md65536 Posted May 22, 2013 Share Posted May 22, 2013 Are the length contraction and redshift factors the same in GR? Say there are 2 observers A and B, with A on the surface of a planet and B hovering high above. If A sends a signal to B it loses energy climbing out of the gravitational well, and appears red-shifted to B. Also, B's meter stick appears longer than a meter to A, and A's appears shorter than a meter to B. Suppose A sends a signal with a wavelength of 1 m, and B receives it with a wavelength of 1.1 m, does that mean that A sees B's meter stick measuring 1.1 m long, and B sees A's measuring 0.909 m long? I think I must have this wrong just because the factors aren't the same in SR. I guess time dilation must also affect the redshift, but not the length contraction??? Would the difference in redshift and length contraction factors be fully accounted for with time dilation? Are there any examples online that calculate the two factors in a simple case? 1 Link to comment Share on other sites More sharing options...

SamBridge Posted May 22, 2013 Share Posted May 22, 2013 (edited) Actually now that I read about it more, rulers should shrink. But at the same time in order to redshift, B's meter stick would have to expand in order to create a redshift like you said. Length contraction with motion is like a relativistic rotation of a coordinate system, but if the relative distance between points that observer "A" measures near B decreased, it should create a blue shift because it would create an acceleration in the amount of length that contracts between points as you get closer to the source of gravity due to the strength of gravity being proportional to the inverse square of the distance, this is getting confusing. If two objects head towards each other at .99 the speed of light, classically the measured speed should be 1.98, but it's not, and if length contracted it would make it even faster, meter sticks must be able to expand just as they can contract. If observer B would observe that as points become further from his source of gravity, that they the contraction decreases, so if B emitted to A, why wouldn't B observe A's stick being longer just as A observe's B being longer? You can try this though http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html Edited May 22, 2013 by SamBridge Link to comment Share on other sites More sharing options...

xyzt Posted May 22, 2013 Share Posted May 22, 2013 Are the length contraction and redshift factors the same in GR? Say there are 2 observers A and B, with A on the surface of a planet and B hovering high above. If A sends a signal to B it loses energy climbing out of the gravitational well, and appears red-shifted to B. Also, B's meter stick appears longer than a meter to A, and A's appears shorter than a meter to B. Suppose A sends a signal with a wavelength of 1 m, and B receives it with a wavelength of 1.1 m, does that mean that A sees B's meter stick measuring 1.1 m long, and B sees A's measuring 0.909 m long? I think I must have this wrong just because the factors aren't the same in SR. I guess time dilation must also affect the redshift, but not the length contraction??? Would the difference in redshift and length contraction factors be fully accounted for with time dilation? Are there any examples online that calculate the two factors in a simple case? Yes, there are , you need to start from a known solution for EFE. For example, you can start from the Schwarzschild solution. This is how gravitational redshift is calculated. You can do similar calculations in order to figure out the length contraction. Link to comment Share on other sites More sharing options...

swansont Posted May 22, 2013 Share Posted May 22, 2013 http://www.mth.uct.ac.za/omei/gr/chap8/node7.html http://www.mth.uct.ac.za/omei/gr/chap8/node8.html The redshift IS the time dilation. The factors are the same. AFAIK you get the same kind of compensation you do in SR so that locally c will be a constant. 2 Link to comment Share on other sites More sharing options...

xyzt Posted May 22, 2013 Share Posted May 22, 2013 (edited) http://www.mth.uct.ac.za/omei/gr/chap8/node7.html http://www.mth.uct.ac.za/omei/gr/chap8/node8.html The redshift IS the time dilation. The factors are the same. AFAIK you get the same kind of compensation you do in SR so that locally c will be a constant. The time dilation link explains the outcome of the Pound Rebka experiment. Indeed: [math]d\tau=dt \sqrt{1-r_s/r}[/math] So, for two different locations [math]r_1[/math] and [math]r_2[/math] one gets the "time dilation": [math]\frac{d\tau_1}{d \tau_2}=\frac{\sqrt{1-r_s/r_1}}{\sqrt{1-r_s/r_2}}[/math] A perfectly similar calculation produces the "length contraction": [math]\frac{ds_2}{ds_1}=\frac{ \sqrt{1-r_s/r_1}}{\sqrt{1-r_s/r_2}}[/math] Lastly: [math]\frac{\lambda_1}{\lambda_2}=\frac{cd\tau_1}{cd \tau_2}=\frac{\sqrt{1-r_s/r_1}}{\sqrt{1-r_s/r_2}}[/math] Edited May 22, 2013 by xyzt Link to comment Share on other sites More sharing options...

