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I understand the precepts of relativistic time dilation, but I'm looking to nail down the local perception of velocity in each frame. My question is threefold:

  1. Given two observers $A$ (rest frame) and $B$, if $A$ observes $B$ traveling away from starting point $A$ at $V=2c/5$, does $B$ perceive a different value for $V$ based on dilation, or the same, based on it's local frame?
  2. If it is the same (which is my assumption), how does this integrate with the predicted experience of $B$ from the point of view of $A$ (which I also assume would seem to act as a multiplier on perceived velocity due to dilation)?
  3. As a further silly notion, does it change if at some point $B$ stops and proceeds to return to $A$ at the same previous velocity? My assumption is it does not.
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3 Answers 3

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  1. No, $B$ will perceive the same value of $v$ for $A$. According to special relativity, it is impossible for $A$ and $B$ to determine which one of them is moving from their reference frames.

  2. Both will have the same experience from their perspectives. $A$ will see $B$ contracting along the direction of motion and its clock moving slower, and vice versa.

  3. If $B$ starts going in the opposite direction with velocity $v$, nothing will change except the perceived direction of motion from both of their reference frames and hence the direction of contraction.

I am not sure if this is all what you were asking. Let me know if I misinterpreted something.

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  • $\begingroup$ in 2, based on the dilation of B as observed from A, would A's perception predict B to observe V as .4c/dilation_fraction_of_A? I'm not trying to violate special relativity, just trying to explore the discontinuity of perception.. I realize that considering from each frame individually gives the correct answer, just trying to nail down loose ends from a single pespective, and see how they fit into the overal scheme. $\endgroup$ Mar 29, 2014 at 11:43
  • $\begingroup$ Their perceptions would be the same. I do not understand you completely. $\endgroup$
    – user42733
    Mar 30, 2014 at 12:14
  • $\begingroup$ working only from A's perspective, My assumption is it would seem that it would view V as (.4c/observered_dilation) if A were to make a linear guess at B's perspective without the benefit of the theorem. Obviously that's an incorrect answer based on the theory, which treats both perspectives as equal. is there another transform that describes this discontinuity, or rectifies it aside from treating both perspectives as equal? $\endgroup$ Mar 30, 2014 at 17:45
  • $\begingroup$ No, both observers will perceive V to be the same. There is no discontinuity, your assumption is wrong. I found this rafimoor.com/english/SRE.htm, give a read to the part about reference frames. Also, if this answer solved your problem, please accept it. $\endgroup$
    – user42733
    Mar 30, 2014 at 19:05
  • $\begingroup$ I don't appear to be explaning myself well. the discontinuity I'm aiming at isn't between both observers, but rather between the observation and the reality... the theory predicts the actual reality (both observers see the same thing)... taking only the single observer A in point 2, they observe dilation of B ergo their observation would also predict a difference in velocity (in the time component) local to B (even though none exists from B's perspective). The difference seems to be born out by the linked page w/ differing perception of event order re: woman on train & man on the ground... $\endgroup$ Apr 1, 2014 at 10:00
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The value B sees for V depends on which combination of frames they use for their reference. If they look entirely in their own frame and are moving at constant velocity, they can't tell what velocity they're moving at. If they observe A, they will see A moving with a velocity V.

If however, they look at preset distance markers in A's frame and use the B frame measurement of time, they will see something called the proper velocity. Brehme referred to this as speedometer velocity[1]. It's the velocity that corresponds to how far B will have travelled in A's frame over the amount of time B experienced. The expression for proper velocity is $v_\tau = \gamma v = sinh(\phi)$, where $\gamma = \sqrt{1-V^2/c^2}$, and $\phi$ is the rapidity of B's frame which is related to V by $V/c = tanh(phi)$.

This brings us to the third question. What happens if B stops and returns to A. Let's say that B slowed to a stop and started back up again in the opposite direction with a constant acceleration. The change in proper velocity of B while accelerating is $\Delta v_\tau = \gamma v = c sinh(\frac{a\tau}{c})$, where $\tau$ is known as the proper time as measured in B's frame.

References

  1. Flynn, R.W., Spacecraft navigation and relativity, AJP, 53, (1985), 113 http://dx.doi.org/10.1119/1.14092

  2. Rindler, W., "Hyperbolic Motion in Curved Space Time", Phys. Rev. 119 2082{2089 (1960).

  3. Rindler, W, "Special Relativity", Interscience Publishers, 43 (1944)

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  • $\begingroup$ The answer I gave to this question is the conventional interpretation. Personally, however, I believe you are absolutely right - local measurement of the velocity does differ from measurement made by a stationary observer (because v' is not the measurement of stationary observer's relative movement, but the measurement of the object's movement made by the moving observer). Therefore I up-vote your answer. $\endgroup$ Apr 28, 2014 at 19:47
  • $\begingroup$ apologies for the late comment. Am I correct in the assumption that it also holds true that if the point of observation changes, but not the distance markers or time referent that the above formulae remain unchanged? $\endgroup$ Apr 30, 2014 at 10:54
  • $\begingroup$ @VoidSinger, I'm not entirely sure I understand the full intent of your question. If the point of observation changes within a frame, you should be fine, and the formulas should be unchanged. $\endgroup$
    – dolphus333
    Apr 30, 2014 at 14:27
  • $\begingroup$ I should have specified from B to A, but working through it it look like it does. the important part was the perception of others remote experience rather than the actual experience $\endgroup$ May 1, 2014 at 16:56
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The Special Theory of Relativity gives us no direct transform equation for velocity for strictly two-body situations. There is only the velocity addition formula $s = \frac{u + v}{1 + (vu/c^2)}$, which can be "simplified" by assuming that $s$ equals zero.

However, the sub-section Principle of relativity in the entry on derivation of Lorentz transforms in Wikipedia states that there is no privileged frame of reference, and explicitly says that you should assume that the apparent velocities perceived by persons A and B have the same value, yet opposing directions. Therefore:$$v = - v'$$ (which is obviously quite puzzling though, since the variables x and t change under the transforms in inverse proportion - hence time dilatation and length contraction - and therefore it seems the values for v and v' should differ).

And yes, you are right, reversing the direction of the movement changes only the directions of the perceived velocity vectors.

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  • $\begingroup$ "The Special Theory of Relativity gives us no transform equation for velocity (at least I did not find any)." — It does. It's known under the (IMHO unfortunate) name "relativistic addition of velocities" (the term is because under Galilean transformations velocities transform by addition). $\endgroup$
    – celtschk
    Apr 27, 2014 at 9:29
  • $\begingroup$ Addition of velocities is about two movements in respect of the third, considered stationary, frame of reference. This case is completely different, as only one of the reference frames is moving, while the other is considered stationary. Therefore we are not adding velocities here, because there is only one velocity - either the A person/object is moving in respect of B or B is moving in respect of A. $\endgroup$ Apr 27, 2014 at 9:40
  • $\begingroup$ Being stationary just means moving at velocity $0$. $\endgroup$
    – celtschk
    Apr 27, 2014 at 9:47
  • $\begingroup$ Being stationary means not moving at all. $\endgroup$ Apr 27, 2014 at 9:55
  • $\begingroup$ "Not moving at all" makes no sense in relativity, since there's no such thing as absolute rest. The "stationary frame" is just the frame of some inertial observer which you have chosen to call "stationary" (another unfortunate convention, as your comments clearly demonstrate). Especially, given that observer A is not accelerated, it is a perfect choice for the "stationary" observer. $\endgroup$
    – celtschk
    Apr 27, 2014 at 9:59

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