Special Relativity 4-vector problem - Doing my head in

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Astrofiend
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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:

[tex] <br /> \frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}<br /> [/tex]

Homework Equations



[tex] <br /> E_o = -p_o.u_o<br /> [/tex]

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:

[tex] <br /> u^\alpha_a = (\gamma_a,\gamma__a V_a ,0,0)<br /> [/tex]
[tex] <br /> u^\alpha_b = (\gamma_b,\gamma__b V_b ,0,0)<br /> [/tex]

and the 4-momentum of the photon as:

[tex] <br /> p^\alpha = (p^t,p^x, 0,0)<br /> [/tex]

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:

[tex] <br /> E_o = -p_\alpha .u^\alpha = - \eta_\alpha_\beta p^\alpha .u^\alpha[/tex]

where [tex]\eta_\alpha_\beta[/tex] is the metric.

so

[tex] E_a = \gamma_a p_t + \gamma_a V_a p_x \\[/tex]

[tex] E_b = \gamma_b p_t + \gamma_b V_b p_x[/tex]

Then, I used the fact that for a photon,

[tex] p^\alpha.p^\alpha = 0 \\[/tex]

[tex] i.e. -p_t^2+ p_x^2 = 0 \\[/tex]

so

[tex] p_t = p_x[/tex]

and we get:

[tex]E_a = \gamma_a p_t + \gamma_a V_a p_t = \gamma_a p_t (1+V_a) \\[/tex]

[tex] E_b = \gamma_b p_t + \gamma_b V_b p_t = \gamma_b p_t (1+V_b) \\[/tex]

so

[tex] \frac{E_a}{E_b} = \frac{\sqrt{1-V_b^2}(1+V_a)}{\sqrt{1-V_a^2}(1+V_b)}[/tex]

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

[tex] <br /> \frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}<br /> [/tex]

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

[tex] E_a = \gamma_a p_t - \gamma_a V_a p_t = \gamma_a p_t (1-V_a) \\[/tex]

[tex] E_b = \gamma_b p_t - \gamma_b V_b p_t = \gamma_b p_t (1-V_b) \\[/tex]

as opposed to:

[tex] E_a = \gamma_a p_t + \gamma_a V_a p_t = \gamma_a p_t (1+V_a) \\[/tex]

[tex] E_b = \gamma_b p_t + \gamma_b V_b p_t = \gamma_b p_t (1+V_b) \\[/tex]

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!
 
Astrofiend said:
[tex] \frac{E_a}{E_b} = \frac{\sqrt{1-V_b^2}(1+V_a)}{\sqrt{1-V_a^2}(1+V_b)}[/tex]

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

[tex] <br /> \frac{E_a}{E_b} = \sqrt{\frac{1+V_a}{1+V_b}.\frac{1-V_b}{1-V_a}}<br /> [/tex]

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 [itex]1-V^2 = (1-V)(1+V)[/itex].
 
Ah - thanks heaps. I'll work through it!

Cheers.