Magnetic monopoles and the quantisation of charge

[Just some guy]

Hi all,

I'm trying to follow somebody's derivation for why the existence of magnetic monopoles would demand electric charge was quantised (from http://forum.physorg.com/quantization-of-electric-charge_385.html . However I'm having problems with one of their steps.

Halfway through it gets to the action of the electron being equal to e multiplied by the integral of the langrangian * the magnetic vector potential. But then it says that this is equal to e multiplied by the integral of the magnetic vector potential across the boundary of the circle encompassing the Dirac string. How does this follow?:grumpy:

Cheers,
Just some guy

 

[marlon]

I cannot access the link you provided. However, i suggest you consult paragraf 2.9 on page 26 in http://arxiv.org/PS_cache/hep-ph/pdf/0310/0310102.pdf

I do not get the explanation that you provide but normally, as is done in the link, one just calculates the induced difference in vector potential if one deformes the Dirac string that stems from a magnetic monopole. More specifically, one calculates the contour integral between two points A and B, along a deformed segment of the Dirac string. The clue is to show that a deformation of the Dirac string is equivalent to a gauge transformation. So the delta A in the link expresses how the vector potential changes when deforming the Dirac String. When doing so, the new vector potential becomes A + delta A. This is a gauge transformation.

The quantization comes from the fact that the electron wavefunctions has a phase difference if you compare the wavefunction of an electron above and beneath the surface subtended by the deformed Dirac String. This implies that one can detect the Dirac string whren looking at the e-wavefunction. This is impossible because the Dirac String has nothing to do with electric fields and magnetic fields. Both E and B are independent of the Dirac String.


PS : Give the pdf-file some time to load.

regards
marlon

 

[Just some guy]

Hmm, the method you've mentioned is completely different to the one I was following, and my knowledge of gague transformations is a bit limited (so why am I doing this? I wish I knew ).

Anyway, I've fixed the link and thought about it some more. I think what it's trying to say is that if one takes an infinately small disk around the Dirac string then the Lagrangian dissapears?:yuck:

I know the integral of Lds is nothing because the circumference of the disk around the dirac string is zero, so the action of an electrically charged relativistic particle (in the presence of a magnetic monopole) around the Dirac string would then be e*int(LA). But it goes on to say this is equal to e*int{circumference of the circle around the dirac string}A. This is the bit that completely evades me.

If anybody has an answer to why this is the case it would help me a lot

Cheers,
Just some guy

 

[marlon]

so the action of an electrically charged relativistic particle (in the presence of a magnetic monopole) around the Dirac string would then be e*int(LA). But it goes on to say this is equal to e*int{circumference of the circle around the dirac string}A.

arrggh, Urs Schreiber , the guy who turns something relatively simple into a world class mystery.

Anyhow,


No, it says that the action of an electrically charged relativistic particle (in the presence of a magnetic monopole) around the Dirac string would then be e*int_LA. You integrate A along the path L...But L is @(S-D).

Note, there is an error:
S = e int_L A = e int_{@(S-A)} A . The A must be D !

The clue is to see that this expression is equal to magnetic charge. The proof for that is the very foundation as to why magnetic monopoles are actually used.

regards
marlon

ps actually, a much better explanation can be found http://hcs.harvard.edu/~jus/0302/song.pdf. this is on undergraduate level. A masterpiece, really.

 

[Just some guy]

arrggh, Urs Schreiber , the guy who turns something relatively simple into a world class mystery.
Anyhow,
No, it says that the action of an electrically charged relativistic particle (in the presence of a magnetic monopole) around the Dirac string would then be e*int_LA. You integrate A along the path L...But L is @(S-D).
Note, there is an error:
S = e int_L A = e int_{@(S-A)} A . The A must be D !
The clue is to see that this expression is equal to magnetic charge. The proof for that is the very foundation as to why magnetic monopoles are actually used.
regards
marlon
ps actually, a much better explanation can be found http://hcs.harvard.edu/~jus/0302/song.pdf. this is on undergraduate level. A masterpiece, really.

Thanks, that lcears a lot up. However I'm still perplexed as to how something inside the integral becomes a boundary condition...

 

[marlon]

However I'm still perplexed as to how something inside the integral becomes a boundary condition...

That is because of the Divergence Theorem. Look i really am too lazy to write this all down but i urge you to look at the Harvard Journal of Undergraduate Sciences (the link in my second post). It is ALL in there.

regards
marlon

 

[divus]

Dirac Strings

Dear Marlon

I would just like to clarfy a point. Is the presence of the string merely a consequnce of globally defining a single potential ie a poor definition of coordinate system. Is this overcome by defining 2 potentials. And is this analogous to the case of trying to use a single global coordinate system on (for example) a sphere- where one would run into problems at the equator of poles depending on which coordinate system used.
Regards