De-excitation of a moving atom with photon emission

[Frostman]

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

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?

 

[anuttarasammyak]

Why do not you proceed to get formula of $v'(\theta,M,m,v)$ first then investigate it ?

 

[Frostman]

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.

 

[anuttarasammyak]

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
$$p=-\frac{\sin\beta}{\sin\theta}P$$$$k=\frac{\sin\alpha}{\sin\theta}P$$
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.

 

[Steve4Physics]

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.

 

[TSny]

(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.

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 $v'$. )

 

[vela]

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.