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

  1. Jul 10, 2012 #1
    Ok, some background:

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

    [tex]{\bf{E}}' = {\bf{E}} + \frac{1}{c}{\bf{v}} \wedge {\bf{B}}[/tex].​

    Whereas [itex]{\bf{B}}[/itex] 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.)
     
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  3. Jul 10, 2012 #2

    Ben Niehoff

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    The Maxwell equations are linear in curved spacetime, too.
     
  4. Jul 11, 2012 #3

    atyy

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    Last edited: Jul 11, 2012
  5. Jul 11, 2012 #4
    Last edited: Jul 11, 2012
  6. Jul 12, 2012 #5
    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.
     
  7. Jul 16, 2012 #6

    bcrowell

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    It may be helpful to look at the example here http://www.lightandmatter.com/html_books/me/ch14/ch14.html#Section14.7 [Broken] 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.
     
    Last edited by a moderator: May 6, 2017
  8. Jul 16, 2012 #7

    tom.stoer

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    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).
     
  9. Jul 16, 2012 #8

    bcrowell

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    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 [itex]k_1 \ne k_2[/itex] and equal numbers of cycles in each wave-train. In the center of mass frame, this is a head-on collision with [itex]k_1' = -k_2'[/itex] 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.
     
  10. Jul 16, 2012 #9

    Ben Niehoff

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    I reiterate my earlier point:

    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.
     
  11. Jul 16, 2012 #10

    bcrowell

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    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.

    Is this a reason for the fact, or is it just a restatement of the fact?
     
  12. Jul 16, 2012 #11

    bcrowell

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    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.
     
    Last edited: Jul 16, 2012
  13. Jul 16, 2012 #12

    Ben Niehoff

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    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.

    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.
     
  14. Jul 16, 2012 #13
    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.

    Notice the word preservation.

    Anyway that's my understanding of the OP, maybe I'm wrong.
     
  15. Jul 16, 2012 #14
    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?
     
  16. Jul 16, 2012 #15

    tom.stoer

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    U(1) is exact in QED b/c the Ward–Takahashi identities are anomaly free at all loops
     
  17. Jul 16, 2012 #16
    I see, is QED considered linear then? Even the Dirac-Maxwell couplings equations?
     
  18. Jul 16, 2012 #17

    Ben Niehoff

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    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.
     
  19. Jul 16, 2012 #18

    Ben Niehoff

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    These two facts have nothing to do with each other, trivially or otherwise.

    Maxwell's equations are

    [tex]dF = 0, \qquad d \ast F = \ast J,[/tex]
    which are linear on ANY background whatsoever.
     
  20. Jul 16, 2012 #19
    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 [itex]Q[/itex] is equal to [itex]4\pi Q\sqrt g [/itex]. Locally, I believe that this change cannot effect the linearity of the differential form of Maxwell's equations (in which you switch from [itex]{\partial ^\mu }[/itex] to [itex]{\nabla ^\mu }[/itex] 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?
     
  21. Jul 16, 2012 #20
    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 [itex]F[/itex]). 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.
     
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