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

• I
• JD_PM
In summary: 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.
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
Dr.AbeNikIanEdL said:
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...

Dr.AbeNikIanEdL said:
.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)

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

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

JD_PM said:
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.

## 1. What is Compton Scattering?

Compton Scattering is a phenomenon in which a photon (electromagnetic radiation) collides with an electron, resulting in a change in the photon's wavelength and direction. This process is important in understanding the interactions between light and matter.

## 2. What is the polarized cross section for Compton Scattering?

The polarized cross section for Compton Scattering is a measure of the probability that a photon will scatter at a particular angle and with a particular polarization state when it interacts with an electron. It is an important factor in understanding the behavior of light and electrons in this process.

## 3. How is the polarized cross section for Compton Scattering calculated?

The polarized cross section for Compton Scattering is calculated using quantum electrodynamics, which is a theoretical framework that describes the interactions between electromagnetic radiation and charged particles. This calculation involves considering the energy and momentum of the photon and electron before and after the scattering event.

## 4. What factors can affect the polarized cross section for Compton Scattering?

Several factors can affect the polarized cross section for Compton Scattering, including the energy and polarization state of the incident photon, the energy of the electron, and the angle at which the photon and electron collide. Additionally, the presence of external fields or other particles can also influence the polarized cross section.

## 5. Why is understanding the polarized cross section for Compton Scattering important?

Understanding the polarized cross section for Compton Scattering is important because it allows us to predict and analyze the behavior of light and matter in this process. This information is crucial in various fields, such as particle physics, astrophysics, and medical imaging, where Compton Scattering plays a significant role. It also helps us gain a deeper understanding of the fundamental interactions between light and matter.

• High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
15
Views
2K
• High Energy, Nuclear, Particle Physics
Replies
14
Views
2K
• High Energy, Nuclear, Particle Physics
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
391
• High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
536
• Introductory Physics Homework Help
Replies
2
Views
734
• Introductory Physics Homework Help
Replies
3
Views
325
• High Energy, Nuclear, Particle Physics
Replies
3
Views
1K