Special Relativity: Photon emission by a moving atom

1. Mar 11, 2008

Helgi

1. The problem statement, all variables and given/known data
The problem involves an atom (Said to be in an excited state of energy Q_0) traveling towards a scintillation counter with speed v. The atom then emits a photon of energy Q and stops completely. The rest mass of the atom is m. I'm supposed to show that
$$Q = Q_0(1+\frac{Q_0}{2mc^2})$$

2. Relevant equations
It's mostly just conservation of energy and momentum stuff.
The kinetic energy of the atom is
$$K = (\gamma -1)m_0 c^2$$
I have a feeling this formula
$$E^2 = (cp)^2+m_0^2c^4$$
has to be used but I don't see where.

3. The attempt at a solution
What I figured I should do was to use conservation of energy and momentum. So I set up 2 equations

$$Q_0 + (\gamma-1)m_0c^2 = m_0 c^2+Q$$ Which is the energy conservation
$$\gamma\cdot m_0v = \frac{Q}{C}$$ and the conservation of momentum

Which I'm pretty sure is set up alright but when I try to solve for Q I don't get anything that is simplifies to what it's supposed to be. The $$\gamma$$ and v usually get in the way and they're not supposed to be in the answer.

Now what I tried was to just simplify the first formula, and I ended up with
$$2Q_0 = 2m_c^2 + Q$$
this looks pretty reasonable but it's missing the second power on the $$Q_0$$.

Last edited: Mar 11, 2008
2. Mar 11, 2008

Shooting Star

The key point to note here that the excited state of energy Q0 is in the rest frame of the atom, not the lab frame as you have considered. This energy Q0 makes the rest energy of the atom (Q0+mc^2). So, the initial energy of the system in labframe is $$(Q0+mc^2)\gamma$$, and the initial momentum is $$(Q0/c^2+m)\gamma v$$.

The rest is as you have done. It's a two liner. Put c=1 to keep it neat, and put it back in the end to make it dimensionally correct.