Can length contraction really be derived from time dilation? Does speed equal speed?

Question

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.

Answer 1

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 $v$.

Answer 2

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.

Answer 3

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.