# A Manifestly covariant Maxwell's equations

1. Oct 21, 2016

### spaghetti3451

Consider the following Maxwell's equation in tensor notation:

$\partial_{k}F_{ij}=0$

$-\partial_{k}\epsilon_{ijm}B_{m}=0$

$\partial_{k}\epsilon_{ijm}B_{m}=0$

$\partial_{k}B_{k}=0$

I wonder how you go from the third line to the fourth line.

2. Oct 22, 2016

### vanhees71

Since the first line is wrong, I don't know, what should follow from it. The Maxwell equations in covariant notation are (in Heaviside-Lorentz units)
$$\partial_{\mu} {^\dagger}F^{\mu \nu}=0, \quad \partial_{\mu} F^{mu \nu} = \frac{1}{c} j^{\nu}.$$
Here the Hodge dual of the Faraday tensor is defined with help of the Levi-Civita tensor
$${^\dagger}F^{\mu \nu}=\epsilon^{\mu\nu\rho\sigma} F_{\rho \sigma}.$$
Now you should be able to get the (1+3)-notation by expressing the Faraday tensor in terms of the usual components in the (1+3) notation, $\vec{E}$ and $\vec{B}$.

3. Oct 22, 2016

### spaghetti3451

Well, you can also write Maxwell's equations in differential forms as well:

$dF=0 \qquad\qquad\qquad d*F=0$

Consider the equation $dF=0$.

$(dx^{\mu}\partial_{\mu})\wedge (F_{0i}\ dx^{0}\wedge dx^{i}+F_{ij}\ dx^{i}\wedge dx^{j}) = 0$

$(\mathrm{d}x^{0}\partial_{0}+\mathrm{d}x^{k}\partial_{k})\wedge (F_{0i}\ \mathrm{d}x^{0}\wedge \mathrm{d}x^{i}+F_{ij}\ \mathrm{d}x^{i}\wedge \mathrm{d}x^{j}) = 0$

$(\partial_{0}F_{0i})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{0}\wedge\mathrm{d}x^{i}+(\partial_{0}F_{ij})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} \qquad +(\partial_{k}F_{0i})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{0}\wedge\mathrm{d}x^{i}+(\partial_{k}F_{ij})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} = 0$

$(\partial_{0}F_{ij})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j}+(\partial_{k}F_{0i})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{0}\wedge\mathrm{d}x^{i} \qquad\\ +(\partial_{k}F_{ij})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} = 0$

$(\partial_{0}F_{ij})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j}-(\partial_{k}F_{0i})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{k}\wedge\mathrm{d}x^{i} \qquad \\ +(\partial_{k}F_{ij})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} = 0$

$(\partial_{0}F_{ij})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j}-(\partial_{i}F_{0j})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} \qquad \\ +(\partial_{k}F_{ij})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} = 0$

$(\partial_{0}F_{ij}-\partial_{i}F_{0j})\ \mathrm{d}x^{0}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j}+(\partial_{k}F_{ij})\ \mathrm{d}x^{k}\wedge\mathrm{d}x^{i}\wedge\mathrm{d}x^{j} = 0$

Now, you get

$(\partial_{0}F_{ij}-\partial_{i}F_{0j})$ and $(\partial_{k}F_{ij})$.

The second equation is the one I quoted in my first post.

4. Oct 22, 2016

### vanhees71

Before you use Cartan calculus you should understand the notation! The wedge product is antisymmetric!

5. Oct 22, 2016

### spaghetti3451

I know that, but I don't know where's my mistake.

6. Oct 22, 2016

### vanhees71

You have to antisymmetrize the expressions in front of the wedge products when you want to get the correct equations for the components. That's achieved in the Ricci calculus with help of the Levi-Civita symbol (which in pseudo-Cartesian coordinates provides the Levi-Civita tensor components).

7. Oct 22, 2016

### spaghetti3451

Can you write down $\mathrm{d}F=0$ in tensor notation?

It will help me understand how to proceed.

8. Oct 22, 2016

### vanhees71

Yes, it's the first equation in #2. The Cartan calculus is completely equivalent to the tensor notation. Alternating multilinear forms are a special case of more general multilinear forms, and multilinear forms are nothing else than tensors. The only advantage of Cartan calculus is a more compact notation. You have to pay the prize when you want to get the equations in terms of the tensor components, i.e., you must not forget to antisymmetrize, which is achieved with the Levi-Civita symbol, which is there in the Ricci calculus from the very beginning. For practical purposes the Cartan calculus is not much more elegant than the Ricci calculus.

