Benish 3 Posted December 19, 2013 Share Posted December 19, 2013 (edited) I think a lot of confusion is avoided about this question by bearing in mind two things: 1) The equivalence principle (EP) and arguments based on accelerating frames of reference have a limited range of applicability. That's because the acceleration side of the "equivalence" uses kinematic arguments; whereas the gravitational side deals with a circumstance that is at least stationary, and usually regarded as static. A simple example to illustrate the limited range of the EP is that, in an accelerating rocket, the universe will gradually appear hotter and hotter in the direction of motion. On the surface of a gravitating body no such effect occurs. 2) Owing to the staticness of the Schwarzschild field, the path of a light ray within is often treated as a pure optical problem, the field being treated as a medium of variable refractive index, n = 1 / (1 - 2GM / rc^2)^.5. For references to the latter approach see the classic book by C. Moller, Relativity (Clarendon, 1972) p. 394; or the more recent paper by Ashby and Bertotti: http://arxiv.org/abs/0912.2705 In accordance with the optical analogy, the coordinate speed of light depends only on position (slower than c in both radial directions the greater the argument of the coefficient). Whereas the proper speed of light is always c. Edited December 19, 2013 by Benish 3 Link to post Share on other sites

xyzt 81 Posted December 19, 2013 Share Posted December 19, 2013 I think a lot of confusion is avoided about this question by bearing in mind two things: 1) The equivalence principle (EP) and arguments based on accelerating frames of reference have a limited range of applicability. That's because the acceleration side of the "equivalence" uses kinematic arguments; whereas the gravitational side deals with a circumstance that is at least stationary, and usually regarded as static. A simple example to illustrate the limited range of the EP is that, in an accelerating rocket, the universe will gradually appear hotter and hotter in the direction of motion. On the surface of a gravitating body no such effect occurs. 2) Owing to the staticness of the Schwarzschild field, the path of a light ray within is often treated as a pure optical problem, the field being treated as a medium of variable refractive index, n = 1 / (1 - 2GM / rc^2)^.5. For references to the latter approach see the classic book by C. Moller, Relativity (Clarendon, 1972) p. 394; or the more recent paper by Ashby and Bertotti: http://arxiv.org/abs/0912.2705 In accordance with the optical analogy, the coordinate speed of light depends only on position (slower than c in both radial directions the greater the argument of the coefficient). Whereas the proper speed of light is always c. Post 3 says essentially the same thing (with more detailed math). Link to post Share on other sites

md65536 383 Posted December 20, 2013 Share Posted December 20, 2013 In accordance with the optical analogy, the coordinate speed of light depends only on position (slower than c in both radial directions the greater the argument of the coefficient). Whereas the proper speed of light is always c.What is proper speed? Link to post Share on other sites

xyzt 81 Posted December 20, 2013 Share Posted December 20, 2013 (edited) What is proper speed? [math]\frac{dr}{d \tau}[/math] Edited December 20, 2013 by xyzt 1 Link to post Share on other sites

