Special Relativity 4-vector problem - Doing my head in

[Astrofiend]

Homework Statement

OK - the problem is thus:

In an inertial frame two observers (called a & b) travel along the positive x-axis with velocities Va and Vb. They encounter a photon traveling in the opposite x-direction. Without using the Lorentz transformations, show that the ratio of the energies of the photon observed by observers a & b is given by:

$$\frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}$$

Homework Equations

$$E_o = -p_o.u_o$$

where E_o is the observed energy of a photon with 4-momentum p_0, by a given observer moving with 4-velocity u_o.

I set up the 4-velocities of observers a & b as:

$$u^\alpha_a = (\gamma_a,\gamma__a V_a ,0,0)$$
$$u^\alpha_b = (\gamma_b,\gamma__b V_b ,0,0)$$

and the 4-momentum of the photon as:

$$p^\alpha = (p^t,p^x, 0,0)$$

The Attempt at a Solution

With these 4-vectors set up, the energies of the photon for each observer should just be the dot product of each 4-velocity with the negative of the 4-momentum of the photon - i.e:

$$E_o = -p_\alpha .u^\alpha = - \eta_\alpha_\beta p^\alpha .u^\alpha$$

where $$\eta_\alpha_\beta$$ is the metric.

so

$$E_a = \gamma_a p_t + \gamma_a V_a p_x \\$$

$$E_b = \gamma_b p_t + \gamma_b V_b p_x$$

Then, I used the fact that for a photon,

$$p^\alpha.p^\alpha = 0 \\$$

$$i.e. -p_t^2+ p_x^2 = 0 \\$$

so

$$p_t = p_x$$

and we get:

$$E_a = \gamma_a p_t + \gamma_a V_a p_t = \gamma_a p_t (1+V_a) \\$$

$$E_b = \gamma_b p_t + \gamma_b V_b p_t = \gamma_b p_t (1+V_b) \\$$

so

$$\frac{E_a}{E_b} = \frac{\sqrt{1-V_b^2}(1+V_a)}{\sqrt{1-V_a^2}(1+V_b)}$$

...which is where I'm falling down. As I said before, I'm after the relation

$$\frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}$$

Can anyone see what I'm doing wrong? Have I made a mistake somewhere, or is there some mathematical trick to take me further from where I am to the required answer? I've stared at this for a while now and can't work out why it's falling down.

Any help would be greatly appreciated.

 

[Astrofiend]

Actually, I think I botched the dot product earlier between the 4-vels and the 4-mom. I think it should give me:

$$E_a = \gamma_a p_t - \gamma_a V_a p_t = \gamma_a p_t (1-V_a) \\$$

$$E_b = \gamma_b p_t - \gamma_b V_b p_t = \gamma_b p_t (1-V_b) \\$$

as opposed to:

$$E_a = \gamma_a p_t + \gamma_a V_a p_t = \gamma_a p_t (1+V_a) \\$$

$$E_b = \gamma_b p_t + \gamma_b V_b p_t = \gamma_b p_t (1+V_b) \\$$

i.e. with minus signs, not the plusses I have before. But this seems to take me even further away from where I want to be. Arrg!

Help!

 

[Avodyne]

$$\frac{E_a}{E_b} = \frac{\sqrt{1-V_b^2}(1+V_a)}{\sqrt{1-V_a^2}(1+V_b)}$$

...which is where I'm falling down. As I said before, I'm after the relation

$$\frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}$$

Can anyone see what I'm doing wrong?

Nothing is wrong. These two expressions are equal. It's easier to see this by squaring each expression, then using $1-V^2 = (1-V)(1+V)$.

 

[Astrofiend]

Ah - thanks heaps. I'll work through it!

Cheers.