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I am trying to understand the proof for Earnshaw's theorem. Though the theorem states

that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges (Wikipedia)

In the proof, Gauss's Law in free space is being used (namely that the charge density $\rho$ is zero). How is that correct if we're looking at a collection of point charges? I feel I am being wrong on a very fundamental level.

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  • $\begingroup$ The point charges are considered to be in free space. This means that there are no other charges other than those you are considering for the theorem. To what point of the wiki article are you referring to exactly? $\endgroup$
    – glS
    Commented Feb 7, 2015 at 17:15
  • $\begingroup$ But how does that mean $\rho=0$? When using Gauss's Law, e.g. for calculating the potential of a charged sphere at point x, one usually considers the charges within the area up to that point. $\endgroup$
    – ari
    Commented Feb 7, 2015 at 17:18
  • $\begingroup$ It says it satifies Laplace's equation which is derived from the Maxwell equation $div \vec{E} = \frac{\rho}{\epsilon_0} $ by stating that $\rho=0$. $\endgroup$
    – ari
    Commented Feb 7, 2015 at 17:24
  • $\begingroup$ Sorry. Here it is (last sentence before equation) en.wikipedia.org/wiki/Earnshaw%27s_theorem#Explanation $\endgroup$
    – ari
    Commented Feb 7, 2015 at 17:39

2 Answers 2

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Consider a single charge $q$ (out of a given system of charges) at a point $\textbf x$ in free space.

The charge $q$ will feel electrostatic forces due to all the other charges. To analyze how $q$ will behave, we have to consider the force that acts on it, which is $$ \textbf F = q \textbf E,$$ where $\textbf E$ is the electrostatic field generated by all the other charges in the system. To remark this point: $q$ is not considered as a source for $\textbf E$.

The divergence of $\textbf E$ at $\textbf x$ must then be zero, because there are no sources of $\textbf E$ at this point.

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  • $\begingroup$ Why are there no sources of $E$? If $E$ is induced by the other charges, why are those not considered sources? $\endgroup$
    – ari
    Commented Feb 7, 2015 at 19:03
  • $\begingroup$ @ari there are no source at the point $\textbf x$. Simply because we are considering point charges, and the charge $q$ at the point $\textbf x$ is not considered as source for $\textbf E$. $\endgroup$
    – glS
    Commented Feb 8, 2015 at 0:39
  • $\begingroup$ I think I am starting to see your point. But why can that equation ever not be zero? I am thinking about a charged sphere again. There are no sources at a point $r>R$ ($R$ - Radius of sphere) and still one uses the mentioned Maxwell equation with $\rho \neq 0$. Or is that because by using Gauss we are now looking at a Volume? If so, why don't we in the Earnshaw problem as well? $\endgroup$
    – ari
    Commented Feb 8, 2015 at 10:03
  • $\begingroup$ Is it because we are only looking at the divergence at exactly that point? $\endgroup$
    – ari
    Commented Feb 8, 2015 at 10:34
  • $\begingroup$ 1 comment up: it is because you are looking at the divergence of the field where the charge is (in my notation this is $\textbf x$). 2 comments up: It depends on what you are calculating. If you want the electric field generated by a charged sphere, then you have to consider the charge density $\rho$ of that sphere as a source. If you instead want to know how a charge behaves in the electric field generated by the other charges, only these other charges must be considered as a source. Otherwise you would be analyzing the behaviour of a charge due to its own field. $\endgroup$
    – glS
    Commented Feb 8, 2015 at 13:57
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Assume that Earnshaw's theorem is not correct and there exists a static system of $N$ entities subject to an inverse square law of force, such as

$$F=G \ m_1 m_2/d_{12}^2$$

Then since The theorem is not correct, there must be a point in the force field of these ($N$) entities where the net force vector of all of the $N$ entities is zero; at which point one may place an entity $N+1$ having any value of the force sensitive parameter $m(n+1)$, which creates a static system of $n+1$ entities.

But since this is a static system, at any other element ($i$) we must also have zero net vector field, so any value of $m(i)$ is stable at that point.

That includes the case of $m(i) = 0$, so any element ( $i$ ) may be removed, without upsetting the equilibrium. So the system of $N-1$ entities is also stable and has a point where another entity may be placed.

This argument simply repeats, and the minimum value of ($n$) must be zero. This is absurd, so clearly out original conjecture that Earnshaw's theorem is false, cannot be true.

The theorem comes from a general theory of orbital mechanics, where objects might be subject to any arbitrary force law, such as a $d_{12}^{-5}$ force law.

In the real classical universe outside of the atomic nucleus, there are only two known forces, and both are inverse square laws, so $m$ and $q$ are the force sensitive parameters. Inside the atomic nucleus, the strong force is not inverse square, so Earnshaw's theorem would not apply.

Of course Heisenberg would invalidate the theorem, so it is a classical result only. Magnetism is a result of moving charges, not static charges, so Earnshaw's theorem doesn't apply to magnetism.

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