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Old Apr 2nd 2009, 06:28 AM   #1
C.E
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Relativity- inertial frames

Suppose the space time co-ordinates of two events in inertial frame S are as follows:
Event 1:
x1=x0, t1=x0/c y1=z1=0
Event 2:
x2=2x0, t2= x0/2c y2=z2=0
Show that there exists an inertial frame s' in which these events occur at the same time (i.e t1'=t2') and find the value of time for which these events occur in this reference frame.


You may assume that: (Δx)^2-c^2Δt^2=(Δx’)^2 - c^2(Δt’)^2


I think I have a solution but annoyingly have not used the above asssumption which I want to do as it is given in the question. Anyway I did the following:

Firstly I set t1'=t2' from this it follows that γ(x0/c - ux0/c^2)=γ(x0/2c-2ux0/c^2) (by the lorentz x co-ordinate transformations) rearranging gives that u=-0.5c. Indeed u=-0.5c seems to work giving t1'=t2'=√3x0/c. However, I am unhappy with my answer in that firstly I seem to have assumed it rather than proved it and secondly I did not use the above assumption. Can somebody please show me how this question should be done?.(By the way this is not assesed work merely revision so feel free to give as much help as you deem appropriate).

Last edited by C.E; Apr 3rd 2009 at 08:46 AM.
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Old Apr 3rd 2009, 12:57 AM   #2
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You may assume that: (Δx)^2+c^2Δt^2=(Δx)^2 + (Δt)^2

Plug Δt = 0 in the above and see if you can prove the rest using the transformation equations.
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Old Apr 6th 2009, 08:08 AM   #3
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I am probably missing something really obvious but why am I using?
(Δx)^2+c^2Δt^2=(Δx)^2 + (Δt)^2
when the question says to assume that:
(Δx)^2-c^2Δt^2=(Δx)^2 - c^2(Δt)^2
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Old Apr 7th 2009, 12:01 AM   #4
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In relativity what is invariant is only the interval and that is given by
(Δx)^2-c^2Δt^2=(Δx)^2 - c^2(Δt)^2 which is what you should be using as that is how the interval is defined.

Just as in coordinate geometry the coordinates of the end points of the same line segment taken from different sets of axes will be different, but both will give the same value for length, i.e the length is invariant there, so is the case here.
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Old Apr 9th 2009, 08:04 AM   #5
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Ok when I set Δt=0 and substitute in Δx=x0, Δt=x0/2c gammaΔx'=Δx. I get that v=0.5c but when I put this into the lorentz x co-ordinate transformation formula it gives me that one of the events occurs at x0/csqrt(3) and the other occurs at -x0/csqrt(3), but the question specifies they should be the same. Where have I gone wrong?
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Old Apr 9th 2009, 11:09 AM   #6
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Since you have used Δt = 0 you should'nt be getting different values for time. The space oordinates however can be different. Pl check again.
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Old Apr 9th 2009, 11:43 AM   #7
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Ok I have checked about 3 times and still can't find a problem (sorry), here are my workings, can you see what is going wrong?

I will denote gamma by g.

(Δx)^2-c^2Δt^2=(Δx’)^2 - c^2(Δt’)^2

Δx=x0, Δx'= Δx/g =x0/g (by length contraction).
Δt’=0. Δt=x0/2c -x0/c=-x0/2c
Therefore
(x0)^2-c^2(x0/2c)^2=(x0/g)^2
so 0.75x0^2=x0/(g^2)
g=2/sqrt(3)
Hence 1-(v/c)^2=0.75 so v=0.5c.

However now using the lorentz coordinate transformations
t'=(t-ux/c^2)g (t denotes the time of an event in S and t' in S').
for event 1 time in S' is given by
t1'= g(x0/c -0.5cx0/c^2)=((x0-0.5x0)/c)2/sqrt(3)=2(0.5x0)/csqrt(3)
t1'=x0/csqrt(3)
Now considering event 2 in S'
t2'= g(x0/2c -0.5c2x0/c^2)=((0.5x0-x0)/c)2/sqrt(3)=2(-0.5x0)/csqrt(3)
t2'=-x0/csqrt(3)
Oddly enough when you try v=-0.5c it works fine but I don't see where I have gone wrong using v=0.5c.

Last edited by C.E; Apr 9th 2009 at 12:47 PM.
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Old Apr 10th 2009, 10:15 AM   #8
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Why were the last 2 posts deleted? I think I see what is happening, I am finding gamma correctly but as it depends only on u^2 and not on the sign of u, u=0.5c or u=-0.5c (by taking positive or negative square roots in finding u from gamma), in selecting 0.5c I clearly chose the wrong option. Anyway how could I know that I should have selected u=-0.5c and not 0.5c? Do you just have to try both and see what works?
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Old Apr 11th 2009, 04:22 AM   #9
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That is because when you view S' from S the velocity should be -ve.
I had posted and it had got posted but now has disappeared!
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Old Apr 14th 2009, 07:04 AM   #10
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How do you know that?
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