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I’m aware that similar questions have been asked numerous times but none of them have helped thus far. When I work through the logic of deriving the formula for length contraction,I keep making an error which I cannot seem to locate. My logic goes like this:

I stand on a platform in the rest frame while and train moving at relative velocity, v, in an inertial frame travels past. The train is at rest with a ruler in its inertial frame.

From my perspective on the platform, the length of the ruler is L = vt where v is the relative velocity of the train and t is the time the I perceive that the ruler takes to pass me.

From the perspective of an observer in the train’s frame, the length of the ruler is L’ = vt’ where v is the relative velocity of me on the platform and t’ is the time that the train observer perceived the ruler required to get past me.

Since, by time dilation, $t = t’/\sqrt{1-v^2/c^2}$, therefore $L = vt’/\sqrt{1-v^2/c^2}$. So, from the perspective of the observer in the rest frame, the length of the train seems to have dilated rather than contracted? Where have I gone wrong? Many thanks in advance.

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  • $\begingroup$ I think it’s that you assume, in a sense, that you agree about “when” the ruler reaches you. Im not positive but I think something in your method is assuming something like that. $\endgroup$
    – Al Brown
    Aug 18 at 4:05
  • $\begingroup$ In other words it’s not just different clock speeds. Youre assessing like you have different speed clocks, thats good, but they are starting up at the same moment in a sense and running. If the ruler is shorter for one guy, then the length also affects the when you start the clocks. Something off there. Im about 80-90% sure there is $\endgroup$
    – Al Brown
    Aug 18 at 4:07
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    $\begingroup$ $L=L'/\sqrt{1-\frac{v^2}{c^2}}$, so the length is dilated as seen by the rest frame. I think this is what the OP intends to say. $\endgroup$
    – Kksen
    Aug 18 at 4:21
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    $\begingroup$ @AlBrown: I of course agree. One thing about drawing the diagram is that it forces you to confront the question of whether your trainbound observer sits at the front of the ruler, the back of the ruler, or somewhere else. $\endgroup$
    – WillO
    Aug 18 at 4:44
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    $\begingroup$ @POW8J6 Keep in mind also that the measurement of the length of an object has to be taken simultaneously at both ends. $\endgroup$
    – Kksen
    Aug 18 at 6:13
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You are talking about 2 time intervals
t = the time the I perceive that the ruler takes to pass me
t' = time that the train observer perceived the ruler required to get past me.

While, you seemed to have correctly assumed that these 2 time intervals wont generally be the same, as these 2 time intervals are in different frames.
Your use of the t = gamma * t' formula , should be the other way round.
The 2 events in question are the head of the ruler crossing you and the tail of the ruler crossing you.

So, in this example, those 2 events happen in the same position , in your frame and they happen in different positions in the train observer frame.
So, the time interval measured by you (t) is the proper time interval .
So, the relation will be t' = gamma * t

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  • $\begingroup$ I understand the general idea of your answer and I had originally thought that an incorrect substitution of the time dilation formula might have been the error. I still don’t understand entirely, however, how the platform observer isn’t the one whose time is dilated relative to the train observer, given that the platform observer is in the rest frame. $\endgroup$
    – P0W8J6
    Aug 18 at 12:38
  • $\begingroup$ @P0W8J6 " given that the platform observer is in the rest frame. " Why do you think this is the case? There is no such thing as absolute rest frame. Strictly speaking, The platform observer is in no more of a rest frame than the train observer. You might consider it rest frame in your example. But, both are at rest in their respective frames, and both are in motion in each other's frames. To know which direction the formula goes, always carefully study the 2 events between which the 2 time intervals are being considered. That will tell you which time interval is the "proper" time interval. $\endgroup$ Aug 18 at 12:43
  • $\begingroup$ @P0W8J6 The time interval in the frame, where the 2 events happen at the same position is the proper time interval. In SR, proper time is the shortest, proper length is the longest. So, the equation will be gamma * (proper time interval) = ( time interval in the other frame ) $\endgroup$ Aug 18 at 12:43
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The answer given by @silverrahul is the right explanation. One way to remember how to apply the time dilation formula is that time dilation is a phenomenon that occurs in SR when one clock moves between two others that are stationary relative to each other. The dilation becomes apparent when you compare the time that has elapsed on the one clock with the difference between the readings on the two others that it has moved between. That's the key asymmetry of the set-up- you have to have two separate clocks in the other frame for the effect to show up.

