Relativistic dynamics problem

[Pedro de la Torre]

Homework Statement

A stationary excited nucleus decays to its grpund state by emitting a γ-ray photon of energy E_γ. The grpund state nucleus recoils in the opposite direction at speed v. Show that when v<<c the change of mass of the nucleus is approximately:

m_e - m_g = E_γ/c² [1+ 1/2(v/c)]

Homework Equations

v/c=pc/E
P^μP_μ = constant = (mc²)²
E² = (mc²)² + (pc)²
P_γc = E_γ

The Attempt at a Solution

I used both, the cuadrimomentum conservation and the energy conservation. I think the solution should be directly found from the cuadrimomentum conservation, but I can not solve it.

 

[TSny]

The Attempt at a Solution

I used both, the cuadrimomentum conservation and the energy conservation.

Conservation of 4-momentum implies conservation of both relativistic energy and relativistic 3-momentum.

I think the solution should be directly found from the cuadrimomentum conservation, but I can not solve it.

I'm not following what you mean here. It would be helpful if you showed some detail of your attempt to solve it this way.

 

[Pedro de la Torre]

Ok, here I attach how I did this problem. (If you can not see correctly, the image is in: https://drive.google.com/open?id=0B6AVsjwLIVdgSDlwa2NYc29zRWM)

I obtained the same as the problem asks, but with a - instead of a +.

Am I wrong?

 

[TSny]

Your work looks good, except I don't understand why you suddenly put in a negative sign at the very end. $\frac{E_{\gamma}}{E_g}$ is positive. $E_{\gamma} = |p_{\gamma}|c$.
So, $1 + \frac{E_{\gamma}}{2 E_g} = 1+ \frac{v}{2c}$.

 

[Pedro de la Torre]

I did that because I considered there P_g = - P_γ. But I think I am realising that this is only for the vectorial form, Isn´t it?

 

[TSny]

I did that because I considered there P_g = - P_γ. But I think I am realising that this is only for the vectorial form, Isn´t it?

Yes.