Iggy 344 Posted December 20, 2013 Share Posted December 20, 2013 (edited) Apologies for the delay in responding. I've been away from the site awhile. Hectic Holidays. There is nothing special about "the v=0.6c frame". Agreed. Besides making the gamma factor an easy rational number, there's nothing significant about 0.6 Also what I said, is that you tried to demonstrate that the distance stayed constant in the launcher frame (and you failed). [your bold] But your attempt at proving that the "distance is constant" used length contraction, so you are contradicting yourself or you do not realize that you have actually (as you admitted at the end of your post) actually calculated the length in the lab frame. See your [math]L=L_0 \sqrt {1-v^2}[/math] [your bold] I didn't try (and fail) to calculate the v=.6 distance relative to the lab frame, nor did I actually calculate the v=.6 distance relative to the lab frame. I very clearly showed (with JVNY's diagram), calculated (with equations), and explained (with words) that the *proper distance* remained constant, not the spatial distance in the lab frame. Let me find the post... Given JVNY's derivation (and illustration) of his thought experiment: The proper distance at t=0 is given along the x axis as: [math]\sigma = 1 - 0.5 = 0.5[/math] The proper distance at the dotted line between the accelerated hyperbolic worldlines is given from the numbers in this frame by length contraction: [math]\sigma = L_{0}\sqrt{1-v^{2}} = (1.25-0.625)(\sqrt{1-0.6^{2}}) = 0.5[/math] So, I'm calculating the *proper distance* (an invariant quantity) between the fore and aft observers. The proper distance is what an at-rest observer measures between two simultaneous events. The first equation calculates the proper distance by nothing more than taking the difference in x coordinates in the lab frame since the rest distance is the proper distance at t=0 in this frame. The answer is 0.5. One could more formally use the Minkowski interval to get the same answer: [math]\sigma = \sqrt{\Delta x^2 - \Delta t^2} = \sqrt{0.5^2 - 0^2} = 0.5[/math] The second equation finds the proper distance between v=0.6 events (along the dotted line). The events are simultaneous from the accelerating observers' perspective (i.e. in an inertial frame with v=0.6c rightward velocity relative to the launch frame). It is done by length contraction, [math]\sigma = L_{0}\sqrt{1-v^{2}} = (1.25-0.625)(\sqrt{1-0.6^{2}}) = 0.5[/math] where [math]L_{0}[/math] is the spatial distance between events in the lab frame (where the events are not simultaneous), and [math]\sqrt{1-v^{2}}[/math] is the inverse of the v=0.6c gamma factor. Multiplying them gives [math]\sigma[/math] (the proper distance where the events are simultaneous). The answer is again [math]\sigma = 0.5[/math] meaning the *proper distance* is constant at the two times we sampled. And, considering the objection to length contraction above, we can again use the spacetime interval to verify: [math]\sigma = \sqrt{\Delta x^2 - \Delta t^2} = \sqrt{0.625^2 - 0.375^2} = 0.5[/math] Sorry to go on and on, but it's worth explaining in detail the actual purpose and successful results above. If you are still saying that the proper distance between uniformly accelerating observers can't be constant if the observers have different accelerations then it seems like you would have to explain why these observers with unequal accelerations have constant proper distance at two arbitrary times. That is to say, why does the lead observer measure the distance to the aft observer as 0.5 at both tau=0 and tau=0.6931? The same calculation can be done, and the same question asked, at any velocity shy of c. The motion is not Born rigid and this problem has nothing to do with Born rigidity, you are dealing with three particles that all have different speeds in any frame you consider, so, there is relative motion between them. As such, their relative distances, contrary to your claims, cannot be constant. Quite the opposite, it is variable. In the launcher frame, the speeds are [math]v_i=\frac{a_i}{\sqrt{1+(a_it/c)^2}}[/math], i.e. they are all different (and they do not depend in any fashion of the initial positions [math]x_{0i}[/math]). It looks like you're missing a variable in the numerator of that equation. The equation finds the velocities of the rockets relative to the launch frame, so it can only say when things have the same velocity in that one frame. The start of the thought experiment (t=0) should be the only time that is true. In other words, that equation proves to us that... ...the red dot and the blue dot are not horizontally aligned (neither is any other velocity we could mark on the hyperbolic worldlines except v=0 at t=0). However, the green line is the line of simultaneity relative to the hyperbolic observers at those events, not a horizontal line. You wouldn't expect the events to be simultaneous in the lab frame -- you expect it in the 'momentarily comoving inertial frame' As far as born rigidity... The uniform acceleration of Born rigid particles means the particles maintain their proper distance as they accelerate. You can imagine tying a string between the rockets/particles and having each accelerate such that the string doesn't strain. The OP imagines just such a system, and it is well-published that the following is true in such a system, Trailing sections of the rod must undergo a greater acceleration in order to maintain Born rigidity with the leading end, and the required acceleration is inversely proportional to the distance from the pivot event Born Rigidity, Acceleration, and Inertia On the other hand, accelerating rockets that do maintain equivalent velocities in the lab frame would typically be called 'Bell's spaceships'. The rockets maintain equivalent accelerations and constant distance in the lab frame, but do not maintain a constant proper distance as they accelerate. In that case a string tied between them would stretch and break. In the frame co-moving with any particle, the calculation is a little more complicated but the conclusion is the same, there is relative motion between the particles because, their speeds, as calculated from any such frames are variable and not equal. Indeed, the proper speed of the particle i=1 is [math]v_1=c * tanh (a_1 \tau/c)[/math]. That equation will tell you the velocity of the launch frame relative to the observer (it does it with tau rather than t is the only difference between it and the equation you just gave). By the principle of reciprocity, it gives numerically equal results. It is problematic for the same reasons I just was saying. It does not tell you the speed of one particle in a frame comoving with another. The proper speed of particle i=2 is [math]v_2=c * tanh (a_2 \tau'/c)[/math]. [math]\tau'[/math] and [math]\tau[/math] are connected by the coordinate time [math]t[/math]: [math]t=\frac{c}{a_1} sinh (a_1 \tau/c)[/math] Again, the equation you have for 'coordinate time' is relative to the launch frame. I've said many times that the distance doesn't remain constant in any particular inertial frame. Because they're accelerating, the frame of measurement has to change as their velocity changes. The easiest way to do this is to set up an accelerating coordinate system (e.g. Rindler coordinates) The solution in which we're interested is derived just above the paragraph and link I'll copy here: This is the magnitude of the constant proper acceleration which the leading end of the rod must undergo in order to meet the stated conditions. Also, the reciprocal of this value represents the distance of the leading end of the rod from the pivot event. Trailing sections of the rod must undergo a greater acceleration in order to maintain Born rigidity with the leading end, and the required acceleration is inversely proportional to the distance from the pivot event http://www.mathpages.com/home/kmath422/kmath422.htm Edited December 20, 2013 by Iggy 2 Link to post Share on other sites

