Help me through a simple SR problem

[adimare]

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 \beta speed to the right relative to the laboratory frame the space separation divided by the time separation of the events would be -\beta (that is dx'/dt'=-\beta)

dt^2-dx^2 = dt'^2-dx'^2

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


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

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.

 

[George Jones]

Your last equation does lead to your desired result. First factor out dt'^2 from both sides of the last equation, and then solve for dx'/dt'.

 

[adimare]

I must suck at this, so far I have
( \frac{dt' + dx'\beta} { \sqrt{1-\beta^2}} )^2 = dt'^2-dx'^2


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


(dt' + dx'\beta)^2 = (dt'^2-dx'^2) (1-\beta^2)


dt'^2 + 2dt'dx'\beta + dx'^2\beta^2 = dt'^2 - dt'^2\beta^2 -dx'^2 +dx'^2\beta^2


2dt'dx'\beta = -dt'^2\beta^2 -dx'^2


dx'^2 + dt'^2\beta^2 + 2dt'dx'\beta = 0


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

Where am I screwing up?

 

[Fredrik]

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 \frac{dx'}{dt'} = -\beta

 

[adimare]

dx'^2 + 2dt'dx'\beta + dt'^2\beta^2 = 0

a = dt'\beta

thus
dx'^2 + 2adx' + a^2 = 0

(dx' + a)^2 = 0

dx' + a = 0

dx' = -a

dx' = -dt'\beta

dx'/dt' = -\beta

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