9. Oct 22, 2016

### spaghetti3451

I'm actually asking you how to write $dF=0$ as a wedge product in Cartan calculus, not in Ricci calculus.

10. Oct 22, 2016

### vanhees71

But that you did (as far as I can see correctly) yourself!

11. Oct 22, 2016

### spaghetti3451

So, are the equations $(\partial_{0}F_{ij}-\partial_{i}F_{0j})=0$ and $(\partial_{k}F_{ij})=0$ correct, in that case?

12. Oct 22, 2016

### vanhees71

The first one looks good except that you forgot a factor 1/2 in the very beginning:
$$F=\frac{1}{2} F_{\mu \nu} \mathrm{d} x^{\mu} \wedge \mathrm{d} x^{\nu}=F_{0 j} \mathrm{d} x^0 \wedge \mathrm{d} x^j + \frac{1}{2} F_{ij} \mathrm{d} x^{i} \wedge \mathrm{d} x^j.$$
So what you get is
$$\frac{1}{2} \partial_0 F_{ij}-\partial_i F_{0j}=0$$
Let's see, what it means. First of all we have
$$F_{ij}=\partial_i A_j -\partial_j A_i = -\partial_i A^j + \partial_j A^i =\epsilon_{jik} B^k$$
and
$$F_{0j}=\partial_0 A_j-\partial_j A_0=-\partial_0 A^j-\partial_j A^0=E^j.$$
$$\partial_0 \epsilon_{jik} B^k-\partial_i E^j=0.$$
This is not in the usual form, but you can contract it with $\epsilon_{jil}$, using
$$\epsilon_{jik} \epsilon_{jil}=\delta_{ii} \delta_{kl}-\delta_{il} \delta_{ki}=2 \delta_{kl}$$
and
$$\epsilon_{jil} \partial_i E^j=-(\vec{\nabla} \times \vec{E})_l.$$
So (including the correct factor 1/2) your equation is equivalent to
$$\partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0,$$
which is Faraday's Law of induction, which is indeed one of the Maxwell equations.

The 2nd equation lacks symmetrization. Correct is that from the Cartan calculus follows
$$\epsilon_{ijk} \partial_i F_{jk}=0,$$
but from the above it's clear that
$$\epsilon_{ijk} F_{jk}=B^i,$$
and thus the equation simply means Gauss's Law for the magnetic field ("no magnetic monopoles")
$$\vec{\nabla} \cdot \vec{B}=0.$$

13. Oct 22, 2016

### spaghetti3451

Ah! I see!

So, is a $p$-form $F$ in $k+1$-dimensions defined as $\displaystyle{F = \frac{1}{p!}\epsilon_{i_{1},\dots,i_{p},j_{0},j_{1},j_{k}}F_{i_{1},\dots,i_{p}}\mathrm{d}x^{0}\wedge\mathrm{d}x^{1}\wedge\mathrm{d}x^{k}}$?

14. Oct 22, 2016

### vanhees71

If $F_{i_1,\ldots,i_p}$ are components of the alternating form (which implies that it's completely antisymmetric under interchange of the indices) with respect to the co-basis $\mathrm{d} x^j$, then the alternating form is given by
$$F=\frac{1}{p!} F_{i_1,\ldots,i_p} \mathrm{d} x^{i_1} \wedge \cdots \wedge \mathrm{d} x^{i_p}.$$

15. Oct 22, 2016

### spaghetti3451

So, why does the epsilon tensor come out from this Cartan-calculus expression when we switch to Ricci calculus?

