# Compton effect does the electron radiate?

In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.

## Answers and Replies

jtbell
Mentor
In the Compton Effect, we normally treat the interaction as a billiard ball collision.
Not really. We simply assume that conservation of energy and conservation of momentum apply, using the relativistic equations for both. Those principles don't depend on the details of the interaction.

Either use classical or quantum mechanics, not both.

According to classical mechanics, the electron is accelerated by an EM wave, and this does indeed cause it to radiate: and that radiation is what cancels out part of the original EM wave and constitutes the colour-shifted wave.

In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.
Textbook calculations are usually limited to the lowest order of the perturbation theory. If you take into account also higher orders you'll see that the Compton scattering is accompanied by the emission of "soft" photons, i.e., additional energy losses.

Not really. We simply assume that conservation of energy and conservation of momentum apply...
Isn't that the way we treat billiard balls?

jtbell
Mentor
Textbook calculations are usually limited to the lowest order of the perturbation theory. If you take into account also higher orders you'll see that the Compton scattering is accompanied by the emission of "soft" photons, i.e., additional energy losses.
Not to dispute this, but simply to get an idea of the size of this effect... consider for example a 662 keV gamma-ray photon Compton-scattering off an electron (which is commonly studied in simple undergraduate laboratories). Roughly how much energy do these soft photons carry?

Staff Emeritus
Roughly how much energy do these soft photons carry?
There's a range of energies, but an estimate of the size would be of order $\alpha/\pi$, probably somewhat less. Perhaps hundreds of eV.

There's a range of energies, but an estimate of the size would be of order $\alpha/\pi$, probably somewhat less. Perhaps hundreds of eV.
The "range" of energies is what I'd like to understand. In this kind of motion there is no sustained oscillation giving a single frequency, but rather a spectrum of frequencies which combines to account for the motion. The energy density within this spectrum is far from equalling a single quantum of energy for any specific frequency. So what does it mean if you detect a soft photon? Does the whole frequency spectrum "collapse"? Because there is no way all the soft energy could have gone into a single frequency at the moment of collision.

Not to dispute this, but simply to get an idea of the size of this effect... consider for example a 662 keV gamma-ray photon Compton-scattering off an electron (which is commonly studied in simple undergraduate laboratories). Roughly how much energy do these soft photons carry?
Vanadium50 gives a ballpark of several hundred eV for the soft photons. This seems really high to me. Lets just ballpark...

The 662 keV photon has a wavelength on the order of 10 ^-12 meters. So in the COM frame, the electron has the same wavelength. This corresponds to an electron energy of approx. 1000 eV. So Vanadium's "soft photons" would carry of a very significant part of the electron energy. How would the undergraduate laboratory demonstration give the right numbers in that case?

jtbell
Mentor
Perhaps hundreds of eV.
I'm just looking for an order of magnitude figure for the total energy radiated. So it's hundreds of eV compared to hundreds of keV for the incoming and scattered photon, which is on the order of 0.1% of the total energy involved in this process.

You'd surely need pretty sensitive experiments to detect this effect, which is one reason why it's not brought up in a standard intro modern physics course. Also the theoretical analysis is a bit above the level of that kind of course.

jtbell
Mentor
The 662 keV photon has a wavelength on the order of 10 ^-12 meters. So in the COM frame, the electron has the same wavelength. This corresponds to an electron energy of approx. 1000 eV. So Vanadium's "soft photons" would carry of a very significant part of the electron energy. How would the undergraduate laboratory demonstration give the right numbers in that case?
Suppose the photon scatters at 45 degrees. The usual Compton-scattering energy equation

$$\frac {1}{E^{\prime}} - \frac {1}{E} = \frac {1}{mc^2} (1 - cos \theta)$$

gives the energy of the scattered photon as E' = 480 keV. Therefore the recoiling electron has kinetic energy 662 - 480 = 182 keV.

A few hundred eV for the additional soft photons is a few times 0.1 keV.

