Ideally I would like to have an $x$-$y$ graph of ($x$) speed relative to centre of mass of the body (star or planet) against ($y$) the number or percentage of electrons having that speed at a given moment in time. For simplicity I can ignore the small proportion of electrons in the atmosphere. Also please ignore effects on electron speed caused either by speed of the body rotating around its axis or by orbital revolution. I do not need high precision just a rough picture.

UPDATE 20160707 I have accepted CR Drost's answer as it is very helpful. But I am not confident that I have got the necessary understanding to answer my question.

A useful related question with a respected answer is here: how-fast-do-electrons-travel-in-an-atomic-orbital.

My (IMPERFECT!) Understanding at present (added following Chris Drost's answer).

Originally I inquired about electron speeds in the Sun and the Earth but following input from Chris Drost's answer this seems too ambitious (for my high-school level physics ability). So I will focus on a simpler imaginary model comprising a white dwarf star and a planet consisting of just a metallic iron core.

Relevant Information

According to wikipedia (Quantum and classical regimes) the classical Maxwell-Boltzmann formula for particle speeds ($v$)

$$f(v) = \left( \frac{m}{2\pi kT} \right)^{3/2} 4\pi v^2 \mathrm{e}^{-\frac{mv^2}{2kT} }$$

applies if the concentration of particles corresponds to an average interparticle separation $\bar{R}$ that is much greater than the average de Broglie wavelength $\bar{\lambda}$ of the particles. Here $\bar{\lambda} = \frac{h}{\sqrt{3mkT}}$ with $h$ the Plank constant ($h=6.62606957\times 10^{-34} \,\mathrm{J\cdot s}$) and $m$ the particle mass ($m_e = 9.11 \times 10^{-31} \,\mathrm{kg}$ for an electron).

But when $\bar{R}$ is not "much greater" than $\bar{\lambda}$ the Fermi-Dirac equation for energy (applicable to Fermions e.g. electrons, protons, neutrons) applies: $$f(E) = (e^{(E-E_f)/kT}+1)^{-1}.$$ where $k$ is the Boltzmann constant ($k=1.3806488 \times 10^{-23} \,\mathrm{J/K}$), $T$ is temperature and $E_f$ is the Fermi Energy wikipedia. The section "Typical Fermi energies" gives example of calculations of the Fermi energy for (i) conduction electrons in metals ($\sim 10\,\mathrm{eV}$ for a density of $10^{28}$ to $10^{29}$ electrons$/m^3$), (ii) for degenerate electrons in white dwarfs (~ $3 \times 10^5 \,\mathrm{eV}$) and (iii) for nucleons in an atom.

To obtain the distribution of electron speeds I would simply apply the relation: $E_i = \frac{1}{2} m V_i^2$ to the electron energies.

White Dwarf

The temperature of a white dwarf is (ignoring its thin outer shell) fairly uniform at about $10^7 \,\mathrm{K}$. The Fermi energy for degenerate electrons in white dwarfs is $\sim 3 \times 10^5 \,\mathrm{eV}$. The Fermi-Dirac formula becomes (using MKS units): $$f(E) = (e^{(E-(3 \times 10^5 \,\mathrm{eV}))/(1.3806488 \times 10^{-23}\, \mathrm{J/K}\cdot 10^7 \,\mathrm{K})}+1)^{-1}.$$ $$f(V) = (e^{((\frac{1}{2}m V^2)-(3 \times 10^5\,\mathrm{eV} \cdot 1.6 \times 10^{-19}\,\mathrm{J/eV}))/(1.3806488 \times 10^{-23}\,\mathrm{J/K}\cdot 10^7\, \mathrm{K})}+1)^{-1}.$$

Initial calculations indicate that $F(V) =1$ for $V = 0$ to $320,000 \,\mathrm{km/s}$ and $F(V) = 0$ for $V \geq 330,000 \, \mathrm{km/s}$ using electron rest mass.

Using the relativistic electron mass instead ($m' = 1/\sqrt{(1-v^2/c^2)}$) indicates that $F(V) =1$ for $V = 0$ to $244,000\, \mathrm{km/s}$ and $F(V) = 0$ for $V = \geq 248,000\, \mathrm{km/s}$.