Iggy 344 Posted December 21, 2013 Share Posted December 21, 2013 Here is a recap and a reopening of the question of the coordinate speed of light in the accelerating frame as a function of the distance from the rear (which should be like the distance from a gravitational mass). I think that the result utilizes parts of the posts from each of md65536, Iggy and xyzt. As md65536 suggests, consider a row of separate ships at rest in an inertial frame that are then simultaneously accelerated (rather than trying to accelerate a single object that has the same length as the row). Each ship accelerates at a proper rate inversely proportional to its distance from the origin. For example, the ship that starts at x=0.5 accelerates at a=1/0.5, or a=2. The ship that starts at a=1 accelerates at a=1/1, or a=1. As Iggy states, under these conditions the ships undergo "Born rigid" motion: they remain together in a shared accelerating reference frame; they keep their same proper distance in that accelerating frame; they agree on simultaneity; they always have the same velocity relative to the inertial frame; and their worldlines in the inertial frame are hyperbolas. Their clocks do not run at the same rate in their frame. The farther forward a ship is, the faster its clock runs. However, the rate is proportional to distance, so each ship behind the front can program its clock to automatically scale up its elapsed time proportionately and thus keep its clock synchronized with the front clock. Iggy advises that the front clock's time is the conventional measure of time in the accelerating frame; it is called the coordinate time of the frame. The time that we measure in that frame when we determine the speed of light in the frame is coordinate time, and so I think that the speed of light in that frame is its coordinate speed. As the ships start to accelerate, the front ship flashes a light toward the rear. The Minkowski diagram for the inertial frame is as follows (ships' worldlines are hyperbolas; light flash's worldline is the arrow): inertial.png Because the ships are all in the same (accelerating) reference frame, we can Lorentz transform the light's path into the ships' frame. The light's worldline is no longer a straight 45 degree line. Rather, it is curved: worldline.png To make the effect clearer, we can diagram the marginal amount of coordinate time that it takes for the light to travel successive equal segments in the ships' frame (here done using distance increments of 0.02). It takes successively greater coordinate time to travel equal segments the farther toward the rear of the row (left in the diagram) that the light is traveling. In coordinate time, light is always traveling less than c, and increasingly less as it moves rearward: marginal.png Notice that the curve flattens out toward the front of the row (toward the right). This suggests that the coordinate speed of light over successive equal increments is higher toward the front (the right), but at a decreasing rate. It suggests that the coordinate speed will not exceed c. This agrees with xyzt, who states that the speed of light is not proportional to the distance from the origin, and more specifically that the speed of light will not exceed c at distances farther away from the rear. Xyzt prefers not to use calculations, but Iggy you can confirm that the coordinate speed of light does not exceed c as follows. Start with two ships, one at x=99.98 and the other at x=100. They accelerate simultaneously and flash a light at each other. Their acceleration rates are 1/99.98 and 1/100, respectively. Next, as you did before use the relativistic rocket formulas to determine the coordinate time that each light flash takes to cross the 0.02 proper distance and strike the other ship. You should find it to be 0.020002 (same in both directions). The coordinate speed of light is still less than c (0.02/0.020002), but it is getting very close to c. This is just an example, but I expect that those of you with better math skills than I have can prove that at any distance away from the origin the coordinate speed will remain below c (I think that it will asymptotically approach c, but I cannot be sure). I think the thing is that accelerated coordinates usually transform away from the horizontal Cartesian-like time coordinate. My understanding is that they should look more like... from this site. Or... From wiki. I think everything you're saying about the spatial coordinate is good, but the time coordinate would spread out as it fans right to reflect how clocks tick faster further from the origin. A light ray at 45 degrees around the middle of these diagrams would cross about one unit of t for every unit of x, but if you imagine traveling far from the origin where the temporal lines are really spread out, a light ray could cross a lot of hyperbolic spatial lines between each temporal coordinate line. Hopefully I'm making sense. I found, also, an interesting site comparing the speed of light in a uniformly accelerating frame, and a uniform gravitational field... Speed of Light in a Gravitational Field Actually, this site: http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/general_relativity_pathway/uniform_acceleration.html about the diagrams, for sure 2 Link to post Share on other sites

Iggy 344 Posted December 22, 2013 Share Posted December 22, 2013 Start with two ships, one at x=99.98 and the other at x=100. They accelerate simultaneously and flash a light at each other. Their acceleration rates are 1/99.98 and 1/100, respectively. Next, as you did before use the relativistic rocket formulas to determine the coordinate time that each light flash takes to cross the 0.02 proper distance and strike the other ship. You should find it to be 0.020002 (same in both directions). The coordinate speed of light is still less than c (0.02/0.020002), but it is getting very close to c. This is just an example, but I expect that those of you with better math skills than I have can prove that at any distance away from the origin the coordinate speed will remain below c (I think that it will asymptotically approach c, but I cannot be sure). I forgot to answer this part, sorry. It seems you're finding the coordinate speed of light where the t axis is made by the worldlines of the accelerating ships, and the x axis is made by the lines of simultaneity in the inertial frame. The problem I tried to touch on in a rush yesterday is that time dilation isn't accounted for in that case. As an analogy, if someone sat near the event horizon of a black hole with rockets firing at enormous acceleration in order to hold their position then they would see the universe above them unfold at an extremely rapid rate. Light would move from one side of a galaxy to the other in an instant (from their perspective). If they set themselves up as the origin of a coordinate system then they'd need to account for time dilation to account for the increased coordinate speed of light above them. The natural way to do this in the case of acceleration is rindler coordinates in which the speed of light is [math]v_x = x \cdot g / c[/math]. In your example, an observer at x=1 is considering the speed of light of someone at x=100. The speed of light in that case is, [math]v_x = 100 \cdot 1 = 100[/math] In other words, the speed of light is one hundred times less for someone one hundred times closer to the rindler horizon. Let's do that in real-world numbers... Let's say I am in a rocket accelerating at 100,000 km/s^{2}. To my right is a line of rockets with less and less proper acceleration, and to my left is a line with greater and greater such that we maintain proper distance. I calculate my distance to the Rindler horizon... [math]x_0 = \frac{c^2}{g_0} = \frac{(299792 \ km/s)^2}{100000 \ km/s^2} = 898752.4 \ km[/math] 100 times this distance is 89,875,243 km, and the coordinate speed of light there would be... [math]v_x = \frac{xg}{c} = \frac{89,875,243 \cdot 100,000}{299792} = 29,979,200 \ km/s[/math] one hundred times the local speed of a ray of light. 2 Link to post Share on other sites

