I am referring to Wikipedia: "Length contraction can also be derived from time dilation." with the following proof which seems to be the result of a circular reasoning.

The proof uses only one and the same velocity v for the point of view of the observer and for the point of view of the reference frame of the observed object:

$\frac{L'}{L} = \frac{T'v}{Tv} = \frac{1}{\gamma}$

It seems that this is a circular reasoning, because we must first proof that both v of the formula are equal, and for this we need the same formula with v' and v:

$\frac{L'}{L} = \frac{T'v'}{Tv} = \frac{1}{\gamma}$

Example: A spaceship is travelling a distance of 8 Lmin (Earth reference frame) within 10 minutes (Earth reference frame), yielding a velocity of 0,8 c (Earth reference frame).

The point of view of the space ship is following from the proper time formula and the Lorentz contraction formula, both are multiplying with the reciprocal Lorentz factor 1/γ. The reciprocal Lorentz factor for v= 0,8 c is 0,6. We get 10 min. x 0,6 = 6 minutes and 8 Lmin x 0,6 = 4,8 Lmin. From the point of view of the spaceship it is travelling 4,8 Lmin in 6 minutes, yielding a velocity of 0,8 c (q.e.d.), so we proved that

v' = v.

Or am I wrong, and it is evident and without need of further proof that v' = v?

For the answer see the answer of Ben Crowell:

"v" is the relative velocity taken into account by the Lorentz factor between inertial reference frames which is the same for both frames. Thus v = v' can be derived directly from the SR postulates. And thus it seems that effectively length contraction can be derived from time dilation, as Wikipedia says.

  • $\begingroup$ another horror from wikipedia! $\endgroup$ – Wolphram jonny Nov 12 '14 at 9:10
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    $\begingroup$ I am a fan of Wikipedia. But it is no religious truth, it is up to us to read it with a convenient distance. $\endgroup$ – Moonraker Nov 12 '14 at 9:26
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    $\begingroup$ I really hate these ad hoc arguments for showing time dilation and length contraction. While they work in simple cases they just lead students astray as soon as things get complicated. Arguments about time dilation and length contraction should be based on the metric, or if that's a bit too advanced derived by working through the Lorentz transformations. $\endgroup$ – John Rennie Nov 12 '14 at 10:12
  • $\begingroup$ @john rennie, as the time clock thought experiment shows, the Lorentz factor, time dilation and length contraction are both deriving directly from the SR postulates. So there is no need to make simple things complicated. This kind of argumenting and this kind of formula (i.e. without Lorentz transforms and metrical considerations) is complying with highest standards, without any physical ambiguity. $\endgroup$ – Moonraker Nov 12 '14 at 10:59
  • $\begingroup$ @john rennie: Two mistakes: First, I suppose that you are the observer, and you are remaining in t=0 x=0. An observed spaceship ("the moving frame", v=0,9c) will let off two firecrackers, but not you. Second: Time in the moving (observed) frame is not γT but 1/γ T. $\endgroup$ – Moonraker Nov 12 '14 at 11:17

Or am I wrong, and it is evident and without need of further proof that v' = v?

It may not be obvious, and it does require proof, but it is true, and it's not anything terribly deep or mysterious.

Consider two observers, Alice and Bob, moving away from each other. Alice says she's at rest and Bob is moving. Bob says the opposite. If they want to find out how fast the motion is, they can do it by sending signals back and forth and measuring how the time lag grows. The details of how they analyze the data are not important. What matters is that due to rotational symmetry (or reflection symmetry) they will both get the same data, and therefore arrive at the same result for $v$.

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  • $\begingroup$ Great answer, thank you! Does everybody agree? $\endgroup$ – Moonraker Nov 12 '14 at 17:05
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    $\begingroup$ It's definitely correct, the question is whether you want to try to prove symmetric velocities using only the basic postulates of SR (along with the idea that different frames always agree about local events like what an observer's clock reads at the moment a light signal reaches her), or if you're OK with invoking other ideas like rotation symmetry. I think it is actually possible to adapt the idea of one observer sending signals to the other to prove velocities are symmetrical without invoking rotation/reflection symmetry, requires more math to derive though so not as conceptually simple. $\endgroup$ – Hypnosifl Nov 13 '14 at 4:38

I think for time dilation you have to divide the time by lorentz factor. Also for calculating velocity you have to use lorentz transformation equations for finding both change in distance traveled and time taken to travel that distance.

An alternate proof for calculating length contraction from time dilation is:

To derive length contraction from time dilation, consider a situation where a rod of length L sits along x-axis of a reference frame, with one end of the rod at origin and other end at position $x=L$. Also at position $x=L$, a mirror is placed.

Now in this reference frame to measure the length of rod suppose a photon is fired from the origin along x direction and time is noted when the photon, after hitting the mirror, comes back to the origin. Which gives: $c=2L/T$, where T is the time recorded.

Now if this reference frame is moving with respect to some stationary observer, the time T will be dilated. This can be proved by taking into consideration that T is the proper time, as time measurements in the moving frame was taken at same position(i.e. the origin).

