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Electromagnetism: Electric Potential Under a Gauge

  1. Jun 16, 2015 #1
    Electromagnetism: Multivalued potential voltage?

    It's been my understanding that specifying the electric and magnetic potential at all points in a system determines the electric and magnetic fields--and more, that we can fix a particular gauge, and it all still works.

    Perhaps I've been naive, but the potential seems to be implicitly single valued. In other words, under a given gauge, we can't have both 10 volts at some space-time event and 12 volts at the same event.

    The Maxwell-Faraday equation, in integral form, gives the voltage around a complete loop. In a particular gauge, if point A is at 0 volts, it might also be at 2 volts after a complete revolution---and 4 volts, 6 volts, etc. depending on winding number.

    I need a question, at this point in order to get some thoughtful feedback, but I'm not sure what it should be!

    Maybe this: Picking a particular gauge, is the voltage multi-valued depending on the physical arrangement of the system under scrutiny?
     
  2. jcsd
  3. Jun 16, 2015 #2

    Dale

    Staff: Mentor

    The electric potential is a scalar and the magnetic potential is a vector. So the potential is multi valued, specifically 4 values at each point in space and time. (Plus the gauge freedom)
     
  4. Jun 16, 2015 #3
    Yes, I know that, and thank you! Maybe now I have the right question:

    I was led to believe that, for a particular physical arrangement, given a global gauge and also a potential at any given event, the electric potential, [itex]\phi(x,y,z,t)[/itex] is uniquely specified over the entire system, and also single valued. Is this true?
     
    Last edited: Jun 16, 2015
  5. Jun 16, 2015 #4

    Dale

    Staff: Mentor

    Well, ##\phi## is always single valued, by definition.
     
  6. Jun 16, 2015 #5
    Say we have a static magnetic field in the z direction, confined about the origin, within a radius r. Define the potential, [itex]\phi[/itex] at some point on the x-axis, x>r. The potential at any other point seems to be path-dependent. To keep things uncluttered, we can constrain our paths to lie outside of the radius, r.

    The potential at any other position at the end of the path depends on how many turns the path takes about the origin, doesn't it?
     
  7. Jun 16, 2015 #6

    Dale

    Staff: Mentor

    If I am understanding your geometry then ##\phi=0## in your scenario, for the Lorentz gauge.
     
  8. Jun 16, 2015 #7
    Yes. At the starting point, call it ##(x_0, y_0)## the gauge is ##\phi=0##. But any value would do. By the way, I should have written a constantly changing magnetic field rather than a static field. Oops.

    The total magnetic flux is ##\Phi_M##. Would the potential at ##(x_0, y_0)## be ##\phi - n\frac{d\Phi_M}{dt}##, where ##n## is the number of turns around the origin, for path from ##(x_0, y_0)## to ##(x_0, y_0)##?
     
  9. Jun 17, 2015 #8

    Dale

    Staff: Mentor

    You are confusing the electric or scalar potential ##\phi## with the magnetic or vector potential ##A##. In this case ##\phi=0##, as I said, but don't forget ##E=-\nabla \phi-\partial A/\partial t##
     
  10. Jun 17, 2015 #9
    I don't think I did. How and where?
     
  11. Jun 17, 2015 #10

    Dale

    Staff: Mentor

    Why else would you think that a changing magnetic field had any relevance to the scalar potential?

    In the Lorentz gauge the scalar potential is just the DAlembertian of the charge density. In the scenario you described that is identically 0. Only the vector potential is non zero.

    Please write down maxwells equations in the potential formulation. This should help you see. Use the Lorentz gauge for simplicity.
     
    Last edited: Jun 17, 2015
  12. Jun 18, 2015 #11
    Dale. This could be interesting, rather than pedestrian. What source do you have that says (phi, A) is single valued over spacetime?

    ddA=0 and *dA=J: maxwells equ in terms of A.
     
  13. Jun 18, 2015 #12

    Dale

    Staff: Mentor

    None. The scalar potential, ##\phi## like all scalar fields, is single valued at each point in spacetime. The vector potential, ##A## like all vector fields, is triple valued at each point in spacetime.
     
    Last edited: Jun 18, 2015
  14. Jun 18, 2015 #13

    Dale

    Staff: Mentor

    I am not quite sure where you are getting this. Are you trying to express it using differential forms? Look here:
    https://en.m.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Lorenz_gauge
    Note that the scalar potential depends only on the charge density and the vector potential depends only on the current density. And on the same page here: https://en.m.wikipedia.org/wiki/Mat...lectromagnetic_field#Potential_field_approach
    You see that the B field is determined only by the vector potential, but the E field is determined by both the vector and scalar potential.
     
  15. Jun 20, 2015 #14
    we could imagine some spatial slice of space-time with the electric potential ##\phi(r,\theta) = \phi(r,\theta+2\pi)##, continuously differentiable except at the origin. If the origin contains a blackhole, or wormhole, the vector potential needn't be single valued. The potential may be multivalued sans singular points, or multivalued over non-simply connected manifolds.
     
  16. Jun 20, 2015 #15

    Dale

    Staff: Mentor

    Not only needn't the vector potential be single valued, it must be triple valued, by definition.

    I must not be understanding what you are asking. The scalar potential is a single valued function of spacetime. The vector potential is a vector valued (3D) function of spacetime. Together the potential has exactly 4 values at each event in spacetime. Those values are not unique due to gauge transformations, but not being unique doesn't change how many they are.

    I don't know what you think changing magnetic fields or wormholes have to do with the number of values. It is ALWAYS one value for the scalar potential and three for the magnetic. By definition.
     
  17. Jun 20, 2015 #16
    I don't necessarily agree, and yes I know there are four components to each 4-vector value. If you recall from studying contour integrals on the complex plan, a particular field may be conservative if and only it does not wrap around the origin. Electromagnetic fields have similar conditions, E.g.: the electromotive force developed around closed loop, in the vicinity of a tube of magnetic field, is dependent upon how many times the path wraps around the tube. Gauss law is another.

    A white hole that connects one part of spacetime with another is an example of non-simply connected manifold. Taking for example, only the electric potential and ignoring the rest, it's possible construct a multivalued electric potential, consistent with maxwell's equations for all paths except, possibly those that wrap the world line of the white hole.
     
    Last edited: Jun 20, 2015
  18. Jun 20, 2015 #17

    Dale

    Staff: Mentor

    The EMF is not the same as the potential. Having multiple loops with different EMFs in no way implies multiple values for the potentials.
     
    Last edited: Jun 20, 2015
  19. Jun 20, 2015 #18
    Note, I was making a comparison. Nevermind
     
    Last edited: Jun 20, 2015
  20. Jun 20, 2015 #19

    Dale

    Staff: Mentor

    Then I have no idea what you have been getting at. Nothing you are saying seems to make any sense.

    The potentials have four values (one from the scalar potential, ##\phi##, and three from the vector potential, ##A##), and those four values are not unique because you can choose a gauge. You don't seem to be referring to either of those facts.

    There isn't any more "multivaluedness" beyond that, and your descriptions of why you think there might be are completely confusing to me.
     
  21. Jun 20, 2015 #20
    sorry dale. If I could well-phrase the concept I would have the answer. Maybe next time.
     
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