# Why Compton scattering probability for gamma ray increases and reaches a maxima and finally decreases with energy? Non mathematical answer please

It is seen from mass attenuation vs energy plot thar compton scattering cross section goes through a maximum. It is low at low energy and high at high energy. Why? I want a non mathematical answer.

Related:

• Compton scattering is defined as scattering of a photon by a free electron, but normally we're interested in Compton scattering that occurs in matter, where the electrons are bound. At the keV and MeV energies we normally have in mind, the eV binding energies are negligible, so we can approximate the electron as free. But at the very lowest energies on this graph, I think that is no longer a good approximation. In the limit of low energies, the Compton cross-section approaches a constant, the Thomson scattering cross-section, of about 0.67 barn. It doesn't fall off as shown in this graph.
– user4552
Jan 5 '18 at 19:54
• I'm not pretending that this is an answer, but note that the Compton cross-section peaks at an energy about equal to the rest mass of the electron. If we assume that the cross-section has a peak, then it kind of makes sense that this is where it peaks. For a free electron, there is no other parameter that could determine where the peak would be. When you integrate the Klein-Nishina cross-section over all angles, you get a complicated expression in terms of the dimensionless variable $x=E_\gamma/mc^2$.
– user4552
Jan 5 '18 at 19:57
• @BenCrowell Despite the complicated formula, the punchline is that the Compton scattering (Klein-Nishina) cross section decreases with energy for energies above the electron rest mass energy, because the photons have a greater chance of being "forward scattered." Put this all together and I think we have an answer Jan 5 '18 at 20:10
• Sir any complete explanation?
– ggs
Jan 6 '18 at 9:17
• You write in your question It is low at low energy and high at high energy. Don't you mean that it is low at high energy? Jan 12 '18 at 21:58

## 2 Answers

This is not a complete answer, but I think I partially understand why the cross-section falls off roughly like $1/E_\gamma$ at high energies. Note that the title of the question refers to probabilities, but the graph is of cross-sections. The probabilities are affected by competition from pair production.

High-energy behavior

In terms of relativistic transformation properties, we expect $\sigma=A/F$ to be invariant under longitudinal boosts, where $A$ is a Lorentz scalar and $F=mE_\gamma$ is the Moller flux factor . The only dimensionless Lorentz scalars we can form are $x=p_\gamma\cdot p_e/p_e\cdot p_e=E_\gamma/m$ and functions of $x$. The high-energy limit of the integrated Klein-Nishina formula is $\sigma\approx \pi r_0^2\ln x/x$, where $r_0$ is the classical electron radius, so $A\approx \pi/\ln x$.

So it seems that the high-energy behavior of the cross-section is mainly just kinematic, except for the rather gentle logarithmic dependence on energy that is present as well. Of course this is not really a fundamental answer, since I don't have any physical reason to offer as to why $A$ is only logarithmic in $x$ at high energies.

Low energies, free electrons

In the limit of low energies, the Compton scattering cross-section for a truly free electron is just the Thomson cross-section, which is classical and constant. I assume the Klein-Nishina cross-section falls off as you go down in energy from $mc^2$ and then approaches that limit. It would be interesting to understand why it falls off, which I don't understand. Mathematically, if you look at the integrated Klein-Nishina cross-section as a function of $x$, a bunch of terms cancel to first order for small $x$.

Low energies, in matter

In matter, it's not really clear to me whether it makes sense to talk about Compton scattering in a low-energy limit, and I don't know how this was defined in the graph the OP posted. Compton scattering is by definition scattering of a photon by a free electron. At very low energies, it will not make sense to describe any electrons (except maybe conduction electrons in a metal?) as free. I would think the cross-section would depend on the condensed matter physics. In a non-metal, below the threshold for the photoelectric effect, I would expect the cross-section for scattering to be zero.

• Is “unavailability of free electrons at low energy” and argument one could make? Jan 9 '18 at 14:08
• @Floris: Hm...an electron is either free or not free, regardless of the photon's energy. I'll rewrite the end of my non-answer a little.
– user4552
Jan 9 '18 at 15:31
• Thanks. Does thermal excitation give you the “occasional free electron” - I realize I am nitpicking but I have a hard time believing there is a hard cutoff energy for Compton. Jan 9 '18 at 18:21
• @Floris: I guess near the Fermi level, occupation numbers will be smeared out thermally, so at low energies you would have bound states available to scatter into. Seems like a bit of a stretch to call that Compton scattering, though.
– user4552
Jan 11 '18 at 15:23

My 25 cents. Note that the diagram, your question is illustrated with, is for Pb.

There is a general factor that decreases the cross sections with increasing energy: The photon field amplitude with increasing energy as any quantized vibration in the first excited state.

The photoelectric effect is decreasing at increasing photon energy is related to that a free electron cannot absorb a photon as it would violate momentum an energy conservation. Bound electrons can absorb a photon due to it can exchange momentum (and energy) with the nucleus. (You may view it as inverse Bremsstrahlung.) When photon energy increases, ultimately the exchange of momentum between electron and nucleus become relatively small, which makes the cross section to go down.

For small photon energies, the Compton scattering is prohibited as the electrons are bound to the nucleus. The scattering off the atom as a whole is Rayleigh scattering which is shown as a separate curve in the diagram. As photon energy increases the Rayleigh scattering is less likely as the electron binding is not sufficient to stop them being torn off. This implies that the Compton scattering is starting to play a rôle. At energies higher than the electron rest energy $m_ec^2$, the forced vibration of the electron, which sends of the scattered photon, gets smaller in analogy with that of a forced vibration above its resonance frequency.

Obviously, the pair creation cross section is zero for $E_{\gamma} < 2m_ec^2$. There it takes off, but also here you have the need to exchange momentum and energy with the nucleus, which is a factor that lowers the cross section with energy.

• The third paragraph is about the photoelectric effect, and the fifth paragraph is about pair production. This doesn't seem directly relevant, since the graph we're trying to explain is a graph of cross-section, not probability, so competition from the other processes is not relevant. (The title of the question does refer to probability rather than cross-section, which doesn't match up with the graph.)
– user4552
Jan 12 '18 at 23:29
• At energies higher than the electron rest energy $m_ec^2$, the forced vibration of the electron, which sends of the scattered photon, gets smaller in analogy with that of a forced vibration above its resonance frequency. This seems like the directly relevant part of the answer, and it would be helpful to have some explanation of why you think this analogy works, and why the resonance would occur at a frequency corresponding to $m_ec^2$. Certainly if we believe there should be a resonance, then there is only one possible energy at which the resonance could occur, but why is there a resonance?
– user4552
Jan 12 '18 at 23:33