# Understanding the (polarized) cross section for Compton Scattering by electrons

• I
• JD_PM
You are only considering the derivative of the first term in the square root. There is also the second one, ##\omega^\prime##, which gives the additional ##1/\omega^\prime## factor.In summary, The conversation is about Compton scattering by electrons and the differential cross section for it. It involves equations and calculations to determine the cross section in the LAB frame. Some of the equations and calculations are incorrect, leading to questions about the derivation. These questions are addressed and corrected by the expert summarizer.f

#### JD_PM

TL;DR Summary
I am studying the derivation of the polarized cross-section for Compton Scattering by electrons (i.e. both initial and final electron and photon states are set) from Mandl & Shaw (M&S)'s QFT book (chapter 8, section 8.6) and vanhees71's Manuscript (6.5.1 Compton Scattering).
I have specific questions, but let's first give context.

Initially we have an electron with momentum ##p=(E, \vec p)## and spin state ##u_r (\vec p)## and a photon with momentum ##k=(\omega, \vec k)## and polarization state ##\epsilon_s (\vec k)##.
Finally we have ##p'=(E', \vec p')##, ##u_r' (\vec p')##, ##k'=(\omega', \vec k')##, ##\epsilon_s' (\vec k')##

I understand that the differential-cross section for Compton Scattering by electrons is given by (more details here):

$$\frac{d \sigma}{d \Omega} = \frac{m^2 \omega'}{16 \pi^2 E E' \omega v_{rel}} \Big[ \Big( \frac{\partial(E'+\omega')}{\partial \omega'}\Big)_{\theta \phi} \Big]^{-1} |\mathscr{M}|^2 \ \ \ \ (1)$$

Where ##(\theta, \phi)## are the polar angles of ##\vec k'## and ##d \Omega = \sin \theta d \theta d \phi## is the corresponding infinitesimal solid angle. We take ##\vec k## as the polar coordinate axis, so that ##\vec k \cdot \vec k' = \omega \omega' \cos \theta## (where ##\cos \theta## arises due to the definition of the dot product).

Conservation of momentum gives

$$p+k=p'+k' \ \ \ \ (2)$$

The Feynman Amplitudes associated to these two figures are as follows $$\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b \ \ \ \ (3)$$

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \gamma^{\mu} \epsilon_{\mu}' (\gamma^{\mu} f_{\mu}+m) \gamma^{\mu} \epsilon_{\mu} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \gamma^{\mu} \epsilon_{\mu}' (\gamma^{\mu} g_{\mu}+m) \gamma^{\mu} \epsilon_{\mu}' u}{2(pk')} \ \ \ \ (4)$$

Where

$$f:=p+k, \ \ \ \ g:=p-k'$$

Eq. (3) is manifestly covariant and must be equivalent (I did not check it out myself though) to vanhees71's equation (6.128).

Experimentally the photon beam (usually) aims at a target of nearly stationary electrons. We now move to the LAB frame, which means that we have

$$p=(m, 0, 0, 0), \ \ \ \ \vec p' = \vec k - \vec k' \ \ \ \ (5)$$

Thus the energy-momentum equation yields

$$E' = \Big( m^2 + (\vec k - \vec k')^2 \Big)^{1/2} = \Big( m^2 + \omega^2 + \omega'^2 - 2\omega\omega' \cos\theta \Big)^{1/2} \ \ \ \ (6)$$

Mandl & Shaw asserted that from equation (2) we can get

$$pk=p'k+k'k=pk'+k'k \ \ \ \ (7)$$

And that Eq. (7) reduces to

$$\omega' = \frac{m \omega}{m+\omega(1-\cos\theta)} \ \ \ \ (8)$$

From Eq. (6) we get

$$\Big( \frac{\partial(E'+\omega')}{\partial \omega'}\Big)_{\theta \phi} = \frac{m \omega}{E' \omega'} \ \ \ \ (9)$$

Then, plugging back into (1) we get the differential cross section in the LAB frame

$$\Big( \frac{d \sigma}{d \Omega}\Big)_{LAB} = \frac{1}{(4 \pi)^2} \Big( \frac{\omega'}{\omega}\Big)^2 |\mathscr{M}|^2 \ \ \ \ (10)$$

These are my questions:

1) Where's Eq. (7) coming from?

M&S stated it comes from equation (2). However, I solve for ##p## and multiply (on the right) by ##k## but don't get (7)

$$pk = p'k+k'k-kk \neq p'k+k'k$$

$$pk = p'k+k'k-kk \neq pk'+k'k$$

2) Where's Eq. (8) coming from?

Let's assume Eq. (7); thus we have (recalling the definition of the 4 inner product ##p_1 \cdot p_2 = E_1 E_2 - \vec p_1 \cdot \vec p_2##)

$$(m\omega, 0)=(m \omega + \omega' \omega, -(\vec k - \vec k') \cdot \vec k - \vec k' \cdot \vec k)=(m \omega' + \omega' \omega, \vec k' \cdot \vec k)$$

Simplifying a bit we get

$$(m\omega, 0)=(m \omega + \omega' \omega, -\vec k \cdot \vec k )=(m \omega' + \omega' \omega, \vec k' \cdot \vec k)$$

Recalling that ##\vec k \cdot \vec k' = \omega \omega' \cos \theta## we get

$$(m\omega, 0)=(m \omega + \omega' \omega, -\omega^2 \cos \theta )=(m \omega' + \omega' \omega, \omega' \omega \cos \theta)$$

So we end up with these two equations

$$m\omega=m\omega+\omega'\omega=m\omega'+\omega'\omega \ \ \ \ (*)$$

$$\omega' \cos \theta=-\omega \cos \theta \ \ \ \ (**)$$

Mmm Eq. (*) is trivial so it tells us nothing interesting, but what can we say about ##(\omega'+\omega)\cos\theta=0##? Does it lead to Eq. (8)?