SamBridge Posted May 23, 2013 Share Posted May 23, 2013 (edited) What about length "Expansion" and not contraction? Unless that's just negative contraction. "http://www.mth.uct.ac.za/omei/gr/chap8/node7.html http://www.mth.uct.ac.za/omei/gr/chap8/node8.html The redshift IS the time dilation. The factors are the same. AFAIK you get the same kind of compensation you do in SR so that locally c will be a constant." How exactly does that work that a photon experiences time dilation? If you measured light as having a slower clock, wouldn't it no longer travel at the speed of light? Thus, why isn't the answer some kind of length transformation? It doesn't seem to make sense that a meter stick get's slower, if you contract the length over an observed distance of observer A, you would in effect be saying the same thing as giving it a higher velocity because it would be traveling a greater measured distance relative to observer A... It would make much more sense if a relative meter stick increased, which in a way would stretch out light's wavelength, while the clock of observer B relative to observer A slowed down due to gravity in such a way that they measured light taking the proper time to reach them so that as they would not measure light traveling fast than "c" as it headed right towards B them. Edited May 23, 2013 by SamBridge Link to comment Share on other sites More sharing options...

md65536 Posted May 23, 2013 Author Share Posted May 23, 2013 How exactly does that work that a photon experiences time dilation? If you measured light as having a slower clock, wouldn't it no longer travel at the speed of light? Thus, why isn't the answer some kind of length transformation? It doesn't seem to make sense that a meter stick get's slower, if you contract the length over an observed distance of observer A, you would in effect be saying the same thing as giving it a higher velocity because it would be traveling a greater measured distance relative to observer A... It would make much more sense if a relative meter stick increased, which in a way would stretch out light's wavelength, while the clock of observer B relative to observer A slowed down due to gravity in such a way that they measured light taking the proper time to reach them so that as they would not measure light traveling fast than "c" as it headed right towards B them. I've been trying to think about this in similar terms. Here's what I have so far... Some mistakes in what you wrote: Light doesn't have a slower clock. It doesn't have its own clock at all. The answer *is* length contraction, and corresponding time dilation and redshift... they all fit together so that velocity of light is c, locally according to any observer. One explanation of redshift comes from the equivalence principle. Here I have A with a higher gravitational potential relative to B, and they're relatively at rest; this is equivalent to A and B at the respective bottom and top of a box that is accelerating upward. So the explanation is, if A sends a signal to B, it takes time for the signal to cross the box, during which the box accelerates. So, B at reception is traveling faster away* from the source event (A at transmission), so the signal appears redshifted. *or something like that, I may have screwed up describing this correctly. I think a similar thing can be said about length contraction with the equivalence principle. If B views A's ruler end-to-end, the far end is farther away and appears slightly older. If A and B are in an accelerating box, then... uh... Okay I haven't figured this out, and anyway the link says that length contraction can't be deduced from the equivalence principle... :/ But in terms of redshift and length contraction: Similar to what I wrote in post #1... Suppose B receives a signal with wavelength of 1 m, redshifted from a source wavelength of 0.909 m. The same wave that is 1 m at B, is 0.909 m at A (measured by B?), so any length??? of 1 m at B is 0.909 m at A (measured by B). I guess this length contraction only applies in the direction of the line AB? Similarly in terms of redshift and time dilation: If A sends a signal with frequency 1.1 Hz (1 wavelength every 0.909 units of time) and B receives it as 1 Hz, then A's clock ticks 0.909 times for every tick of B's clock. This is just a sloppy interpretation of what I think are the right answers, forcefully made to fit into a hand-wavey explanation. I don't think I've got it yet. I agree, intuitively it seems like a shrunk ruler means less time for light to cross means a faster ticking clock, and I've always guessed wrong about it. I think it can be counter-productive to try to figure it out intuitively, because it's way too easy to come up with intuitive explanations for incorrect physics! 1 Link to comment Share on other sites More sharing options...

swansont Posted May 23, 2013 Share Posted May 23, 2013 A climbs out of a potential well and arrives where B is. B thinks A's clock was running slow, but from A's perspective the distance was smaller, and so B still thinks c has the right value. Exactly the same as a SR example. Link to comment Share on other sites More sharing options...

