# Magnetic Monopoles and Gravity

1. Sep 9, 2010

### Anamitra

Maxwell's equations in the covariant form are the most general relations expressing electromagnetic phenomena in the presence of gravity.
Now $$\nabla$$B=0
does not follow in a general way as a necessary condition from the covariant form of the Maxwell's equations.
We can always calculate the classical three divergence[does't matter what Maxwell's equations are saying] and if
$$\nabla$$B$$\neq$$0
B has sources which may be identified with magnetic monopoles, created by the action of gravity!

2. Sep 10, 2010

The general relativistic expression for non-existence of monopoles is exactly the same as before:

$$dF=0$$

where $$F$$ is the electromagnetic tensor (two-form). Now, write it down in terms of E and B and see what you get.

3. Sep 10, 2010

### JustinLevy

When we look at Maxwell's equations (in good old inertial frames):
$$\nabla \cdot \mathbf{E} = \rho_e / \epsilon_0$$
$$\nabla \cdot \mathbf{B} = 0$$
$$\nabla \times \mathbf{E} = -\frac{\partial}{\partial t} \mathbf{B}$$
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e + \mu_0 \epsilon_0\frac{\partial}{\partial t} \mathbf{E}$$
Some people feel it is "begging" for magnetic monopoles, to add to more terms and make it look more symmetric between E and B fields.

However, that feeling is in some respect due to the way the fields were chosen (E and B fields). As soon as you try to write electrodynamics as an action principle, such as in a Lagrangian, it becomes clear the root piece is the four-vector potential.

Written that way, it does not beg for monopoles. In fact, it is extremely difficult to add in magnetic monopoles in a local consistent way. There's a thread discussing this around here somewhere, but I suggest you play with it yourself to convince yourself of this. You'd have to add in another vector potential, and then somehow get them to couple correctly so that there is a single set of E and B fields. I've seen some papers try, but nothing that really solved it. It's kind of fun to play with.

...

anyway, getting back on track:
Yes, del.B can be != 0 in some cases if you consider del to be the usual coordinate derivatives. As arkajad points out though, with covariant derivates it still holds.

Heck, even in just a Rindler frame (accelerating observer, but still just SR -- flat spacetime), del.B != 0 if you are only considering coordinate derivatives. (Maxwell's equations have strange source terms if written that way.)

4. Sep 10, 2010

Well, to make it easier one can write the handy $$dF=0$$ in components

$$\partial_\mu F_{\nu\sigma}+\partial_\nu F_{\sigma\mu}+\partial_\sigma F_{\mu\nu}=0$$

where

$$\partial_\mu F_{\nu\sigma}$$ stands for

$$\frac{\partial F_{\nu\sigma}}{\partial x^\mu}$$

etc.

This is covariant even without covariant derivateives, due to the fact that $$F_{\mu\nu}$$ is antisyymetric.

Last edited: Sep 10, 2010
5. Sep 10, 2010

### Anamitra

Now let's see:
Maxwell's equations in Flat space-time:

$$\frac{\partial {F^{\alpha\beta}}}{\partial {x^{\alpha}}}$$=-$${J^{\beta}}$$ --------------------- (1)

$$\frac{\partial {F_{\beta\gamma}}}{\partial {x^{\alpha}}}$$+$$\frac{\partial {F_{\gamma\alpha}}}{\partial {x^{\beta}}}$$+$$\frac{\partial {F_{\alpha \beta}}}{\partial {x^{\gamma}}}$$=0
--------- (2)
Here,
$${F_{\gamma\delta}}$$=$$\eta_{\gamma\alpha}\eta_{\delta\beta}{F^{\alpha\beta}$$

$${F_{\gamma\delta}}$$=$${F^{\gamma\delta}}$$
Noting $$\eta$$ represents the flat space-time matrix.
One gets $$\nabla$$B=0 from equation (2)

Maxwell's equations in curved space-time:

$${F^{\mu\nu}}$$;$${\mu}$$=-$${J^{\nu}} ---------- (3) [tex]{{F^{\mu\nu}}{;}}_{\lambda}$$+$${{F^{\lambda\mu}}{;}}_{\nu}$$+$${{F^{\nu\lambda}}{;}}_{\mu}$$=0 ----------- (4)

