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The use of gauges in physics

  1. Oct 20, 2008 #1
    I posted this here because gauges seem to play a far more significant role in modern physics than in classical (bar the obvious exceptions).

    I want to ask a general question: I've been reading about gauges and gauges theories recently, but nowhere have I actually found solid justification for/reasoning as to why these gauges were introduced.

    For example, is the use of the Lorenz gauge in (classical) electromagnetism merely a device so that the vector and scalar potentials can be expressed in the form of homogenous wave equations?

    This question basically summarises what I am wondering - why a particular gauge?

    If anyone could shed some more light on what appears in most of the literature I have consulted to have been introduced in an ad hoc manner with no justification, I would be grateful; thank you.
  2. jcsd
  3. Oct 20, 2008 #2


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    The use of Gauge in particle physics, the standard model, is that the phase of the field is not an observable (just as the phase of the wave function in nonrelativistic quantum mechanics is not an observable). Since it is not an observable, any observer may choose his/her own value on that phase - maybe to simplify calculations or whatever (compare with phase conventions made in nonrelativistic quantum mechanics).

    So now we require that any an observer can choose his/her own phase value (gauge), we postulate that the equations of motion must be invariant under a LOCAL phase transformation (Local since we require that observers can be anywhere in space-time)

    [tex] \phi (x) \rightarrow \phi (x) e^{i \Theta (x)} [/tex]

    Since I am not an expert on Lorentz Gauge, I'll leave that topic to someone else;-)
  4. Oct 20, 2008 #3


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    The answer to that is very simple: to make calculations easier. I think the really important question is: why any gauge at all? Again, one could say the answer is: to make calculations easier. Let me explain in loose terms what I mean by that. In general, one tries to write down a mathematical model for some physical system. Often, such a model contains more freedom than there actually is. For example, it is convenient to describe a photon with a vector, which has 4 (complex) components. We do this because we have a nice formalism to work with vectors and stuff like that; however, in principle two components would be enough to describe the photon completely. This means that there are certain operations we can apply which will change the vector, but when you try to calculate anything physical from it, the result will not be affected (you can compare this to a coordinate transformation: in a different basis the vector looks completely different, but it describes the same point). This is called gauge freedom, these operations are called gauge transformations and if we take away the extra degrees of freedom this is called gauge fixing. Sometimes, it may be convenient to work in a gauge where the "extra" non-physical degrees of freedom vanish, to bring out the physical content of the theory. In other cases, one may choose some other gauge which obscures it more, but eventually makes calculations easier.
  5. Oct 21, 2008 #4
    Thanks, malawi_glenn & CompuChip.

    So in the case of the Coulomb gauge, would it be correct to say that it is acceptable to set divA = 0 because whether or not we consider the components of A as being (spatially) constant, we do not lose the E generated by a time-varying A?

    And, following a similiar line of thought to this, the Lorenz gauge sets:


    So, would it be correct to say that because from the definition of E, both the spatial dependence of the vector potential and the time dependence of the scalar potential do not manifest themselves physically (i.e. as a measurable change in E), and hence it is okay to set them equal to zero?

    I appreciate your insight, the books I have been reading just introduce these gauges with absolutly no justification or even explanation of what a gauge is.
  6. Oct 21, 2008 #5


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    Sorry about the unreadable formula's, it's a server problem which is being worked on. Since I am using LaTeX quite heavily in this post, I have also put it http://mathbin.net/1644 [Broken] so you can read it properly until Greg and chroot have fixed the problem.

    Indeed, applying my post to the case of electromagnetism: our mathematical formulation consists of the potentials [itex]\phi, \vec A[/itex] because one scalar and a vector is easier to work with than 2 vectors. The physical quantities, however, are the electric field [itex]\vec E = - \nabla\phi[/itex] and magnetic field [itex]\vec B = \nabla \times \vec A[/itex]. One can easily check by elementary vector calculus, that we can take any function [itex]\xi(x, t)[/itex] and work with
    [tex]\phi' = \phi - \frac{\partial \xi}{\partial t}, \vec A' = \vec A + \nabla\xi[/tex]
    instead, then the electric and magnetic fields generated by these potentials are the same as those of the original potentials. In other words, you have a gauge freedom. You can use this to make the divergence of A vanish:
    [tex]\nabla \cdot \vec A' = \nabla \cdot \vec A + \nabla \cdot \nabla \xi = \nabla \cdot \vec A + \nabla^2\xi[/tex]
    so you can choose [itex]\xi[/itex] such that its Laplacian [itex]\nabla^2[/itex] is the divergence of the original potential [itex]\vec A[/itex] and work with [itex]A'[/itex] instead, but that means that your scalar potential will be slightly more complicated:
    [tex]\phi' = \phi - \frac{\partial \xi}{\partial t}[/tex] instead of just [tex]\phi[/tex].

    You can also use your gauge freedom to make the divergence of [itex]\vec A[/itex] not equal to zero, but to the time derivative of phi. In first instance, you might wonder why someone would like to do that. However, in the notation of special relativity, the Lorentz gauge is just
    [tex]\partial_\mu \partial^\mu A^\nu = 0[/tex]
    where [itex]A^\mu[/itex] contains both the scalar and vector potential in one, and so this is somewhat the analog of "div A = 0" when describing electromagnetism in a Lorentz-invariant ("special-relativistic", if you want) way.

    I wouldn't say
    because the spatial dependence of the vector potential does manifest itself; not in the electric field but in the magnetic field. However, it is not the "full" spatial dependence, so to speak -- it's just the part of it described by the curl of the vector potential (I suppose you could think of it this way: to measure of how much it rotates around a point we can deform the "flow" to a circle -- but I don't know if I am making sense to you now :smile:).
    Last edited by a moderator: May 3, 2017
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