Electromagnetic Curr. in Curved Spacetime: Questions Answered

[gerald V]

I assume this forum to be the appropriate one, since the real problem is about covariance rather than electromagnetism.

In electrodynamics in a curved background, the relation $F^{\mu \nu} = A^{\mu , \nu} - A^{\mu , \nu}$ stays in terms of ordinary derivatives. So, in particular $F_{,\mu \nu}^{\mu \nu}$ holds in terms of ordinary derivatives. One can define $F_{,\mu}^{\mu \nu} = - \hat{j}^\nu$, which is ordinarily conserved.

In contrast, the field equation $F_{;\nu}^{\mu \nu} = \frac{1}{\sqrt{- \det g}} (\sqrt{- \det g} F^{\mu \nu})_{,\nu} = -j^\nu$ and the conservation law for the current $j_{;\nu}^\nu = \frac{1}{\sqrt{- \det g}} (\sqrt{- \det g} j^\nu)_{,\nu} = 0$ are in terms of covariant derivatives.

Starting from the identiy $(\sqrt{- \det g} \; F^{\alpha \beta})_{, \beta} = (\sqrt{- \det g})_{, \beta} F^{\alpha \beta} + \sqrt{- \det g} \; F^{\alpha \beta}_{, \beta}$ and using the field equation, one gets $- \sqrt{- \det g} \; j^\alpha = (\sqrt{- \det g})_{, \beta} F^{\alpha \beta} + \sqrt{- \det g} \;( -\hat{j}^\alpha)$ or $j^\alpha = - (\ln \sqrt{- \det g})_{, \beta} F^{\alpha \beta} + \hat{j}^\alpha = - \Gamma_{\mu \beta}^\mu F^{\alpha \beta} + \hat{j}^\alpha$, where the current $j$ is covariantly conserved, while the current $\hat{j}$ is ordinarily conserved.

Questions:

- Is my analysis correct?
- If so, what does it mean that there co-exist a covariantly as well as an ordinarily conserved current? Is $\hat{j}$ observable?
- Is there anything comparable for gravitation, that means the co-existence of a covariantly as well as an ordinarily conserved energy-momentum Tensor?

Thank you very much in advance.

 

[Orodruin]

So, in particular Fμν,μνF,μνμνF_{,\mu \nu}^{\mu \nu} holds in terms of ordinary derivatives.

What do you mean by ”holds”? What you wrote down is not an equation. That F in terms of A can be written without the covariant derivative has to do with its anti-symmetry. This does not mean you can take arbitrary derivatives and expect to get something covariant.

 

[gerald V]

Sorry, that is just a clerical error. I intended to say $F_{,\mu \nu}^{\mu \nu} = 0$ holds in terms of ordinary derivatives. I think this follows from performing the ordinary derivatives as usual. But this is not a covariant equation, what exactly is the reason for my confusion.

 

[sweet springs]

n electrodynamics in a curved background, the relation Fμν=Aμ,ν−Aμ,νF^{\mu \nu} = A^{\mu , \nu} - A^{\mu , \nu} stays in terms of ordinary derivatives.

In GR electromagnetic tensor is derived from covariant derivative of A thus
$$F_{\mu\nu}=A_{\mu:\nu}-A_{\nu:\mu}$$
where : means covariant derivative.

 

[Orodruin]

Sorry, that is just a clerical error. I intended to say $F_{,\mu \nu}^{\mu \nu} = 0$ holds in terms of ordinary derivatives. I think this follows from performing the ordinary derivatives as usual. But this is not a covariant equation, what exactly is the reason for my confusion.

It is identically zero because of the anti-symmetry of the field tensor. You cannot draw the conclusion that $\partial_\mu F^{\mu\nu}$ holds any sort of physical meaning from this. The coordinate invariant way is to use the covariant derivative and the quantity you defined as $\hat j$ is coordinate dependent. I would therefore not use it.

In GR electromagnetic tensor is derived from covariant derivative of A thus
$$F_{\mu\nu}=A_{\mu:\nu}-A_{\nu:\mu}$$
where : means covariant derivative.

