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Consider two electrical point charges $q_1$ and $q_2$ described by the total charge distribution $\rho = \rho_1 + \rho_2 = q_1 \delta(\vec{r} - \vec{r_1}) + q_2 \delta(\vec{r} - \vec{r_2})$. The total electrical potential could then be calculated by $$ \phi(\vec{r}) = \frac{1}{4 \pi \varepsilon_0} \int_V{\frac{\rho(\vec{r'})}{|\vec{r} - \vec{r'}|} d^3r'} = \frac{1}{4 \pi \varepsilon_0} \int_V{\frac{\rho_1(\vec{r'})}{|\vec{r} - \vec{r'}|} d^3r'} + \frac{1}{4 \pi \varepsilon_0} \int_V{\frac{\rho_2(\vec{r'})}{|\vec{r} - \vec{r'}|} d^3r'} = \phi_1 + \phi_2 $$ as solution of the Poisson's equation, which can be derived from Maxwell's equations.

Since $\vec{E} = - \nabla \phi$ and the gradient operator is linear we have $$\vec{E} = - \nabla \phi = - \nabla (\phi_1 + \phi_2) = - \nabla \phi_1 - \nabla \phi_2 = \vec{E_1} + \vec{E_2}$$ what the superposition principle is claiming. The superposition principle is sometimes called a postulate in electrostatics. But couldn't it be "derived" in that way?

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  • $\begingroup$ One can't derive anything in physics, one can, at best, give different but equivalent formulations that are describing the observations equally well. $\endgroup$ – CuriousOne Aug 8 '16 at 4:58
  • $\begingroup$ There are a lot of different choices of postulates you can start with. As you found, Maxwell's equations are enough to give you everything else. You can also start with a larger number of simpler postulates instead, possibly including the superposition principle. $\endgroup$ – knzhou Aug 8 '16 at 4:59
  • $\begingroup$ But I'm not aware of any modern textbook that does this. Can you give an example of your claim? $\endgroup$ – knzhou Aug 8 '16 at 5:00
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    $\begingroup$ Yes, it is an empirical postulate one starts with, that is eventually confirmed when writing down the full set of Maxwell's equations and realising they are linear come what may. $\endgroup$ – gented Aug 8 '16 at 7:43
  • $\begingroup$ @knzhou: the line of arguments above wasn't taken from a textbook, it has been rather my own idea. So if you don't assume Maxwell's equations as postulate, do you have to include the superposition principle in that set of simpler postulates? $\endgroup$ – Ben Aug 9 '16 at 4:31
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But couldn't it be "derived" in that way?

No, because your starting statement $\rho = \rho_1 + \rho_2$, assumes the superposition principle. Thus using your derivation to justify the superposition principle would be circular reasoning.

Principles, by their very definition, are not derived. Instead we empirically evaluate them with experiments. Theories are then built on these principles. There is no theoretical reason charge densities need add linearly (at least, not that I'm aware of), perhaps in a different universe they don't.

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  • $\begingroup$ Is $\rho = \rho_1 + \rho_2$ not like a definition for the "total charge density" as sum of the individual densities? Or the other way round is the superposition principle of electric fields equivalent to the fact, that the total charge density equals the sum of the single charge densities? $\endgroup$ – Ben Aug 9 '16 at 4:36
  • $\begingroup$ The fact that the charges sum in the first place assumes the super-position principle. I'm not sure which is more fundamental and thus I'm not sure which way round it's best to think about it. On the one hand, it's conventional to think of charge producing electric fields, but on the other hand we only know charge exists because we measure its electric field. If you grant, via principle, that one is true then the other follows suit closely after, as you excellent derivation demonstrates :) $\endgroup$ – Judge Aug 9 '16 at 9:49
  • $\begingroup$ How does summing two charges require the superposition principle? The latter is a property of fields propagation, the former is just summing two things to have the total. $\endgroup$ – gented Aug 9 '16 at 9:55
  • $\begingroup$ Good question. Perhaps I've misunderstood the precise definition of the superposition principle. But given that charge/charge densities add linearly, and using Maxwell's equations you can get to fields with a linear transformation. Doesn't one lead to the other? If so, then it's a question of where exactly along the process does the super-position principle come in. Finally, in a more general sense the superposition principle just states that for linear systems, the resultant quantity is the sum of the individual quantities. Surely $\rho = \rho_1 + \rho_2$ is an example of this? $\endgroup$ – Judge Aug 9 '16 at 10:11
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The Maxwell equations contain electrostatics as special case and are linear, so electrostatic's linearity in the electric field is already contained in the Maxwell equations. I see the linearity in the electric field as consequence of $\vec F = q \cdot \vec E$ (where $\vec F$ is a force, $q$ the charge and $\vec E$ the electric field) and the fact that forces (force vectors) are simply added up to give the total force.

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Maxwell equations for electrostatics are $\nabla\cdot\vec D = \rho, \nabla\times \vec E = 0$. The second one tells you, that there is a potential field $\phi$ generating electric field as $\vec E = -\nabla \phi$.

Both equations are linear, it means if there is $\rho_1, \rho_2$ satisfying $\nabla\cdot \vec D_1 = \rho_1$ and $\nabla\cdot \vec D_2 = \rho_2$, then $\rho = \rho_1 + \rho_2$ must hold $\nabla\cdot (\vec D_1 + \vec D_2) = \rho$. Equation with circulation is trivial.

The important fact here is, that theory is constructed on Maxwell equations, not on their consequence $\vec E = - \nabla \phi$. The superposition is then another consequency of linearity of differential operators in equations (divergence and circulation). Realize at this point that superposition would hold even if you didn't construct potentials.

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