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Let's all agree that the formula $ \frac{1}{2} \iiint \rho_{macro} V_{macro} d\tau $ is a decent approximation to the work done to build this system of charges (basically a considerably large charge distribution).

Let's say I have a charge distribution, I calculate the work done to build this (approximately), by considering it to be made of macroscopically small but microscopically large blobs of charges, over which we average the true charge densities to get the macroscopic charge densities. The formula $ \frac{1}{2} \iiint \rho_{{macro}_{1}} V_{{macro}_{1}} d\tau $ would give us the work it takes to bring each one of these blobs from infinity to its prescribed point.

Now let's reconfigure this system into one which has a totally different charge distribution, shape, etc. The work it takes to freshly build this new configuration by bringing each one of these blobs from infinity to their new prescribed points would be $\frac{1}{2}\iiint \rho_{{macro}_{2}} V_{{macro}_{2}} d\tau $. Can we simply say that the work it takes to reconfigure the system would be the difference between the two integrals $\frac{1}{2}\iiint\rho_{{macro}_{2}} V_{{macro}_{2}} d\tau - \frac{1}{2} \iiint \rho_{{macro}_{1}} V_{{macro}_{1}} d\tau $?

I ask the above question because when we calculate the work it takes to go from one configuration to another, we must also consider the difference between the total sum of the self energies of these blobs but the above integrals (by definition) don't take into consideration the self energies of these blobs.

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  • $\begingroup$ The half is missing in the latter case. And no, self-energies do not matter because you have already correctly brought the charges in from infinity, and that act allows you to ignore self-energies as being givens. $\endgroup$
    – naturallyInconsistent
    Commented May 18, 2023 at 14:16
  • $\begingroup$ @naturallyInconsistent , I am fine with this when I am simply moving around these blobs into different positions and calling it a new config, without altering the inherent structure of each one of these blobs. Here, the self energies do cancel. But consider the case when I fundamentally change the microscopic charge configuration, and as a result, changing the structure of each of these blobs. In such a case, we cant really say that the self energies cancel, as they would be different. How do we handle such a situation? $\endgroup$
    – nickbros123
    Commented May 18, 2023 at 15:12
  • $\begingroup$ Please use more descriptive titles $\endgroup$
    – Brian
    Commented May 18, 2023 at 15:27

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The answer is yes. You are right that in general, the potential energy integral might contain self-energies that are not zero even when all the "blobs" are at infinity, but when you calculate the potential energy difference between two configurations, they cancel out. Because of this, you can also write the potential energy as $$W = \frac{1}{2}\iiint\epsilon_0 E^2 d\tau.$$

With this definition, the self-energies will be different in general, but the work you must do to go from one configuration to another can still be calculated as the potential energy difference.

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  • $\begingroup$ In the case when we are not changing the structure of these macroscopically small blobs and just moving them around, I am fine with the result. But consider the case when I fundamentally change the microscopic structure of the configuration, we are also changing the inherent structure of the blobs, and we can no longer say that the self energies cancel since the self energies of the blobs in each case would now be different. Do we have a method to account for this, or do we not consider it to be a physical problem? $\endgroup$
    – nickbros123
    Commented May 18, 2023 at 15:08
  • $\begingroup$ @nickbros123 If in changing the structures the only work you are doing is against electrostatic forces, the work you must do in this process is already accounted for in the potential energy integral so you should still be able to calculate the work that must be done to go from one configuration to the other in the same way. If you are doing work against another type of conservative force (e.g. a spring), the appropriate potential energy term should also be included. If you are doing work against non-conservative forces, in general I don't think you can meaningfully speak of potential energies. $\endgroup$
    – Puk
    Commented May 18, 2023 at 15:17
  • $\begingroup$ But the $\rho$ and $V$ are averaged out quantities, so when we write the macroscopic integrals, we don't account for the self energies of these blobs, over which we average the field. So we cant really say the integral takes care of the self energy term. In the case when we weren't changing the structure, the self energy terms existed, and the integral didn't account for it, but still it simply got cancelled and we could simply subtract the two integrals. here they wont cancel, so subtracting the integrals wouldn't give us the total change in energy. $\endgroup$
    – nickbros123
    Commented May 18, 2023 at 15:22
  • $\begingroup$ @nickbros123 Perhaps it would be easier to discuss this if you had an example in mind for such a change in structure. $\endgroup$
    – Puk
    Commented May 18, 2023 at 15:24
  • $\begingroup$ consider the case when I have a bunch of solid balls that make up my system that somehow interact. If I move these around without changing their internal structures, ie, they have the same internal self energies, I can say that difference of these integrals would give me the work done in the process of moving these balls. The integrals, inherently being done on macroscopic quantities, only account for the interaction energy, not the self energy. $\endgroup$
    – nickbros123
    Commented May 18, 2023 at 15:29

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