Lorentz Transformation/Time Dilation


by bon
Tags: dilation, lorentz
bon
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#1
Mar27-10, 05:16 AM
P: 569
1. The problem statement, all variables and given/known data

Ok so I know how to prove Time Dilation using the Lorentz transformation...

My problem is that I don't see why one couldn't use the Inverse Lorentz Transform and come to the opposite result as follows:

Time period in rest frame S: T = t2 - t1

ILT: t = $(t1' + B/c x') where $ is gamma

So: t2-t1 = $(t2'-t1') so T' = T/$ which is the wrong result...

I.e. how does one know that one must use the LT rather than ILT?

Thanks


2. Relevant equations



3. The attempt at a solution
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vela
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#2
Mar27-10, 05:46 AM
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Why is that wrong? γ is greater than 1, so T' < T, as expected. The time elapsed in the moving frame is shorter than the time elapsed in the rest frame.
bon
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#3
Mar27-10, 05:50 AM
P: 569
No it should be the other way round. "Time Dilation" - time measured in any frame moving wrt rest frame should be dilated (i.e. longer time period) than in rest frame..

CompuChip
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#4
Mar27-10, 05:51 AM
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Lorentz Transformation/Time Dilation


I didn't look at it too closely, but I suspect that by using ILT you have implicitly inverted the rest frame and the moving frame.
Basically, you have relabelled T to T' and vice versa.
bon
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#5
Mar27-10, 05:53 AM
P: 569
Thanks, yes I agree. I just can't see where the inconsistency comes in in the maths... :S
bon
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#6
Mar27-10, 06:22 AM
P: 569
anyone?
vela
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#7
Mar27-10, 06:24 AM
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Quote Quote by bon View Post
No it should be the other way round. "Time Dilation" - time measured in any frame moving wrt rest frame should be dilated (i.e. longer time period) than in rest frame..
OK, so you're assuming the clock is in the rest frame and trying to calculate what someone in the moving frame would see?
vela
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#8
Mar27-10, 06:33 AM
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Your inconsistency stems from ignoring x' in the Lorentz transformation. It looks like you just erased it going from the ILT to your conclusion.
bon
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#9
Mar27-10, 06:42 AM
P: 569
but i though x1' = x2' since the two times were measured at the same position in S'?
vela
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#10
Mar27-10, 06:48 AM
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If the clock is at rest in S, it's moving in S', so x1' and x2' won't be equal.
bon
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#11
Mar27-10, 06:56 AM
P: 569
Ahh this is getting confusing now. Ok so its measured in S at a position x to have time interval t2-t1

Now what happens when the measurement is made in S' - what stays constant? My understanding was that the time interval is measured to be t2'-t1' made at a fixed position in S' - call this x'..

What have i don wrong?

thanks for your help
bon
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#12
Mar27-10, 07:01 AM
P: 569
Basically what i don't understand is the asymmetry:

using your above expression

time dilation: clock at rest in S. moving in S' therefore x1' and x2' aren't equal..
length contraction: clock at rest in S. moving in S'. Therefore t1' and t2' aren't equal..

but for length contraction t1' and t2' are equal - we use the inverse LT:

x1 = γ(x1'+Bct')
x2 = γ(x2' + Bct')

i.e. t1'=t2'=t
which gives the right result: x2-x1 = γ(x2'-x1')
vela
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#13
Mar27-10, 07:08 AM
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For time dilation, you're comparing the time elapsed between the same two events in different frames.

When measuring length, you're finding the spatial separation of, say, the two ends of a stick at the same instant. Since simultaneity is frame dependent, you're looking at different pairs of events in different frames.
bon
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#14
Mar27-10, 07:10 AM
P: 569
Thanks ok - so what went wrong in my last post? (no. 12)..

Thanks
bon
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#15
Mar27-10, 07:28 AM
P: 569
?
thanks
vela
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#16
Mar27-10, 07:34 AM
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Just when you said this: "length contraction: clock at rest in S. moving in S'. Therefore t1' and t2' aren't equal." When you measure the length in S', you have to have t1'=t2', right? You have to measure where the ends are at the same time (in S').
bon
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#17
Mar27-10, 07:36 AM
P: 569
but then by parity of argument, when we measure time in S', we have to measure at x1'=x2', right? You have to measure the time at the same point in space?
vela
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#18
Mar27-10, 07:48 AM
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No. The reason you have to measure where the endpoints are simultaneously is just an operational requirement. If the stick is moving, you can't measure where one end is at one time, measure where the other end is a second later, and subtract the two to get the length. You have to see where both ends are at the same time, and only then can you subtract those numbers to get the length.

With time dilation, you have two spacetime events, and different observers will say different amounts of time separate them, depending on their relative motion.

The symmetry you're looking for isn't there. Time dilation compares the same two events in different frames. Length contraction compares a pair of events in one frame to a different pair in another.


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