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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.

Señor O
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P0W8J6
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4 Answers4

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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

silverrahul
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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.

Professor Sushing
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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.

Señor O
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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.

JEB
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