Wolfram Alpha:

```Electromagnetic 4 - Potential
-----------------------------

Aμ = (φ/c, A) where φ is the electric potential and A
is the magnetic vector potential.  Therefore,

Aμ = ημνAν = (φ/c, -A)  η is the Minkowski metric of
the form +---

The corresponding 4 - gradient operators are,

∂μ = ∂/∂xμ = {(1/c)∂/∂t, ∇}

and

∂ν = ημν∂μ = {(1/c)∂/∂t, -∇}

Now,

E = -∂A/∂t - ∇φ

Consider only x dimension,

Ex = -∂A1/∂t - ∂φ/∂x

Now, from before,

∂0A1 = (1/c)∂A1/∂t + ∂(φ/c)/∂x

Therefore, in order to get the RHS to equal Ex, we
have to multiply by -c.

Ex = -c{∂A1/∂t - ∂φ/∂x}

Similarly, for the B field we can find the Bx component
of the curl.

Bx = ∇ x A1 = ∂2A3 - ∂3A2

So E and B have the same form.  We can now define an
asymmetric tensor, Fμν such that,

Fμν = ∂μAν - ∂νAμ

where,

F00 = ∂0A0 - ∂0A0 = 0

F01 = ∂0A1 - ∂1A0 = -Ex/c

F02 = ∂0A2 - ∂2A0 = -Ey/c

F03 = ∂0A3 - ∂3A0 = -Ez/c

F10 = ∂1A0 - ∂0A1 = Ex/c

F11 = ∂1A1 - ∂1A1 = 0

F12 = ∂1A2 - ∂2A1 = Bz

F13 = ∂1A3 - ∂3A1 = -By

F20 = ∂2A0 - ∂0A2 = Ey/c

F21 = ∂2A1 - ∂1A2 = -Bz/c

F22 = ∂2A2 - ∂2A2 = 0

F23 = ∂2A3 - ∂3A2 = Bx

F30 = ∂3A0 - ∂0A3 = Ez/c

F31 = ∂3A1 - ∂1A3 = By/c

F32 = ∂3A2 - ∂2A3 = -Bx

F33 = ∂3A3 - ∂3A3 = 0

so we can write (note by convention the B components
are written as the -ve of the actual values).

-                         -
| 0      Ex/c   Ey/c   Ez/c |
|-Ex/c   0     -Bz     By   |  = Fμν ... The COVARIANT
|-Ey/c   Bz     0     -Bx   |            ELECTROMAGNETIC
|-Ez/c  -By     Bx     0    |            TENSOR
-                         -

-                         -
| 0     -Ex/c  -Ey/c  -Ez/c |
| Ex/c   0     -Bz     By   |  = Fμν ... The CONTRAVARIANT
| Ey/c   Bz     0     -Bx   |            ELECTROMAGNETIC
| Ez/c  -By     Bx     0    |            TENSOR
-                         -

Note:  Fμν = ημαηνβFαβ and Fμν = ημαηνβFαβ

Properties:

Fμν = -Fνμ  ... antisymmetric

Fμν is gauge invariant but not Lorentz invariant (see note
on Gauge Theory).

Inner product:

FμνFμν = 2(B2 - E2/c2) which is invariant under a Lorentz
transformation.

Maxwell's Equation in Flat Spacetime
------------------------------------

∇.E = ρ/ε and ∇ x B = μ0J + (1/c2)∂E/∂t reduce to,

∂Fμν/∂xμ = ∂μFμν = μ0jν

where jν = ρ∂xν/∂t = (cρ, j) ... the 4-current

∇.B = 0 and ∇ x E = -∂B/∂t reduce to,

∂Fμν/∂xγ + ∂Fνγ/∂xμ + ∂Fγμ/∂xμ = ∂γFμν + ∂μFνγ + ∂νFγμ = 0

Maxwell's Equation in Curved Spacetime
--------------------------------------

Replace the partial derivative with the covariant
derivative to get,

∇μFμν = μ0Jν

∇γFμν + ∇μFνγ + ∇νFγμ = 0

This is equal to,

(∂γFμν - ΓσγμFσν - ΓσγνFμσ) + (∂μFνγ - ΓσμνFσγ - ΓσμγFνσ)
+ (∂νFγμ - ΓσνγFσμ - ΓσνμFγσ) = 0

We can use the antisymmetric properties of Fab = -Fba to get,

(∂γFμν - ΓσγμFσν - ΓσγνFμσ) + (∂μFνγ - ΓσμνFσγ + ΓσμγFσν) +
+ (∂νFγμ + ΓσνγFμσ ΓσνμFσγ) = 0

We can use the symmetry of the Christofell symbols
Γσαβ = Γσβα to get,

∂γFμν + ∂μFνγ + ∂νFγμ - ΓσγμFσν + ΓσγμFσν - ΓσγνFμσ
+ ΓσγνFμσ - ΓσμνFσγ + ΓσμνFσγ = 0

Which simplifies to,

∂γFμν + ∂μFνγ + ∂νFγμ = 0```