This is equal to $F_{\mu\nu} = A_{\mu,\nu} - A_{\nu,\mu}$ because of the anti-symmetry - as already mentioned in #2. Also note that the notation typically used for the covariant derivative is ; and not :.

 

[Dale]

It is identically zero because of the anti-symmetry of the field tensor. You cannot draw the conclusion that $\partial_\mu F^{\mu\nu}$ holds any sort of physical meaning from this.

I agree here. For an antisymmetric tensor $\partial_\mu F^{\mu\nu}$ =$\nabla_\mu F^{\mu\nu}$. So the physically significant derivative is still the covariant derivative, and the ordinary derivative is simply a computational shortcut available for antisymmetric tensors. Personally, I prefer to write the covariant derivative to make the quantity manifestly covariant.

 

[Orodruin]

I agree here. For an antisymmetric tensor $\partial_\mu F^{\mu\nu}$ =$\nabla_\mu F^{\mu\nu}$. So the physically significant derivative is still the covariant derivative, and the ordinary derivative is simply a computational shortcut available for antisymmetric tensors. Personally, I prefer to write the covariant derivative to make the quantity manifestly covariant.

This is not generally true,
$$
\nabla_\mu F^{\mu\nu} = \partial_\mu F^{\mu\nu} + \Gamma_{\mu\alpha}^\mu F^{\alpha\nu} + \Gamma_{\mu\alpha}^\nu F^{\mu\alpha}.
$$
Note that the two terms with Christoffel symbols do not generally cancel.

 

[Dale]

Note that the two terms with Christoffel symbols do not generally cancel.

They do for antisymmetric tensors, right? I thought they did.

 

[kent davidge]

They do for antisymmetric tensors, right? I thought they did.

Only in a symmetric connection.

 

[Orodruin]

Only in a symmetric connection.

No, this is not true either. The last term in the expression in #7 disappears for a symmetric connection, the middle one does not.

 

[Orodruin]

They do for antisymmetric tensors, right? I thought they did.

What cancels out are the terms involving Christoffel symbols in the expressions for $F_{\mu\nu}$.

 

[gerald V]

Thank you very much. I think, I understand what was said here. I should have mentioned that I assume General Relativity, thus a symmetric connection.

Maybe my point is not very relevant. I just was surprised that, in the special case of an asymetric field an a symmetric connection, one can construct an "animal" $\hat{j}$ which is conserved in terms of ordinary derivatives. I am aware that $\hat{j}$ does not emerge as the covariant derivative of $F$. Nevertheless I wondered whether it has any physical meaning. But the common opinion seems to be that the answer is "no".

 

[Orodruin]

Well, in the end, it is a coordinate current density. The point with conserved (physical) currents is that you should have
$$
0 = \int_V (\nabla_\mu j^\mu) dV = \oint_S j^\mu dS_\mu,
$$
where $S$ is the boundary of $V$, $dV$ is the physical volume element ($\sqrt{|g|} d^4x$ in a 4D spacetime), and $dS_\mu$ the corresponding hypersurface element. This is because you can make nice conservation interpretations of the right-hand side.

However, note that if you look at the coordinates, the $\sqrt{|g|}$ ends up outside of the divergence of the current so you cannot just apply the divergence theorem out of the blue. Instead, you can look at the coordinates and (it is relatively easy to show that)
$$
\sqrt{|g|} \nabla_\mu j^\mu = \partial_\mu(\sqrt{|g|} j^\mu),
$$
leading to
$$
\int_V (\nabla_\mu j^\mu) dV = \int_V (\partial_\mu \tilde j^\mu) d^4 x,
$$
where $\tilde j^\mu = \sqrt{|g|}j^\mu$. Note that $d^4x$ here is the coordinate volume element, not the physical volume element $dV$. In the end, it is a question of whether you want to deal with the physical current density $j^\mu$ (a vector field) or the coordinate current density $\tilde j^\mu$ (which is not a vector field, but a vector density field).