3) I am not getting Eq. 9

It has to be simply about taking the partial derivative but I am making a mistake I do not see...

$$\frac{\partial}{\partial \omega'} (E'+\omega') = \frac{\partial}{\partial \omega'} \Big[ \Big( m^2 + \omega^2 + \omega'^2 - 2\omega\omega' \cos\theta \Big)^{1/2}+\omega' \Big] = \frac{\omega'-\omega \cos \theta}{\sqrt{m^2 + \omega^2 + \omega'^2 - 2\omega\omega' \cos\theta}}+1=\frac{\omega'-\omega \cos \theta+E}{E} \neq \frac{m \omega}{E' \omega'}$$

Any help is appreciated.

Thank you Last edited:
For the first two:

1) ##k## should be an on-shell photon momentum, so ##k^2=0=(k^\prime)^2##. Similarly ##p^2=m^2=(p^\prime)^2##. From energy-momentum conservation (2) it then also follows that ##p^\prime k = pk^\prime##.

2) Why do you have vector equations there? Some of them look definitely wrong as they imply things like ##\vec{k}\vec{k}=0##, but I don't understand where they come from anyway. Eq. (7) is a relation between the scalars ##p\cdot k##, ##p\cdot k^\prime## and ##k\cdot k^\prime##. Calculating each of them in the rest frame of the initial electron, i.e. where ##p=(m,\vec{0})##, ##k=(\omega,\vec{k})##, ##k^\prime=(\omega^\prime,\vec{k}^\prime)## and using ##\vec{k}\vec{k}^\prime=\omega\omega^\prime \cos\theta## should almost immediately give (8) from the second version of (7).

Last edited:
• JD_PM
For the first two:

1) ##k## should be an on-shell photon momentum, so ##k^2=0=(k^\prime)^2##. Similarly ##p^2=m^2=(p^\prime)^2##. From energy-momentum conservation (2) it then also follows that ##p^\prime k = pk^\prime##.

Oh I see so we take Eq. (2), solve it for ##p## and multiply it (on the right side) by ##k##

$$p \cdot k = p' \cdot k + k' \cdot k - k \cdot k$$

Due to the on-shell condition ##k^2=0=(k^\prime)^2##, ##k \cdot k=0##. Thus I get the first equality on Eq. (7)

$$p \cdot k = p' \cdot k + k' \cdot k$$

I still have to proof the second equality on Eq. (7).

$$p \cdot k = p \cdot k' + k' \cdot k$$

This is my reasoning:

We now solve Eq. (2) for ##p'##, multiply (on the right side) by ##k'##

$$p' \cdot k' = p \cdot k' + k \cdot k' - k' \cdot k'$$

Due to the on-shell condition ##k^2=0=(k^\prime)^2##, ##k' \cdot k'=0##. Thus I get

$$p' \cdot k' = p \cdot k' + k \cdot k'$$

Mmm but I am not really convinced because to show Eq. (7) I need ##p' \cdot k'=p \cdot k## to hold and I think this is not true...

.2) Why do you have vector equations there? Some of them look definitely wrong as they imply things like ##\vec{k}\vec{k}=0##, but I don't understand where they come from anyway. Eq. (7) is a relation between the scalars ##p\cdot k##, ##p\cdot k^\prime## and ##k\cdot k^\prime##. Calculating each of them in the rest frame of the initial electron, i.e. where ##p=(m,\vec{0})##, ##k=(\omega,\vec{k})##, ##k^\prime=(\omega^\prime,\vec{k}^\prime)## and using ##\vec{k}\vec{k}^\prime=\omega\omega^\prime \cos\theta## should almost immediately give (8) from the second version of (7).

Ahhh big mistake of mine! Now I got it, thanks Eq. (7) tells us that

$$p \cdot k = p \cdot k' + k \cdot k'$$

Where

$$p \cdot k=m\omega$$

$$p \cdot k'=m\omega'$$

$$k \cdot k'=\omega\omega'-\vec k \cdot \vec k'$$

Thus we get Eq. (8)

$$m\omega=m\omega'+\omega\omega'(1-\cos \theta)$$

$$\omega'=\frac{m\omega}{m+\omega(1-\cos \theta)}$$

Mmm but I am not really convinced because to show Eq. (7) I need p′⋅k′=p⋅kp′⋅k′=p⋅kp' \cdot k'=p \cdot k to hold and I think this is not true...

Try squaring both sides of (2)...

Or first subtract ##k## and ##k^\prime## from both sides and then square to immediately get the second expression of (7) from the first.

• JD_PM
Alright so indeed we have ##p' \cdot k'=p \cdot k##, thank you.

About 3). There has to be a naive mistake, if I find it I'll post it.