Does "magnetic charge" fit into vector potential?

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  • Thread starter MichPod
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Disclaimer: I am not a physicist, just trying to learn some parts of it in my free time. And I do not mean to propose any kind of "new-theory" with my question.

I always thought that Maxwell equations in their differential form for B and E may be reformulated/updated to include a magnetic charge, even a hypothetical one. On the other hand, when a vector potential is introduced, and the Maxwell equations are reformulated in its terms, it looks like that only allows for no magnetic charge as B is defined as a rotor of the vector field and such a rotor may have only zero divergence.

Is that right? Or can a vector potential of electromagnetic field be used or updated to incorporate a magnetic charge?
 
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  • #2
Charles Link
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The pole model of magnetism does include magnetic charges, and they always come in pairs. For a uniformly magnetized cylinder of magnetization ## \vec{M} ## along its axis, there are, according to the pole method, magnetic surface charge densities of ## \sigma_m=\vec{M} \cdot \hat{n} ## on the end faces, with the dot product giving them signs of "plus" and "minus". The ## \vec{H} ## is computed from the magnetic charges/poles using the inverse square law just like ## \vec{E} ## in electrostatics, with ## \epsilon_o ## replaced by ## \mu_o##. The magnetic field ## \vec{B} ## is then computed as ## \vec{B}=\mu_o \vec{H}+\vec{M} ## , and the resulting ## \vec{B} ## does obey ## \nabla \cdot \vec{B}=0 ## everywhere. ## \\ ## An alternative calculation can be done that considers the problem to be one of magnetic surface currents and computing ## \vec{B} ## from Biot-Savart. The magnetic surface current per unit length ## \vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o} ##. The surface current method gets the exact same answer for ## \vec{B} ## everywhere as the pole method. The vector potential ## \vec{A} ## can readily be computed from these magnetic surface currents, or alternatively, from the magnetization ## \vec{M} ## itself. ## \\ ## The magnetic surface current method offers a much better explanation for the underlying physics than the pole method. The vector potential can not be computed directly from the magnetic charges, which appear to be the result of a mathematical construction and are actually "fictitious". Even though the "pole" method does get the correct answer for the magnetic field ## \vec{B} ## everywhere, the calculation, including the magnetic charges, appears to be the result of a mathematical construction, rather than the result of the actual existence of any magnetic charges.
 
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So the answer to your question is no, you cannot have magnetic charges if you are using a vector potential.
 
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  • #4
samalkhaiat
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can a vector potential of electromagnetic field be used or updated to incorporate a magnetic charge?
Yes. The quantum description of monopole is based on the existence of a vector potential. However, the potential must be singular at at-least one point on every closed surface surrounding a monopole. For example, the magnetic field of static monopole [tex]\vec{H} = \frac{g}{4 \pi} \frac{\hat{r}}{| \vec{r} |^{2}} ,[/tex] can be derived from the potential [tex]A_{x} = \frac{g y }{4 \pi r (r + z)} , \ A_{y} = \frac{ - g x }{4 \pi r (r + z)} , \ \ A_{z} = 0 ,[/tex] which is singular all along the negative z-axis. This line of singularities, first noted by Dirac, is called Dirac string. Of course there is nothing special about the negative z-axis. Indeed, we could have chosen to derive the magnetic field [itex]\vec{H}[/itex] from the potential [tex]\bar{A}_{x} = \frac{- g y }{4 \pi r (r - z)} , \ \bar{A}_{y} = \frac{ g x }{4 \pi r (r - z)} , \ \ \bar{A}_{z} = 0 ,[/tex] placing the Dirac string along the positive z-axis. We notice, however, that the difference between the potentials is just a gauge transformation: [tex]\bar{A}_{i} - A_{i} = \frac{g}{2 \pi} \partial_{i}\tan^{-1}(y/x) .[/tex] This shows the Dirac string is not a physical singularity but a “coordinate” singularity.
 
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  • #5
Charles Link
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@samalkhaiat Very interesting, but very abstract. The prediction of things like the possible existence of magnetic monopoles takes me beyond what I am able to readily infer from the equations at hand. I leave the subject to the theoreticians in that field who seem to be able to somehow make "heads" and "tails" out of what are some rather abstract computations. In any case, very interesting.
 
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samalkhaiat
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@samalkhaiat Very interesting, but very abstract. The prediction of things like the possible existence of magnetic monopoles takes me beyond what I am able to readily infer from the equations at hand.
Monopole type solution in the Standard Model is just as abstract, bizarre and important as Black Hole type solution in General Relativity. Would you suggest ignoring Black Hole solutions in GR?
 
