It is said that whole of electromagnetism can be completely described by the Maxwell's equations.

The thing that intrigues me is that how does superposition principle follow? First, I take an example, where $2$ charges $q_i$ are placed at a distance $d_i$ from a point, $i=1,2$. Now to calculate the field at that at point, we simply add the electric field due to individual charges at that point if the other charge was not there (Field calculated using Gauss' law). But how do we know that now, after the charges are placed together, the electric field due to both of them there is unaffected and also, that it adds up?

Edit: For all those who are saying that since Gauss' Law is a linear equation, that is why the SP follows. Please read the question again. I am asking the reasoning behind it, why SP follows from linearity (Or how).

It would be appreciated if someone explains it in the simple example I chose above. (2 point charges)

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    $\begingroup$ Ignoring constant factors, Gauss' law is $\nabla \cdot \mathbf E = \rho$. So if both $\nabla \cdot \mathbf E_1 = \rho_1$ and $\nabla \cdot \mathbf E_2 = \rho_2$ are solutions, then by the linearity of Gauss' law $\nabla \cdot \mathbf (\mathbf E_1+\mathbf E_2) = (\rho_1+\rho_2)$ is also a solution. $\endgroup$ – Cuspy Code Jan 9 '19 at 19:59
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    $\begingroup$ @CuspyCode, in case of point charges, the above equations become a bit unclear. So lets use integral forms for the sake of simplicity. Now for both, individually, the equations hold (in absence of any other charges). But when they are placed together, how do they add up? How does the field at the point follow from Gauss' Law? $\endgroup$ – Aditya Agarwal Jan 10 '19 at 14:52
  • $\begingroup$ @AdityaAgarwal, are you really asking how to use Gauss' flux law to calculate E using a Gaussian surface? If so this is neither practical or logical. The failure to be able to do this does not negate the principle of superposition. This could be a cause of confusion. $\endgroup$ – ggcg Jan 10 '19 at 15:28
  • $\begingroup$ In the above I meant E total, not simply E. $\endgroup$ – ggcg Jan 10 '19 at 15:34
  • $\begingroup$ @AdityaAgarwal, with all due respect I refer you to my answer. You edit doesn't change anything. I am explaining how SP follows from linearity, the proof is almost trivial. What I offer is a generalization of Cuspy Code's original comment. $\endgroup$ – ggcg Jan 10 '19 at 17:24

To put it simply superposition does NOT follow from Maxwell's equations. It is a mathematical principle and it holds for any linear equation or system of equations. Maxwell's equations are just one of many that fall into this category. The general form is:

D(Field) = Source

where D() is some abstract differential linear operator. Linearity means that if F1 is a solution for S1, and F2 is a solution for S2 then F1 + F2 is a solution for S1 + S2. The proof can be found in most math texts on ordinary or partial differential equations, also linear algebra or operator theory. Once you "prove" that a set of equations is in fact linear you can immediately apply this result.

You state using Gauss' law to calculate the field. The more fundamental equation is Coulomb's law, which was discovered as an empirical law, along with its linearity, by experiments conducted by Coulomb. This is how we know that the field behaved linearly. When we learn all this physics today we sometime forget the history, how our knowledge is connected to history, and what comes first (i.e. what piece of knowledge is fundamental rather than derived).

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  • $\begingroup$ Regarding your final part, I agree that history is fundamental. But finally, when the theory is established, then we kind of forget the history. I am told that the most fundamental thing in electromagnetism is the Maxwell's equations, not the Coulomb's law. (Whole of electromagnetism can be described using these 4 equations, and nowhere, the need of superposition principle is needed.) $\endgroup$ – Aditya Agarwal Jan 10 '19 at 14:49
  • $\begingroup$ Now as far as the linearity part is concerned. Can you explain it in the context of the example I used? Like, the field of $q_1$ at point in absence of all other charges, by Gauss Law equation, is say $E_1$ and similarly for $q_2$. But when both are placed at a finite distance from each other, how do we know that the fields (the ones which were calculated in absence of each other) would be added up? $\endgroup$ – Aditya Agarwal Jan 10 '19 at 14:50
  • $\begingroup$ (1) Maxwell's equations might be proved wrong some day but whatever replaces them better give Coulomb's law, (2) "...electromagnetism can be described using these 4..." is true, but "...and nowhere, the need of superposition principle is needed..." is both wrong and misleading. Maxwell's equations are in fact linear therefore the "superposition principle" is ensured (not needed). $\endgroup$ – ggcg Jan 10 '19 at 15:19
  • $\begingroup$ @AdityaAgarwal, I am not sure how to address your second comment as I think you are not reading my answer. You seem to be using "Gauss' Law" as a motivation to find E and that's okay but again, this equation in both differential and integral form is mathematically linear hence superposition is ensured. In my answer q1 and q2 would be source1 and source2, etc. $\endgroup$ – ggcg Jan 10 '19 at 15:22
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    $\begingroup$ @AdityaAgarwal, another way to look at things, from a PDE theory perspective, is that Coulomb's law is proportional to the Green's function of the field equation, e.g. solution for a Dirac delta distribution = point source. From the general theory the result for an arbitrary distribution is the integral of Green's function over the source distribution which is once again the principle of linear superposition. $\endgroup$ – ggcg Jan 10 '19 at 15:25

Superposition principle (SP) holds for every physical phenomenon which may be modeled by linear equations.

A frequently neglected consequence is that, in the case of macroscopic electromagnetic fields, SP is not unconditionally true, even if Maxwell's equations are linear, due to the completely general (non-linear) relation between the fields $\mathbf{E}$, $\mathbf{D}$, $\mathbf{B}$, $\mathbf{H}$. It is only in the cases of vacuum and linear materials that the full set of Maxwell's equations and constitutive equations is globally linear and SP holds.