In your example, if a clock on the train moves between two clock on the platform, then the train clock will seem time dilated compared to the readings on the two platform clocks. However, if you compare just one of the platform clocks against two clocks at either end of the train, then the platform clock will appear time dilated against the train clocks.

That's how time dilation is seen from both sides, because they are not comparing like with like. In each case a pair of clocks in one frame is being used to assess a time interval on a single clock in the other. The time dilation effect arises because the planes of simultaneity of the two frames are tilted relative to each other, so the clocks in one frame seem systematically out of synch to the clocks in the other.

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  • $\begingroup$ I’ve just been studying simultaneity in more depth and I’ve looked back to your answer which makes much more sense now and is something that I have never considered before. To confirm that I’ve understood correctly, would I be correct in saying that time dilation is the result of the fact that distance-separated clocks in a moving frame are, from the rest frame’s perspective, not synchronised... $\endgroup$
    – P0W8J6
    Aug 25 at 2:23
  • $\begingroup$ ....Therefore, time elapsed on the single clock in the rest frame that moves from one clock in the moving frame to the other, does not “account” for the original lack of synchronisation between the first and second clocks and so, when compared with the second moving clock, the single clock shows more time to have elapsed than the time difference between the two moving clocks. $\endgroup$
    – P0W8J6
    Aug 25 at 2:28
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    $\begingroup$ While the effects are two sides of the same coin, I think it is better to think of time-dilation as being the result of a lack of simultaneity than vice versa. Imagine walking down a long corridor where every 10 metres there is a clock on the wall. If those clocks were set so that each was 1 second ahead of the previous one, then as you walked down you would see that the time on your watch was falling a second behind the clock every ten metres. You would not know if your watch was running slow or the clocks were out of synch- the two effects are inter-related. $\endgroup$ Aug 25 at 7:35
  • $\begingroup$ Follow-on question: physics.stackexchange.com/q/661598/123208 $\endgroup$
    – PM 2Ring
    Aug 26 at 17:49
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You can't just willy-nilly throw time dilation around, as the front and back of the rulers at the same time (any time) on train, are not at the same time in a moving frame. It's best just to do the Lorentz transform.

Starting on the train frame ($S'$), the platform observer pass the ruler ends are 2 events:

The front: $$ E'_f = (t'=0,x'=0) $$

and the rear:

$$ E'_r = (L/v, -L) $$

where $L$ is the proper length of the ruler and $v$ is the train's speed.

The length is, as measure by $vt$:

$$ L = v(t_r'-t'_f) = v\frac L v = L $$

Inverse Lorentz transforming that to the platform frame ($S$):

$$ E_f=(t=0,x=0)$$

$$ E_r=\big(\gamma[\frac L v+ (-Lv)],\gamma[(-L+vL/v)]\big)$$ $$ E_r=\big(L\cdot\gamma(\frac 1 v -v)),0\big)$$ $$ E_r=\big(L\cdot\frac{\gamma} v(1 -v^2)),0\big)$$ $$ E_r=\big(\frac{L}{\gamma}/v ,0\big)$$ $$ E_r=\big(\frac{L'}{v} ,0\big)$$

$$ L' = \frac{L}{\gamma} = v\times(t_r-t_f)$$

showing that the ruler is Lorentz contracted.

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  • $\begingroup$ Just worked through this and it’s all very clear and easy to understand. Thanks- the utility of space time diagrams is now very evident to me. $\endgroup$
    – P0W8J6
    Aug 19 at 2:57
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This is essentially a slightly different version of the ladder paradox, a well known thought experiment in relativity.

Both this and the ladder paradox arrive from the false assumption that the front and back end of the ruler can agree on when they are directly across from the observer. When the back of the ruler sees that the front is crossing the platform's observer, it is actually already well past the observer. Thus the equation $L' = vt'$ is not the correct length of the ruler in the train's reference frame.

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  • $\begingroup$ Thank you all for your quick and clear responses $\endgroup$
    – P0W8J6
    Aug 18 at 5:01
  • $\begingroup$ Ok so vote his answer up please POW $\endgroup$
    – Al Brown
    Aug 18 at 5:08

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