SamBridge Posted May 23, 2013 Share Posted May 23, 2013 A climbs out of a potential well and arrives where B is. B thinks A's clock was running slow, but from A's perspective the distance was smaller, and so B still thinks c has the right value. Exactly the same as a SR example. But if B's clock wasn't slowed down to counter the length contraction, reducing the relative distance from B to A would be the same thing as saying that light traveled a greater distance than one light second per second. A is the one in the gravitational potential, to A it's cocks would run normally, but because B is in a lesser gravitational potential or a greater height off the ground, why wouldn't that make B's clock run faster relative to A? Link to comment Share on other sites More sharing options...

swansont Posted May 23, 2013 Share Posted May 23, 2013 But if B's clock wasn't slowed down to counter the length contraction, reducing the relative distance from B to A would be the same thing as saying that light traveled a greater distance than one light second per second. This translates to "If relativity didn't apply, things would be different", at least in my reading of it. A is the one in the gravitational potential, to A it's cocks would run normally, but because B is in a lesser gravitational potential or a greater height off the ground, why wouldn't that make B's clock run faster relative to A? B's clocks do run faster than A's. But since A also experiences a length contraction, the distance/time for the light is still the same. Link to comment Share on other sites More sharing options...

md65536 Posted May 23, 2013 Author Share Posted May 23, 2013 B's clocks do run faster than A's. But since A also experiences a length contraction, the distance/time for the light is still the same.Does this mean that the distance A measures to B is smaller than the distance B measures to A? (I always got this backward, guess I still don't get it.) In terms of a twin paradox, if A and B are together, and in negligible time A enters a gravity well and spends considerable time there, then returns in negligible time, A has aged less. If A and B are continuously bouncing signals off each other, A might say "each signal takes 1 second round trip to travel, so the distance is 1 light second" and ages 100 seconds per 100 signals, while B might say "each signal takes 1.1 seconds, the distance is 1.1 light seconds," and ages 110 seconds per the same 100 signals. Can they apply the local speed of light universally like that (neglecting inflation)? It still doesn't make sense to me, because... say you place metersticks between A and B, say 1000 of them, I would think that each measures a local ruler as 1 meter, and both agree that the rulers at B are bigger than the rulers at A ("rulers shrink in a gravitational field"???). Then A would measure the distance as being greater than 1000 m and B would measure it as less than 1000 m. Am I mistaking the meaning of rulers shrinking in a gravitational field? Is it A who observes a meterstick brought into the field as smaller than a meter, while B observes it as 1 m? 1 Link to comment Share on other sites More sharing options...

xyzt Posted May 23, 2013 Share Posted May 23, 2013 Does this mean that the distance A measures to B is smaller than the distance B measures to A? (I always got this backward, guess I still don't get it.) Do you think that the radar distance Earth-Moon is different from the radar distance Moon-Earth? Link to comment Share on other sites More sharing options...

md65536 Posted May 23, 2013 Author Share Posted May 23, 2013 Do you think that the radar distance Earth-Moon is different from the radar distance Moon-Earth?Not if both are measured from one place. Measured from Earth vs from moon, yes... isn't that what length contraction implies? 1 Link to comment Share on other sites More sharing options...

xyzt Posted May 23, 2013 Share Posted May 23, 2013 Not if both are measured from one place. Measured from Earth vs from moon, yes... isn't that what length contraction implies? Let's try again: Radar_Distance_1: From Earth->Moon->Earth Radar_Distance_2: Moon->Earth->Moon Obviously, the two distances are not measured "from one place". Are they equal or not? Link to comment Share on other sites More sharing options...

swansont Posted May 23, 2013 Share Posted May 23, 2013 Does this mean that the distance A measures to B is smaller than the distance B measures to A? (I always got this backward, guess I still don't get it.) I think so, yes. I'm not well-versed in GR beyond a few of the "simple" effects. But AFAICT it's the same as if A were moving toward B. A would measure a shorter distance than B measures. In terms of a twin paradox, if A and B are together, and in negligible time A enters a gravity well and spends considerable time there, then returns in negligible time, A has aged less. Right. Sitting stationary in a gravitational potential (i.e. a situation where you feel an acceleration) is not an inertial frame. Let's try again: Radar_Distance_1: From Earth->Moon->Earth Radar_Distance_2: Moon->Earth->Moon Obviously, the two distances are not measured "from one place". Are they equal or not? No, just as a moving rocket ship measures a shorter distance of travel than the stationary observer. Length depends on your frame of reference. Link to comment Share on other sites More sharing options...

