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I understands how the time dilation work. The derivation of time dilation perfectly understandable for me. However the thought that two observers observes that time dilated for each other do puzzles me. For instance, observer A in his inertial frame of reference observed observer B in his own inertial frame of reference moving away at a certain velocity. Observer A observed that B's time is slowed (by whatever feasible measure), whereas B observed A's time slowed. What if B changes his direction of motion, heading back toward A. According to relativity, their time is still dilated.

Both observe one another having a slower time. What would happen when they come together?

tryst with freedom
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Johnson Zhou
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1 Answers1

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Yes. If B suddenly reversed direction at some point (or if we consider another frame C going in the opposite direction at the same speed and synchronized to B at the reversal point), his/her time would still appear time dilated to A. A and B would no longer be synchronized in time at their 2nd rendezvous.

The usual argument goes like this: Say A sees B reversing direction at time $t_A^0$ and position $x_A^0 = v t_A^0$. B's time at that moment is $t_B^0 = \gamma(t_A^0 - \frac{v}{c^2}x_A^0) = t_A^0/\gamma$ and he observes $A$ at position $x_B(A) = -v t_B^0 = -x_A^0/\gamma$. From the point of view of A, when A and B meet again in $x_A=0$, A's time is $t_A^0 + x_A^0/v = 2t_A^0$, while B's time is $t_B^0 + |x_B(A)|/v = 2t_B^0 = 2t_A^0/\gamma$. So A sees B as time dilated by a factor of $\gamma$.

But we still have a couple of problems: After B reverses direction, does s/he also see that A is time dilated relative to him? And how did A and B loose synchronization of their origins, even though they were synchronized in the beginning?

The answer to the first problem is again yes, B does observe A as time dilated. In order to prove this we need the answer to the 2nd problem, which is that B's reversal of direction implies a change in synchronization through a shift in coordinates, despite the fact that he does not change the coordinates of his location at the moment of reversal. The Lorentz transforms between A and B after B's reversal read $$ \begin{eqnarray} x_A &=& \gamma (x_B - vt_B) + 2vt_A^0 \\ t_A &=& \gamma (t_B - \frac{v}{c^2}x_B) \end{eqnarray} $$ and $$ \begin{eqnarray} x_B &=& \gamma (x_A + vt_A) - 2\gamma v t_A^0\\ t_B &=& \gamma (t_A + \frac{v}{c^2}x_A) - 2\gamma \left(\frac{v}{c}\right)^2t_A^0 \end{eqnarray} $$ (All that is needed to derive these is to account for a shift in coordinate origins in the usual form of the Lorentz transforms. I am skipping the details, but we can easily verify that all coordinates verify their correct values for $t_A = t_A^0$.)

What happens as a consequence is that B sees A at a different time after his direction reversal: Right before the reversal B observed A at position $x_B(A) = - x_A^0/\gamma$, corresponding to A's time $t_A = \gamma(t_B^0 + \frac{v}{c^2}x_B(A)) = t_A^0/\gamma^2$. Right after the reversal the new Lorentz transforms show that B still observes A at position $x_B(A) = - x_A^0/\gamma$, but now this corresponds to A's time $\bar{t}_A = \left(1+\left(\frac{v}{c}\right)^2\right)t_A^0$.

On the other hand, the 2nd rendezvous still takes place at time $t_B^0 + |x_B(A)|/v = 2t_B^0 = 2t_A^0/\gamma$ for B, and at $t_A^0 + x_A^0/v = 2t_A^0$ for A. So from B's point of view, s/he meets A again after a time $\Delta t_B = 2t_B^0 -t_B^0 = t_A^0/\gamma$, while A only observes a time lapse $\Delta t_A = 2t_A^0 - \left(1+\left(\frac{v}{c}\right)^2\right)t_A^0 = t_A^0/\gamma^2 = \Delta t_B/\gamma$. In other words, B sees A undergoing time dilation, as expected.