Presumably because the electrons are degenerate their speed does not depend on the speed of local nucleons. Is this true?

Metal Planet

The surface temperature of Mercury is between $100$ and $700\,\mathrm{K}$. The temperature of Earth's inner core is about $6000\,\mathrm{K}$.

Having assumed suitable core and outer temperatures for our imaginary metal planet, next we can sub-divide the planet into a nested sequence of spherical shells of some arbitrary thickness and apply a temperature gradient radially across the shells and calculate a speed distribution for the electrons in each shell based on shell temperature. The mass density of each shell would be required to calculate the number of electrons in each shell. And the elemental composition can be used to determine the proportion of conduction to bound electrons in each atom.

In the metal there are two sets of electrons to consider (a) conduction electrons and (b) electrons bound to the nucleus.

Can "non-thermal" motions of electrons in their orbitals be ignored?

The speeds of (a) conduction electrons can be obtained using Fermi-Dirac. The Fermi energy of $E_f \approx 10\,\mathrm{eV}$ quoted earlier applies to metals in lab conditions, so a different figure may be required according to pressure (function of depth of burial below planet surface).

For (b) bound electrons in the metal. I am unclear whether I need to give different treatments according to position of electron in the atom. I do not know how to obtain the Fermi-energies of these different electrons. Or can I just use Maxwell-Bolzman for these electrons?

Presumably I need to determine the speed distribution for the atomic nucleus at each depth and then convolve (smear) the nucleus velocity distribution onto the raw electron distribution in each shell. This presumably is required for bound electrons but not for conduction electrons. Am I right?


1 Answer 1


Something analogous to the Fermi-Dirac distribution function will probably work pretty well for the electrons in the Sun; see for example this PDF. You may need to put in some "fudge factors" for their interaction energy with the rest of the proton soup, but you're in some luck, because the interaction energy in the Sun is actually mostly lower than the thermal energy, so this is not going to be a huge effect.

In the Earth, well, you'd have to assume that there are basically two chunks, one is orbiting a non-conductive nucleus (Silicon etc.) which needs to be treated as electrons-on-a-semiconductor plus the motion of the nucleus; the other chunk is probably in a metal state, especially the metal core of the Earth. You would probably want to find the various mass-percentages of different elements in the Earth, weight these by the number of electrons in that element, and sum their individual Maxwell velocity distributions together--then you'd want to look at estimates of what the electrons in the core are doing. It may be worth your while to also figure out, for each element, the average electron velocity tends to be, and try to "add" this number assuming that the direction of an electron's orbital velocity about the nucleus is totally uncorrelated to the velocity of the nucleus itself. This might smear those curves a little higher. Then you just smooth out the curve a bit because some electrons are conduction electrons, others are in chemical bonds, etc. If you're lucky, the electrons in the core predominate and you just have two Fermi-Dirac velocity distributions.

  • $\begingroup$ Thanks for the thoughtful reply. It will take me some time to digest. Following on I read in wikipedia that Fermi-Dirac is required for a White Dwarf due to high electron concentration. So I guess that classic Maxwell-Bolzman is OK for the much less dense Sun, even in the core? $\endgroup$
    – steveOw
    Jul 9, 2015 at 14:23
  • $\begingroup$ Honestly, I'd use Fermi-Dirac whenever I can when talking about electrons that are only thermally excited. The difference in the statistics comes from the identicalness of the particles, not necessarily their density. However, it is totally 100% true that at low densities (and the high $T$ of the sun may effectively give it a "low" density), the Fermi-Dirac statistics will mirror Maxwell-Boltzmann statistics. $\endgroup$
    – CR Drost
    Jul 10, 2015 at 14:50
  • $\begingroup$ Ah, thanks. I think I will simplify my question by addressing a simpler (hypothetical) model comprising a white dwarf star and a planet consisting exclusively of a metallic core. This will make it easier for me to estimate the appropriate Fermi Energy values for input to the Fermi-Dirac equation. $\endgroup$
    – steveOw
    Jul 10, 2015 at 15:59
  • $\begingroup$ Why "for each element" add "the average electron velocity"? Should this be the "average nucleus velocity?" $\endgroup$
    – steveOw
    Jul 10, 2015 at 16:30

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