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## Linearity of Maxwell's equations as a result of special relativity.

 Quote by TrickyDicky The fact that in curved spacetime Maxwell eq. are also linear trivially reflects IMO the fact that the spatial part of curved spacetimes can be flat.
These two facts have nothing to do with each other, trivially or otherwise.

Maxwell's equations are

$$dF = 0, \qquad d \ast F = \ast J,$$
which are linear on ANY background whatsoever.

 Blog Entries: 1 This thread has gone a bit beyond what I was expecting, and far beyond my knowledge of physics (I'm an incoming college freshman, so "Yang-Mills field theory" is a bit out of my grasp). I was hoping to create a simpler discussion based on the question "Does the linear way photons propogate in space-time have anything to do with how electric and magnetic fields add up?" It may be my fault for being unclear. Still, here's my go at a point that has been raised: Maxwell's equations are linear in curved spacetime. I believe that the generalization pf the electric Gauss' law for general relativity would be this: The electric flux through the surface of an infinitesimal enclosure around the charge $Q$ is equal to $4\pi Q\sqrt g$. Locally, I believe that this change cannot effect the linearity of the differential form of Maxwell's equations (in which you switch from ${\partial ^\mu }$ to ${\nabla ^\mu }$ in the covariant formulation; both operates being linear). However, personally, I feel that the covariant formulation of Maxwell's equation obfuscates some otherwise simple concepts so that the linearity --> superposition implication becomes less clear. So I suppose I ought to ask simply: Do electric fields and magnetic fields superimpose themselves linearly in curved spacetime? Or, given a system of moving point-charges in curved spacetime, is it possible to determine each contribution of electric and magnetic field at some arbitrary point?
 Are you familiar with the Faraday field? I really wouldn't treat the electric and magnetic fields as separate. They're part of one object (Ben just alluded to it; it's usually denoted $F$). The Faraday field has six components, and so, just as you can find the component in a given direction for a vector field, the Faraday field always allows you to split it into electric and magnetic parts. You just have to choose a specific vector as the direction of time--time is what distinguishes electric from magnetic.

Blog Entries: 1
 Quote by Muphrid Are you familiar with the Faraday field? I really wouldn't treat the electric and magnetic fields as separate. They're part of one object (Ben just alluded to it; it's usually denoted $F$). The Faraday field has six components, and so, just as you can find the component in a given direction for a vector field, the Faraday field always allows you to split it into electric and magnetic parts. You just have to choose a specific vector as the direction of time--time is what distinguishes electric from magnetic.
I know about that, but that is not really the issue (I assume the electromagnetic tensor ${F^{\mu \nu }}$ I alluded to is the Faraday field you are talking about). That is why I only mentioned the electric field in my OP, as you can view the magnetic field as a relativistic correction of the electrostatic field (equivalently, you can view the electric field as a relativistic correction of the magnetostatic field). They are indeed one and the same.

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 Quote by Snicker I believe that the generalization pf the electric Gauss' law for general relativity would be this: The electric flux through the surface of an infinitesimal enclosure around the charge $Q$ is equal to $4\pi Q\sqrt g$.
In curved spacetime you should use the differential form; volume integrals do not always make sense

The Maxwell equations can be generalized to the Einstein-Maxwell equations:

http://en.wikipedia.org/wiki/Maxwell...rved_spacetime

Note that these are no longer linear due to the coupling of the el.-mag. field with the metric, which is a solution of the Einstein equation with the el.-mag. stress-energy tensor on the r.h.s. So formally the el.-mag. part seems to be linear (w.r.t. the el.-mag. fields) but the full equations aren't.

 Quote by Snicker So I suppose I ought to ask simply: Do electric fields and magnetic fields superimpose themselves linearly in curved spacetime?
Yes, as long as you can neglect the back-reaction of el.-mag. fields to spacetime; so if you fix a space-time background you can still use the linear equations for the el.-mag. field.

 Quote by Snicker Or, given a system of moving point-charges in curved spacetime, is it possible to determine each contribution of electric and magnetic field at some arbitrary point?
Yes, but again you have to fix spacetime and neglect back-reaction of el.-mag. fields; otherwise I don't think that using point-like charges will make sense.

 Quote by Ben Niehoff These two facts have nothing to do with each other, trivially or otherwise. Maxwell's equations are $$dF = 0, \qquad d \ast F = \ast J,$$ which are linear on ANY background whatsoever.
You are right, of course.
I was thinking only in terms of the context in which certain equations could be derived, but obviously my statement is misleading, equation's linearity or nonlinearity per se is independent of any background.

 Quote by Ben Niehoff No, QED includes non-linear effects (due to the fermion fields). However, the OP was about classical E&M, and we have been talking about classical gauge theories.
Ah, ok, so your statement about the link between linearity and abelian gauge symmetry holds only for classical theories?

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 Quote by TrickyDicky Ah, ok, so your statement about the link between linearity and abelian gauge symmetry holds only for classical theories?
Yes and no. The operator equations look identical to the classical ones, but already classically you get a non-linearity when treating the current correctly.

