# Maxwell's equations in curved space-time

1. Apr 1, 2013

### ash64449

what changes will take place in maxwell's equations if the space-time was curved?

2. Apr 1, 2013

### Staff: Mentor

The local equations stay the same, as far as I know (locally, space always looks like special relativity). The integrated versions can get problematic.

3. Apr 1, 2013

### ash64449

Ok. What change can arrive is space was like dynamical,like in General Theory of relativity?..

4. Apr 1, 2013

### Staff: Mentor

I don't understand that question.

The integrated versions require a global time where the integrals are evaluated, that is not given in GR.
Actually, I am not even sure if they are true in special relativity, as local changes need some time to propagate.

5. Apr 1, 2013

### Staff: Mentor

All laws of physics (expressed in their differential form) have basically the same changes. Express them in terms of tensors and replace any derivatives with covariant derivatives. That is it.

EDIT: this is incorrect, see WannabeNewton's correction below.

Last edited: Apr 1, 2013
6. Apr 1, 2013

### ash64449

Well,yes friend. But i have heard that by adding 5th dimension,Electromagnetic phenomena and gravity was unified. But it said that this theory had some inconsistencies because According to GR,Space is dynamical.i.e It can curve. But Maxwell's equation is only consistent if space is flat.So i heard that they have to freeze the tensor to make it consistent. That is the reason why i asked. Why maxwell's equation break out if space is curved? i wanted to know that. That is why i started the discussion

7. Apr 1, 2013

### WannabeNewton

Let's be careful here. Maxwell's equations are a perfect example of why naively replacing $\partial _{a}\rightarrow \nabla_{a}$ doesn't always work. Consider the following form of maxwell's equations in flat spacetime: $\partial ^{a}F_{ab} = -4\pi j_{b},\partial _{[a}F_{bc]} = 0$. If we make the replacement of partials with covariant derivatives in order to make the equations covariant in curved space-time then we simply get $\nabla ^{a}F_{ab} = -4\pi j_{b},\nabla _{[a}F_{bc]} = 0$. One can show (if you want me to I can show you the calculation right here on this thread) that the inhomogenous Maxwell equations do indeed give us local charge conservation i.e. $\nabla ^{a}F_{ab} = -4\pi j_{b}\Rightarrow \nabla^{a}j_{a} = 0$.

Consider now the fact that the poincare lemma allows us to claim that locally there exists a one-form $A_{a}$ such that $F_{ab} = \partial _{a}A_{b} - \partial _{b}A_{a}$ (of course $A_{a}$ is just the 4-potential). Writing the inhomogenous Maxwell equations in flat space-time in terms of the 4-potential, after fixing the Lorenz gauge, we simply have $\partial ^{a}\partial _{a}A_{b} = -4\pi j_{b}$. If we now naively replace the partials with covariant derivatives then we find that $\nabla^{a}\nabla_{a}A_{b} = -4\pi j_{b}$ but this will not give us local charge conservation. By simply replacing partials with covariant derivatives we overlooked the facts that covariant derivatives do NOT commute and that their commutator gives us a curvature term. In fact, the correct form should be $\nabla^{a}\nabla_{a}A_{b} - R^{c}{}{}_{b}A_{c} = -4\pi j_{b}$. Again you can show with a calculation that this does in fact imply $\nabla^{a}j_{a} = 0$.

8. Apr 1, 2013

### Staff: Mentor

Interesting. I was not aware of your good counter-example. I will have to not say that in the future.

9. Apr 1, 2013

### WannabeNewton

As far as I know, it almost always works but there are examples where it doesn't. It is sort of sad that it doesn't work for the 4-potential form of Maxwell's equations , in the Lorenz gauge, when you think about it, consider they are the most awesome equations known to man (yes I am a Maxwell fanboy).

10. Apr 1, 2013

### Staff: Mentor

I guess that you could take the approach that the examples where it doesn't work are not "laws of nature", but in the counter-example you provided I don't like that approach.

11. Apr 1, 2013

### dextercioby

Actually field equations always follow from variational principles. So write the variational principle for electromagnetism in curved space-time and find what you're supposed to find. See the 75 page brochure by Dirac.

12. Apr 2, 2013

### TrickyDicky

I remember a thread where I was trying to make a distinction between equations in tensorial form(tensor field equations like the EFE for instance) and equations in differential form(covector field equations like the Maxwell eq. discussed in this thread). I was told by Dalespam and others that since vectors are tensors there was no distinction wrt general coordinate transformation invariance.(And a quite silly exchange about this followed IIRC).
My point was precisely that one can't replace in general partial derivatives with covariant derivatives but I didn't addressed it that way nor came up with any counter-example so I'm not trying to vindicate here my position then just see if I can understand this issue better.
So my understanding from WB post is that to handle curvature and and keep the general (not only for inertial coordinates) coordinate transformations invariance one needs a tensor field equation, not just a covector field equation that would be inertial transformation invariant, would this be right?
I'm actually not completely sure this translates well to the pseudoRiemannian case since the minimal coupling principle states that: "No terms explicitly containing the curvature tensor should be added in making the transition from SR to GR" as stated in D'inverno (page 131).
It also says this principle is vague and should be used with care, maybe it is referring to cases like the one WB showed.

13. Apr 2, 2013

### Bill_K

Well, I think that's right. Often the same equation can be written in more than one way, e.g. by permuting derivatives, and this will cause the curvature to appear explicitly.

In the example given by WannabeNewton, we're led to a wave equation for Aμ in the form ◻Aμ - Aν;μν = 0, and it's only because we want to impose the Lorenz condition that we need to permute the indices, bringing in the Ricci tensor term.