xyzt 81 Posted December 24, 2013 Share Posted December 24, 2013 (edited) Apologies for the delay in responding. I've been away from the site awhile. Hectic Holidays. Agreed. Besides making the gamma factor an easy rational number, there's nothing significant about 0.6 I didn't try (and fail) to calculate the v=.6 distance relative to the lab frame, nor did I actually calculate the v=.6 distance relative to the lab frame. I very clearly showed (with JVNY's diagram), calculated (with equations), and explained (with words) that the *proper distance* remained constant, not the spatial distance in the lab frame. Let me find the post... So, I'm calculating the *proper distance* (an invariant quantity) between the fore and aft observers. The proper distance is what an at-rest observer measures between two simultaneous events. The first equation calculates the proper distance by nothing more than taking the difference in x coordinates in the lab frame since the rest distance is the proper distance at t=0 in this frame. The answer is 0.5. One could more formally use the Minkowski interval to get the same answer: [math]\sigma = \sqrt{\Delta x^2 - \Delta t^2} = \sqrt{0.5^2 - 0^2} = 0.5[/math] The second equation finds the proper distance between v=0.6 events (along the dotted line). The events are simultaneous from the accelerating observers' perspective (i.e. in an inertial frame with v=0.6c rightward velocity relative to the launch frame). It is done by length contraction, [math]\sigma = L_{0}\sqrt{1-v^{2}} = (1.25-0.625)(\sqrt{1-0.6^{2}}) = 0.5[/math] where [math]L_{0}[/math] is the spatial distance between events in the lab frame (where the events are not simultaneous), and [math]\sqrt{1-v^{2}}[/math] is the inverse of the v=0.6c gamma factor. Multiplying them gives [math]\sigma[/math] (the proper distance where the events are simultaneous). The answer is again [math]\sigma = 0.5[/math] meaning the *proper distance* is constant at the two times we sampled. And, considering the objection to length contraction above, we can again use the spacetime interval to verify: [math]\sigma = \sqrt{\Delta x^2 - \Delta t^2} = \sqrt{0.625^2 - 0.375^2} = 0.5[/math] Sorry to go on and on, but it's worth explaining in detail the actual purpose and successful results above. If you are still saying that the proper distance between uniformly accelerating observers can't be constant if the observers have different accelerations then it seems like you would have to explain why these observers with unequal accelerations have constant proper distance at two arbitrary times. That is to say, why does the lead observer measure the distance to the aft observer as 0.5 at both tau=0 and tau=0.6931? The simple answer is that you are trying to calculate the proper length using the Minkowski metric [math]ds^2=dx^2-(cdt)^2[/math], when , in reality, the metric is Rindler: [math]ds^2=(d \zeta)^2-(1+\frac{g \zeta}{c^2})^2 (c d \tau)^2[/math] where [math]\tau, \zeta[/math] are respecively, the proper time, proper distance measured by one of the accelerated observers. I have pointed out your mistake repeatedly in this thread, in different ways, this is yet another way to point out why your claims are wrong. They are just as wrong as your even more serious claim that the Shapiro delay would somehow be explained by light "slowing down" in the presence of a gravitating body. I mentioned before that I will not even respond anymore to your calculations on the JVNY exercise until you retract the more egregious claim about the Shapiro delay but I broke my promise since you persist in the claim that the distance between the rockets is constant. I forgot to answer this part, sorry. It seems you're finding the coordinate speed of light where the t axis is made by the worldlines of the accelerating ships, and the x axis is made by the lines of simultaneity in the inertial frame. The problem I tried to touch on in a rush yesterday is that time dilation isn't accounted for in that case. The natural way to do this in the case of acceleration is rindler coordinates in which the speed of light is [math]v_x = x \cdot g / c[/math]. In your example, an observer at x=1 is considering the speed of light of someone at x=100. The speed of light in that case is, Once again, this is the coordinate speed of light, an entity devoid of any physical meaning. As explained already in post 3, this entity can take any value (in GR, it is dependent on the coordinates [math]r, \theta[/math] ), in accelerated frames, is dependent on the acceleration and on the coordinate x. The elegant way of showing that is to remember that light follows null geodesics and to make [math]ds=0[/math] in the Rindler metric above. You would get [math]d \zeta=(1+g \zeta/c^2)(c d \tau)[/math] That is, the correct formula of the coordinate speed of light for the accelerated observer is [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math]. The natural way to do this in the case of acceleration is rindler coordinates in which the speed of light is [math]v_x = x \cdot g / c[/math]. In your example, an observer at x=1 is considering the speed of light of someone at x=100. According to your above logic, the observer at [math]x=0[/math] will measure the speed of light to be [math]v_x=0[/math], which is clearly absurd. The reason is that your formula, wherever you picked it up from (since you show no derivation) is incorrect. The correct formula is, as derived above , [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math]. This means that an observer at [math]\zeta=0[/math] measures the speed of light to be [math]c[/math] , as one would expect. As the observer accelerates away from the light source, the coordinate speed of light decreases, following the rule [math]\frac{d \zeta}{d \tau}=c(1-g | \zeta |/c^2)[/math]. At the Rindler horizon [math]\zeta=\frac{c^2}{g}[/math] so [math]c=0[/math]. A pulse of light emitted when the accelerated observer is already past the Rindler horizon, will never "catch up" with the accelerated observer. On the other hand, if the observer accelerates towards the light source, then [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math] teaches you that the coordinate speed of light "increases" linearly, as the observer "speeds up" towards the light source (and away from the launch pad). Notice how similar are these formulas with the ones derived from the Schwarzschild metric in post 3. Edited December 24, 2013 by xyzt Link to post Share on other sites

DimaMazin 93 Posted December 24, 2013 Share Posted December 24, 2013 I think a lot of confusion is avoided about this question by bearing in mind two things: 1) The equivalence principle (EP) and arguments based on accelerating frames of reference have a limited range of applicability. That's because the acceleration side of the "equivalence" uses kinematic arguments; whereas the gravitational side deals with a circumstance that is at least stationary, and usually regarded as static. A simple example to illustrate the limited range of the EP is that, in an accelerating rocket, the universe will gradually appear hotter and hotter in the direction of motion. On the surface of a gravitating body no such effect occurs. 2) Owing to the staticness of the Schwarzschild field, the path of a light ray within is often treated as a pure optical problem, the field being treated as a medium of variable refractive index, n = 1 / (1 - 2GM / rc^2)^.5. I think variable refractive index should be = 1/(1-2GM/rc^{2})^{1 } Link to post Share on other sites