16. Oct 22, 2016

### vanhees71

Let's take a very simple example. The four-potential can be taken as a one-form
$$A=A_{\mu} \mathrm{d} x^{\mu}.$$
It's "exterior derivative" is by definition the Faraday two-form
$$F=\mathrm{d} A=\partial_{\mu} A_{\nu} \mathrm{d} x^{\mu} \wedge \mathrm{d} x^{\nu}.$$
Since by definition the wedge product is skew symmetric, the unique way to map this to its components is to completely antisymmetrize them, i.e., in this case you have
$$F=\frac{1}{2} (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}) \mathrm{d} x^{\mu} \wedge \mathrm{d} x^{\nu}.$$
Now take the exterior derivative of this, which is a three-form
$$\mathrm{d} F = \frac{1}{2} \partial_{\rho} F_{\mu \nu} \mathrm{d} x^{\rho} \wedge \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
This maps one-to-one to the totally antisymmetrized components
$$G_{\rho \mu \nu}=\partial_{\rho} F_{\mu \nu} -\partial_{\mu} F_{\rho \nu} - \partial_{\nu} F_{\mu \rho}=\partial_{\rho} F_{\mu \nu} + \partial_{\mu} F_{\nu \rho} + \partial_{\nu} F_{\rho \mu},$$
i.e., to the "cylcic sum", and so on.

Now the totally antisymmetric components of the three-form, can be mapped to 1st-rank-tensor components
$$G_{\rho \mu \nu} = \epsilon_{\rho \mu \nu \sigma} (^\dagger G)^{\sigma}.$$
Obviously we have
$$(^\dagger G)^{\sigma} = \det \hat{\eta} \frac{1}{3!} \epsilon^{\rho \mu \nu \sigma} G_{\rho \mu \nu}=-\frac{1}{3!} \epsilon^{\rho \mu \nu \sigma} G_{\rho \mu \nu}.$$
The somewhat cumbersome factor $\det \hat{\eta}=-1$ comes from the pseudometric $\hat{\eta}=\mathrm{diag}(1,-1,-1,-1)$, and one defines by convention $\epsilon^{\mu \nu \rho \sigma}=\text{sign}(\mu,\nu,\rho\,sigma)$, i.e., as the totally antisymmetric Levi-Civita symbol with $\epsilon^{0123}$. Since then lowering the indices is done with $\eta_{\mu \nu}$ as for any other tensor components you have $\epsilon_{\mu \nu \rho \sigma}=-\epsilon^{\mu \nu \rho \sigma}$, and that's where the minus sign comes from in the Hodge dual.

Now we can write the homogeneous Maxwell equations as
$$G_{\rho \mu \nu}=0$$
or equivalently as
$$(^{\dagger} G)^{\sigma} = -\frac{1}{6!} G_{\rho \mu \nu} \epsilon^{\rho \mu \nu \sigma}=0.$$
This means that
$$\epsilon^{\rho \mu \nu \sigma} \partial_{\rho} F_{\mu \nu}=+\epsilon^{\mu \nu \rho \sigma} \partial_{\rho} F_{\mu \nu} \partial_{\rho} (^\dagger F)^{\rho \sigma}=0$$
is a more compact form of $\mathrm{d} F=0$ when using the index notation than writing out the cyclic form $G_{\rho \mu \nu}$.
In the same way you can show that the inhomogeneous Maxwell equations
$$\partial_{\mu} F^{\mu \nu}=\frac{1}{c} j^{\nu}$$
can be written as
$$\mathrm{d} ^{\dagger} F=^{\dagger} j.$$

17. Oct 22, 2016

### robphy

This should read $\partial_{[k}F_{ij]}=0$--the left hand side is the exterior derivative of a two form.

18. Oct 22, 2016

### vanhees71

One should add that the square brackets in this notation denote the total antisymmetrization of the tensor components, which I expressed above with help of the Levi-Civita symbol.

19. Oct 22, 2016

### robphy

Yes, totally-antisymmetrized, thanks.

Are you referring to
Note that $\partial_{[k}F_{ij]}=0$ is a three-form (or a total-antisymmetric three-index tensor) for Minkowski spacetime.
However, $\epsilon_{ijk} \partial_i F_{jk}=0$ looks like a scalar equation. Is it in Minkowski spacetime... or is it in a spatial 3D slice, with $\epsilon_{ijk}$ as the symbol associated with that slice?

20. Oct 22, 2016

### vanhees71

This is the result of the split in temporal and spatial components with respect to some fixed inertial frame. As shown above, it's just Gauss's Law for the magnetic field $\vec{\nabla} \cdot \vec{B}=0$ ("no magnetic poles").

21. Oct 23, 2016

### spaghetti3451

How does the action $\displaystyle{\int F \wedge *F}$ become the action $\displaystyle{\int d^{4}x\ F_{\mu\nu}F^{\mu\nu}}$?