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A What's Your Answer To Dyson's Challenge

  1. Jun 11, 2017 #1

    bhobba

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    I had been meaning to go into Feynman's derivation of Maxwell's Equations for a while now. I finally got around to it:
    http://signallake.com/innovation/DysonMaxwell041989.pdf

    He didn't make use of gauge invariance which Schwinger showed is its real basis and I know that derivation, as well as from the Klein Gordon Equation with mass zero. But Feynmans is a new one to me - interesting.

    Now for the real interesting bit. Feynman, and even those that base it on quantum commutators, are not making use of relativity, yet it immediately follows from the equations. Dyson's challenge is how can this be?

    I have wracked my brain about that and think I know the answer.

    But first what do others think - maybe I am on entirely the wrong track.

    Thanks
    Bill
     
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  3. Jun 11, 2017 #2

    vanhees71

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    That's interesting, but the first point is that of course the proof doesn't give the full closed set of Maxwell's equations + equations of motion for matter. Dyson just mentions the inhomogeneous Maxwell equations in passing. So the question is, whether a system of equations consisting only of the homogeneous Maxwell equations and the non-relativistic approximation of the equation of motion of a point particle can somehow be made Galileo covariant.

    Also Feynman's assumptions make neither use of the commutation relations from Poincare nor Galileo symmetry, but he introduces ##m \dot{\vec{x}}## instead of the (canonical) momentum as the commutation relations. So it's not so clear to me at a first glance, in which sense this is incompatible with relativistic physics of a massive particle. If you interpret ##\dot{\vec{x}}## as the derivative wrt. the proper time, it should be well compatible with relativistic physics (although leading to the well-known problems with the first-quantization formalism in relativistic QT).
     
  4. Jun 11, 2017 #3

    stevendaryl

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    Thanks! It's a great read. Presumably, though, there is nothing particularly quantum-mechanical about this, right? If you did it classically, and used Poisson brackets instead of commutators, it would have worked out the same (or maybe not--the quantum commutator is bilinear in the two arguments, but that's not true in general for Poisson brackets).

    Of course, the assumption that [itex]m \{\dot{x_j}, x_k\} = - \delta_{jk}[/itex] is only true for certain types of Lagrangians.
     
  5. Jun 11, 2017 #4

    bhobba

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    Now that's true. One must make an assumption that the p can somehow be 'attached' to a particle then set up the most reasonable full Lagrangian. The gauge approach doesn't require that - the full Lagrangian falls out immediately and is easily generalized to SU(2) and SU(3).

    My view is the definition of the 4 current used in the last two equations ie defining it as J(v) = ∂(u)F(uv) implicitly makes it covariant. All you can really do is define ∇E = p then providing you assume p obeys the equation of continuity you have ∇.(J + ∂E/∂t) =0 hence some B' exists ∇XB' = J + ∂E/∂t. Of course its only natural to associate B' with B but there is nothing forcing you to. My suspicion is its that assumption that is at the heart of it being relativistic.

    Thanks
    Bill
     
  6. Jun 11, 2017 #5

    bhobba

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    Two approaches were given - one involving QM and Feynmans that didn't.

    I was in two minds because of that where to post it but chose here just because I mostly post here.

    Thanks
    Bill
     
  7. Jun 11, 2017 #6
    That paper assumes units of c=1. Thus the velocity of light is a constant independent of any space-time co-ordinates. If you translate from one space-time point to another, this is equivalent to the reverse translation on the frame of reference. Even two frames related by a boost are also related by translation at any given time if one accepts Einstein's linearity assumption. Thus c must be the same in all frames of reference related by translation or boost or any other linear transformation.

    In fact as soon as you assume anything has a constant velocity independent of its frame of reference (and therefore of its dynamics too), then you end up with SR for precisely this reason. (A constant velocity implies time is a spatial measurement of whatever has that velocity and so time must transform like a spatial co-ordinate. That light has that property is because photons are massless.)

    In the case of Dyson's telling of Feynman's argument (assuming Feynman also assumed a space-time independent c) then SR is implicit from the start.
     
  8. Jun 12, 2017 #7
    This looks like an important insight. Could how and where exactly this assumption is made in Feynman's premises be spelled out more clearly? Or you just mean the modern interpretation of the paper?
    I mean, you are not suggesting that Newtonian mechanics assumes c=1, are you?
     
    Last edited: Jun 12, 2017
  9. Jun 12, 2017 #8

    kith

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    If you scroll down the pdf-file of the OP, you will find a number of interesting comments. In the last comment, Farquhar elaborates on the significance of [itex]c[/itex].

    By looking at the Maxwell equations in SI units, we see that Feynman only derives the two Maxwell equations which don't depend on the electromagnetic constants (cgs muddles this by including [itex]c[/itex] in the definition of [itex]\vec B[/itex]). So for these two equations, there's no difference between a speed of light of [itex]c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} = \text{const}[/itex] (Lorentz transformation) and of [itex]c=\infty[/itex] (Galilean transformation as limiting case of the Lorentz transformation).

    So the derivation by Feynman is fine but incomplete. The remaining two equations are not merely defining the charge and current densities (as Dyson states) but also determine the type of transformation and the physics. For example, there's no displacement current if we insist on the Galilean transformation. (The first comment -by Dombey- cites research by Le Bellac and Levy-Leblond about how Galilean-invariant electromagnetism would look like.)
     
    Last edited: Jun 12, 2017
  10. Jun 12, 2017 #9

    vanhees71

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  11. Jun 12, 2017 #10

    bhobba

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    As others have pointed out the remaining two equations is more than a definition - it has physical significance. To be specific the definition is you, from the first two equations, define the vector potential which allows you to define the EM tensor F(uv). Then it's simple to define the 4 current as J(v) = ∂(u)F(uv) and you get the last 2 equations. So the paper is correct - its simply a matter of definition. BUT what, in the units used do these equation predict? EM radiation at the speed of 1. However what is it in another frame classically - not 1 - hence the equations are not true in another frame, classically so are not relativistic. Note the definition of J I gave is from the EM tensor - the key point is its a tensor so you have already put in relativity.

    To be honest Dyson should have picked it up - or maybe he did and simply wanted people to think about it a bit.

    Thanks
    Bill
     
  12. Jun 12, 2017 #11

    vanhees71

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    Well, the two inhomogeneous equations are the dynamical equations for the em. field. The homogeneous equations are constraints, usually eliminated by the introduction of the potentials (at the cost of introducing the gauge problem, which however turns out as a key feature in building the quantum version of electrodynamics, QED).
     
  13. Jun 12, 2017 #12

    stevendaryl

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    I don't know where to find the reference, but I read a paper once that claimed that Maxwell's equations can be made Galilean-invariant if you allow for the constants [itex]\epsilon[/itex] and [itex]\mu[/itex] be quantities that vary from frame to frame, rather than constants.
     
  14. Jun 12, 2017 #13
    Ok, I had missed the comments.
    But the way I understand what Dyson is saying he is more puzzled by the fact that there are equations of electrodynamics where Galilean and Lorentzian invariance coexist, just depending on whether one assumes c equals a limit value tending to infinity(infinity itsel is not a number) or to 1. Because in purely mathematical terms in the presence of scale invariance both assumptions seem equivalent.
     
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