JVNY 8 Posted December 24, 2013 Author Share Posted December 24, 2013 As an analogy, if someone sat near the event horizon of a black hole with rockets firing at enormous acceleration in order to hold their position then they would see the universe above them unfold at an extremely rapid rate. Light would move from one side of a galaxy to the other in an instant (from their perspective). If they set themselves up as the origin of a coordinate system then they'd need to account for time dilation to account for the increased coordinate speed of light above them. I agree with this. What I did in the other exercises was to proportionately adjust all of the clocks' tick rates to the same rate (the farther rearward, the more elapsed ticks show on the clock than actual ticks of time for that rocket's clock). I think that you would call this putting all of the clocks on the same coordinate time. That eliminates the time dilation effect. It leaves only the distance effect. If you calculate the speed of light using the proper distance, then you get the speed of light slowing the closer the rocket is to the rear of the accelerating row (or the deeper in a gravitational well). I think that xyzt objects to that description: Once again, this is the coordinate speed of light, an entity devoid of any physical meaning. As explained already in post 3, this entity can take any value (in GR, it is dependent on the coordinates Having heard from a lot of members who have been kind enough to respond, and done a lot of reading, I think that it is reasonable to consider two descriptions to be true. First, the speed of light is constant. In gravity, the distance that one is measuring when doing a Shapiro delay experiment is greater when the sun is between the earth and the other planet than when the the sun is not. The greater time that the experimenter on earth measures in the former case is due to the fact that the distance is greater. Second, in the earth experimenter's frame the speed of light is not constant; the speed slows when the sun is between the earth and the other planet. Call this the coordinate speed of light; that is fine. But it is not "devoid of any physical meaning." It is measurable by the earth experimenter. We cannot simply change our coordinates to get another value -- we live on earth, we send signals out into space, and we need to take this effect into account when doing calculations that depend on the amount of our proper time that passes before the signals return (or before the signals reach a spacecraft that we have sent into space). Here is an excellent example of the two views, both from the same person. Neil Ashby is responsible for the GR corrections in the GPS system: http://en.wikipedia.org/wiki/Neil_Ashby Here is a piece that he coauthored about GPS corrections: http://www.gps.gov/cgsic/meetings/2012/weiss2.pdf This notes that there are relativity terms that are becoming important in the GPS system. Slide 2 describes two effects that are not currently used in the GPS corrections: "Signal Propagation Delay" caused by "Earth's Gravitational Field." The first of the two is expressly called the Shapiro delay on slide 15. So, one way of looking at the effect is that gravity causes signal propagation delay, which I think can fairly be interpreted only one way -- that gravity slows down the signal (delaying something is slowing it). If we want to make the GPS system more precise, we would need to correct for this signal delay. It has physical meaning; we live on the earth, drive our cars here, shoot guided missiles at each other here, etc. So we cannot deny the physical reality or importance of this signal slowing. On the other hand, Ashby also says (from the other perspective) that the effect is one of greater distance: the distance is the coordinate distance plus the "so-called" Shapiro delay (slide 3). So it seems to be fair to view the question from both perspectives, at least if you want to be able to both understand GR and build an accurate GPS system. 1 Link to post Share on other sites

xyzt 81 Posted December 25, 2013 Share Posted December 25, 2013 (edited) This is a complicated question, the answer has two parts: 1. Vesselin Petkov is a crank, this is known in the physics community. 2. Your question is answered trivially in most textbooks: 2.1 LOCALLY, the speed of light in a gravitational field is CONSTANT, if it weren't the whole GR theory would fall apart. 2.2. Nevertheless, the COORDINATE speed of light is variable. Indeed, start with the reduced Schwarzschild metric: [math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)[/math] Light follows null geodesics, therefore: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)[/math] giving: [math]\frac{dr}{dt}=c(1-r_s/r)[/math] Th LHS represents the COORDINATE speed of light, i.e., the speed of light measured by a distant observer. As one can easily see, it approaches 0 when the light approaches the event horizon , i.e. [math]r \to \infty[/math]. 2.3. When starting from the complete metric, things get even more complicated, resulting into what is taught as "photon orbits" in GR classes. For example: [math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] Since light follows null geodesics: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] 2.3,1. If there is no radial motion (i.e [math]dr=0[/math]) then we obtain the so-called spherical orbits": [math]0=(1-r_s/r)(cdt)^2-r^2 (d\theta)^2[/math] The tangential speed is: [math]r \omega=\pm c \sqrt{1-r_s/r}[/math] 2.3.2 If there is both radial and tangential motion, then: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] so: [math]\frac{dr}{dt}= \sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}[/math] JVNY, What Ashby and Weiss are telling you (without spelling out the exact formulas) is that from [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] one gets: [math]\frac{dt}{dr}= \frac{1}{\sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}}[/math] so, the one way total delay time can be obtained via an integration wrt the radial coordinate: [math]T_{total}=\int_{r_1}^{r_2}{\frac{dr}{\sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}}}[/math] The above integral doesn't have a closed form (best I know), so, a series of approximations needs to be done. There is no "slowing down of light speed", the effect can be calculated rigorously by a simple integration. If you neglect the rotational motion of the Earth, the integral is even simpler: [math]c*T_{total}=\int_{r_1}^{r_2}{\frac{dr}{1-r_s/r}} \approx \int_{r_1}^{r_2}{dr}+r_s \int_{r_1}^{r_2}{\frac{dr}{r}}=r_2-r_1+r_s ln \frac{r_2}{r_1}[/math] where [math]r_s=\frac{2GM}{c^2}[/math] is the Schwarzschild radius of the Earth. Post 3 contains all the mathematical formalism that you needed in order to answer your questions. BTW, the second term is what Ashby calls "the Shapiro delay" on slide 3. On the other hand, Ashby also says (from the other perspective) that the effect is one of greater distance: the distance is the coordinate distance plus the "so-called" Shapiro delay (slide 3). So it seems to be fair to view the question from both perspectives, at least if you want to be able to both understand GR and build an accurate GPS system. The "two perspectives" is only one, GPS was designed based on the GR machinery. Edited December 25, 2013 by xyzt Link to post Share on other sites

