De-excitation of a moving atom with photon emission

Frostman
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Homework Statement
An excited atom of mass ##M## moving with velocity ##v## in the laboratory system decays into a state of mass ##m## by emitting a photon.
Calculate the relationship between the final speed ##v'## of the atom and the angle ##\theta##, formed by the directions of flight of the final particles.
Relevant Equations
Conservation of the four-momentum
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The information I have are the following:

##p^\mu=(E, p, 0, 0)##
##p'^\mu=(E', p'\cos\beta, -p'\sin\beta,0)##
##k^\mu=\tilde{E}(1, \cos\alpha, \sin\alpha, 0)##

Where:

##E=\sqrt{M^2+p^2}##
##E'=\sqrt{m^2+p'^2}##

Using the conservation of the four-momentum

##p^\mu=p'^\mu+k^\mu##
##(p^\mu)^2=(p'^\mu+k^\mu)^2##
##M^2=m^2+0+2(E'\tilde{E}-p'\tilde{E}\cos\theta)##
##\frac{M^2-m^2}{2(E-E')}=E'-p'\cos\theta##
##\frac{M^2-m^2}{2E'(E-E')}=1-v'\cos\theta##

Now I should replace the values of ##E'## and ##E##, but I don't know if this is the correct way, after that I would have all the roots of their definitions and I would have to rework the formula to write ##v'## as a function of ##\theta##, ##M##, ##m## and ##v##.

What do you think?
 
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Why do not you proceed to get formula of ##v'(\theta,M,m,v)## first then investigate it ?
 
I tried, but the more I go I get lost in the accounts. I'm not even rewriting what I wrote here on the forum because it seems to me just a waste of time, but to show that I tried to do the math I am attaching the spreadsheets.

As this is an exam to be done along with 2 exercises in 2 hours, it's strange that I have to lose myself in so many accounts.

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A tough calculation. I just try an introduction.
Energy conservation
\sqrt{P^2+(Mc)^2}=\sqrt{p^2+(mc)^2}+k...(1)
Momentum conservation
P=p \cos \alpha+k \cos\beta
0=p \sin \alpha + k \sin \beta
\theta=\alpha-\beta
so
p=-\frac{\sin\beta}{\sin\theta}Pk=\frac{\sin\alpha}{\sin\theta}P
with ##\beta<0<\alpha<\theta##. The formula
sin\theta=\sin\alpha \cos\beta - \cos \alpha \sin\beta would give
1=\frac{k}{P}\sqrt{1-\frac{p^2}{P^2}\sin^2\theta}+\frac{p}{P}\sqrt{1-\frac{k^2}{P^2}\sin^2\theta}
\frac{1}{\rho\kappa}=\sqrt{\frac{1}{\rho^2}-\sin^2\theta}+\sqrt{\frac{1}{\kappa^2}-\sin^2\theta}
where
\rho=\frac{p}{P}=\frac{mv}{MV}\sqrt{\frac{1-\frac{V^2}{c^2}}{1-\frac{v^2}{c^2}}}
and from (1)
\kappa=\frac{k}{P}=\frac{\sqrt{P^2+(Mc)^2}-\sqrt{p^2+(mc)^2}}{P}=\frac{c}{V}\{1-\frac{m}{M}\sqrt{\frac{1-\frac{V^2}{c^2}}{1-\frac{v^2}{c^2}}}\}
I hope it is correct and may help you check validation of your calculation.
 
Last edited:
I’m not too familiar with this stuff, but I’ll take a shot…

M and m are rest masses and we take c=1.
Let γ = 1/√(1 -v²) and let γ’ = 1/√(1 -v’²).
In the lab’ frame:
- the (initial) energy of M is γM
- the (final) energy of m is γ’m.
So the photon has energy γM - γ’m (and therefore its momentum is γM - γ’m).

In the lab’ frame, use xy axes oriented so that the photon is moving in the +x direction; this means that m is moving at angle θ anticlockwise with respect to the +x axis.

Therefore the photon’s momentum is γM - γ’ in the +x direction.
I.e. the photon’s 4-momentum is k = (γM - γ’m, γM - γ’m, 0, 0)

And m’s 4-momentum is: p = (γ’m, γ’mv’cosθ, γ’mv’sinθ, 0)
________________

Total 4-momentum after emission is P = k + p

P
= (γM – γ’m + γ’m, γM – γ’m + γ’mv’cosθ, γ’mv’sinθ, 0)
= (γM, γM + γ’m(v’cosθ -1), γ’mv’sinθ, 0)

Since the length-squared of the 4-momentun is Lorentz invariant:
P•P = M²
(γM, γM + γ’m(v’cosθ -1), γ’mv’sinθ, 0)•(γM, γM + γ’m(v’cosθ -1), γ’mv’sinθ, 0) = M²
(γM )² – [(γM + γ’m(v’cosθ -1))² + (γ’mv’sinθ)² + 0²] = M²

The question states “Calculate the relationship between the final speed v’ of the atom and the angle θ…” which is a bit ambiguous/vague. Technically, once we replace γ and γ’ by their definitions, the last equation (above) answers the question. If you want an equation for v', then it's now a (very nasty) exercise in algebra.

If I've misunderstood something, I would be more than happy if someone could correct me.
 
Steve4Physics said:
(γM )² – [(γM + γ’m(v’cosθ -1))² + (γ’mv’sinθ)² + 0²] = M²
I think this result is equivalent to the OP's result if ##E## and ##E'## in the OP's result are replaced by ##\gamma M## and ##\gamma' m##, respectively.

Steve4Physics said:
The question states “Calculate the relationship between the final speed v’ of the atom and the angle θ…” which is a bit ambiguous/vague. Technically, once we replace γ and γ’ by their definitions, the last equation (above) answers the question. If you want an equation for v', then it's now a (very nasty) exercise in algebra.
I tend to agree. Finding a relationship between ##v'## and ##\theta## does not necessarily imply that you have to express ##v'## explicitly as a function of ##\theta##.

( It wouldn't be hard to do the reverse; i.e. express ##\theta## as a function of ##v'##. )
 
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An alternative approach would be to find the energy and momentum in the rest frame of the (initial) atom and then transform back to the lab frame.
 
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