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For Length Contraction using Lorentz Transformation, we use the condition that from the 'Fixed frame' (in the following image) the length of the object was measured simultaneously (i.e. $t_1 = t_2$ ). I understand that if this condition is not fulfilled, an observer at 'Fixed frame' will make error in measuring the object length (since the object is moving with respect to the observer in 'Fixed frame').

Now for time dilation proof, a similar condition is used. Here,for the 'Moving frame', the time difference has been measured for such two events that took place in the same position (i.e. $x_1' = x_2'$). But I cannot understand why this condition must be fulfilled.

enter image description here

Would you please explain this? Thanks.

(image source: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html)

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2 Answers 2

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Now for time dilation proof, a similar condition is used. Here,for the 'Moving frame', the time difference has been measured for such two events that took place in the same position (i.e. x1' = x2'). But I cannot understand why this condition must be fulfilled.

They're assuming that the "moving frame" is the frame following the clock itself. That is, it is the frame where the clock isn't moving at all, so $x_1' = x_2'$.

We consider that frame, because time dilation is about comparing the time in your (unprimed) frame with the time that elapses in the clock's frame itself, i.e. the frame where the clock doesn't move.

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  • $\begingroup$ Thanks for your reply. But still, I have another confusion. You have clarified that in the clock's frame, the clock is stationary with respect to that frame, so two events take place at same position. But couldn't we measure the time difference between two ticks of two different clocks at different position?(or there might be two guns positioned at x1' and x2'. They fire at time t1' and t2' respectively.Now we want to express the time difference observed by an observer from Fixed frame between two guns' firing i.e. time difference between t1 and t2 with respect to t1',t2',x1' and x2'.) $\endgroup$ Commented Feb 19, 2020 at 13:34
  • $\begingroup$ @NafisSadik You could definitely do that. That would be a totally valid and possibly useful thing to calculate, it just wouldn't be the same thing as the simple "time dilation" effect. It would end up being a combination of a time dilation, length contraction, and loss of simultaneity effect. $\endgroup$
    – knzhou
    Commented Feb 19, 2020 at 20:34
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    $\begingroup$ @NafisSadik Textbooks focus on the case $x_1' = x_2'$ because it's simpler, so it's less confusing for students if they introduce one new effect at a time. $\endgroup$
    – knzhou
    Commented Feb 19, 2020 at 20:35
  • $\begingroup$ Thanks for your help, knzhou. Your comments have cleared my confusion. $\endgroup$ Commented Feb 20, 2020 at 15:54
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@NafisSadik Another way to address the two guns scenario could be to realise that the two guns are at rest in the moving frame, and there their firing delta time can be measured using two synchronised clocks in the two positions, and calculating the time difference in one of the two, for example in x_1', through a simple time recording message exchange from x_2' to x_1' when the gun in x_2' fires. This gives in x_1' a dt' = t_1' - t'_2, and due to the "static" scenario in this frame, i.e. all clocks at rest, this is still a proper time measure. Applying then the transformation rule for proper time it would give the corresponding delta time in the fixed frame, i.e. dt = gamma*dt' The mentioned length contraction would enter in the game in the fixed frame to calculate the gun distance x_1 - x_2 there, if compared with the one in the moving frame x_1' - x_2', and imposing on both frames constant speed of light.

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