Staff Emeritus
I'm just looking for an order of magnitude figure for the total energy radiated. So it's hundreds of eV compared to hundreds of keV for the incoming and scattered photon, which is on the order of 0.1% of the total energy involved in this process.
I'm also probably on the high side here as well. The other scales in the problem are all smaller, and I know that the interferences are destructive, not constructive.

You can visualize the soft photon emission as follows: the photon scatters off the electron, sending it flying. Now that the electron has experienced an acceleration, it radiates.

jtbell
Mentor
The 662 keV photon has a wavelength on the order of 10 ^-12 meters. So in the COM frame, the electron has the same wavelength. This corresponds to an electron energy of approx. 1000 eV.
A wavelength of $10^{-12}$ m correponds to a momentum of 1240 keV/c. For an electron, this means a (total) energy of $E = \sqrt {(pc)^2 + (mc^2)^2}$ = 1342 keV.

For my example, with a scattering angle of 45 degrees, the wavelength is $3.55 \times 10^{-12}$ m, which leads to a (total) electron energy of 619 keV.

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Here is a web reference that discusses scattering of 1-eV laser photons off a 4 to 8 GeV electron beam. references to published papers on Compton scattering vertex corrections are included.
The vertex corrections are less than 1%, Do a google search on "SLAC Compton Scattering" to find other similar papers

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Staff Emeritus
Um...no, that's a link to an Alnico data sheet.

The 1-loop vertex corrections are not the same as this. Indeed, this particular piece of the calculation is IR-divergent.

In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.
As meopemuk said, there is always soft radiation with a continuous spectrum.

The Compton scattering kinematics is similar to billiard ball one if one considers photons as particles with specific energy-momentum (quantum) relationship and the electrons as relativistic particles. But the Compton wavelength Λ=ћ/mc is written for electron as if it were a quantum wave (not a particle) and the photon were a classical wave of a given frequency (not a particle). In QED both are de Broglie waves, so their scattering is due to non linearity of QED's wave equations. Of course, the other photon modes are also excited (soft radiation). Normally their energy is within the hard photon source/detector and electron detector accuracy range (i.e. included experimentally events). It corresponds to the inclusive theoretical QED cross section. See my "Atom as a "dressed" nucleus" for the inclusive cross section physics under large-angle scattering.

Bob_for_short.

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A wavelength of $10^{-12}$ m correponds to a momentum of 1240 keV/c. For an electron, this means a (total) energy of $E = \sqrt {(pc)^2 + (mc^2)^2}$ = 1342 keV.

For my example, with a scattering angle of 45 degrees, the wavelength is $3.55 \times 10^{-12}$ m, which leads to a (total) electron energy of 619 keV.
Thanks for checking my numbers. I see what I did wrong. (I ballparked a 10eV electron wavelength as 10^{-10}m based on the hydrogen atom, then multiplied the momentum by 100 to equate to the wavelength of the photon. Forgot to square 100 to get the energy multiplier. Still out by a factor of 10 but what the heck.)

... consider for example a 662 keV gamma-ray photon Compton-scattering off an electron (which is commonly studied in simple undergraduate laboratories). Roughly how much energy do these soft photons carry?
What is the incident gamma-ray energy uncertainty in such experiments?
What are the gamma-detector and electron detector uncertainties?

Bob.

What is the incident gamma-ray energy uncertainty in such experiments?
What are the gamma-detector and electron detector uncertainties?
Bob.
The 662 KeV gamma from Cs-137 is very precise, because the half-life of Cs-137 is many years. A gamma detector has statistical uncetainties on the energy resolution, which depends on what fraction of the photon energy is absorbed, and/or electron/holes (silicon) or photoelectrons (for NaI(Tl)) are created for each individual incident photon. In both cases (silicon and NaI(Tl)), the incident photon can create a Compton electron, and the recoil secondary photon escapes the detector. This will show up in the energy spectrum as a Compton backscatter peak at about 184 KeV.
My first "exposure" to the 662 KeV cesium spectrum was when we received a pulse height analyzer (PHA) from the vendor in 1958, and the spectrum was still stored in the PHA core memory.

Hans de Vries
Gold Member
In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.

Technically, there is an interference pattern between the initial and final
state of the electron. This sinusoidal interference pattern of charge and
spin density is responsible for the radiation.

