# Why does Gauss's Law for Magnetism prohibit monopoles?

1. Nov 15, 2014

### QuantumCurt

I'm currently in an introductory electricity and magnetism course, and I've been pondering magnetic monopoles. We didn't go in depth with them at all, but the professor made a comment when discussing Gauss's Law for Magnetism, $\oint \vec B \cdot d \vec A=0$ (can't figure out how to type a closed surface integral, as it should be), that this is a mathematically equivalent way of saying that there are no magnetic monopoles.

Why exactly is this? As Gauss's Law states, the sum of the flux across a closed surface must equal zero. Is it as simple as the idea that the net flux of a magnetic monopole would not equal zero? I'm a bit confused by this.

Flux is essentially the 'number' of field lines passing through a given surface. In a normal dipole, magnetic field lines essentially take the form of loops passing through the point from which they are emanating. With a magnetic monopole, magnetic field lines are radial and emanating straight out in every direction, correct? Using that fact, we can say that net flux is -not- equal to zero because field lines are not passing back through the surface?

I'm still a bit confused on this idea, and I haven't been able to find much in the searching that I've done. Could anyone provide any insight?

2. Nov 15, 2014

### ZetaOfThree

Yep, it's about as simple as that.

3. Nov 15, 2014

### QuantumCurt

Thanks for the response. I always seem to think these things are more complicated than I understand them to be...lol

Now, that being the case, I still have some questions. Magnetic monopoles are theorized to exist, and one of the necessary conditions for them to exist is the quantization of all charges in the universe. Given that all charges are quantized, it's said that they are predicted to exist. Why exactly is charge quantization a necessary condition for their existence? What other conditions are necessary for them to exist? I've sifted through a great deal of articles on this topic, and I can't really find any kind of explanation.

4. Nov 15, 2014

### ZetaOfThree

If you calculate the motion of an electric charge $q$ in a field produced by a magnetic monopole, an angular momentum term arises that is proportional to $q$. Since quantum mechanics says that angular momentum is quantized, this means that $q$ must also be quantized. So if only one monopole exists in the whole universe, then all $q$'s in the universe must be quantized. That's how magnetic monopoles lead to quantization of charge, in a nutshell.

5. Nov 19, 2014

### Matterwave

I think you have the converse of the argument given by Dirac. Dirac's argument says IF there is at least 1 magnetic monopole, then charges must be quantized. This gives a good motivation for quantization of charge. However, I don't believe the argument works the other way that IF charges are quantized, then there must be a magnetic monopole. You can quantize charges simply by fiat.

6. Nov 19, 2014

### ZetaOfThree

I think QuantumCurt got it right. He/she said quantization of charge is necessary for magnetic monopoles, not that it is sufficient.

7. Nov 19, 2014

### Matterwave

I read the statement "given that all charges are quantized, it is said that [magnetic monopoles] are predicted to exist" as the converse of Dirac's argument. I guess I read "are predicted to exist" as "they must exist theoretically".

8. Nov 20, 2014

### ChrisVer

Isn't it easier to understand that equation by just comparing it to the electric field?
In any case it means that there are no monopole sources of the magnetic field, but the sources appear at higher order in the n-pole expansion (for n>1)...

9. Nov 20, 2014

### Staff: Mentor

Professor Susskind had a better way to put it.

$\oint \vec B \cdot d \vec A=R$

Where R is the magnetic charge density in the volume. If there are no monopoles, then R=0.

Sou you could say that Gauss' Law doesn't require that there be no monopoles, it assumes that there aren't any.

Actually, Susskind's way of writing it demonstrates the symmetry between E and B in Maxwell's equations. It is lack of monopoles, not the equations that break that symmetry. I think it would be cool if we always wrote Maxwell,s equations that way, with a note thst R=0 in all case we know so far.

10. Nov 23, 2014

### vanhees71

The existence of a magnetic monopole is a possible reason for charges to be discrete but it's not necessary. In the Standard Model there are no magnetic monopoles (at least no elementary ones) but the electric charges all come in integer multiples of $e/3$, where $e$ is the electric charge of a proton. However, this is just by putting the empirical charge values by hand. The only restriction to the possible charge pattern of the quarks and leptons in the Standard Model is that the fundamental gauge $\mathrm{SU}(2)_{\text{flavor}} \times \mathrm{U}(1)_{\text{hypercharge}}$ local symmetry is free of anomalies. The empirical values, where each family of quarks and leptons has the charges $2/3 e$, $-e/3$, $-e$ and $0$. Together with the three colors for each quark this makes the electroweak sector of the standard model free of anomalies (separately for each family). As far as I know, 't Hooft once thought about the possibility that the observed charge pattern is determined by this necessity of freedom from anomalies, but that's not the case. He found other possible charge patterns. In this sense the charge pattern of the so far observed elementary particles (quarks and leptons) is empirical input and not yet understood from a more fundamental symmetry principle.

11. Nov 28, 2014