# Energy of charge distributions,advantage of one equation vs the other?

1. May 6, 2014

### iScience

here are the two equations i'm dealing with:

Eq. 1: $$W=\frac{1}{2}\sum_{i=1}^n q_{i}V(r_{i})$$

Eq. 2: $$W=\frac{ε_{0}}{2}\int{E^{2}dT}$$

both equations describe the energy required to assemble a distribution of charges. (should go without saying that the first one is for point charge distributions, and the second one for continuous distributions)

i'm reading griffith's intro to electrodynamics and i don't understand the following:

An excerpt from ch. 2; he states: "Eq. 1 does not take into account the work necessary to make the point charge in the first place; we started with the point charges and simply found the work required to bring them together. This is a wise policy, since Eq. 2 indicates that the energy of a point charge is in fact infinite. Eq. 2 is more complete in the sense that it tells you the total energy stored in the charge configuration, but Eq. 1 is more appropriate when you're dealing with point charges, because we prefer to leave out that portion of the total energy that is attributable to the fabrication of the point charges themselves. In practice, after all, the point charges (electrons say) are given to us ready-made; all we do is move them around. Since we did not put them together and cannot take them apart it is immaterial how much work the process would involve. Still, the infinite energy of a point charge is a recurring source of embarrassment for electromagnetic theory, afflicting the quantum version as well as the classical."

I don't understand; are we referring to the energy required to actually create the electron? the energy stored in matter? i see no 'pc' terms, or anything of the like, anywhere in equation 2. it only has the work expressed in terms of the electric field, and volume. Where in the world did matter energy come into play?

Is this the only advantage eq. 2 has over eq. 1? that eq. 2 includes the 'total energy'?

Last edited: May 7, 2014
2. May 7, 2014

### Matterwave

Notice in equation one what happens if you choose the particle itself in $V(r_i)$. The potential due to a point charge located at the origin is:

$$V(r)=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$$

If I look at what happens AT where this point charge is located, r=0, my potential term diverges. In other words, what is the potential energy of the point charge with respect to its own electric field? It's infinite. It's good to talk about the potential energy of OTHER charges in this potential, but it's no good to talk about the potential energy of the point charge itself.

3. May 8, 2014

### iScience

I see, but still, how does this make equation 2 more informative than equation 1?

4. May 8, 2014

### iScience

i'm still having trouble with identifying what/where the problem is.

is this the problem?: where, because the distances between charges in a conductor is so small, the work diverges to infinity?

(but that wouldn't make sense!)

5. May 8, 2014

### Matterwave

The work only diverges to infinity for the charge itself. It's self-interaction is infinite.

6. May 8, 2014

### vanhees71

Let's start from Eq. (2.41) in Griffiths's book:
$$V_{\text{pot, interactions}}=\frac{1}{8 \pi \epsilon_0} \sum_{i \neq j} \frac{q_i q_j}{|\vec{x}_i-\vec{x}_j|}. \qquad (1)$$
To that end let $\Phi$ be the static potential of the total electric field, which is given by
$$\Phi(\vec{x})=\sum_{j} \frac{q_j}{4 \pi \epsilon_0 |\vec{x}-\vec{x}_j|}. \qquad (2)$$
Now we have
$$\vec{E}=-\vec{\nabla} \cdot \vec{\Phi} \qquad (3)$$
and thus for the total field energy
$$\mathcal{E}_{\text{em}}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \vec{E}^2(\vec{x})=\frac{1}{2} \int_{\mathbb{R}^3} (\vec{\nabla} \cdot \vec{E}) \Phi(\vec{x}). \qquad (4)$$
In the last step I've used (3) and an integration by parts (assuming that the integrand vanishes quickly enough at infinity). Now according to Gauß's Law (one of Maxwell's Equations) we have
$$\vec{\nabla} \cdot \vec{E}=\rho/\epsilon_0. \qquad (5)$$
This finally gives from (4)
$$\mathcal{E}_{\text{em}} = \frac{1}{\epsilon_0} \int_{\mathbb{R}^3} \rho(\vec{x}) \Phi(\vec{x}). \qquad (5)$$
But now in our case of point particles we have
$$\rho(\vec{x})=\sum_{j} q_j \delta^{(3)}(\vec{x}-\vec{x}_j). \qquad (6)$$
Then you see where the trouble comes from! Plugging also in (2) into (5) according to the $\delta$ distributions describing point charges you have to evaluate the potential also at the points where the charges are sitting, which means nothing else to include the interaction of the point particles with their own electric field, and this of course diverges.

For most purposes this is not a big desaster. It depends on what you want to achieve with your description. If, e.g., you want to describe the motion of one of the charges in the field produced by all the other charges, you just need expression (1) for the interaction potential, i.e., the force on particle 1 due to the field produced by all other charges is of course
$$\vec{F}=-\vec{\nabla} \vec{U} \quad \text{with} \quad U(\vec{x})=q_1 \sum_{j=2}^{N} \frac{q_j}{4 \pi \epsilon_0 |\vec{x}-\vec{x}_j|}. \qquad (7)$$
This explicitly takes out the contribution of particle 1 to the electric potential and also the electric field, and that's the right thing to do in an approximate point of view.

