When I ask this, I mean it as in when a test charge $q$ is placed in a region that contains two fixed charges $q_1$ and $q_2$, the force acting on it is the vector sum of the forces it would experience when placed alone with either of the two charges. In essence, I'm asking about the principle of superposition. Why is it that charges don't have effects on each other's fields? Is there some sort of symmetry argument that can be used to show that?

I'm also trying not to simply accept the argument that Maxwell's equations are linear, as from my understanding their linearity is more postulated than anything else.

EDIT: So many quantum explanations! Those are all fine and good, but I was hoping for something that did not have to rely on more advanced and fundamental particle physics, just a classical argument.


From a Classical perspective, this is because (as you mention) Maxwell's equations are linear. That is to say that the theory is linear (an abstract mathematical property) because it is represented by an abstract mathematical formalism which has that property.

That said, perhaps you will find this more satisfying: Photons have no charge. We know that the electromagnetic force is delivered via photons, so if the fields of two charges were to interfere with one another, then that means that photons are interacting with and disrupting the behavior of other photons. But only particles with charge (i.e. "charges") experience the electromagnetic force. Photons carry no charge; ergo, photons do not affect the behavior of other photons.

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    $\begingroup$ Just a minor comment: The OP can relate your second paragraph with what happens in GR and see that it really reflects on why physically classical EM is linear. In GR, the gravitational field (or graviton) itself has the "gravitational mass" which makes them interact with themselves gravitationally. This makes Einstein equations non-linear. Whereas, as you mention, since the electromagnetic field (or photon) has no charge, a photon doesn't self-interact electromagnetically - making the EM linear. $\endgroup$ – Dvij D.C. Oct 13 '17 at 23:43
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    $\begingroup$ Your first paragraph is backwards. Maxwell's equations are linear because E and M fields are observed to behave linearly. Nature doesn't decide what to do on the basis of a theory. The theory is constructed to describe nature. $\endgroup$ – garyp Oct 14 '17 at 4:16

Electromagnetism is a gauge theory with the gauge group U(1), which is abelian. This means there are no terms in the EM Lagrangian where the electromagnetic field interacts with itself.

$L = \frac{-1}{4}F^2 + \bar{\Psi}(i\gamma^{\mu}D_{\mu} - m)\Psi$, where D is the covariant derivative containing the gauge field $A_{\mu}$, at tree (classical) level there are no photon-photon interactions. This carries over to the classical theory.

Of course there are higher order effects with photons interacting with each other via. loops of fermions, but this is no longer classical.

Hope this is what you were looking for.

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    $\begingroup$ I don't like this answer. It looks like a super awesome explanation using really high powered math, but it just restates that EM is linear. $\endgroup$ – Javier Oct 13 '17 at 23:33
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    $\begingroup$ I don't understand the down vote. You asked for something beyond, "Maxwell's equations are linear", so I gave an answer from which that can be derived. Electromagnetism is a U(1) gauge theory, there is nothing beyond that behind experimental fact. $\endgroup$ – CStarAlgebra Oct 13 '17 at 23:35
  • $\begingroup$ @Javier U(1) is a fundamental symmetry of nature. How is it not exactly the symmetry the OP was asking about? $\endgroup$ – safesphere Oct 13 '17 at 23:44
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    $\begingroup$ @cstaralgebra Well, I've thought about it a little better and I don't think a downvote is deserved anymore (though I can't take it back unless the answer is edited). I still think the answer is needlessly fancy, though. Your comments are a much better explanation IMO. $\endgroup$ – Javier Oct 13 '17 at 23:44
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    $\begingroup$ @Javier I think that there is no way to answer OP's question in Classical Electromagnetism itself except saying that the electromagnetic field has no charge and doesn't self-interact. But this answer provides a further reason as to why the photons (electromagnetic field) don't self-interact in the classical theory. I don't think this further explanation can be provided without invoking at least somewhat fancy machinery because this further explanation lies in QFT. Anything related to QFT is bound to look somewhat fancy I guess! ;-) $\endgroup$ – Dvij D.C. Oct 13 '17 at 23:47

The Newtonian limit of general relativity is also linear, even though there is nothing inherently linear about general relativity. Why should it be?