Iggy 344 Posted December 26, 2013 Share Posted December 26, 2013 The simple answer is that you are trying to calculate the proper length using the Minkowski metric [math]ds^2=dx^2-(cdt)^2[/math], when , in reality, the metric is Rindler: [math]ds^2=(d \zeta)^2-(1+\frac{g \zeta}{c^2})^2 (c d \tau)^2[/math] where [math]\tau, \zeta[/math] are respecively, the proper time, proper distance measured by one of the accelerated observers. I have pointed out your mistake repeatedly in this thread, in different ways, this is yet another way to point out why your claims are wrong. I appreciate your feedback, unfortunately we are no closer to a common understanding. Both your metric and mine are Rindler, and they are physically equivalent. I will explain in a moment. I should first say that the t variable on the right hand side of your equation is not proper time, but rather coordinate time. This is common to all metrics (the t and r on the right hand side of the Schwarzschild metric, for example, are coordinate time and radial distance). In fact, you rightly solve the metric above to get the following: ...null geodesics and to make [math]ds=0[/math] in the Rindler metric above. You would get... the correct formula of the coordinate speed of light for the accelerated observer is [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math] If [math]d \zeta[/math] and [math]d \tau[/math] were proper distance and time then that wouldn't be a coordinate speed (which it should indeed be). It would be a proper speed. More important to your point, however... ...is dependent on the acceleration and on the coordinate x. The elegant way of showing that is to remember that light follows null geodesics and to make [math]ds=0[/math] in the Rindler metric above. You would get [math]d \zeta=(1+g \zeta/c^2)(c d \tau)[/math] That is, the correct formula of the coordinate speed of light for the accelerated observer is [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math]. According to your above logic, the observer at [math]x=0[/math] will measure the speed of light to be [math]v_x=0[/math], which is clearly absurd. The reason is that your formula, wherever you picked it up from (since you show no derivation) is incorrect. The correct formula is, as derived above , [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math]. This means that an observer at [math]\zeta=0[/math] measures the speed of light to be [math]c[/math] , as one would expect. As the observer accelerates away from the light source, the coordinate speed of light decreases, following the rule [math]\frac{d \zeta}{d \tau}=c(1-g | \zeta |/c^2)[/math]. At the Rindler horizon [math]\zeta=\frac{c^2}{g}[/math] so [math]c=0[/math]. A pulse of light emitted when the accelerated observer is already past the Rindler horizon, will never "catch up" with the accelerated observer. The Rindler metric I gave is more common than yours (it is found on wiki you'll notice). Yours can be found on the last link I gave to JVNY (here). Mine: [math]ds^2 = (ax)^2(dt)^2 - dx^2[/math] Yours: [math]ds^2 = (1+ax_i)^2 dt^2 - {dx_i}^2[/math] The difference between them is nothing more than a shifting of the x-axis origin by 1/a. In geometry it is called a coordinate translation. In other words, plug [math]x = x_i + 1/a[/math] into my line element and you will get yours. The x axis has been relabeled -- it has been shifted (which makes no physical difference). The rindler horizon in my metric is at x = 0 and proper acceleration, a, starts at x = 1/a The rindler horizon in your metric is at x_{i} = -(1/a) and proper acceleration, a, starts at x_{i} = 0 Of course, this means that the observer at x in my line element has a different speed of light from the observer at x_{i} in yours, because we are talking about two different observers with a distance 1/a between them. As an analogy, if I said "Mt. Everest is 4,650 meters tall from its base", it makes no sense for you to object by saying "No, Mt. Everest is 8,848 meters tall from sea level". It is an obfuscation, not a disagreement. My speed of light: [math]v_c = ax[/math] where my x is the same as your [math]x_i + 1/a[/math]... therefore my speed of light rewritten in your relabeled x-axis: [math]v_c = a(x_i + \frac{1}{a})[/math] [math]v_c = ax_i + 1[/math] My speed of light is the same as your speed of light. You've just found a line element where the x axis is shifted by 1/a. Mt. Everest can be both one height from its base and another from sea level. This absolutely does not change the conclusion I made -- that the coordinate speed of light is 100 times greater 100 times further from the horizon. According to your above logic, the observer at [math]x=0[/math] will measure the speed of light to be [math]v_x=0[/math], which is clearly absurd. No, that's either a typo or you're misunderstanding how to use the metric. The observer at x = 1/a will measure the speed of light to be zero at x = 0. The observer at x = 1/a does all of the measuring to determine the coordinate speed [math]v_c = xa[/math]. In my metric, the Rindler horizon is at x=0 which is why the coordinate speed of light is zero there (just like it would be zero at the Schwarzschild horizon). In your metric the Rindler horizon is at x = -1/a (remember, the x axis has been shifted by 1/a between them), and the observer at x = 0 does all the measuring. That is why plugging -1/a into your speed of light equation gives zero. The observer at x=0 measures the coordinate speed of light to be zero at the horizon. I understand that you are trying to show that coordinate speeds have no physical meaning, but if you understand what the coordinates mean I think there is a good amount of meaning. There is, after all, no way to discuss non-local speeds in GR except via coordinate speeds. Denying physical meaning because GR works with any coordinate system is a bit like saying "this building could be measured in feet or meters and they give different values therefore the height of the building has no physical meaning". I agree with this. What I did in the other exercises was to proportionately adjust all of the clocks' tick rates to the same rate (the farther rearward, the more elapsed ticks show on the clock than actual ticks of time for that rocket's clock). I think that you would call this putting all of the clocks on the same coordinate time. That eliminates the time dilation effect. It leaves only the distance effect. If you calculate the speed of light using the proper distance, then you get the speed of light slowing the closer the rocket is to the rear of the accelerating row (or the deeper in a gravitational well). Ok, I think I gotcha. Link to post Share on other sites