Hence, the time measured in this stationary reference frame will be $T'=T/\gamma$. According to the observer in this reference frame, the distance traveled by photon will be $L'=1/2*c*T' = 1/2*c*T/\gamma = L/\gamma$.

This proves length contraction.

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  • $\begingroup$ I agree with your proof which seems to be similar to the light clock thought experiment. Your are proving that both time dilation and length contraction are deriving directly from the postulates of SR (without need of Lorentz transforms), but you are not proving that length contraction is deriving from time dilation! $\endgroup$ – Moonraker Nov 12 '14 at 9:56
  • $\begingroup$ you have to notice that all we are doing is measuring two coordinates of space time in a moving frame from a stationary reference frame. I will us the same geometry as i have used in my answer. Now suppose at x'=0,t'=0 we have t=0 ans x=0. And at a certain instant t=t an event occur at x=L in moving frame. Now the observer in the stationary frame, will say that this event has occurred at an earlier time in moving frame than what he has recorded in his clock. SO if he take a rod of Length L from his frame and put in moving frame the length will appear to be shorter. $\endgroup$ – ruskin23 Nov 12 '14 at 10:23
  • $\begingroup$ This derivation is wrong. Just by looking at the last part, we see that L' is related to L in the same way as T' is related to T, whereas one of them should be multiplied by $\gamma$ and the other one divided. It's not a simple typo, but I am not totally sure where the mistake is. $\endgroup$ – Heterotic Feb 4 '18 at 18:03

Imagine that two different inertial observers , one sitting on a train moving through a station with uniform velocity v and the other at rest in the station, want to measure the length to be L and claims that the passenger covered this distance in a time L/v. This time, ∆t, is a nonpropre time, for the events observed occur at two different places in the ground(S) frame and are timed by two different clocks. The passenger, however, observes the platform approach and recede and finds the two events to occour at the same place in his (S’) frame. Now we know the proper time interval $\Delta {t^{'}}$ and nonpropre time interval(∆t) are related by the equation$ \Delta{t}^{'}=∆t{\sqrt{(1-v^2/c^2 )}}$ . But$ \Delta{t}=L/v $ ,so that$\Delta {t}^{'}=L {\sqrt{(1-v^2/c^2 )}⁄v } $ . The passenger claims that the platform moves with the same speed v relative to him so that he would measure the distance from back to front of the platform as $ v\Delta {t}^{'}$. Hence the length of the platform to him is $ L'=v \Delta {t}^{'}=L√(1-v^2/c^2 )$

And yes, in this case the velocity of the train to the ground observer is equal to the velocity of the platform to the 2nd observer sitting on the train. And if want to know why the both observer claim the same velocity then see my comment on this ans.

  • $\begingroup$ Aren't you assuming without proof that if the train passenger is moving at v in the station frame, the station is also moving at v in the passenger frame? ("The passenger claims that the platform moves with the same speed v relative to him...") This is the main issue the original question was asking about. Of course this assumption is true in SR, and the rest of your argument is correct given the assumption, but without explicitly giving an argument as to why it's true, I don't think you're really answering the question. $\endgroup$ – Hypnosifl Nov 13 '14 at 4:47
  • $\begingroup$ Here both the frame are inertial frame, and the observer on platform think he is at rest again the observer sitting on the train also think he is at rest. Now according to spacial theory of relativity there is no preferred inertial system exist. Hence if the rest observer claim that the train moving with velocity v then the observer on the train(who think he is at rest) must also claim that the platform moving with velocity v. $\endgroup$ – Rajesh Sardar Nov 13 '14 at 5:04
  • $\begingroup$ Why would asymmetric velocities automatically imply a preferred frame? One could imagine that even if Alice sees Bob moving at speed v and contracted by L1 whereas Bob sees Alice moving at speed v' and contracted by L2, it could still be true that if Alice saw some different observer moving at speed v' she would see their length contracted by L2, and likewise if Bob saw some different observer moving at speed v he would see their length contracted by L1--in other words they would both see the same function for length contraction (and other properties) as a function of velocity. $\endgroup$ – Hypnosifl Nov 13 '14 at 13:34
  • $\begingroup$ Yes, I do agree that they are observe different length of the platform and this is because observer on the platform measures the length form two positions( i.e. at the start of the platform and at the end of the platform) but observer sitting on the train is measuring length from the same place of the moving train. For this difference of measurement the observed length becomes different. $\endgroup$ – Rajesh Sardar Nov 13 '14 at 14:28
  • $\begingroup$ OK, but you're just talking about what's true in SR, I'm talking about what we might imagine to be true if we knew the two postulates of SR and had seen a derivation of time dilation, but didn't yet see how these facts necessarily implied symmetric velocities. I'm saying that you can't just rule out asymmetric velocities based on "no preferred frame", since the description I gave above (which doesn't match what SR predicts about velocities or length contraction) still seems to allow for the possibility that each frame could use the same equations for length contraction as a function of v. $\endgroup$ – Hypnosifl Nov 13 '14 at 14:39

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