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

[Snicker]

Ok, some background:

In the static case, the force on a charge is the multiplication of the charge into the electric field ${\bf{E}}$, defined by Gauss' law, the force on a moving charge with velocity ${\bf{v}}$ is given by the multiplication of the charge (which is Lorentz-invariant) into the "Lorentz transformed" electric field ${\bf{E}}'$, which is given by

$${\bf{E}}' = {\bf{E}} + \frac{1}{c}{\bf{v}} \wedge {\bf{B}}$$.​

Whereas ${\bf{B}}$ denotes the so-called "magnetic field," and that Maxwell's equations apply.
_____________________________________________________________

That is, electromagnetism may be viewed as the application of special relativity to electrostatics whilst postulating that electric charge is Lorentz-invariant. Cool, right?

Well, there's this property of Maxwell's equations that I think is too under-appreciated: It is linear in the electric and magnetic fields. The linearity is pretty neat because it implies the principle of superposition for both electric and magnetic fields, which, in turn allows for Lorentz forces to be linearly combined.

So I was thinking: Particularly which features of special relativity imply the preservation of linearity? I think it might be because space-time is an affine space, which is basically linear (an affine space is a vector space that is also invariant under the group of translations). Can a simple, pretty argument explain this connection, if any?

I mean, I'm pretty sure it works. It works for Newtonian gravity (Galilean space-time is also affine and Galilean transformations are identity mappings, which are trivially linear, when applied to a Newtonian gravitational field.)

 

[Ben Niehoff]

The Maxwell equations are linear in curved spacetime, too.

 

[atyy]

http://en.wikipedia.org/wiki/Relativistic_wave_equations

Also, a nonlinear scalar equation
http://arxiv.org/abs/gr-qc/0405030 (Eq 10)

 

[PhilDSP]

But there is a price to be payed for that linearity. Considered rigorously, Minkowski space requires that geometric rules concerning space and time be modified from their classic Euclidean basis. That means, for one thing, that the inner product operation must be redefined.

http://en.wikipedia.org/wiki/Minkowski_space#The_Minkowski_inner_product

 

[lugita15]

Particularly which features of special relativity imply the preservation of linearity?

I suppose it arises from the linearity of the Lorentz transformations of the electric field and the magnetic field, which in turn arises from the linearity of the ordinary Lorentz transformations.

 

[bcrowell]

It may be helpful to look at the example here http://www.lightandmatter.com/html_books/me/ch14/ch14.html#Section14.7 titled "Light rays don't interact." This argument rules out scattering of light by light unless you consider either polarization or quantum mechanics. I would be interested in knowing whether this argument can be strengthened by showing that something bad or implausible is required to happen in the classical case where the polarizations are affected but not the energies or the momenta.

[EDIT] "Polarization" above should really be "wave properties." See #7-8.

 

[tom.stoer]

... which features of special relativity imply the preservation of linearity?

There can't be such principle in SR b/c there are non-linear wave equations consistent with Lorentz invariance as well, e.g. Yang-Mills equations. The linearity is something that has to be specified on the level of the specific field equations, not in the context of SR, which serves as a kinematic framework only. That means that for field theories no specific dynamics follows from SR (in contrast to point particles where SR specifies the dynamics as well).

 

[bcrowell]

There can't be such principle in SR b/c there are non-linear wave equations consistent with Lorentz invariance as well, e.g. Yang-Mills equations. The linearity is something that has to be specified on the level of the specific field equations, not in the context of SR, which serves as a kinematic framework only. That means that for field theories no specific dynamics follows from SR (in contrast to point particles where SR specifies the dynamics as well).

Yes, my symmetry argument linked to from my #6 is really an argument about particles, not waves. I think my logic about polarization should really be replaced with logic about any degrees of freedom not describable in terms of point particles.

I think we can gain some more insight by looking at the phenomenology of classical Yang-Mills wave packets. This paper shows some numerical simulations: Hu et al., Wavepacket Dynamics in Yang-Mills Theory, http://arxiv.org/abs/hep-ph/9502276

I believe such packets propagate at c, so they have no rest frame, and this means that they can't disperse, since there is no natural frame in which to set a time-scale for the dispersion process. This seems to be the behavior shown in figure 1 from t=400 to t=600.

Suppose we collide two sinusoidal pulses with wave-vectors $k_1 \ne k_2$ and equal numbers of cycles in each wave-train. In the center of mass frame, this is a head-on collision with $k_1' = -k_2'$ and identical wave envelopes. What nontrivial scattering process can we then have without violating symmetry? Just a polarization swap? I think this may be right. Hu's figure 1 shows a simulation of "an elastic collision of wavepackets with no relative isospin polarization." Although the field during the collision is presumably not a linear superposition of the two envelopes, the pulses do reemerge unchanged. Figure 2 shows a case where nontrivial spin interactions are possible, and the pulses emerge distorted.

 

[Ben Niehoff]

I reiterate my earlier point:

The Maxwell equations are linear in curved spacetime, too.

Hence Lorentz symmetry has nothing to do with linearity of Maxwell's equations.

