and pick you up again. In your analogy my car has driven 30 blocks (10 from your house to home, 10 for you going to the store and another 10 coming back to pick you up), where in reality it drove 20 (10 from your house to mine, and 10 the next morning to pick you up).

captain swoop · #23 (**permalink**) 21-May-2009, 06:52 PM

Why would it say 20 x C? light only travels at C

tommac · #24 (**permalink**) 21-May-2009, 06:54 PM

So would you agree with the statement then that light has travelled 4 light years in 1/5th of a year based on Bs observation?

(BTW I only semi-agree with the statement and will tell why after you reply )

Quote:

Originally Posted by rommel543

Regardless of B leaving and coming back, both photon streams were released from the same point. Thats what is important.

It's like watching someone play with a basket ball. They throw the ball up 10 ft and it comes down where they catch the ball. The ball traveled a distance of 20 ft, 10 up and 10 down. If the person throws the ball up and runs in a circle it doesn't effect the distance that the ball travels, only the person. The same for your flashlight analogy. The photon stream goes it's 2 light years one way, and 2 light years back from point of origin . It just so happened that the flash light also traveled for a bit in there as well, but it doesn't effect the distance that the photons traveled.

If you and I are driving down the street and I drop you off in front of your house, then drive 10 more blocks to my house. That night you walk 5 blocks to the store and back. Next morning I drive 10 blocks to your house and pick you up again. In your analogy my car has driven 30 blocks (10 from your house to home, 10 for you going to the store and another 10 coming back to pick you up), where in reality it drove 20 (10 from your house to mine, and 10 the next morning to pick you up).

tommac · #25 (**permalink**) 21-May-2009, 07:01 PM

Quote:

Originally Posted by captain swoop

Why would it say 20 x C? light only travels at C

It has a hard coded 4 light years into the program.

grant hutchison · #26 (**permalink**) 21-May-2009, 07:03 PM

Quote:

Originally Posted by tommac

Then they try it again release the photon streams ... but this time after releasing the photons B travels off at near the speed of light perpendicular to the direction of the light ... A percieves B to come back after 364 days, B rests a day and then that next day A + B detect the photons and their calculator comes back with results.

You're jumping between frames again. What do you hope to learn from this?

Quote:

Originally Posted by tommac

It has a hard coded 4 light years into the program.

So it has the wrong value coded into it, as far as B is concerned. He has been issued with misleading equipment. Again, what's the point?

Grant Hutchison

tommac · #27 (**permalink**) 21-May-2009, 07:12 PM

Quote:

Originally Posted by grant hutchison

You're jumping between frames again. What do you hope to learn from this?

So it has the wrong value coded into it, as far as B is concerned. He has been issued with misleading equipment. Again, what's the point?

Grant Hutchison

Agreed all agreed. However if B measured the distance to the mirrors whatever he did he would measure 2 ly ( assuming he now keeps his frame constant ). The point here is that there is an illusion that light has travelled away from B super-luminally.

If we were unaware that we ( b) changed frames we would miscalculate the speed of light.

Grey · #28 (**permalink**) 21-May-2009, 07:20 PM

Quote:

Originally Posted by tommac

Then they try it again release the photon streams ... but this time after releasing the photons B travels off at near the speed of light perpendicular to the direction of the light ... A percieves B to come back after 364 days, B rests a day and then that next day A + B detect the photons and their calculator comes back with results. A says that light travelled at C, but Bs caculator disagrees and computers 20 x C.

I assume that you mean that B comes back after 1,460 days, so he's got one day before the light returns from its trip.

So, here's the thing. If you measure the distance something travels in one reference frame, and the time something travels in a different reference frame, you can get any answer you want for its velocity (by making the right choices for those two reference precisely because you can get any answer that you want. Instead, you need to calculate the distance travelled and the time taken using the same reference frame. If you do that, you'll find that you always determine the speed of light to be the same. Here, you've got the added complication that there are really three reference frames: A, B going out, and B coming back. Essentially, what you have here is the classic twin paradox, with A using a really big light beam clock for the time measurement. There's not a paradox for the same reason that there isn't one in the twin paradox: the change in simultaneity of distant events when he turns around.

Grey · #29 (**permalink**) 21-May-2009, 07:26 PM

Quote:

Originally Posted by tommac

If we were unaware that we ( b) changed frames we would miscalculate the speed of light.

But the change in reference frames isn't something subtle that you might miss. You'll feel an enormous acceleration as you change to moving at near-light-speed, and then another huge acceleration as you turn around halfway through your trip, and a third acceleration when you match speeds with A again at your starting point. If you insist on claiming that you remained in the same spot the whole time and that it was really A that was moving, you'll have to take into account the effects of those big "gravitational fields" that appeared and then vanished.

tommac · #30 (**permalink**) 21-May-2009, 07:28 PM

Quote:

Originally Posted by Grey

I assume that you mean that B comes back after 1,460 days, so he's got one day before the light returns from its trip.

So, here's the thing. If you measure the distance something travels in one reference frame, and the time something travels in a different reference frame, you can get any answer you want for its velocity (by making the right choices for those two reference precisely because you can get any answer that you want. Instead, you need to calculate the distance travelled and the time taken using the same reference frame. If you do that, you'll find that you always determine the speed of light to be the same. Here, you've got the added complication that there are really three reference frames: A, B going out, and B coming back. Essentially, what you have here is the classic twin paradox, with A using a really big light beam clock for the time measurement. There's not a paradox for the same reason that there isn't one in the twin paradox: the change in simultaneity of distant events when he turns around.

I totally agree with everything here. And trust me I have learned oodles with this post and one other.

Now the same could be said using a Black hole right? That if you started deep in a well then came out of the well and did the measuring you could be confused as to why you dont measure the speed of light over the period that you released the photon in a deep well, until you measured the distance ( when you were outside of the well ) right?

It would be a very similar experiment.

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