# Complete relativistic treatment of photon emission recoil

I'm not finding an answer to this. I tried to do it forward, but checking backward gave a residual, presumably to a mistake in the fully relativistic calculation. Edit: I rechecked the calculations and think the below is correct.

Consider every quantity in the lab frame. A nucleus at rest in the lab frame and in an excited nuclear state with a rest mass of $$M$$ emits a photon. The resulting ground state nucleus has a rest mass of $$m$$. The idea is that $$M$$ might be 7 AMU and $$m$$ might be 6 AMU, just as an example to require a full relativistic treatment. I'm not looking for numbers, just formulas (with $$c$$ in them)

What is the momentum of the resulting gamma ray (in the lab frame)? What is its energy in the lab frame? Of course in the lab frame the momentum of the recoiled nuclear is minus that of the gamma ray.

What is the correct relativistic total energy of the recoiled nucleus in the lab frame?

Calculations:

The original nuclear energy is E_0 + dE thus the rest mass is E_0/c^2 + dE/c^2

after emission

photon momentum is p nucleus momentum is -p total nuclear energy, lab frame, is sqrt(E_0^2 + p^2c^2)

E_0 + dE = pc + sqrt(E_0^2 + p^2c^2)

E_0 + dE - pc = sqrt(E_0^2 + p^2c^2)

E_0^2 + dE^2 + p^2c^2 + 2E_0dE -2E_0pc -2dE*pc = E_0^2 + p^2c^2

dE^2 + 2E_0*dE = 2pc(E_0 + dE)

2pc = (dE^2 + 2E_0*dE)/(E_0 + dE) = dE * (2E_0 + dE)/(E_0 + dE) = dE * (2 + dE/E_0)/(1 + dE/E_0)

pc = dE * (1 + dE/(2 E_0))/(1 + dE/E_0)

And the rest of my questions have formulas above.

whew!

and lim as x -> 0 of pc ~= dE * (1 - dE/(2E_0))

• Have you tried the calculation yourself? Ie conserving the four momentum in the center of mass frame and then boosting to the lab frame? If you have you should provide any and all work attempted. Commented Feb 18, 2023 at 22:32
• To Triatticus: no ... I tried in the lab frame. Commented Feb 18, 2023 at 22:47
• To Triatticus: As I said, I have tried several times. But no ... I tried in the lab frame. But wait .... if the nucleus starts at rest in the lab, which I specified, those two frames coincide. I was just thinking ... maybe its impossible (i.e. need a second low energy photon). What bothers me is that I can't find a correctly done calculation anywhere. I want to see somebody else's, worked out. Commented Feb 18, 2023 at 23:01
• Still you need to put what you've tried in the body of the question, that is the actual calculation you attempted. Commented Feb 18, 2023 at 23:04
• Now is also a good time to familiarize yourself with the MathJaX the site uses to make mathematics and equations more readable, I don't normally have the link copied but there is a minor tutorial on it on the site. Commented Feb 19, 2023 at 3:24

Performed efficiently, this calculation should take you at most 10 seconds. You have a particle with mass $$M$$ decaying into a particle with mass $$m$$ and a photon. Energy-momentum conservation is expressed by the relation $$P=Q +p$$ in terms of the energy-momentum $$4$$-vectors $$P$$ (particle in the initial state), $$Q$$ (particle in the final state) and $$p$$ (photon). Because of the energy-momentum relations $$P^2 = M^2 c^2$$, $$Q^2 = m^2 c^2$$ and $$p^2=0$$, the equation $$(P-p)^2= Q^2$$ implies $$M^2 c^2-2 \, P\cdot p = m^2 c^2$$. In the rest frame of the decaying particle, you have $$P= (Mc, \vec{0})$$ with $$P \cdot p= Mc |\vec{p}|$$ for the $$4$$-scalar product (remember $$p=(|\vec{p}|, \vec{p})$$ for a massless particle). Expressed in terms of the masses $$M$$ and $$m$$, the absolute value of the photon momentum is thus given by $$|\vec{p}|=\frac{(M^2 -m^2) c}{2 M}.$$ It is now an easy task to express $$M$$ and $$m$$ in terms of $$E_0$$ and $$\Delta E_0$$ and compare with your result.

• Yes, Hyperion, my expression is easily derived from yours. Its now clear what I was really looking for ... which was an expression not in your M and m, but in m and (M-m) or the equivalent in energies, which I used. (M-m) is the variable you use in expanding in a series, which gives my final answer as (M-m) goes to zero. Using the squares of the masses obscures that. I'd still like to see a published reference to that. Commented Feb 19, 2023 at 15:20