By suitable manipulations[Gravitation and Cosmology,Steven Weinberg,Chapter 5,Section 2,Electrodynamics] we may write (3) and (4) as:

$$\frac{\partial\sqrt{g}{F^{\mu\nu}}}{\partial {x^{\mu}}}$$=-$${J^{\nu}}$$ ---------------- (5)

$$\frac{\partial {F_{\beta\gamma}}}{\partial {x^{\alpha}}}$$+$$\frac{\partial {F_{\gamma\alpha}}}{\partial {x^{\beta}}}$$+$$\frac{\partial {F_{\alpha\beta}}}{\partial {x^{\gamma}}}$$=0

----------------------- (6)

Here ,

$${F_{\gamma\delta}}$$=$${g_{\gamma\alpha}}{g_{\delta\beta}}{F^{\alpha\beta}$$

$${g_{\mu\nu}}$$ in the above equation are no more simple constants as we have for flat spacetime.

From equation (6) we do not have

$$\nabla$$B=0

Important to note that in each of the above cases,

$${F^{12}}{=}{B_3}$$, $${F^{23}}{=}{B_1}$$, $${F^{31}}{=}{B_2}$$
$${F^{01}}{=}{E_1}$$, $${F^{02}}{=}{E_2}$$,$${F^{03}}{=}{E_3}$$

$${F^{\alpha\beta}}$$=-$${F^{\beta\alpha}}$$

Last edited: Sep 10, 2010
6. Sep 11, 2010

$$\partial_1 F_{23}+\partial_2 F_{31}+\partial_3 F_{12}=\nabla\cdot \bf{B} =0$$

7. Sep 11, 2010

### Anamitra

$${B_{1}}{\neq}{F_{23}}{;} {B_{2}}{\neq}{F_{31}}{;} {B_{3}}{\neq}{F_{12}}{;}{\partial_1 }{F^{23}}{+}{\partial_2}{ F^{31}}{+}{\partial_3} {F^{12}}{\neq}0$$

It is important to remember that I have been trying to calculate the sources of B[$${B_{1}},{B_{2}},{B_{3}}$$] by using

$$\frac{\partial B_1}{dx^1}{+}\frac{\partial B_2}{dx^2}{+}\frac{\partial B_3}{dx^3}$$

Last edited: Sep 11, 2010
8. Sep 11, 2010

From http://arxiv.org/abs/gr-qc/9712019" [Broken] by Sean M. Carroll:

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9. Sep 11, 2010

### Anamitra

That holds in flat space time but not in curved spacetime

In flat spacetime:
$${F_{\gamma\delta}}$$=$$\eta_{\gamma\alpha}\eta_{\delta\beta}{F^{\alpha\beta}$$
This corresponds to what has been given in the thumbnail.

In curved space time:
$${F_{\gamma\delta}}$$=$${g_{\gamma\alpha}}{g_{\delta\beta}}{F^{\alpha\beta}$$

The metric coefficients can be complicated functions of the coordinates in curved spacetime.

Last edited: Sep 11, 2010
10. Sep 11, 2010

There is a more difficult way:

Landau, Lifgarbagez, "The Classical Theory of Fields", Vol 2, Fourth Revised English Edition, pp. 276-277

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11. Sep 11, 2010

### Staff: Mentor

That is only true where the metric is the Minkowski metric. In other metrics it is not even clear to me that the electromagnetic field tensor can reasonably be decomposed into electric field components and magnetic field components.

Last edited: Sep 11, 2010
12. Sep 12, 2010

### Anamitra

The approach by Landau and Lifgarbagez,as indicated by Arkajad:
$${B^{\alpha}}{=}{-}{1/2}{\frac{1}{\sqrt{\gamma}}{\epsilon}^{\alpha\beta\gamma}}{F_{\beta\gamma}}$$
$${B^{\alpha}}{=}{-}{1/2}{\frac{1}{\sqrt{\frac{-g}{g(0,0)}}}{\epsilon}^{\alpha\beta\gamma}}{F_{\beta\gamma}{\neq}{g^{\beta\mu}{g^{\gamma\nu}}{F_{\mu\nu}{=}{F^{\beta\gamma}$$
[ $${\gamma}{=}{-}\frac{g}{g(0,0)}}$$] ----------- eq 84.10 in Landau and Lifgarbagez
For such a definition we have
$${\nabla}{B}{=}{0}$$
The definition has been aimed to produce the above result. In fact the authors have remarked-----We introduce the vectors B and E………
[Please see attachment in Thread #10 ]
We continue to have in curved spacetime,
$$\frac{\partial {F^{\alpha\beta}}}{\partial{x^{\gamma}}}{+}{\frac{\partial {F^{\beta\gamma}}}{\partial{x^{\alpha}}}}{+}{\frac{\partial {F^{\gamma\alpha}}}{\partial {x^{\beta}}}}{\neq}{0}$$

