Energy of electrons
Update: Dani Fong in the comments has demonstrated that I’m wrong in the case Scott is talking about of electrons in metals. Obviously I’ve been growing bacteria for too long.
Scott Aaronson has a basic article on electromagnetism which makes one of the great basic mistakes:
if you have a bunch of electrons going through a wire, then the energy scales like the number of electrons times the speed of the electrons squared.
Unfortunately not. Consider a superconducting ring with some current flowing in it. Each charge carrier is in the same state in the wire, and produces some potential in space. We’ll take the limit of many, many charge carriers so we can regard the number 
as a continuous variable. You don’t have to, but it makes the property we want really simple to see.
Say I have charge carriers, and each of them produces a potential which, when felt by a charge carrier in that state in the ring has energy 
(where 
is the increment of one charge carrier. Then as I add a charge to a system of charges, each pre-existing charge exerts a potential, and the change in energy from the new charge id 
. If we say that the energy with zero charges is 0, then 
. The energy of a system of electrons scales as the square of the number of electrons.
It’s not that the electrons don’t have kinetic energy, but it only scales with . For any significant number of charges, it is absolutely swamped by the interaction effect.
If you try to do electromagnetism as point particles flowing through pipes, you are doomed to horrendous difficulties. The theory only becomes clear if you work in terms of waves. I didn’t begin to have a clue what was going on in E&M until after I had already fought through Jackson. Then I found Carver Meade’s Collective Electrodynamics, and everything suddenly snapped into place.
Dani Fong:
I don’t think that this is right. Most currents take place in a quasi-neutral free electron gas (e.g. a metal), provided the gas is non-relativistic. This means that the E field is not a function of current at all, it’s nearly zero.
One can account for the intrinsic energy of the current by accounting for the kinetic energy of the electrons — this scales as v squared, and also the magnetic field squared, which also scales as v squared. So Scott is asymptotically correct, which is the best that he ever hopes for.
8 May 2007, 6:38 pmmadhadron:
Electrodynamics doesn’t have just a scalar potential. If we take fixed positive charge carriers and flowing negative charge carriers which gives us net zero charge, we can still have a 4-potential given by
for a current flowing in the positive 
-direction. The other components are zero. This gives us a nonzero energy for each charge carrier.
If we were in a plasma, and had the positive charge carriers flowing in the opposite direction from the negative ones, and both having roughly comparable mass, then we would get a net zero 4-potential, and there would be no such effects, but even in plasmas, one charge carrier usually has much more mass.
What’s actually happening is that most people write down the energy as
, then tack on the rest as the energy of the electromagnetic field (
, and definitely scaling as 
) elsewhere.
8 May 2007, 7:03 pmDani Fong:
Why do you think that B^2 scales as n^2? B^2 scales as J^2, the product of n^2 and v^2. But in a free electron gas, like nearly all conductors, n doesn’t vary — new charges don’t simply pop up or disappear out of/into nowhere. This occurs in superconductors and semiconductors, true, but Scott explicitly excluded this from the discussion.
So B^2 scales as v^2 in a conductor, and not as n^2, since the number of charge carriers doesn’t vary. And even if it did vary, B^2 would scale as J^2, not n^2!
The only energy sensitive to n quadratically is the electrostatic energy, E^2. But this depends not on the number of charge carriers but rather the charge imbalance, which is entirely not mentioned. And even (especially) in the case of semi/super conductors, there is not a simple linear relationship between the number of charge carriers and the charge imbalance, as there is Debye shielding in the valence bands and conduction bands and polarization in the core electrons.
8 May 2007, 9:19 pmmadhadron:
So, we agree that
. Thus the mathematical form of the energy definitely scales as 
. It doesn’t matter if the number of charge carriers doesn’t change. In a purely mathematical sense, it does scale as 
, which is all I every hoped to imply.
If the electrostatic energy scales as
, then the magnetic energy must as well, because we can just transform into a moving coordinate frame, and the two mix, but we haven’t changed the number of particles.
If you have a wire with current going down it but no change in the number of charge carriers, then the part of the energy that depends on
is a constant, and in the nonrelativistic case we can ignore it — it’s just a gauge transformation — but you need to explicitly say so. If the wire interacts with anything, then you need to worry about it. I know that the approximation in circuits is that wires don’t do anything and everything happens in localized black boxes, but I think it really confuses the issue to start with that rather than arrive at that.
9 May 2007, 12:33 amDani Fong:
No,
a) B is a function of J in general, and v in conductors in the realm Scott was discussing. It is not a function of n. This may be confusing — the point is that B(n) would be multivalent unless there’s a specific law constraining the system, like conservation of charge carriers for example. If B is a function of variables x1 through xn, then dB should be a perfect differential. dB/dn dn is not a perfect differential. Ok?
This is why we can’t just arbitrarily say that this scales with that — the relation changes as the constraints change, and the constraints are important. This should remind you of thermo.
b) This argument does not hold, since field energy is not Lorentz invariant. Here is another example: quasineutral current along the Z axis. There is no electric field. Giving a Lorentz boost along Z now does not create an electric field, but it does modify the magnetic field.
c) One can show thermodynamically that any buildup of charge in a wire will be almost entirely eliminated due to Debye shielding. It is not merely that the charge carriers do not change in number, because even if they change in number, in most circumstances the charge buildup does not scale as n^2 but in fact scales as one over the 3/2 th’s of the charge carrier density! (Debye length cubed). This is slightly changed in the case of semiconductors with strange doping layers, but it is almost always less than n^2 due to shielding except in perfect vacuums.
It is slightly more significant near the surface, but still negligible, and in particular the potential on such a surface is nearly constant so it does not effect the physics of the wire.
d) Here’s another way to look at it. You say “It’s not that the electrons don’t have kinetic energy, but it only scales with n. For any significant number of charges, it is absolutely swamped by the interaction effect.”
I say that the physics of a steady state flow can be examined thermodynamically, ok? Now, in a metal, electrons are almost entirely in momentum eigenstates. The interaction between these electrons is almost entirely limited to Pauli exclusion, then, and since is also tiny, we have close to no interaction effect.
Want classical E&M? Ok, let’s give electrons each a position. Calculating the partition function we get an equilibrium with electrons filling up the whole allowed space in a way nearly uniform. Okay, so that gives an electron pressure -> constant energy. Now perhaps we’d like to add electrons — how do we do this? Charge is conserved — we need to create ions. If we don’t, we’re limited by the Child-Langmuir law — again, the energy isn’t scaling as n^2 but as linearly with the chemical potential of the electron supply and the electron skin depth. But if we do, we lose exactly as much electrostatic energy to shielding by ions as we gain — it is fruitless.
Alright, we’ve looked now at a bunch of systems. Absolutely none have an internal energy that has a simple relationship to n^2. We only get this in a system at constant volume and constant entropy, and if we adjust N up directly we need to balance the chemical potential somehow (charge must be conserved!), so the system is not closed.
In almost all large systems the interaction energy does -not- swamp the kinetic energy. Okay?
9 May 2007, 7:05 ammadhadron:
For (a), I think we’ve been using different definitions. Once we start using quantities which aren’t just the result of adding up every particle in sight but actually have constraints, you’re absolutely right.
For (b), alright. Any time our electrons are localized and shielded, the
dependence is moot.
And today’s lesson is: don’t try to nitpick on things you haven’t thought about in four years.
9 May 2007, 11:26 am