Help me through a simple SR problem

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  • #1
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I'm reading Wheeler's spacetime physics and have been doing some newbie SR problems.
I thought up what shouldd be an extremely simple problem but am having trouble with the math, I'm sure one of you guys can probably help me out with it.

Events A and B occur with a time separation in the laboratory frame but no space separation, I thought it'd be easy to prove that in a rocket frame moving with a [tex]\beta[/tex] speed to the right relative to the laboratory frame the space separation divided by the time separation of the events would be [tex]-\beta[/tex] (that is [tex]dx'/dt'=-\beta[/tex])

[tex]dt^2-dx^2 = dt'^2-dx'^2[/tex]

Since the events occur in the same place in the laboratory frame
[tex]dt^2 = dt'^2-dx'^2[/tex]


After Lorentz transformation
[tex] ( \frac{dt' + dx'\beta} { \sqrt{1-\beta^2}} )^2 = dt'^2-dx'^2 [/tex]

However, I've been unable to derive this properly, is it just lack of math skills or did I set the equations inproperly? Any help is appreciated.
 

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  • #2
George Jones
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Your last equation does lead to your desired result. First factor out [itex]dt'^2[/itex] from both sides of the last equation, and then solve for [itex]dx'/dt'[/itex].
 
  • #3
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I must suck at this, so far I have
[tex] ( \frac{dt' + dx'\beta} { \sqrt{1-\beta^2}} )^2 = dt'^2-dx'^2 [/tex]


[tex] \frac{(dt' + dx'\beta)^2} {1-\beta^2} = dt'^2-dx'^2 [/tex]


[tex] (dt' + dx'\beta)^2 = (dt'^2-dx'^2) (1-\beta^2) [/tex]


[tex] dt'^2 + 2dt'dx'\beta + dx'^2\beta^2 = dt'^2 - dt'^2\beta^2 -dx'^2 +dx'^2\beta^2 [/tex]


[tex] 2dt'dx'\beta = -dt'^2\beta^2 -dx'^2[/tex]


[tex] dx'^2 + dt'^2\beta^2 + 2dt'dx'\beta = 0[/tex]


This does not seem to hold up when [tex]\frac{dx'}{dt'} = -\beta[/tex]

Where am I screwing up?
 
  • #4
Fredrik
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I posted this once already and deleted it because I incorrectly thought I had made a mistake, and the forum won't let me post a duplicate, so I had to add this pointless sentence.

Where am I screwing up?
Here:
This does not seem to hold up when [tex]\frac{dx'}{dt'} = -\beta[/tex]
 
  • #5
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[tex]dx'^2 + 2dt'dx'\beta + dt'^2\beta^2 = 0[/tex]

[tex]a = dt'\beta[/tex]

thus
[tex]dx'^2 + 2adx' + a^2 = 0[/tex]

[tex](dx' + a)^2 = 0[/tex]

[tex]dx' + a = 0[/tex]

[tex]dx' = -a[/tex]

[tex]dx' = -dt'\beta[/tex]

[tex]dx'/dt' = -\beta[/tex]

Thanks everyone, I will proceed and slap myself in the head so you won't have to
 

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