xyzt Posted May 23, 2013 Share Posted May 23, 2013 (edited) No, just as a moving rocket ship measures a shorter distance of travel than the stationary observer. Length depends on your frame of reference. This is false, just as the first answer you gave in this entry, the two radar distances are the same. If they were different, the whole GR would be falsified and lunar ranging method would be invalid, since the results would depend on where the signal originated. Edited May 23, 2013 by xyzt Link to comment Share on other sites More sharing options...

swansont Posted May 23, 2013 Share Posted May 23, 2013 This is false, just as the first answer you gave in this entry, the two radar distances are the same. If they were different, the whole GR would be falsified and lunar ranging method would be invalid, since the results would depend on where the signal originated. How so? The radar systems are in different gravitational potentials. There's no symmetry that demands they get the same answer. A clock on earth and a clock on the moon will not keep the same time. AFAIK there's no legitimate reason to think they should measure the same distances. Lunar ranging gives the distance from the earth to the moon. It's valid for anyone on the earth. Not that the difference would be a big deal; the length contraction would be small. In GPS, for example, it is ignored. IIRC it's about 5 cm, so much smaller than the distance errors from the timing signals. But it's not zero. Link to comment Share on other sites More sharing options...

xyzt Posted May 23, 2013 Share Posted May 23, 2013 How so? The radar systems are in different gravitational potentials. There's no symmetry that demands they get the same answer. A clock on earth and a clock on the moon will not keep the same time. AFAIK there's no legitimate reason to think they should measure the same distances. Lunar ranging gives the distance from the earth to the moon. It's valid for anyone on the earth. Not that the difference would be a big deal; the length contraction would be small. In GPS, for example, it is ignored. IIRC it's about 5 cm, so much smaller than the distance errors from the timing signals. But it's not zero. The "difference in gravitational potential" has nothing to do with it. Actually, the gravitational potential has nothing to do with it. Imagine an observer floating somewhere between the Earth and the Moon, such an observer would be inertial. For "it" the distance Earth-Moon is equal to the distance Moon-Earth: [math]d(EM)=d(ME)=D[/math]. A radar beam sent from the Earth will return to Earth after bouncing off the Moon in the time :[math]\tau_{EME}=\frac{2D}{c}[/math]. A radar beam sent from the Moon will return to the Moon after bouncing off the Earth in the time :[math]\tau_{MEM}=\frac{2D}{c}[/math]. Even if you considered for a second that (coordinate) light speed is not isotropic, the effects of such "anisotropy" cancel out due to the fact that light speed "climbs out" of Earth and Moon gravitational potentials for both trips. But light speed is not anisotropic and we are not using coordinate light speed anyways, so, [math]\tau_{EME}=\tau_{MEM}[/math] and this ill there is to it. Link to comment Share on other sites More sharing options...

swansont Posted May 23, 2013 Share Posted May 23, 2013 The "difference in gravitational potential" has nothing to do with it. Actually, the gravitational potential has nothing to do with it. We are talking about general relativity, right? A radar beam sent from the Earth will return to Earth after bouncing off the Moon in the time :[math]\tau_{EME}=\frac{2D}{c}[/math]. A radar beam sent from the Moon will return to the Moon after bouncing off the Earth in the time :[math]\tau_{MEM}=\frac{2D}{c}[/math]. The clock on Earth used to measure this value is ticking slower/faster than the clock on the moon, as it is deeper/shallower in a gravitational potential. Thus, the Earth observer will measure a shorter time and conclude that D is smaller. The only way they come to the same answer is if by some accident the two observers have the same gravitational potential. Link to comment Share on other sites More sharing options...