tryst with freedom
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udrv
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  • Thank you for all your incredible derivation for my questions. My brain is just obtuse facing the equations. It took me a while to go through all your explanation. I think I get it now, and thank you for all the help. – Johnson Zhou Dec 29 '15 at 13:13
  • I wish I can handle the Lorentz transformation as handy as you can. Can you suggest some ways to strengthen my math and the physical implications systematically? (since I am a freshman in high school and exploring the glamorous realm of physics on my own) – Johnson Zhou Dec 29 '15 at 13:15
  • Plus, how to determine quickly if it is to multiply γ or divide by γ. It seems confusing to me – Johnson Zhou Dec 29 '15 at 13:34
  • You are very welcome. Please let me know if you'd like more detailed intermediate steps anywhere in the derivation. As to how to handle $\gamma$, the fast rules are: length contraction = division by $\gamma$; time dilation and Lorentz transformations = multiplication by $\gamma$. When in doubt always use the Lorentz transforms to see how events relate between different frames, taking into account both space and time coordinates. – udrv Dec 29 '15 at 21:34
  • Use Lorentz transforms to find coordinates in one frame from coordinates in another frame, but also use with mixed coordinates: knowing space coordinate in frame 1 and time in frame 2, you can find time in frame 1 and space coordinate in frame 2. One other important thing to always keep in mind, because it easily gets forgotten or overlooked, is relativity of simultaneity: whatever is seen at the same time in one frame is not occurring at the same time in another frame. – udrv Dec 29 '15 at 21:34
  • This means, for instance, that different points of a moving frame along the direction of motion, as seen at the same time in the stationary frame, correspond to different times in the moving frame. Hope this helps :) – udrv Dec 29 '15 at 21:34
  • Thank you. In fact, the realisation of the relativity of simultaneity was one of the very first concepts I encountered during the study of the theory. There is one question I have recently. When Lorentz or Einstein thought about the formula for the transformation, how did they actually derived the formula? I found a lot of different ways of derivations and found a majority of them features an assumption of a constant between the distance I observe and the distance the other observer perceive. (In this case gamma) – Johnson Zhou Dec 31 '15 at 04:34
  • It seems to me not very plausible that they assumed the existence of gamma and derived it. The length contraction was easy to derive from the Michelson-Morley experiment. What about the time dilation formula? And also, the time dilation formula t'=γ(t-ux/c^2) seems counter intuitive to me. The ux/c^2 part really baffles me. It would be great if you can explain its physical implications to me. – Johnson Zhou Dec 31 '15 at 04:39
  • Einstein actually rederived the Lorentz transformations with a constant at first, by assuming what we call the linearity of the time transformation and by making use of wisely selected thought experiments with the moving rod. But the constant was not $\gamma$ (or $\beta$ as he called it), it was a multiple of $\gamma$ of the form $\phi(v)\gamma(v)$, where the form of $\gamma$ was already evident at this point. By considering what must happen to a light front under the new transformations he then showed that $\phi(v) = 1$. – udrv Dec 31 '15 at 12:07
  • The time dilation formula is easiest to obtain from the "light clock" thought experiment: laser pulse in tube with mirrors at the ends, the tube oriented along $y$ or $z$ and moving at velocity $v$ along $x$. Say the half period of the clock in its own frame is $T$. Take two snapshots of the tube in the stationary frame: one when the pulse is at the bottom, the next after half period $t$, when the pulse reaches the top. – udrv Dec 31 '15 at 12:08
  • In the triangle formed by the first position of the pulse (bottom) and the whole length of the tube in the 2nd snapshot, the actual trajectory of the pulse is the hypothenuse, its speed along it is $c$, and its length must be $ct$. The leg along $x$ must be $vt$, and the length of the tube is obviously the same as in its own frame, so $cT$. Pithogora's theorem now gives $(ct)^2 - (vt)^2 = (cT)^2$, wherefrom you get $T = t/\gamma$. – udrv Dec 31 '15 at 12:09
  • As for the $vx/c^2$ part in $t' = \gamma(t-vx/c^2)$, that is exactly what accounts for relativity of simultaneity. It tells you that, as seen from another frame, the proper time of the moving frame slides along the direction of motion proportional to the x coordinate. Basically from the stationary frame each point along the direction of motion (or plane perpendicular to it) is observed at a different proper time. – udrv Dec 31 '15 at 12:10