xyzt 81 Posted December 26, 2013 Share Posted December 26, 2013 (edited) Of course we are not any closer until you retract yur erroneous claim thath the Shapiro delay would somehow confirm the "slowing of light speed". This absolutely does not change the conclusion I made -- that the coordinate speed of light is 100 times greater 100 times further from ...The point I made was that your proof that the proper distance between the rockets is constant is incorrect. The thing with the linear dependence on distance for the coordinate speed is not the point of disagreement, the validity of your proof of constant distance is.You did not use the Rindler metric in computing the distance, aas you should have, you used the Minkowski metric. You never addressed this criticsm. Edited December 26, 2013 by xyzt Link to post Share on other sites

Iggy 344 Posted December 26, 2013 Share Posted December 26, 2013 You did not use the Rindler metric in computing the distance, aas you should have, you used the Minkowski metric. Distance is constant in the Rindler metric a priori. A rindler observer at x=1 is always at x=1, and the same with an observer at x=2. The distance between them is constant one by definition (it's assumed by the metric). Constant distance is proven in the derivation of the metric, not with the metric. It's like asking someone to use a meter stick to prove that the meter stick is one meter. It's a confused question. 1 Link to post Share on other sites

xyzt 81 Posted December 26, 2013 Share Posted December 26, 2013 The point was that you used the MINKOWSKI metric in your attempt to show that the distance between rockets is constant. When do you plan to retract your false statement that the Shapiro delay somehow proves the "slowing" of light speed? Link to post Share on other sites

Iggy 344 Posted December 30, 2013 Share Posted December 30, 2013 The point was that you used the MINKOWSKI metric in your attempt to show that the distance between rockets is constant. Yes, I'm perfectly capable of doing that. When do you plan to retract your false statement that the Shapiro delay somehow proves the "slowing" of light speed? If, The answer in terms of Schwarzschild coordinates is the standard, mainstream answer. then the radial coordinate velocity of light is, [math]v_c = \pm \left(1- \frac{r_s}{r} \right)[/math] and in terms of acceleration of gravity, g, and two radial points, A and B... [math]v_{BA} = \left(1- \frac{gr}{2c^2} \right)[/math] (where A is further than B and [math]v_{BA}[/math] is the average velocity of light between the two as measured by A) If true then we get Shapiro delay for a round trip A->B->A... [math]\Delta \tau = \Delta t_{flat} \left(1 + \frac{gr}{2c^2} \right)[/math] and then light is slowed relative to flat spacetime, and many experiments have confirmed this, therefore, Shapiro delay confirms it. You have to decide if you're the sort of person that says "the answer in terms of Schwarzschild coordinates is the standard" or decide if you're the sort of person that denies that Shapiro delay originates in a variable speed of light. Declaring both is incongruent. Are you more interested in proving yourself (and post #3) correct, or proving me wrong? 2 Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) Yes, I'm perfectly capable of doing that Minkowski metric applies only to inertial observers, the observers in JVNY's exercise are accelerated, so your attempt at applying the Minkowski metric in order to calculate distances is clearly incorrect. [math]v_c = \pm \left(1- \frac{r_s}{r} \right)[/math] and in terms of acceleration of gravity, g, and two radial points, A and B... [math]v_{BA} = \left(1- \frac{gr}{2c^2} \right)[/math] (where A is further than B and [math]v_{BA}[/math] is the average velocity of light between the two as measured by A) If true then we get Shapiro delay for a round trip A->B->A... [math]\Delta \tau = \Delta t_{flat} \left(1 + \frac{gr}{2c^2} \right)[/math] There are multiple mistakes in the above answer: 1. The Shapiro delay is not calculated using the radial speed of light. 2. The light ray in the Shapiro delay doesn't move radially, it moves on a geodesic that grazes the gravitating object 3. The formula that you posted for the Shapiro delay is incorrect, the correct formula has been given early in this thread. You have to decide if you're the sort of person that says "the answer in terms of Schwarzschild coordinates is the standard" or decide if you're the sort of person that denies that Shapiro delay originates in a variable speed of light. Declaring both is incongruent. The Shapiro delay requires a simple understanding of how to calculate the elapsed time of the photon traveling along a null geodesic. Nowhere in the calculation one uses "variable speed of light". The use of the Schwarzschild metric in figuring out the delay does not validate in any form the fringe concept of "variable light speed". Are you more interested in proving yourself (and post #3) correct, or proving me wrong? Post 3 is textbook stuff, therefore it is correct. Your posts, on the other hand, are not. Edited December 31, 2013 by xyzt 1 Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 The answer in terms of Schwarzschild coordinates is the standard, mainstream answer. The use of the Schwarzschild metric in figuring out the delay does not validate in any form the fringe concept of "variable light speed". Tell me about it. The speed of light is variable in Schwarzschild coordinates. How wrong can one person be? Which of the two above quotes would you like to retract? I'll be happy with either, but you still haven't answered. Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) Tell me about it. The speed of light is variable in Schwarzschild coordinates. Correction: the coordinate speed of light is variable in Schwarzschild coordinates. As it is in Riessner coordinates, as it is in Rindler coordinates, as it is in Kerr coordinates. So what? This unphysical effect is not used in the derivation of the Shapiro delay. Why do you persist in pushing your fringe ideas? This is supposed to be a science forum. Which of the two above quotes would you like to retract? I'll be happy with either, but you still haven't answered. Both my statements are correct, so there isn't going to be any retraction. As I explained to you (repeatedly), the Schwarzschild solution is used in deriving the correct prediction for the Shapiro delay (not the incorrect one that you posted). The Schwarzschild solution also predicts a variable coordinate speed of light. You cannot infer from this that a variable speed of light explains the Shapiro delay, you are committing a basic fallacy: A->B and A->C doesn't mean C->B. Edited December 31, 2013 by xyzt 1 Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 (edited) Tell me about it. The speed of light is variable in Schwarzschild coordinates. Correction: the coordinate speed of light is variable in Schwarzschild coordinates. That isn't a correction. That is something logically equivalent to what I said. I'm sure Euclid made some point about things logically equivalent to a mutual party are logically equivalent to themselves. The speed of light in a certain set of coordinates doesn't need corrected as the certain coordinate speed of light in coordinates. That is just you skipping the issue as always, I hope. I hope for you as always. This unphysical effect is not used in the derivation of the Shapiro delay. Please derive Shapiro delay in a way that reflects nothing of Schwarzschild coordinates. While you're at it, please say that post 3 has nothing to do with anything real or physical. Thank you. Edited December 31, 2013 by Iggy 1 Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 That isn't a correction. Sure it is, it points out that the variability is applied to a unphysical entity, to the coordinate speed of light , not to the local speed of light. This has been explained early on, in post 3, I recommend that you re-read it and try to understand it. Please derive Shapiro delay in a way that reflects nothing of Schwarzschild coordinates. This demonstrates that you STILL do not understand the issue: using the Schwarzschild solution doesn't mean that one uses the "variable" speed of (coordinate) light as a means of explaining the Shapiro delay. While you're at it, please say that post 3 has nothing to do with anything real or physical. Thank you. I did. It is right there, in post 3, has been there from the beginning. I suggest that you re-read it and try to understand it. Link to post Share on other sites

