# Help me through a simple SR problem

1. Jun 9, 2008

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.

2. Jun 9, 2008

### George Jones

Staff Emeritus
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 $dx'/dt'$.

3. Jun 9, 2008

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?

4. Jun 9, 2008

### Fredrik

Staff Emeritus
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.

Here:

5. Jun 9, 2008

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

$$a = dt'\beta$$

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