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In the question A charged sphere with pulsing radius the answer says that the charged sphere does not radiate. However, compressing a sphere of charge to a smaller radius requires work, so where does the electromagnetic energy needed to compress the charged sphere periodically go to?

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The energy required to compress the sphere of charges goes into the electric field which has an associated energy density of $\frac{1}{2}\epsilon_0 \mathbf E^2$. As the radius of the sphere decreases, the integral of this quantity increases. Per Poynting's theorem the amount of this increase is equal to the electrical work done compressing the sphere.

Note, the energy remains in the local field because the monopole field does not radiate. So the energy is completely recoverable if the sphere later expands by the opposite process, i.e. the energy of the field decreases doing work on the charges.

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  • $\begingroup$ Thanks, are there sources for potential wave solutions for $\phi$ and $\overrightarrow{A}$ for which $\overrightarrow{E}=\overrightarrow{B}=0$ globally? So that $$ 0=\overrightarrow{E}=-\nabla\phi-\partial_{t}\overrightarrow{A}$$ and $$0=\overrightarrow{B}=\nabla\times\overrightarrow{A}$$ $\endgroup$ – v217 Aug 5 at 13:44
  • $\begingroup$ Yes, you can just start with $\phi = 0$ and $A=0$ and then do any gauge transformation to get a solution for which $E=B=0$ globally. For instance, $\phi = -\sin(t-x-y-z)$ and $A=\left(\sin(t-x-y-z),\sin(t-x-y-z),\sin(t-x-y-z)\right)$ is a potential wave with no E or B fields. $\endgroup$ – Dale Aug 5 at 17:48
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If it was a real physical experiment, the repulsion of the charges on the sphere would have to be compensated by some tension of the surface, eg. a rubber based balloon. Then the pulsation of the sphere would be nothing but a harmonic oscillation around the equilibrium, where the tension is equal to the repulsion. For larger radii, the tension exceeds the repulsion and vice versa for smaller radii. Once started and damping neglected, the pulsation can go on forever.

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  • $\begingroup$ See the update to my question. Of course I am interested in the electromagnetic energy! $\endgroup$ – v217 Aug 5 at 13:34

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