Unfortunately that's not the end of our trouble! Your particle 1 will be accelerated, because the force (7) is acting on it. As is well known, according to Maxwell electromagnetics, an accelerated charge radiates off electromagnetic waves that carry energy and momentum, and due to the conservation of total energy and momentum this must somehow react back on the particle. This radiation back reaction can be interpreted indeed as an interaction of the charged particle with it's own electromagnetic field (indeed, a moving charge has both an electric and a magnetic field around it). This interaction of pointlike particles with their own radiation field has been a problem since the early days of classical electron theory. The most famous players where Lorentz and Abraham in the early 1900s, and a lot of famous physicists tried to solve this problem.

The final solution, if you wish to call it a solution, is that a point particle in the strict mathematical sense is a singular object that doesn't really fit into classical electrodynamics, which in its very nature is a continuum theory. What can be done, however, is the description of the point particle as a finite quasi-rigid object within the full relativistic theory (which is the only consistent scheme of classical electrodynamics, which is the paradigmatic example for a local classical relativistic field theory) carrying some charge distribution making up the charge of the "particle". Of course such a system alone is not stable, because the like-sign charges contained somehow in a compact object will repell each other and the whole construction is mechanically instable. Thus, as has been found out by Poincare in 1906, one should take additional mechanical stresses into account that hold the like-sign charges together within the compact extended object. Working this idea out, leads to the conclusion that you can have stable configurations which might serve as a classical model for a charged particle. It also turns out that there is a term which exactly behaves as an additional mass of the object due to the inertia of the electromagnetic field created by this charged object. This "self-mass" can be lumped together with the "bare" mass of the particle to give the total physical mass of this object. This is a thing reminiscent to what's known as "mass renormalization" in relativistic quantum field theory (e.g., quantum electrodynamics).

Now the trouble hits back when you want to let the extension of the finite charged object go to 0 to get a real mathematical point charge. This makes the "self-mass" infinite. You can of course argue that you make the bare mass also infinite such as that the total mass stays the mass of your point particle (say an electron). Nevertheless the mechanical stability gets lost when making the extension of the charged object smaller than a certain critical value, and the equations of motion of the particle in its own em. field also become ambigous. A fully satisfactory theory of a classical point charge in classical electrodynamics is thus not yet found.

In quantum electrodynamics it's somewhat better, because at least in the sense of the usual perturbation theory, you can get a concistent theory of particles including the self-masses (here it's known by the better name self-energy) due to their em. field at any order of perturbation theory via the now well-developed renormalization theory. Quantum electrodynamics is one of the most accurate theories ever discovered, but there's still not an exact solution beyond perturbation theory, and it's not even proven whether such a solution can exist in principle nor whether QED is a consistent mathematical theory at all!

If you are interested in the very fascinating ideas about classical charged particles, have a look at

F. Rohrlich, Classical Charged Particles, World Scientific Publishers

or, a bit more convenient, to the many papers cited and commented in Griffiths's recent Resource Letter in the American Journal of Physics

Griffiths, D. J. (2011). Resource letter EM-1: Electromagnetic momentum. American Journal of Physics, 80(1), 7-18.
http://dx.doi.org/10.1119/1.3641979

7. May 10, 2014

### Jano L.

The first formula is written somewhat incorrectly. There is no one function $V$ that could be used in the Eq. 1; the potential is different function of position for each particle. Correct formula for potential energy of point charges is
$$W = \sum_i \frac{1}{2}q_i V_{-i}(\mathbf r_i) ~~~Eq. 1^*$$
where
$$V_{-i}(\mathbf x) = \sum_j{}^{'} \frac{1}{4\pi\epsilon_0} \frac{q_j}{|\mathbf x - \mathbf r_j|}$$
is Coulomb potential due to point particles other than $i$ and is different function of position for each particle $i$.

Equation Eq. 1* and Eq. 2 are different formulae for different systems; Eq. 1 gives potential energy of arrangement of point particles and cannot be applied to continuous charge distribution. Eq. 2 gives potential energy for non-singular charge distribution and cannot be applied to point charges. These are mathematically two different cases, there is no "one is more general than the other".

8. May 10, 2014

### Jano L.

Only in macroscopic theory, where the Poynting expressions for energy can be derived for finite source densities. For point particles, the derivation of the Poynting theorem is invalid and the theorem is not valid either - some terms are infinite, some are meaningless.

Alternatively to the resolution with extended particles you mention, one can stick with point particles, consider only action of one particle on another and discard self-interaction. Then accelerated charged point particle is associated with EM radiation, but not necessarily with positive EM energy, for the EM energy is not given by integral of $E^2$ but by integral of sums of products of fields $\mathbf E_a \cdot \mathbf E_b + \mathbf B_a \cdot \mathbf B_b$ where $a$ is different from $b$. Check out these papers:

J. Frenkel, Zur Elektrodynamik punktf¨ ormiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534.
http://dx.doi.org/10.1007/BF01331692

J. A. Wheeler, R. P. Feynman, Classical Electrodynamics in Terms of Direct
Interparticle Interaction, Rev. Mod. Phys., 21, 3, (1949), p. 425-433.
http://dx.doi.org/10.1103/RevModPhys.21.425

In addition to non-presence of self-interaction, these assumed half-advanced, half-retarded fields and other conditions (Feynman and Wheeler assumed the absorber condition), but this is not necessary. The theory of point particles without self-force seems to work well whether one takes retarded or other solutions to the Maxwell equations - which one is best for microscopic theory is another question.