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Gauge transformations and the coulomb gauge

  1. Oct 9, 2012 #1
    My book has introduced the idea of gauge invariance in terms of classical electrodynamics (attached file). However, I am not sure I completely understand how it works. On the one hand they use a lot of time on specifying how you can add to the vector potential the gradient of any scalar, whilst at the same time subtracting the time derivative of this scalar from the generalized electrical potential. On the other hand when it comes to the use of this property, as in the section "The coloumb Gauge" they just PICK the divergence of A to be zero. I know that is always possible but shouldn't V be adjusted at the same time? I think I'm getting something wrong. I don't think I understand what the original potentials always are... Because gauge transformations speaks of adjusting these, but can you really say that there are original potentials?
     

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  3. Oct 9, 2012 #2

    vanhees71

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    The point of gauge invariance in classical electromagnetics is that you can choose the scalar and vector potential (more appropriate is to say you have one four-vector potential in the sense of special relativity, but unfortunately most textbooks first start with the old-fashioned 1+3-dimensional formalism, so I also use this formalism) with some arbitrariness. Physically meaningful, i.e., measurable quantities are the electric and magnetic components of the electromagnetic field, which are related to the potentials (using Heaviside Lorentz units) via
    [tex]\vec{E}=-\frac{1}{c} \frac{\partial \vec{A}}{\partial t}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.[/tex]
    This ansatz solves the two homogeneous Maxwell equations, and vice versa these Maxwell equations also guarantee the existence of such potentials:
    [tex]\vec{\nabla} \cdot \vec{B}=0, \quad \frac{1}{c} \frac{\partial \vec{B}}{\partial t}+\vec{\nabla} \times \vec{E}=0.[/tex]
    Any other set of potentials, given by
    [tex]\Phi'=\Phi+\frac{1}{c} \partial \chi, \quad \vec{A}'=\vec{a}-\vec{\nabla} \chi[/tex]
    obviously gives the same em. field [itex](\vec{E},\vec{B})[/itex], no matter which scalar field [itex]\chi[/itex] you take.

    Now, usually the task is to solve Maxwell's equations for given charge and current distributions and maybe subject to some boundary conditions. This task can be made much easier by "choosing the gauge" appropriately. This is done by imposing one constraint on the potentials. For dynamical problems most convenient is the Lorenz gauge (sometimes still called Lorentz gauge, which is a bit unjust against Lorenz, who found this gauge way earlier than Lorentz)
    [tex]\frac{1}{c} \frac{\partial \Phi}{\partial t}+\vec{\nabla} \cdot \vec{A}=0.[/tex]
    Particularly in static or stationary problems the Coulomb gauge is most convenient:
    [tex]\vec{\nabla} \cdot \vec{A}=0.[/tex]
    There is in principle no way to say the one or the other gauge is more natural than the other.

    With the introduction of the potentials you can forget about the homogeneous Maxwell equations and plug the definitions of the potentials into the inhomogeneous Maxwell equations
    [tex]\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B} - \frac{1}{c} \frac{\partial \vec{E}}{\partial t}=\vec{j}.
    [/tex]
    For an arbitrary gauge you get
    [tex]
    -\frac{1}{c} \frac{\partial \vec{\nabla} \cdot \vec{A}}{\partial t} - \Delta \Phi=\rho, \quad
    \vec{\nabla} \times (\vec{\nabla} \times \vec{A})+\frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2}+\vec{\nabla} \frac{1}{c} \frac{\partial \Phi}{\partial t}=\vec{j}.[/tex]

    Of course, the resulting equations change with the choice of gauge, depending on the chosen gauge condition, and this adapts both the vector and the scalar potential to this choice. For the Lorenz gauge the gauge condition leads to
    [tex]\left (\frac{1}{c^2} \partial_t^2-\Delta \right ) \Phi=\rho, \quad \left (\frac{1}{c^2} \partial_t^2-\Delta \right ) \vec{A}=\vec{j}.[/tex]
    You should figure out, how the equations look in the Coulomb gauge.

    It's a good exercise is also to think about the [itex]\chi[/itex] which leads from Lorenz to Coulomb gauge and vice versa. A very nice didactical paper about this subject by J.D. Jackson (the "textbook Jackson") can be found here:

    J.D. Jackson, From Lorenz to Coulomb and other explicit gauge transformations, Am. J. Phys. <b>70</b>, 917 (2002)
    http://arxiv.org/abs/physics/0204034
     
  4. Oct 9, 2012 #3
    there is an effect called aharnov-bohm effect which verifies the reality of vector potential because magnetic field itself is zero in the region where the setup is used but not A.In this situation still the gauge freedom remains i.e.