14. Apr 2, 2013

### vanhees71

I think there is a subtle problem with the derivation of Maxwell's equations within GR, and the solution is to refer to Hamilton's principle at the end. I have to check it this evening when I'm at home and have my GR books at hand, but I think this is discussed in both Landau/Lifgarbagez Vol. II and Misner/Thorne/Wheeler.

15. Apr 2, 2013

### WannabeNewton

If we instead look at the differential forms version, $dF = 0, d^{\star }F = 4\pi^{\star }j$ and use Stoke's theorem $\int_{\Omega } d\omega = \int_{\partial \omega} \omega$ on the second equation we can show e.g. that Gauss's law still holds in GR. See the remarks under part (b) of problem 2 in Chapter 4 of Wald.

EDIT: if you wanna see how to get the differential forms version from the usual version, I did it on the homework section here a while back: https://www.physicsforums.com/showthread.php?t=674049 (it's just the exercise I referred you to in Wald).

Last edited: Apr 2, 2013
16. Apr 2, 2013

### atyy

I think it's viable to say the examples where it doesn't work aren't "laws of nature".

From one point of view, the EP applies only locally, so it fails for non-local laws, such as those that have double gradients. (If we like this version of the EP, then isn't it good that it fails for Maxwell's equations in Lorenz gauge?)

OTOH, if you write the known fields of special relativity (say the standard model of particle physics) using an action, then only first derivatives of the fields appear, and no derivatives of the metric appear, so you can use the minimal coupling prescription. If we say EP = minimal coupling, then the EP is exact and never fails.

Last edited: Apr 2, 2013
17. Apr 2, 2013

### WannabeNewton

If you wanted to be as natural as possible then going along with dexter, the field equations would invariably come out correctly when you vary the appropriate action for the field theory. If we are talking about a Klein Gordon scalar field then $\mathcal{L}_{KG} = -\frac{1}{2}\sqrt{-g}(g^{ab}\nabla_{a}\varphi \nabla_{b} \varphi + m^{2}\varphi^{2})$ will naturally gives us $\nabla^{a}\nabla_{a}\varphi - m^2\varphi = 0$. Similarly $\mathcal{L}_{EM} = -\sqrt{-g}g^{ac}g^{bd}\nabla_{[a}A_{b]}\nabla_{[c}A_{d]}$ will give us the correct field equations of electromagnetism on curved space-time. We can simply use a variational principle for the respective field theory on curved space-time and avoid the whole minimal coupling thing.

18. Apr 2, 2013

### Ben Niehoff

What I think you really have is an example of how index notation is cumbersome and can lead the unwary astray...

In differential forms notation, one has

$$dF = 0, \qquad d \star F = \star J, \qquad F = dA$$
and hence

$$\star d \star dA = -J$$
Now, the Lorentz gauge condition in curved space is

$$d \star A = 0$$
Noting that the Laplace-Beltrami operator is

$$\Delta = \star d \star d + d \star d \star$$
we can, after imposing the Lorentz condition, write the wave equation

$$\Delta A = - J$$
But for comparison with index notation, we should write out what the gauge condition is:

$$(d \star A)_{abcd} = 4 \nabla_{[a} (\sqrt{-g} \frac{1}{3!} \varepsilon_{bcd]e} g^{ef} A_f) = \sqrt{-g} \varepsilon_{abcd} \nabla_e A^e = 0$$
And then in this gauge, the wave operator is

$$- (\star \Delta A)_{abc} = (d \star dA)_{abc} = 3 \nabla_{[a} (\sqrt{-g} \frac{1}{2! 2!} \varepsilon_{bc]de} g^{dm} g^{en} \nabla_{[m} A_{n]}) = \sqrt{-g} \varepsilon_{abcd} \nabla_e \nabla^{[d} A^{e]}$$
Up to factors of -1 and maybe other constants because I'm being sloppy.

We have already imposed the gauge condition, but if you were to write all of this out, there would be yet another term that you can cancel using $\nabla_a A^a = 0$. However, you have to commute two covariant derivatives to get there, hence the appearance of the Ricci tensor.

But what's really happening is this: It is still true that imposing the Lorentz gauge condition leaves you with the wave equation for A. The problem is assuming that the wave operator for vector fields is just $\nabla_a \nabla^a$. It clearly isn't.

In differential form notation, the wave operator is always $\star d \star d + d \star d \star$ (up to an overall factor of -1 which depends on both the number of dimensions and the signature of the metric). The catch is that when this operator acts on fields with indices, you will get factors of the Ricci tensor.

So the problem is nothing specifically to do with electromagnetic fields at all. It's in making unwarranted assumptions about the Laplace operator (i.e. wave operator). The substitution $\partial_a \partial^a \rightarrow \nabla_a \nabla^a$ only works on scalar fields.

19. Apr 2, 2013

### WannabeNewton

Sure I agree there is nothing remotely deep here. I was just using the EM field as a specific example because making a mistake as simple as assuming $\nabla^{a}\nabla_{a}$ acts on $A_{b}$ like $\partial ^{a}\partial _{a}$ does in the flat space-time case can obstruct one from showing $\nabla^{a}j_{a} = 0$. With $\nabla^{a}F_{ab} = -4\pi j_{b}$ one manages to doge that bullet whether one knew about the pitfalls of the notation or not and $\nabla^{a}j_{a} = 0$ successful comes out of calculations. Ideally, one could just stick to $dF = 0, d^{\star }F = 4\pi ^{\star} j$ because physical results regarding the EM field can be much more elegantly derived in this form and one can avoid the cumbersome index business to boot.

20. Apr 2, 2013

### atyy

Wouldn't one fail to get covariant conservation of energy without minimal coupling? I've seen a claim like that in http://arxiv.org/abs/gr-qc/0505128 (Eq 11) and in http://arxiv.org/abs/0704.1733 .