The reason for linearity is that the gauge group U(1) is abelian.

 

[bcrowell]

The Maxwell equations are linear in curved spacetime, too.

I don't think this addresses the OP's question. S/he conjectured that A implies B, where A is special relativity, and B says that wave equations are linear. Tom's #7 gives a counterexample where A holds and B fails. You're giving an example where A fails and B holds, but that doesn't disprove the conjecture that A implies B.

The reason for linearity is that the gauge group U(1) is abelian.

Is this a reason for the fact, or is it just a restatement of the fact?

 

[bcrowell]

That means that for field theories no specific dynamics follows from SR (in contrast to point particles where SR specifies the dynamics as well).

Hmm...is it really true that SR specifies the dynamics of point particles? Or do they have to be point particles without any degree of freedom besides translation?

Suppose I make a theory T1 of interacting, massless point particles in which each particle has an intrinsic degree of freedom which is that it can be either happy or sad. There are no happy-happy or sad-sad interactions. When a happy and a sad collide, the result is predicted by going into the c.m. frame and saying that the final state has a happy particle with its momentum reversed compared to the momentum of the original happy particle, and similarly for the sad particle.

Now let's also have theory T2 which is the same as T1 except that it's a trivial theory in which there are no interactions at all.

Since both T1 and T2 are consistent with SR, I don't think it's true that SR specifies the dynamics.

I think it's also necessary to assume that it's a deterministic theory, although that is usually understood when physicists talk about a classical theory. (You can certainly have nondeterministic classical theories, e.g., GR with naked singularities.) In the actual QED process of photon-photon scattering, for example, symmetry in the c.m. frame would forbid momentum transfer except that the momentum transfer is nondeterministic. One can easily construct purely classical, but nondeterministic, theories of point particles with this kind of behavior.

 

[Ben Niehoff]

I don't think this addresses the OP's question. S/he conjectured that A implies B, where A is special relativity, and B says that wave equations are linear. Tom's #7 gives a counterexample where A holds and B fails. You're giving an example where A fails and B holds, but that doesn't disprove the conjecture that A implies B.

I see now that the OP used the word "imply". When I first read it, I thought the OP had the impression that flat space was somehow special, and that the linearity of Maxwell's equations was caused by the flatness of space (causation is stronger than implication). I think this interpretation is consistent with the thread title. So I pointed out that flat space is not special.

In fact, flatness of space has nothing to do with linearity of Maxwell's equations, as it is clearly true that neither A implies B nor B implies A.

Is this a reason for the fact, or is it just a restatement of the fact?

If we're talking about gauge theories, a gauge theory is linear if and only if its gauge group is Abelian. So they are equivalent facts, whichever way you cut it. Note also that "Why?" is an ill-defined question in science.

In any case, the base manifold has nothing to do with it.

 

[Dead Boss]

I think you are misreading OP. The problem at hand is not that the electric field in electrostatic case is linear. That's the starting point. The problem is that when you apply Lorentz transformation to the field you get a mixture of E and B fields and they are still linear.

Particularly which features of special relativity imply the preservation of linearity?

Notice the word preservation.

Anyway that's my understanding of the OP, maybe I'm wrong.

 

[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.

All this suggests me a question:Since QED is non-linear at high enough energies,
is U(1) an exact gauge symmetry of QED, or only approximate due to the radiative corrections?

 

[tom.stoer]

U(1) is exact in QED b/c the Ward–Takahashi identities are anomaly free at all loops

 

[TrickyDicky]

U(1) is exact in QED b/c the Ward–Takahashi identities are anomaly free at all loops

I see, is QED considered linear then? Even the Dirac-Maxwell couplings equations?

 

[Ben Niehoff]

I see, is QED considered linear then? Even the Dirac-Maxwell couplings equations?

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.

 

[Ben Niehoff]

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.

 

[Snicker]

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?

 

[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.

 

[Snicker]

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.

 

[tom.stoer]

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's_equations_in_curved_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.

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.

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.

 

[TrickyDicky]

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.

 

[TrickyDicky]

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?

 

[tom.stoer]

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.

 

[PhilDSP]

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)

 

[TrickyDicky]

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.

 

[PhilDSP]

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.

 

[TrickyDicky]

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.

 

[tom.stoer]

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.

 

[TrickyDicky]

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.

 

[Ben Niehoff]

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.

QED is not a pure gauge theory; it is coupled to fermions. The fermions provide the nonlinearity as Tom described above.

Generally, when we solve classical E&M problems, we consider the sources J to be non-dynamical. They are fixed, or are given by a predetermined forcing function (cf. antennas). In this case, the problem we are left to solve is actually pure E&M.

If you allow the sources to be dynamical--i.e., to allow the fields to move the charges in the way that they actually would in nature--then the theory as a whole is no longer linear. At least I think so. This fact may actually depend on the sources having a nonlinear coupling to the fields, as in the case with fermions.

In any case, we should be careful what we mean by "gauge theory". I've been referring to the pure gauge sector, possibly up to including external, non-dynamical sources. QED, QCD, etc. are "gauge theories coupled to matter".