The same quantity was zero in flat space time. This difference seems to indicate he existence of sources of $${F^{\alpha\beta}}$$.[We consider only the spatial indices and keeping in mind the antisymmetry factor we get three components.]

Last edited: Sep 12, 2010
13. Sep 12, 2010

From C. Moller, "The Theory of Relativity", Oxford (1952) pp. 304-305

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14. Sep 12, 2010

### Anamitra

[I am uploading attachment at my earliest]

We consider the relation [from Moller]
$${B^{\alpha}}{=}{\frac{1}{\sqrt \gamma}}{ F_{\beta\gamma}}$$ ------- (1)

In flat spacetime,
$${B(f)^{\alpha}}{=}{F^{\beta\gamma}}$$ ----------------- (2)
Again
$${B(f)^{\alpha}}{=}{g^{\beta\mu}}{g^{\gamma\nu }}{F_{\mu\nu}}$$ ----------------- (3)

Now if we transform from flat spacetime to curved spacetime, the left side of the last two equations transform to the left side of the first equation. Consequently the right sides of the last two should produce the right hand side of (1)
Considering the transformation from equation (3) to (1) we have
$${ \frac{1}{\sqrt \gamma}}{ F_{{'}\beta\gamma}}{=}{ g^{{'}\beta\mu}}{g^{{'}\gamma\nu }}{F_{'\mu\nu}}{=}{F^{'\beta\gamma}}$$
Implies:
$${{{F^{'}}_{\beta\gamma}}{=}{\sqrt\gamma}{F^{'\beta\gamma}}$$
[From (1) we have,
$$B^{'\alpha}{=}{\frac{1}{\sqrt\gamma}}{F_{{'}\beta\gamma}}$$]
This inconsistency means that the quantity defined by equation (1) does not represent physically the transformed magnetic field.
But if we define:
$${B^{\alpha}}{=}{F^{\beta\gamma}}$$ then
1)There is no inconsistency as such.
2)We have $${\nabla}{B}{\neq}{0}$$

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15. Sep 12, 2010

But this is not the relation from Landau-Lifgarbagez. In Landau-Lifgarbagez they use 3-dimensional Levi-Civita symbol, which means it should apply only to the space-like part - it needs to be read from the context.

A good reference (though one needs to know differential forms to appreciate it):

http://mpej.unige.ch/~durrer/papers/threeplusone.pdf", by Durrer and Strauman

Last edited by a moderator: Apr 25, 2017
16. Sep 12, 2010

Another good reference that is using only index notation:

http://arxiv.org/abs/0907.1151" [Broken]
by
Miguel Alcubierre, Juan Carlos Degollado, Marcelo Salgado
(Submitted on 7 Jul 2009 (v1), last revised 29 Nov 2009 (this version, v2))

Also: "Black hole physics: basic concepts and new developments"
by Valeriĭ Pavlovich Frolov,Igorʹ Dmitrievich Novikov" (available on Google books p. 292 and the following pages)

Last edited by a moderator: May 4, 2017
17. Sep 12, 2010

### Anamitra

In my previous thread the indices refer to only to the spatial parts.

Regarding what has been said in Landau and Lifgarbagez:

$${B^{\alpha}}{=}{-}{1/2}{\frac{1}{\sqrt\gamma}}{\epsilon}^{\alpha\beta\gamma}{F_{\beta\gamma}$$

For example,

$${B^{1}}{=}{-}{1/2}{\frac{1}{\sqrt\gamma}}{[}{\epsilon}^{123}{F_{23}}{+}{\epsilon}^{132}{F_{32}{]}$$

Implies that:

$${B^{1}}{=}{-}{\frac{1}{\sqrt\gamma}}{F_{23}}$$

Essentially this is of the same content as that of C Moller

Last edited: Sep 12, 2010