# How long does it take for an electron to reach equilibrium with blackbody radiation?

While teaching a course on electrodynamics, I thought of an interesting question that I think deserves some attention.

Consider an ensemble of electrons all with momentum $$\hbar \mathbf{k}$$ traveling in free space. By considering an ensemble, we are ignoring interactions between electrons, so e.g. each one is in an independent part of space/time. Now, consider free space to have a finite temperature $$T$$, like the CMB or in a vacuum chamber. Then, there is a finite density of photons in space determined by the Bose distribution.

My question is: how long does it take for the electron ensemble to equilibriate with blackbody radiation, if ever on the scale of the universe? Free electrons can't absorb photons, so I would guess Compton-type scattering would dominate the equilibriation process, which is really weak especially at temperatures well below the electron mass (e.g. 1-10,000 Kelvin).

Edit: I meant Compton (inelastic), not Thomson (elastic) scattering.

• You are describing an electron beam. If no effort (applied electric and magnetic fields) is made to keep it together it will disperse due to the electric repulsion between the electrons. They cannot form the ensemble you visualize – anna v Aug 20 '19 at 5:41
• @anna v I know that, but that is irrelevant, the question is still a valid one. One can always choose large enough spacing in space or time between the electrons to ignore "space charge"-type repulsion. This is done regularly in beam optics by using high energy electrons at very low currents. – KF Gauss Aug 20 '19 at 6:16
• Nitpick: Compton scattering is elastic, in the (usual) sense that the total kinetic energy before & after the collision is the same. Saying that it's inelastic might be a bit confusing; unless it's standard terminology in some sub-fields of physics that I'm not familiar with? – Michael Seifert Aug 20 '19 at 14:10
• @MichaelSeifert I think this is standard terminology in experimental particle physics, condensed matter physics, plasma physics, atomic physics, and nuclear physics. Usually in scattering you only measure one of the products, so whether something is elastic or inelastic comes down to whether energy is changed for the subset of particles you can measure. See this question: physics.stackexchange.com/questions/319299/… – KF Gauss Aug 20 '19 at 14:33
• Interesting question, but there is one big problem: how do you define equilibrium state for the ensemble of electrons? The closer the ensemble average energy $\langle E\rangle$ gets to expected average for temperature $T$ (probably $E_{av}(T) = \frac{3}{2}k_BT$?) the slower the rate of change of that average becomes, because when electron gets slow enough, the radiation it experiences can increase and decrease its energy with similar probabilities. This decreases the rate of loss of energy below naive estimate from Larmor's formula, so convergence to expected average may take infinite time. – Ján Lalinský Aug 20 '19 at 21:04

If you are going to ignore electron interactions then the only processes going on are interactions between photons and electrons.

If those interactions are limited to Thomson scattering (which is elastic), then the electrons cannot change their energies and so can never achieve a Maxwell-Boltzmann distribution.

However, even for low electron energies there will be some inelastic processes occurring - namely Compton scattering, whereby a photon loses some of its energy to an electron. For a typical collsion, the change in wavelength is of order $$h/m_e c = 2.4\times 10^{-12}$$ m. For a low-energy blackbody radiation field, say at temperatures below $$10^6$$ K, the cross-section will almost be the same as the Thomson scattering cross-section ($$6.6\times 10^{-29}$$ m$$^2$$).

If the electron energy is small compared with $$k_B T$$, then the number of collisions to thermalise an electron will be of order $$k_B T/\Delta E$$, where $$\Delta E = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \sim \frac{h^2}{\lambda^2 m_e}$$ A typical photon also has energy of around $$3k_BT$$, so we can let $$\lambda \sim hc/3k_BT$$ and so the number of collisions required is $$N \sim \frac{m_e c^2}{9k_BT} \simeq \frac{10^9}{T},$$ where $$T$$ is in Kelvin.

The total number density of photons in a blackbody raditation field is just a function of temperature and is given (in SI units) roughly by $$n \sim 60 \left(\frac{k_B T}{hc}\right)^3$$

The time between collisions will be $$\sim (n \sigma c)^{-1}$$, where $$\sigma$$ is the cross-section. So my final result is that the time taken will be $$\tau \sim \left(\frac{m_e c^2}{9k_BT}\right) \left(\frac{hc}{k_B T}\right)^{3} \left(\frac{1}{60\sigma c}\right) \sim 10^{21}\ T^{-4}\ {\rm seconds}$$

I think this will not be applicable if the electrons are relativistic or if the electron kinetic energies are large compared with $$k_B T$$. In the latter case a means of transferring energy from the electrons to the photons will be required - i.e. the inverse Compton effect and this does become important if the electrons are relativistic. In that case $$\Delta E \sim (\gamma^2 -1) E$$, where $$\gamma$$ is the Lorentz factor of the electrons and $$E$$ is the photon energy (e.g. see https://casper.ssl.berkeley.edu/astrobaki/index.php/Inverse_Compton_Scattering ). To thermalise the electrons requires reducing their kinetic energy from $$(\gamma-1) m_e c^2$$ to $$\sim k_B T \sim 0$$. The number of collisions required is $$N \sim \frac{E}{\Delta E} \sim \frac{m_e c^2}{3(\gamma +1)k_B T},$$ where again I have used $$\sim 3 k_B T$$ as a typical photon energy.

The rate of interaction is still $$(n \sigma c)^{-1}$$ and this then gives me a timescale for thermalisation of $$\tau \sim 5\times10^{21} \left(\frac{1}{\gamma +1}\right) T^{-4}\ {\rm seconds},$$ which for $$\gamma \gg 1$$ is faster than the timescale when the electrons are at much lower energy than the radiation field.