In more detail it's a two step process, the first is the interaction of the
electron with the incoming photon while the second is the interaction
with the radiated photon.

Regards, Hans

Technically, there is an interference pattern between the initial and final
state of the electron. This sinusoidal interference pattern of charge and
spin density is responsible for the radiation.

In more detail it's a two step process, the first is the interaction of the
electron with the incoming photon while the second is the interaction
with the radiated photon.

Regards, Hans
Thanks, Hans. But I'm not sure you're interpreting it exactly the way I do. For me, the incoming and outgoing electron interfere to form a stationary diffraction grating. That's what the light reflects from. It provides a neat semi-classical explanation for the Compton effect without the need to invoke photons.

Hans de Vries
Gold Member
Thanks, Hans. But I'm not sure you're interpreting it exactly the way I do. For me, the incoming and outgoing electron interfere to form a stationary diffraction grating. That's what the light reflects from. It provides a neat semi-classical explanation for the Compton effect without the need to invoke photons.

What I described is the is the first order approximation of QED
which, besides being 99+% accurate, also provides the neat
semi classical explanation you are looking for.

Maxwell's laws stay 100% the same, but we have to consider:

1. The electron field is a continuous charge/current density
and magnetization/polarization distribution ($\bar{\varphi}\gamma^\mu\varphi$ and $\bar{\varphi}\sigma^{\mu\nu}\varphi$)
2. The electron field can interfere resulting in sinusoidal
charge/current density and magnetization terms.
The importance of the intrinsic spin of the electron is illustrated
by the fact that the source of the transverse components of the
em radiation (emitted real photon) stem from the transverse currents
in the interference pattern caused by the magnetization gradients.
(as shown via the so called Gordon decomposition)

Regards, Hans

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What I described is the is the first order approximation of QED
which, besides being 99+% accurate, also provides the neat
semi classical explanation you are looking for.

Maxwell's laws stay 100% the same, but we have to consider:

1. The electron field is a continuous charge/current density
and magnetization/polarization distribution ($\bar{\varphi}\gamma^\mu\varphi$ and $\bar{\varphi}\sigma^{\mu\nu}\varphi$)
2. The electron field can interfere resulting in sinusoidal
charge/current density and magnetization terms.
The importance of the intrinsic spin of the electron is illustrated
by the fact that the source of the transverse components of the
em radiation (emitted real photon) stem from the transverse currents
in the interference pattern caused by the magnetization gradients.
(as shown via the so called Gordon decomposition)

Regards, Hans
I'm sure this is the right picture but your language is different from mine in certain ways and I'm not understanding it. In my superposition of electron states I simply have a sinusoidal charge distribution. Just like the potential well. No current distribution. Why do you have a sinusoidal current distribution?

Yes, you see an oscillating current distribution once you let the incoming radiation fall on these parallel sheets of charge, but that doesn't appear just from the superposition of states, does it?

Hans de Vries
Gold Member
I'm sure this is the right picture but your language is different from mine in certain ways and I'm not understanding it. In my superposition of electron states I simply have a sinusoidal charge distribution. Just like the potential well. No current distribution. Why do you have a sinusoidal current distribution?

Yes, you see an oscillating current distribution once you let the incoming radiation fall on these parallel sheets of charge, but that doesn't appear just from the superposition of states, does it?

The alternating transverse current distribution is the result of the
intrinsic spin of the electron. Each point of the electron field repre-
sents a small magnetic dipole caused by a circular current.

This circular current is described by the axial current density which
is calculated from Dirac's electron field $\varphi$ with the expression $\bar{\varphi}\gamma^\mu\gamma^5\varphi$

A sinusoidal distribution of such a circular current is equivalent, via
Stokes law, to an alternating current distribution. This can be seen
in the attached image below. If the circular currents are equal then
they cancel each other locally and there is no net (vector) current.

If there is a gradient however then they do not cancel each other
and there is an effective (vector) current. The left hand side of the
image shows the circular (axial) currents and the right hand side of
the image shows the effective (vector) currents which are the source
of the transverse components of the emitted photon.

Regards, Hans

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