  • Here is one interesting way to interpret the Lorentz transformations: For $x' = \gamma(x - vt)$ write $x'/\gamma = x - vt$. Then you have: lenth contracted $x'$ coordinate in moving frame = $x$ coordinate in stationary frame - drift $vt$ due to uniform motion. For $t' = \gamma(t - vx/c^2)$ write $t'= t/\gamma - vx'/c$ (substitute $x'$ to check!) and this gives: time $t'$ in moving frame = time $t$ in stationary frame dilated by $\gamma$ - time shift $vx'/c$ due to relativity of simultaneity. – udrv Dec 31 '15 at 12:10
  • This is incredible. So, does the time transformation equation imply that if there are two frames traveling at the same speed according to one stationary observer(me), the farther frame's time move more slowly than the nearer one? – Johnson Zhou Jan 05 '16 at 08:41
  • And how do I substitute to acquire t′=t/γ−vx′/c? It ends unsolvable for me in a strange form – Johnson Zhou Jan 05 '16 at 09:08
  • And I am did a thought experiment recently. The context is the same as the classical astronaut falling into a black hole. According to us time dilation is so extreme at the horizon, the falling astronaut's time literally stopped according to us. (From the astronaut's perspective, the rest of the universe would also dilate, right?) According to Hawking radiation, a black hole could slowly disappear, in a theoretical explosion. At the time of the annihilation of the black hole, the falling astronaut should be still falling in? Then, what would happen to the astronaut? – Johnson Zhou Jan 05 '16 at 09:14
  • And there is a strange loophole of the common explanation to Hawking radiation. The fluctuations could create positive energy particle and negative energy particle. Why would only negative particles get sucked in? If it possess negative energy, won't it try to get away from the black hole by gravity? – Johnson Zhou Jan 05 '16 at 09:17
  • I looked back to the original question I could not understand this step: t0B=γ(t0A−vc2x0A)=t0A/γ. Why did γ became the denominator? – Johnson Zhou Jan 05 '16 at 09:50
  • And I more question: how does the special relativity cope with Newton's bucket? – Johnson Zhou Jan 05 '16 at 09:55
  • Lots of questions? So one at a time: "Does the farther frame's time move more slowly than the nearer one?" Instead of two frames take two clocks at rest and synchronized in the moving frame, say C1 at $x'1=0$, C2 at $x'_2= L>0$. The farther one, C2, runs not slower, but later then the closer one, C1. The stationary frame sees both runinng at the same time dilated _rate, but C2 does show an earlier time then C1. So yes, C2 does lag behind relative to C1. – udrv Jan 05 '16 at 19:35
  • "How do I substitute to acquire $t′=t/\gamma−vx′/c$?" My error, it should be $t′=t/\gamma−vx′/c^2$. Substitute $x' = \gamma(x - vt)$, does it work now? – udrv Jan 05 '16 at 19:36
  • "I could not understand this step: t0B=γ(t0A−vc2x0A)=t0A/γ" You have $x_A^0 = vt_A^0$ and $t_B^0 = \gamma(t_A^0 - vx_A^0/c^2)$, so: $t_B^0 = \gamma(t_A^0 - v(vt_A^0)/c^2) = \gamma (1-v^2/c^2) t_A^0 = \gamma \gamma^{-2} t_A^0 = t_A^0/\gamma$ – udrv Jan 05 '16 at 19:36
  • "At the time of the annihilation of the black hole, the falling astronaut should be still falling in? Then, what would happen to the astronaut?" The astronaut should be shredded by gravitational forces long before he reached the horizon (sorry, "Interstellar" is a fun movie though), but even assuming that he somehow survives that, he would still be caught up in the evaporation and energy release. Would it still look like an explosion to him? Probably much much slower in his time. – udrv Jan 05 '16 at 19:37
  • "Why would only negative particles get sucked in?" Energy would only be "negative" as seen by a distant observer. The answers to these questions explain nicely what this means: http://physics.stackexchange.com/questions/106882/how-does-negative-energy-from-hawking-radiation-cause-a-black-hole-to-shrink, http://physics.stackexchange.com/questions/30597/black-holes-and-positive-negative-energy-particles – udrv Jan 05 '16 at 19:37
  • "Newton's bucket in special relativity" Not a bucket, but a spinning water sphere explained here: http://physics.stackexchange.com/questions/3986/newtons-bucket – udrv Jan 05 '16 at 19:38
  • About the black holes, I read that black holes' mass is proportional to the radius of the event horizon, whereas gravity weakens at a inverse square law(just using it as an approximation, as I barely know how it works). Theoretically, a super-massive black hole can suck one in rather gently. Btw, I did enjoy reading "the science of interstellar by Kip Thorne – Johnson Zhou Jan 05 '16 at 23:18
  • Since I would keep studying physics on the college level, and my high school is not the best suitable mentor for this kind of questions. May I keep a private mean of communication with you like email as you're such an expert? – Johnson Zhou Jan 06 '16 at 02:27