• The time between collisions should be inversely proportional to n sigma. Also intuitively time to thermalization should be longer for colder radiation. – Ján Lalinský Aug 20 '19 at 13:31
• > "The number of collisions to thermalise an electron will be of order $k_BT/\Delta E$." -- Could you motivate this a bit more? I assume by thermalisation you mean increase of kinetic energy of electron until it reaches value $3/2k_BT$. That makes sense only if initial electron kinetic energy is much smaller than that. However there are higher energies that are still far from relativistic regime. – Ján Lalinský Aug 20 '19 at 14:41
• Thanks! There are a number of assumptions here that I have to think about. I have some doubts about assuming a constant $\sigma$ over both very low and high energy electrons. I assume this can be derived explicitly from the Klein-Nishina formula, but I would imagine it would change a lot with $\gamma$. – KF Gauss Aug 20 '19 at 14:50
• > "a means of transferring energy from the photons to the electrons will be required - i.e. the inverse Compton effect" Shouldn't that be the opposite? For fast electron, transfer of energy from electron to radiation is required. Also, could you explain why change of energy in single scattering on relativistic electron varies as $\gamma^2$? – Ján Lalinský Aug 20 '19 at 14:57
• Nice work. The final conclusion about relativistic electrons is surprising to me; the more energy it has, the faster it thermalizes. Is there some support for this curious prediction in literature or experiments? – Ján Lalinský Aug 20 '19 at 17:40

Building off of Rob Jeffries answer, I will try to write out a complete expression for the case of electrons with higher kinetic energy than the Blackbody thermal energy $$\sim k_b T$$.

The starting point is the expression for the total power irradiated by an electron to a photon density $$U_{\mathrm{photon}}$$ through inverse Compton scattering (see the nice reference https://casper.ssl.berkeley.edu/astrobaki/index.php/Inverse_Compton_Scattering).

$$P_{\mathrm{tot}}=\frac{4}{3} \sigma_T U_{\mathrm{photon}} c \beta^2 \gamma^2$$

Where $$\sigma_T$$ is the Thomson cross-section $$\sigma_T = \frac{8\pi}{3}(\frac{\alpha \hbar c}{m c^2})^2 \approx 66.5 \,\mathrm{fm}^2$$, $$U$$ is the photon density, which for blackbody radiation is simply $$U=\frac{4 \sigma T^4}{c}$$, where $$\sigma$$ is the Stefan-Boltzmann constant.

We can write $$P_{\mathrm{tot}}$$ as a function of (kinetic) energy by noting $$KE=(\gamma-1)m c^2$$ and $$\beta^2 = 1-1/\gamma^2$$.

$$P_{\mathrm{tot}}=\frac{4}{3} \sigma_T U c \frac{KE\,(KE+2 mc^2)}{m c^2}$$

Then because $$P_{\mathrm{tot}} dt = dKE$$, the total time taken to radiate from the starting (kinetic) energy $$E_0$$ to the thermal energy $$\frac{3}{2} k_b T$$ is given by the integral:

$$\tau = \int_{E_0}^{\frac{3}{2} k_b T} \frac{1}{P_{\mathrm{tot}}} dKE$$

The end result of this is:

$$\tau = \frac{m c^2}{\frac{8}{3} \sigma_T U c} \mathrm{log}\left[ \frac{E_0+2m c^2}{\frac{3}{2} k_B T + 2m c^2} \frac{\frac{3}{2} k_B T}{E_0}\right]$$

Plugging in all the constants and using units of Kelvin for the temperature $$T$$, we get

$$\tau \approx \frac{1}{T^4} \mathrm{log}\left[ \frac{E_0+2m c^2}{\frac{3}{2} k_B T + 2m c^2} \frac{\frac{3}{2} k_B T}{E_0}\right] \times2\times 10^{21} \,\,\mathrm{Seconds}$$

The weak logarithmic behavior means for practical purposes the logarithm only changes by a factor of about 3 for all electron kinetic energies above $$k_b T$$.

So this means it would take $$\sim 10^{21}$$ seconds for any electron to equilibriate with the CMB blackbody radiation, which is much longer than the age of the universe. On the other hand, if the electron was traveling through an empty region at 10,000 Kelvin, it would equilibriate in $$\sim 10^{6}$$ seconds, which may be of some astrophysical relevance.

If you are ignoring the e-e interaction your system is effectively a free electron. This electron is moving in a background of thermal radiation. Inelastic scattering of photons will eventually decrease the initial electron momentum to a thermal value. I assume a nonrelativistic electron.

The question is then at what timescale will its initial momentum decay to a thermal value $$kT/c$$ by inelastic scattering. The differential cross section can be found on Wikipedia. Now you have to work out the momentum transfer, which will depend on the angle with respect to the electron momentum and on the black body radiation momentum. Integrate over angle and black body spectrum - or simply take the average energy over c - to get a net momentum transfer opposed to the original electron momentum, per time unit. I suspect that you will find an exponential decay law with the desired time constant.

• I think this belongs as a comment, not an answer. – KF Gauss Aug 20 '19 at 11:57
• @KFGauss This is a blueprint of how to work this out theoretically. If you publish please reference me. – my2cts Aug 20 '19 at 16:35
• A minus? There is no pleasing some people. – my2cts Aug 20 '19 at 16:36
• Thomson scattering is elastic. – ProfRob Nov 17 '20 at 16:31