Because most physical laws are linear when the input is close enough to zero. For example, we can model thermal expansion as linear because if we expand $V(T)$ as a Taylor series $V(T + \Delta T) = V(T) + V'(T) \Delta T + \frac{1}{2}V''(T) (\Delta T)^2 + \ldots$ and $\Delta T$ is small enough, then we can neglect all higher order terms.

We can say the same thing about classical electromagnetism. We should expect the electric field intensity at a point $E(x)$ to be expressed as some Taylor series in functional derivatives, $E(x) = E_0 + \int \rho(x') \frac{\delta E(x)}{\delta \rho(x')} \, \mathrm{d}^3x' + \frac{1}{2} \iint \rho(x') \rho(x'') \frac{\delta^2 E(x)}{\delta \rho(x') \delta \rho(x'')} \, \mathrm{d}^3 x' \, \mathrm{d}^3 x'' + \ldots$, where if $\rho$ is small enough then we can neglect higher-order terms and think of $E$ as linear in $\rho$. (Here $E_0$ comes from the boundary conditions.)

We simply happen to find ourselves at a point in time where the universe has cooled enough so that the fields around us are low enough to fall within the linear regime. A fraction of a second after the Big Bang, this was not the case and classical electrodynamics would not have been an accurate description of the electromagnetic interaction.


It is enough to show one counterexample to see that the observed and claimed linearity of electrostatics is a simplification. Your example is

a test charge q is placed in a region that contains two fixed charges q1 and q2, the force acting on it is the vector sum of the forces it would experience when placed alone with either of the two charges.

Let’s move the charge q1 behind q2, so that the three charges are aligned:

enter image description here

In my understanding the charge q1 is now perfectly shielded by the charge q2 and the test charge feel only charge q2.

To round up the answer I dare say that due to the mutual displacement of the fields from q1 and q2 the force between q2 and the test charge should be a little bit stronger than without q1, but of course not the vector sum from q1 and q2 with the test charge.

Since there is a downvote perhaps I have to add a little bit more explanation.

Having the three charges say in a triangle position one would find the force on the test charge as the sum of the two forces from q1 and q2. Otherwise about the shielding of the charge q1 in the case this charge is behind q2 can’t be any doubt.

Now, moving q1 from the triangle position into the shielded position should, thee force between the test charge and q1 has to be a steady function and to be zero in the shielded position.

It is fully ok to be downvoted but in this case I really would be asking to explain what is wrong. The clearness of the above example with the shilded force seems to be without any doubt for me.

  • $\begingroup$ "In my understanding the charge q1 is now perfectly shielded by the charge q2 and the test charge feel only charge q2." No. This is not right. q2 has a field $1/r^2$, and q1 has a field $-1/(r+a)^2$. Your assertion is that $1/r^2 - 1/(r+a)^2= 1/r^2$, which clearly can't be true. Further I imagine the downvotes come from you trying to disprove linearity, using the assumption that you can add the fields linearly. $\endgroup$ – CDCM Oct 14 '17 at 12:57
  • $\begingroup$ @CDCM Thank you for clarifications. So the sketch has to be wrong? According the sketch the field \frac{1}{{(r+a)}^{2}} is completely shilded by q2. $\endgroup$ – HolgerFiedler Oct 14 '17 at 15:53
  • $\begingroup$ the sketch is showing two different scenarios, as labelled (a) and (c), so the two sketches are separate things. $\endgroup$ – CDCM Oct 14 '17 at 16:04
  • $\begingroup$ A@CDCM The sketch I adopted not to draw my one. It shows perfectly what happens between the three charges. Furthermore from Wikipedia “the shielding effect of negatively charged electrons prevents higher orbital electrons from experiencing the full nuclear charge of the nucleus due to the repelling effect of inner-layer electrons.” So the shielding does exist or does not? $\endgroup$ – HolgerFiedler Oct 14 '17 at 16:17
  • $\begingroup$ Yes it reduces it, that is $1/r^2 -1/(r+a)^2 < 1/r^2$, but it's not a "perfect shielding". If you got really really far away, that is $a<<r$, then it would start to look like no charge. I recommend you play with the explicit equations, or ask a separate question about it. $\endgroup$ – CDCM Oct 14 '17 at 16:22

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