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##Ax=\frac{gy}{4πr(r+z)}, Ay=\frac{−gx}{4πr(r+z)}, Az=0##,
Samalkhaiat wrote that this is singular all along the negative z axis. It seems to me that it is only singular where ##z=-r##. Or am I missing something?
 
  • #8
samalkhaiat
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Samalkhaiat wrote that this is singular all along the negative z axis. It seems to me that it is only singular where ##z=-r##. Or am I missing something?
Which line is the line [itex]z = -r[/itex]?
 
  • #9
Charles Link
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Monopole type solution in the Standard Model is just as abstract, bizarre and important as Black Hole type solution in General Relativity. Would you suggest ignoring Black Hole solutions in GR?
Not to be ignored, but the equations and their interpretation, both in QED and GR, is pretty much the territory of experts and some very advanced students. For some of these problems involving second quantized quantum mechanical field operators, some people seem to have a knack for seeing what are some very subtle symmetries. ## \\ ## Additional comment=I don't want to get too far off the subject of the original post, which I think has been addressed reasonably satisfactorily with the latest inputs from @samalkhaiat .
 
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It is a plane parallel to the xy plane, not on the z axis.
 
  • #11
samalkhaiat
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It is a plane parallel to the xy plane, not on the z axis.
Calculate the potential for [itex](x,y,z) = (0,0, - |a|)[/itex] for any non-zero real number [itex]a[/itex].
 
  • #12
Ibix
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It is a plane parallel to the xy plane, not on the z axis.
I think you are interpreting ##r## as a constant (##z=\mathrm{const}## is indeed a plane parallel to xy), where @samalkhaiat intends it as ##\sqrt{x^2+y^2+z^2}##. In the latter case ##z=-r## implies ##x=y=0## and ##z<0##.
 
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But, @samalkhaiat, I would like to calculate the potential. Should I use rectangular coordinates?
 
  • #15
samalkhaiat
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But, @samalkhaiat, I would like to calculate the potential. Should I use rectangular coordinates?
Do you expect to find a number? I said the potential is singular all along the negative z-axis [itex](0,0, - |a|)[/itex].
 
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No, I just want to see it go singular
 
  • #17
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Well, really, I guess I want to calculate the magnetic field
 
  • #18
samalkhaiat
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Well, really, I guess I want to calculate the magnetic field
I have given you the potential. You can calculate the field from [itex]\vec{H} = \vec{\nabla} \times \vec{A}[/itex]. And you can do it in any coordinate system.
 
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  • #19
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Well @samalkhaiat, I computed the field and I got monopole components for ##H_x## and ##H_y##, but for ##H_z## I didn't.
##H_z=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=##
##\frac{-gx}{4\pi}[\frac{-x}{r^2(r+z)}(\frac{1}{r}+\frac{1}{r+z})]+\frac{g}{4\pi r(r+z)}##
+
##\frac{gy}{4\pi}[\frac{y}{r^2(r+z)}(\frac{1}{r}+\frac{1}{r+z})]-\frac{g}{4\pi r(r+z)}##
##=\frac{g}{4\pi (r+z)}(\frac{1}{r}+\frac{1}{r+z})##
 
  • #20
Charles Link
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Well @samalkhaiat, I computed the field and I got monopole components for ##H_x## and ##H_y##, but for ##H_z## I didn't.
##H_z=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=##
##\frac{-gx}{4\pi}[\frac{-x}{r^2(r+z)}(\frac{1}{r}+\frac{1}{r+z})]+\frac{g}{4\pi r(r+z)}##
+
##\frac{gy}{4\pi}[\frac{y}{r^2(r+z)}(\frac{1}{r}+\frac{1}{r+z})]-\frac{g}{4\pi r(r+z)}##
##=\frac{g}{4\pi (r+z)}(\frac{1}{r}+\frac{1}{r+z})##
@Gene Naden When I computed ## H_z ## from @samalkhaiat 's potential, I get ## H_z=-\frac{g}{4 \pi} \frac{z}{r^3} ##. Perhaps I missed a minus sign, but otherwise I get the expected result. ## \\ ## Edit: I also computed ## H_y=-\frac{g}{4 \pi} \frac{y}{r^3} ##. It looks like the ## H ## has a minus sign in it. ## \\ ## Additional editing: The best I can tell, I think both of your ## \frac{g}{4 \pi r(r+z) } ## terms should have a minus sign. In the first one, you have a plus sign.
 
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OK, @Charles Link and @samalkhaiat, thanks for sticking with me on this. I rechecked my derivations and now they agree with you. @MichPod, I was wrong. The vector potential can in fact have magnetic monopoles.
 
