How Does Recoil Affect Photon Emission Frequency in Special Relativity?

[Quattro]

I’m having some trouble with a question from my special relativity assignment.

It relates to the frequency of a photon emitted from an atom which then subsequently recoils. The atom has initial rest mass m0, and loses rest energy ‘e’ in the emission (“e is the difference between the rest energy of the final and initial atom”).

The question asks to prove the frequency of the emitted photon is:

$$f=\frac{e}{h}(1-\frac{e}{2m_0c^2})$$

I really enjoy special relativity (we just had a lecture about how the existence of a magnetic field around a current carrying wire can be explained simply by considering SR, that’s so cool!), but I often find myself getting horrible confused. This is especially true when it comes to relativistic mass type questions like the above (and similarly nuclear reactions equations). I always thought the best thing to do with such problems was to just ensure both energy and momentum are conserved, but I still usually get mixed up.

Anyway, back to this problem, here’s what I tried.

For starters, I figured that since e is the difference in rest mass energies, this value must be equivalent to the gained kinetic energy of the atom plus the energy of the photon:

$$e = E_k + hf$$

I then equated the initial and final energy of the system:

$$E_i=m_0c^2=E_f=hf+mc^2$$

Where $$m$$ is the relativistic mass.

This is where the trouble starts… Is the above line correct? I get the feeling the final energy isn’t right. It has to be the energy of the photon plus the kinetic energy and rest mass energy of the new ‘lighter’ atom (one photon ‘lighter’), but how can this be expressed? Introduce a new rest mass: $$m_02$$?

I get the feeling this is getting unnecessarily complicated…

Someone said the thing to do is use momentum rather than velocity, specifically the formula: $$E^2=p^2c^2+(m_0c^2)^2$$ but I really can't see how that helps.

I will be grateful for any help, tips and advice on this question (or even methods for solving such problems in general). The question is only worth 1 mark so really shouldn’t be so hard, I think I’m just doing something obviously wrong.

Thanks in advance.

(please excuse any tex errors, its my first time using it…)

 

[Dick]

Do yourself a great favor and forget about 'relativistic mass'. Just deal in rest masses. Unless you enjoy being confused. The relation $$E^2=p^2c^2+(m_0c^2)^2$$ is what you need. To use it, equate the initial and final total energies and the initial and final total momenta and use this equation to eliminate extra variables. Eg label the two atoms 0 and 1 and the photon g. Then

E0=E1+Eg
p0=p1+pg

where E's are energies and p's are momenta. Let's get started. p0=0, E0=m0*c^2, right? Since the photon is massless Eg=c*|pg|. Etc etc. Try it out!

 

[Saketh]

I always thought the best thing to do with such problems was to just ensure both energy and momentum are conserved, but I still usually get mixed up.

You thought correctly.

The thing is, you have to be sure that your expressions are conserving total energy. This is one issue in moving from classical to relativistic dynamics -- getting used to dealing with total energy (it's actually much more straightforward). The expression $$E^2 = p^2 c^2 + m_0 c^2$$ has E as the total energy. For a photon, this expression reduces to $pc$, which is equivalent to $h\nu$.

As you notice, momentum is present in both the energy and momentum equations. Therefore, you will set up a system and solve it.

 

[Quattro]

Thanks for the input guys, but I’m still having trouble.

I now see the significance of the $$E^2=p^2c^2+(m_0c^2)^2$$ formula and used it to setup the system of energy and momentum you introduced Dick:

Energy: $$E_0=E_1+E_\gamma$$

$$E_0=m_0c^2$$
$$E_1=\sqrt{(m_1c^2)^2+p_1^2c^2}$$
$$E_\gamma=p_\gamma c$$

$$\rightarrow m_0c^2=\sqrt{(m_1c^2)^2+p_1^2c^2}+ p_\gamma c$$

Momentum: $$p_0=p_1+p_\gamma$$

$$p_0=0$$
$$p_1=-p_\gamma$$
$$p_\gamma=hf/c$$

$$\rightarrow p_1=-p_\gamma=-hf/c$$

Also I took the definition of $$e$$ to be:

$$e=(m_0-m_1)c^2$$

I get an answer for f in terms of e and $$m_0$$ but its not the same as the one I’m supposed to get. Are my above equations correct? If so, I guess ill just keep trying, its just algebra…

 

[Mentz114]

Quattro, I think m1 = m0. There's only one rest mass involved here.

$$E_1=\sqrt{(m_0c^2)^2+p_1^2c^2}$$

 

[Quattro]

Doesn't the rest mass have to decrease slightly to produce the energy for the photon?

 

[George Jones]

Doesn't the rest mass have to decrease slightly to produce the energy for the photon?

Yes, and e is define as defined it in post #4. In some sense, the rest mass of the atom includes contributions from the binding energies of the electrons to the nucleus, as well as from the masses of the electrons and the nucleus. I haven't checked your other work, and I have to go out now, but if you can't get it to work out, I'll try and come back to this thread later today or tomorrow. Yesterday, I did manage to get the formula that you're supposed to derive.

 

[Quattro]

Thanks George.

Im on a bit of a high at the moment after finally solving another similar (but 5 mark) question in this same assignment i was stuck on. I'd say this thread is at least partially resposible, that total relativisic energy formula really is useful!

But ill give this question with the equations i posted another go later and see what happens.

[EDIT]: It worked after all! I sort of cheated by working my way back from the answer, but hey, its the same in the end.

Thanks for all your help.

 

[George Jones]

From your equation in post #4, I get the result. Use $e = m_0 c^2 - m_1 c^2$ to eliminate $m_1$ in your other equation.

A tip: $E^2 - c^2 p^2$, where $E$ and $p$ are the total energy and momentum of system, is conserved. Equating the expression for this quantity before with the expression after avoids the square root.

 

[Quattro]

Yeah, thanks George, i did get their in the end (see my last post).
The process in going from what i got initially to including the defined e was a bit messy, but hey, and answer is an answer.

Thanks again