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  • $\begingroup$ So it should be mentioned that Electromagnetism is described by Maxwell's equations, and the superposition principle, not just the Maxwell's equations? That is what my question finally points out. $\endgroup$ – Aditya Agarwal Jan 10 '19 at 14:56
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    $\begingroup$ @AdityaAgarwal Not completely true. If one would like to look at the EM in the vacuum as a fundamental theory,the constitutive equations are linear and the whole set of equations (Maxwell's equations + contitutive equations) is linear. Thus, superposition pronciple holds. But if one is dealing with macroscopic fields, its important to recall that in general constitutive equations are not linear (for example it is not linear the relation between $\mathbf{B}$ and $\mathbf{H}$ in the case of a permanent magnet). $\endgroup$ – GiorgioP Jan 10 '19 at 15:14
  • $\begingroup$ In those cases, superposition principle doesn't hold even if Mxwell's equations are linear in the four fields. $\endgroup$ – GiorgioP Jan 10 '19 at 15:15
  • $\begingroup$ @GiorgioP, the non-linearity you are referring is duel, presumably to some average coupling between elements to generate a macro effect, as in non-linear optical phenomenon and second harmonic generation in non-linear crystals? Good point +1. $\endgroup$ – ggcg Jan 10 '19 at 15:32
  • $\begingroup$ @ggcg Yes, the non-linearity enters into play as soon as one has to consider macroscopic fields, which may be obtained by averaging the microscopic fields. Non-linear induction effects destroy the linearity of the equations. $\endgroup$ – GiorgioP Jan 10 '19 at 15:42

After a few comments and the last edit of the original question, I think that the original question has been better focused and it is more a question about how to prove SP for linear electrodynamics (Maxwell's equations plus linear constitutive equations) using Gauss' law in the integral form.

Answer to this question is that it cannot be done at the same level of simplicity one is able to derive Coulomb's low from Gauss' law. Indeed, in the case of Coulomb's law, one exploits heavily the symmetry of the field due to a point-like charge. The symmetry allows to obtain from a unique number (the flux through the surface of a sphere with the charge at the centre) the value of the electric field at all the points on the spherical surface.

Missing the simplification offered by symmetry, Gauss' law is not able to provide directly the electric field at one point. It can be done indirectly by using the integral form to derive the differential form and the differential form of Gauss'law plus the condition on the curl of $\mathbf{E}$ and the boundary conditions allow to get the value of electric field everywhere.

As far as the derivation of SP from Maxwell's equations in differential form, it is basically contained in the first comment by Cuspy Code.

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  • $\begingroup$ So, this sounds like a completely new question. Any maybe should be one, closing this. Linearity exists in Gauss' flux law. In fact this is what makes flux calculation possible when there is no symmetry!! But that is derived from other laws, SP is not derived from it. $\endgroup$ – ggcg Jan 10 '19 at 17:23

First, superposition principle and linearity is the same thing, exactly how @Cuspy Code wrote in a comment above.

The answer to why you could add electric fields in the first place, or more precisely, what causes the equations of motion to become linear (instead of something else random), comes from symmetry. Interactions in a theory are mediated by gauge fields. In quantum electrodynamics(QED), there is one gauge field called the photon and is denoted by $A_\mu$. The key point is that the local symmetry group in QED, which is U(1), is Abelian. You can change the order of 2 successive gauge transformations without changing anything. The resulting equations of motion are linear in $A_\mu$: there are no self-interaction terms. If the local symmetry was non-Abelian, you would get self-interactions between the gauge fields, as happens for gluons that describe the strong nuclear force in quantum chromodynamics. In general relativity (GR) too, there are graviton self-interactions. So although for many purposes you can add Newtonian potentials, we really should be taking the non-linearities into account. The non-linear structure of GR explains Mercury's perihelion precession, for instance; see my answer here.

But in all this analysis, it must be noted that we considered an idealized situation where the sources are fixed. In reality, everything is affected by everything else and all backreaction effects (which cause non-linearity) must be taken into account in order to be as precise as possible.

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  • $\begingroup$ Linearity does not come from symmetry. One can easily have non-linear equations with symmetry. How else do you explain non-abelian gauge theories or general relativity? This answer is misleading -1. $\endgroup$ – ggcg Jan 10 '19 at 15:29
  • $\begingroup$ @ggcg You misunderstood the point I was making. I did not state that GR or non-Abelian gauge theories don't have symmetries. That is (untrue and) tangential from the point I was making. It's not about the 'presence' or 'absence' of a symmetry, but what it is, exactly. That makes ALL the difference, to the physics of any theory. To be pedantic, the answer comes from the fact that the gauge group is Abelian. $\endgroup$ – Avantgarde Jan 10 '19 at 16:38
  • $\begingroup$ You statement is simply untrue. Linearity does not in any way come from an abelian gauge symmetry. There are plenty of non-gauge theories that are linear. $\endgroup$ – ggcg Jan 10 '19 at 17:05
  • $\begingroup$ Like, any and every linear equation. Linearity is an older concept than gauge theory and does not require a gauge group to exist. Conversely non-linear theories can have gauge symmetry. The two are mutually exclusive. $\endgroup$ – ggcg Jan 10 '19 at 17:12
  • $\begingroup$ @ggcg Of the 4 forces in the universe, which one is not described by a gauge theory? Exclude gravity for a moment, which has general coordinate invariance. $\endgroup$ – Avantgarde Jan 10 '19 at 17:12

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