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  • #22
Charles Link
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@samalkhaiat There is perhaps one item missing from this, and that is the requirements on the vector potential ## \vec{A} ##. Is the requirement here that ## \nabla^2 \vec{A}=0 ## everywhere except perhaps at ## r=0 ##? Is that how the prediction of the existence of this monopole came about? And does the vector potential ## \vec{A} ## satisfy this? ## \\ ##Edit: I just googled the subject and found this paper: http://www.th.physik.uni-bonn.de/nilles/exercises/DreesNillesSeminar06/DNSeminar06_Monopoles.pdf It looks like it is going to take a lot of work to be able to follow the details of much of this.
 
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  • #23
samalkhaiat
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@Gene Naden When I computed ## H_z ## from @samalkhaiat 's potential, I get ## H_z=-\frac{g}{4 \pi} \frac{z}{r^3} ##. Perhaps I missed a minus sign, but otherwise I get the expected result. ## \\ ## Edit: I also computed ## H_y=-\frac{g}{4 \pi} \frac{y}{r^3} ##. It looks like the ## H ## has a minus sign in it. ## \\ ## Additional editing: The best I can tell, I think both of your ## \frac{g}{4 \pi r(r+z) } ## terms should have a minus sign. In the first one, you have a plus sign.
Okay, to agree with a left-handed coordinate system, set [itex]x \to - x[/itex] and [itex]y \to - y[/itex] in the components of the vector potentials in post #4.
In spherical polar coordinate, the vector potential becomes [tex]\vec{A}(r, \theta , \phi ) = \frac{g}{4 \pi} \frac{1 - \cos \theta}{r \sin \theta} \hat{\phi} ,[/tex] where [itex]\hat{\phi} = - \hat{x} \sin \phi + \hat{y} \cos \phi[/itex] in the left-handed coordinate system. Now

[tex]\nabla \times \vec{A} = \frac{1}{r^{2}\sin \theta } \begin{vmatrix} \hat{r} & r \hat{\theta} & r \sin \theta \hat{\phi} \\

\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\

A_{r} & r A_{\theta} & r \sin \theta A_{\phi}

\end{vmatrix}[/tex] So, if you substitute [itex]A_{r} = A_{\theta} = 0[/itex] and [itex]A_{\phi} = \frac{g (1 - \cos \theta )}{4 \pi r \sin \theta}[/itex], you get [tex]\nabla \times \vec{A} = \frac{g}{4 \pi} \frac{\hat{r}}{r^{2}} .[/tex]
 
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  • #24
samalkhaiat
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@samalkhaiat There is perhaps one item missing from this, and that is the requirements on the vector potential ## \vec{A} ##. Is the requirement here that ## \nabla^2 \vec{A}=0 ## everywhere except perhaps at ## r=0 ##? Is that how the prediction of the existence of this monopole came about? And does the vector potential ## \vec{A} ## satisfy this?
No, this does not lead to monopole. There must be a singular field in the theory. In general, i.e., when the Dirac string is an arbitrary line, one can write [tex]\vec{H}_{mon} = \nabla \times \vec{A} - g \vec{h} ,[/tex] where [itex]\vec{h}[/itex] is a singular field which vanishes everywhere except on a line from the origin to infinity, and satisfies [tex]\nabla \cdot \vec{h}(\vec{r}) = - \delta^{3}( \vec{r}) .[/tex] So, if you take the curl of [itex]\vec{H}_{mon}[/itex], and imposing the gauge condition [itex]\nabla \cdot \vec{A} = 0[/itex] you get [tex]\nabla^{2} \vec{A} (\vec{r}) = -g \nabla \times \vec{h}(\vec{r}) .[/tex] This is the equation which replaces the electrostatic equation [itex]\nabla^{2}\vec{A}(\vec{r}) = 0[/itex].
 
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  • #25
Charles Link
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No, this does not lead to monopole. There must be a singular field in the theory. In general, i.e., when the Dirac string is an arbitrary line, one can write [tex]\vec{H}_{mon} = \nabla \times \vec{A} - g \vec{h} ,[/tex] where [itex]\vec{h}[/itex] is a singular field which vanishes everywhere except on a line from the origin to infinity, and satisfies [tex]\nabla \cdot \vec{h}(\vec{r}) = - \delta^{3}( \vec{r}) .[/tex] So, if you take the curl of [itex]\vec{H}_{mon}[/itex], and imposing the gauge condition [itex]\nabla \cdot \vec{A} = 0[/itex] you get [tex]\nabla^{2} \vec{A} (\vec{r}) = -g \nabla \times \vec{h}(\vec{r}) .[/tex] This is the equation which replaces the electrostatic equation [itex]\nabla^{2}\vec{A}(\vec{r}) = 0[/itex].
It's quite interesting. What these equations say is that ## \nabla^2 \vec{A} =0 ## except along the isolated line, the Dirac string, where ## h(r) ## is non-zero.
 

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