    [tex] \quad \vec{A}'=\vec{a}-\vec{\nabla} \chi[/tex]
    does not effect the situation because of the presence of gradient which does not effect anything.you can see it here
    http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect
    there are good sources on this topic like Sakurai(advanced quantum mechanics) and also an elementary discussion for the involving path integral is given in feynman lectures vol.2.
     
  5. Oct 9, 2012 #4

    vanhees71

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    The phase difference, important for the Aharonov-Bohm effect, is not gauge dependent as it must be. It can be expressed uniquely with the magnetic field. It's more a proof of nonlocal correlations in quantum theory than a proof of the physicality of the vector potential. This is also illuminated by the very nice work

    Wu, T. T., and Yang, C. N. Concept of nonintegrable phase factors and global formulation of gauge fields. Phys. Rev. D 12 (1975), 3845.
     
  6. Oct 9, 2012 #5
    There are two things going on here. Basically, this gauge transformation stuff with potentials is the special relativity equivalent of this:

    [tex]-\nabla \times (\nabla \times T) = \tau[/tex]

    For two vector fields [itex]T, \tau[/itex]. If you have already chosen a gauge, you can transform the potentials to another gauge by adding any potential [itex]T'[/itex] such that [itex]\nabla \times T' = 0[/itex]--hence, it won't alter the differential equation we have. The formulas for the separate transformations of [itex]V,A[/itex] are just breaking it down into components.

    Alternatively, you can rewrite the eqation via vector calculus identities to

    [tex]\nabla^2 T - \nabla (\nabla \cdot T) = \tau[/tex]

    You can choose a gauge by imposing some condition that simplifies the equations to your liking. For instance, we can choose [itex]\nabla \cdot T = 0[/itex], and this reduces the equations to [itex]\nabla^2 T = \tau[/itex], or

    [tex]\frac{\partial^2 T^x}{\partial x^2} = \tau^x \\
    \frac{\partial^2 T^y}{\partial y^2} = \tau^y \\
    \frac{\partial^2 T^z}{\partial z^2} = \tau^z[/tex]

    This is the 3d equivalent of the Lorenz gauge.


    In short, there are two things going on here: you can either pick a gauge and that's the end of it, or you can start with some potentials (maybe you calculated them from some charge/current distribution) and transform them if you know of a convenient transformation.
     
  7. Oct 9, 2012 #6
    But if you have already picked a gauge, say the coloumb gauge where you specify A, how do you then know that the potential in the equations is exactly the one that goes well with this A? I guess it should follow directly from how you defined gauge symmetry, but I'm still a bit confused whether it does or whether you should also define a potential that goes well with your A.
     
  8. Oct 9, 2012 #7

    vanhees71

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    The integrability conditions are of course on the currents, i.e., the four-current must be conserved. In 3D notation that means the continuity equation
    [tex]\frac{\partial{\rho}}{\partial t}+\vec{\nabla} \cdot \vec{j}=0[/tex]
    must hold true. Otherwise you violate gauge invariance of electromagnetism, and you get a contradiction with the gauge condition after solving the equations.

    Let's look at the Coulomb gauge. In this case the inhomogeneous Maxwell equations translate into (again in Heaviside-Lorentz units, see my first posting in this thread)
    [tex]\Delta \Phi=-\rho, \quad \Box \vec{A}=\frac{1}{c} \left (\vec{J}-\frac{\partial \vec{\nabla} \Phi}{\partial t} \right )=:\frac{1}{c} \vec{J}_{\perp}.[/tex]
    The first equation is solved by
    [tex]\Phi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(t,\vec{x}')}{4 \pi \left |\vec{x}-\vec{x}' \right|},[/tex]
    and for the usual retarded condition, describing the physics of radiation off a charge-current distribution, you get
    [tex]\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \; \vec{j}_{\perp}\left (t-\frac{\left |\vec{x}-\vec{x}' \right|}{c},\vec{x}' \right ) \frac{1}{4 \pi \left |\vec{x}-\vec{x}' \right|}.[/tex]
    Now, it's not too difficult to show with help of the continuity equation that indeed this solution fulfills the Coulomb-gauge condition,
    [tex]\vec{\nabla} \cdot \vec{A}=0.[/tex]
     
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