Usually you have $j_\mu A^\mu$ which is linear in A; the current is fixed by hand. But in QED you have a fermionic current $j_\mu = \bar{\psi}\gamma_\mu\psi$ which introduces an interaction $\bar{\psi}\gamma_\mu\psi A^\mu$. w/o this term even QED (w/o fermions!) would be linear, no scattering, no bound states, superposition, ...; but this term adds an interaction between fermions and the electromagnetic field, which e.g. causes a (tiny) photon-photon scattering, i.e. linearity is lost. Two photons will scatter via exchange of virtual electron-positron pairs, whereas in classical electrodynamics nothing like that is present. Note that this is not related to quantization but to the different structure of the current - which in principle could also be studied in classical theories.

However, the theory remains gauge invariant and the fermionic current is still conserved via (the quantum analogue of) Noether's theorem. So non-linear interaction terms are compatible with gauge symmetry.

 Quote by TrickyDicky When talking about Lorentz invariance most people don't include translations, that are included in the broader Poincare group, so certainly the equations of a theory that is Lorentz invariant in the usual sense, don't have to be linear, as Tom stoer's example of Yang-Mills theory shows. Maybe the OP is confused by the fact that Lorentz transformations are indeed linear. In any case Maxwell's equations linearity as previously commented is more related to the fact that it has an abelian gauge symmetry:U(1) in the Minkowskian formulation and also to the fact that it was originally formulated in pre-relativistic (pre-4spacetime) terms and the classical EM vector fields are based in 3D Euclidean geometry with absolute time as a parameter and this conditions the linearity of the EM field equations since the vector fields belong to the Euclidean vector space. The fact that in curved spacetime Maxwell eq. are also linear trivially reflects IMO the fact that the spatial part of curved spacetimes can be flat.
I don't see how U(1) could possibly support the Poincare group which includes the combination of rotations and boosts. Usually the Poincare group is associated with an SL algebra, isn't it? The SL group linearity is geometrically projective rather than Euclidean and that's what separates SL from GL groups, right? (From a Euclidean perspective, continuity relationships between series of projections are non-linear even if each projection is linear)

 Quote by PhilDSP I don't see how U(1) could possibly support the Poincare group which includes the combination of rotations and boosts. Usually the Poincare group is associated with an SL algebra, isn't it? The SL group linearity is geometrically projective rather than Euclidean and that's what separates SL from GL groups, right? (From a Euclidean perspective, continuity relationships between series of projections are non-linear even if each projection is linear)
I 'm not sure how all these questions are related to my post, I only mentioned the Poincare group to stress that the Lorentz group is a subgroup of Poincare's that doesn't include translations.

 Quote by TrickyDicky I 'm not sure how all these questions are related to my post, I only mentioned the Poincare group to stress that the Lorentz group is a subgroup of Poincare's that doesn't include translations.
And that's a good observation, especially when digging deeper into group representations (which I was attempting to do above). One complication with groups that can easily be overlooked is that it may not be sufficient to describe only one transformation at a time, but it may be necessary to account for all possible transformations concurrently and collectively.

 Quote by tom.stoer However, the theory remains gauge invariant and the fermionic current is still conserved via (the quantum analogue of) Noether's theorem. So non-linear interaction terms are compatible with gauge symmetry.
Yes, again, as the gauge symmetry of non-abelian Yang-Mills gauge theories show. But then in those we must include other internal symmetries to form U(1)XSU(2) and the electroweak interaction.
However what I fail to see is how QED contains that non-linear term only with the U(1) abelian gauge symmetry. This might be getting offtopic and not related to relativity, so maybe I should open a new thread.

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 Quote by TrickyDicky However what I fail to see is how QED contains that non-linear term only with the U(1) abelian gauge symmetry. This might be getting offtopic and not related to relativity, so maybe I should open a new thread.
This is a general construction principle for coupling gauge fields to spin 1/2 fields in the Lagrangian in a both Lorentz- and gauge invariant manner

$$\bar{\psi} \gamma^\mu \ldots (\partial_\mu - ig A_\mu^a T^a)\psi$$

where gamma and '...' mean a Clifford algebra, in () we have the covariant derivative (for the gauge group) and a=1..dim(G) labels the generators of the gauge group in the adjount rep. For U(1) we have a=1

This is the minimal coupling scheme and therefore gauge theory + fermions is always non-linear.

In gauge theories w/o fermions the Lagrangian reduces to

$$F_{\mu\nu}^a F_{\mu\nu}^a$$

which is non-linear when using a non-abelian gauge group b/c

$$F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a +gf^{a}_{bc}A_\mu^bA_\nu^c$$

For abelian gauge theories like U(1)*U(1)*... the structure constants f vanish identically, whereas for non-abelian groups already in the pure gauge sector the theory is necessarily non-linear.

 Thanks Tom. I guess I interpreted the phrase "a gauge theory is linear if and only if its gauge group is Abelian" in a previous post as meaning "a gauge theory is non-linear if and only if its gauge group is non-Abelian" which are not exactly equivalent, the latter is wrong while the former is right.

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