xyzt Posted May 23, 2013 Share Posted May 23, 2013 (edited) We are talking about general relativity, right? The clock on Earth used to measure this value is ticking slower/faster than the clock on the moon, as it is deeper/shallower in a gravitational potential. Thus, the Earth observer will measure a shorter time and conclude that D is smaller. The only way they come to the same answer is if by some accident the two observers have the same gravitational potential. Yes, we are talking GR, a subject that I am very familiar with. I can tell you in no uncertain terms that your attempts at calculating are incorrect. Here is the deal: A probe particle will : -climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math] -on the return trip, it will climb out of the gravitational well of the Moon and fall into the gravitational well of the Erath. This will take a time [math]\tau_{ME}[/math] The other probe particle will do the reverse trip: -climb out of the gravitational well of the Moon and fall into the gravitational well of the Earth. This will take a time [math]\tau_{ME}[/math] -climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math] The total time for either particle (the "radar distance") is the same, [math]\tau_{EM}+\tau_{ME}[/math], just the order of addition is reversed. Now, for people interested in the subject, I could show the the expressions for [math]\tau_{ME}[/math] and [math]\tau_{EM}[/math]. For example: [math]c \tau_{ME}=\sqrt{\frac{r_0}{r_E}}(r_0 arctan \sqrt{\frac{r}{r_0-r}}+\sqrt{r(r_0-r)})[/math] where [math]r_E[/math] is the Earth Schwarzshild radius, [math]r_0=D[/math] and [math]r[/math] is the Earth radius. The derivation is quite complicated but I can attach a file that shows how it is done. Likewise: [math]c \tau_{EM}=\sqrt{\frac{r_0}{r_M}}(r_0 arctan \sqrt{\frac{r}{r_0-r}}+\sqrt{r(r_0-r)})[/math] where [math]r_M[/math] is the Moon Schwarzshild radius. The presence of the Schwarzschild radii encapsulates the gravitational effects on clocks. Edited May 24, 2013 by xyzt Link to comment Share on other sites More sharing options...

md65536 Posted May 23, 2013 Author Share Posted May 23, 2013 (edited) The total time for either particle (the "radar distance") is the same, [math]\tau_{EM}+\tau_{ME}[/math], just the order of addition is reversed.How do you go from your example using proper time of a single observer and extend it to the measures of time made by two different clocks, which is what we were talking about here? Edit: By the way, radar time is measured by the radar transceiver observer, not a traveling particle. Edited May 23, 2013 by md65536 1 Link to comment Share on other sites More sharing options...

swansont Posted May 23, 2013 Share Posted May 23, 2013 I am not contending that T_{EM} = T_{ME }for any single observer. You've not addressed my point that the two clocks used to measure the time run at different rates. Why does that not matter? Link to comment Share on other sites More sharing options...

SamBridge Posted May 23, 2013 Share Posted May 23, 2013 This translates to "If relativity didn't apply, things would be different", at least in my reading of it. Um, no, it translates what to based on my knowledge, what the outcome should be. It length contracts in the direction of light while light is heading towards something like say observer A, since time dilation alone isn't the complete picture, light should be measured as traveling faster than C. But it doesn't, it can only mean the relative distance expands, which I imagine would be negative length contraction. Link to comment Share on other sites More sharing options...

md65536 Posted May 23, 2013 Author Share Posted May 23, 2013 Um, no, it translates what to based on my knowledge, what the outcome should be. It length contracts in the direction of light while light is heading towards something like say observer A, since time dilation alone isn't the complete picture, light should be measured as traveling faster than C. But it doesn't, it can only mean the relative distance expands, which I imagine would be negative length contraction.Wouldn't the same faulty reasoning apply to SR? If there's time dilation in SR, wouldn't you predict expanded distances? 1 Link to comment Share on other sites More sharing options...

xyzt Posted May 23, 2013 Share Posted May 23, 2013 I am not contending that T_{EM} = T_{ME }for any single observer. You've not addressed my point that the two clocks used to measure the time run at different rates. Why does that not matter? Very simple, because the formulas that I have given you already correctly incorporates the effect on gravitational fields on clocks (see the presence of the [math]r_E[/math]?). How do you go from your example using proper time of a single observer and extend it to the measures of time made by two different clocks, which is what we were talking about here? Edit: By the way, radar time is measured by the radar transceiver observer, not a traveling particle. [math]\tau[/math] is proper time, this is an invariant, remember? we went over this in the other thread, the one where you claiming that acceleration is not important in the twins paradox. Very simple, because the formulas that I have given you already correctly incorporates the effect on gravitational fields on clocks (see the presence of the [math]r_E[/math]?). [math]\tau[/math] is proper time, this is an invariant, remember? we went over this in the other thread, the one where you claiming that acceleration is not important in the twins paradox. One more thing, the times I have given are valid for radar, you only need to understand that in GR, the test probe is infinitely small, so your contention is invalid. -1 Link to comment Share on other sites More sharing options...

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