md65536 383 Posted December 31, 2013 Share Posted December 31, 2013 Correction: the coordinate speed of light is variable in Schwarzschild coordinates. As it is in Riessner coordinates, as it is in Rindler coordinates, as it is in Kerr coordinates. So what?It is clear that this is what's been discussed all along. It's been pointed out several times (including by you) that it's coordinate speed of light that is being discussed. If it's confusing or bothersome perhaps everyone could repeat "coordinate speed" every time it's used. I'm curious about your correction. What is the difference between speed in Schwarzschild coordinates and coordinate speed in Schwarzschild coordinates? Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 Sure it is, it points out that the variability is applied to a unphysical entity, to the coordinate speed of light I'm sorry... the "coordinate speed of light" is unphysical, or "Schoolchild coordinates are the standard". Which were you saying? It looks above like you just negated something you said earlier. Were you ok with that as long as you're interjecting with me now? It's a bit difficult to tell. Can you just quickly say "the coordinate speed of light is unphysical, and therefore the earlier thing I said about it being the standard answer is just my best attempt at a pail stain on the carpet". Can't you just say that? Or some humerus variation of it? It would end our confrontation immediately. Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) It is clear that this is what's been discussed all along. It's been pointed out several times (including by you) that it's coordinate speed of light that is being discussed. If it's confusing or bothersome perhaps everyone could repeat "coordinate speed" every time it's used. I'm curious about your correction. What is the difference between speed in Schwarzschild coordinates and coordinate speed in Schwarzschild coordinates? Read post 3, it is made clear from the first lines. I'm sorry... the "coordinate speed of light" is unphysical, or "Schoolchild coordinates are the standard". Which were you saying? It looks above like you just negated something you said earlier. Were you ok with that as long as you're interjecting with me now? It's a bit difficult to tell. Can you just quickly say "the coordinate speed of light is unphysical, and therefore the earlier thing I said about it being the standard answer is just my best attempt at a pail stain on the carpet". Can't you just say that? Or some humerus variation of it? It would end our confrontation immediately. Do you really not understand the "variable" speed of light is not the mainstream explanation of the Shapiro delay ? That you are pushing your personal fringe idea? That your "solution" is negated by the mainstream solutions? Let's try again: 1. The Schwarzschild solution is used for deriving the Shapiro delay. 2. The Schwarschild/Kerr/Riessner/etc. solutions happen to predict variable coordinate light speed. 3. This does NOT mean that variable light speed is the valid/mainstream explanation of the Shapiro delay, contrary to your fringe ideas pushed in your posts. There is no physical connection between statements 1 and 2. Both are valid but they aren't connected. Edited December 31, 2013 by xyzt Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 Sure it is, it points out that the variability is applied to a unphysical entity, to the coordinate speed of light I apologize profusely. Is the variable speed of light the unphysical entity, or the "standard answer", both of which you'd said. Could you please just say one is right and one is wrong so I can understand what you mean? You were right before and wrong now, or right now and wrong before? I can't tell until you say it explicitly. Is it that the "variability is applied to an unphysical entity" or is it that "Schwarzschild coordinates are standard"? I'm happy with either. They are physically dielectric. Can't you choose one thing of which you've said? Link to post Share on other sites

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