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A Do the equations of motion simply tell us which degrees of freedom apply?

  1. Oct 1, 2017 #1


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    A massless spin 1 particle has 2 degrees of freedom. However, we usually describe it using four-vectors, which have four components. Hence, somehow we must get rid of the superfluous degrees of freedom. This job is done by the Maxwell equations. To quote from Gilmore's "Lie Groups, Physics, and Geometry"

    “In some sense, Maxwell's equations were a historical accident. Had the discovery of quantum mechanics preceded the unification of electricity and magnetism, Maxwell's equations might not have loomed so large in the history physics. … Since the quantum description has only two independent components associated with each four momentum, there are four dimensions worth of linear combinations of the classical field components that do not describe physically allowed states, for each four momentum. Some mechanism must be derived for annihilating these superpositions. This mechanism is the set of equations discovered by Maxwell. In this sense, Maxwell's equations are an expression of our ignorance.”

    Is there an analogous interpretation for the Dirac equation and possibly other equations of motion?
  2. jcsd
  3. Oct 7, 2017 #2


    Staff: Mentor

  4. Oct 9, 2017 #3


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    Sure, the connection between degrees of freedom and conservation laws is pretty straight-forward. However, as far as I know, the equations of motions, like the Dirac equation, the Klein-Gordon equation etc. do not follow from Noether's theorem.

    However, in some sense we can understand them as "projection equations" that project out the unphysical degrees of freedom from our field description. As mentioned above, a massless spin 1 particle has only 2 physical degrees of freedom. We describe it using a spin $1$ field, which is represented by a four-vector. This four-vector contains too many degrees of freedom. The Maxwell equation tells us which of them are actually physical.

    I'm looking for a similar interpretation of the Dirac and Klein-Gordon equations.
  5. Oct 9, 2017 #4

    king vitamin

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    Yes, there are similar interpretations for all of the EoMs. For spin>0 quantum fields, all individual components satisfy the Klein-Gordon equation. But then there are more complicated equations which project out unphysical modes. There is the Dirac equation for spin-1/2, Maxwell/Proca for massless and massive spin-1, Rarita-Schwinger for spin-3/2, and Einstein/Fierz-Pauli equations for massless/massive spin-2.

    In some sense this is sort of "automatic" when you try to reconcile the "wrong sign" in the Minkowski metric with the positive-definiteness of the spectrum. Let's say you have a massive spin-1 field. You might first try the Lagrangian

    [tex]- \frac{1}{2}\left( \partial_{\mu} A_{\nu} \right)\left( \partial^{\mu} A^{\nu} \right) + \frac{m^2}{2} A_\mu A^\mu [/tex]

    Here, all four of the [itex]A^{\mu}[/itex] fields satisfy the Klein-Gordon equation, but if you compute the Hamiltonian you'll see that one of the four fields comes in with the wrong sign, and the vacuum is unstable. This describes four spin-0 fields, and one of the has energies [itex]E = -\sqrt{p^2 + m^2}[/itex] so the theory is unacceptable. If you instead consider the Proca action, its equations of motion,

    [tex] \partial_{\mu} F^{\mu \nu} + m^2 A^{\nu} = 0 [/tex]

    imply [itex]\partial_{\mu} A^{\mu} = 0[/itex], or the projection to three propagating degrees of freedom.
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