# Special Relativity - Conservation of Energy/Momentum

1. Nov 28, 2013

### Zatman

1. The problem statement, all variables and given/known data
An atom of rest mass M in an excited state of energy Q0 (measured in its rest frame) above the ground state moves towards a counter with speed v. The atom decays to its ground state by emitting a photon of energy Q (as recorded by the counter), coming completely to rest as it does so. Show that

$Q=Q_0[1+\frac{Q_0}{2Mc^2}]$

2. Relevant equations
$E^2=p^2c^2 + m_0^2c^4$
$E = \gamma m_0c^2$
$\gamma = (1-\frac{v^2}{c^2})^{-\frac{1}{2}}$

3. The attempt at a solution
In the lab frame, the atom is initially moving with speed v. When the photon is emitted, the atom is at rest. Hence using conservation of energy:

$\gamma Q_0 = Mc^2 + Q$

By conservation of momentum, the photon's momentum must equal the initial momentum of the atom. This gives:

$\gamma Mv = \frac{Q}{c}$

There is sufficient information here to eliminate v and γ (since γ depends on v). However, when I do so, I get nothing close to the required answer. So I guess my interpretation of the context is wrong. Please could I have a hint?

2. Nov 28, 2013

### TSny

Your expressions for the final energy and final momentum look correct. However, the expressions for the initial energy and initial momentum are not correct.

Think about the rest mass of the excited atom.

3. Nov 28, 2013

### Zatman

Okay, so I think that the initial energy of the atom given as Q0 does not include the rest energy? Which would mean the energy equation would become:

$\gamma Q_0 + Mc^2 = Mc^2 + Q$

$\Rightarrow \gamma Q_0 = Q$

I'm not sure why the momentum is incorrect. I am working in the lab frame, in which the atom is moving with speed v. Hence I must use

$\mathbf{p_{initial}} = \gamma M\mathbf{v}$

$\Rightarrow p_{initial} = \gamma Mv$

4. Nov 28, 2013

### TSny

This is still not correct.

You have the general equations $E = \gamma m_0 c^2$ and $p=\gamma m_0 v$.

You just have to get the $m_0$ correct when the atom is excited. (How does $Q_0$ affect $m_0$?)

5. Nov 29, 2013

### Zatman

Thanks for you replies, TSny, but I'm afraid I still cannot get it.

The rest mass of the atom in its excited state is given by

$Q_0 = m_0 c^2 \Rightarrow m_0 = \frac{Q_0}{c^2}$?

If so, this would change the energy equation to:

$\gamma Q_0 + \frac{Q_0}{c^2}c^2 = Mc^2 + Q$
$\Rightarrow (\gamma + 1)Q_0 = Mc^2 + Q$

and the momentum equation to:

$\gamma\frac{Q_0}{c^2}v = \frac{Q}{c}$
$\Rightarrow\gamma Q_0v=Qc$

6. Nov 29, 2013

### TSny

$Q_0/c^2$ represents the additional rest mass due to the excitation energy. This should be added to $Mc^2$ to get the total rest mass of the excited atom.

A given quantity of hot water has more rest mass than the same water when cold.

7. Nov 29, 2013

### Zatman

I see. This gives the excited rest mass as

$M+\frac{Q_0}{c^2}$

Giving the conservation of energy equation:

$\gamma (M+\frac{Q_0}{c^2})c^2 = Mc^2+Q$

And conservation of momentum:

$\gamma(M+\frac{Q_0}{c^2})v = \frac{Q}{c}$

Which then solve to give the required answer. Thank you for your help!

8. Nov 29, 2013

### TSny

Looks good.

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