The electromagnetic field

Continuing from last time, today I will talk about the electromagnetic field as a gauge theory.

1. The gauge group

In this case the gauge group is $U(1)$ - the phase rotations. This group is commutative. This can be determined if we start from Dirac's equation and we demand that the group leaves the Dirac current of probability density:

$j^\mu = \Psi^\dagger \gamma^0 \gamma^\mu \Psi$

invariant.

2. The covariant derivative giving rise to the gauge group

Here the covariant derivative takes the form:

$D_\mu = \partial_\mu - i A_\mu$

To determine the gauge connection $A_\mu$ we can substitute this expression in Dirac's equation:

$i\gamma^\mu D_\mu \Psi = m\Psi$

and require the equation to be invariant under a gauge transformation:

$\Psi^{'} = e^{i \chi}\Psi$

which yields:

$A^{'}_{\mu} = A_\mu + \partial_\mu \chi$

This shows that:
- the general gauge field for Dirac's equation is an arbitrary vector field $A_\mu (x)$
-The part of the gauge field which compensates for an arbitrary gauge transformation of the Dirac field $\Psi (x)$ is the gradient of on an arbitrary scalar field.

3. The integrability condition

Here we want to extract a physically observable object out of a given vector field $A_\mu (x)$. From above it follows that there is no external potential if $A_\mu = \partial_\mu \chi$ and this is the case if and only if:

$\partial_\mu A_\nu - \partial_\nu A_\mu = 0$

4. The curvature

The curvature measures the amount of failure for the integrability condition and by definition is the left-hand side of the equation from above:

$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$

and this is the electromagnetic field tensor.

5. The algebraic identities

There is only one algebraic identity in this case stemming from the curvature tensor antisymmetry:

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

6. The homogeneous differential equations

If we take the derivative of $F_{\mu\nu}$ and we do a cyclic sum we obtain:

$F_{\mu\nu , \lambda} + F_{\lambda\mu , \nu} + F_{\nu\lambda , \mu} = 0$

which is analogous with the Bianchi identity in general relativity.

This identity can be expressed using the Hodge dual as follows:

$\partial_\rho {* F}^{\rho\mu} = 0$

and this is nothing but two of the Maxwell's equations:

$\nabla \cdot \overrightarrow{B} = 0$
$\nabla \times \overrightarrow{E} + \frac{\partial}{\partial t} \overrightarrow{B} = 0$

7. The inhomogeneous differential equations

If we take the derivative of $\partial_\beta F^{\alpha\beta}$ we get zero because the F is antisymetric and $\partial_{\alpha\beta} = \partial_{\beta\alpha}$. and so the vector  $\beta F^{\alpha\beta}$ is divergenless. We interpret this as a current of a conserved quantity: the source for the electromagnetic field and we write:

$\partial_\rho F^{\mu\rho} = 4\pi J^\mu$

where the constant of proportionality comes from recovering Maxwell's theory (recall that last time $8\pi G$ came from similar arguments.

From this we now get the other two Maxwell's equations:

$\nabla \cdot \overrightarrow{E} = 4\pi \rho$
$\nabla \times \overrightarrow{B} - \frac{\partial}{\partial t} \overrightarrow{E} = 4\pi \overrightarrow{j}$

Now we can compare general relativity with electromagnetism:

Coordinate transformation - Gauge transformation
Affine connection $\Gamma^{\alpha}_{\rho\sigma}$ - Gauge connection $iA_\mu$
Gravitational potential $\Gamma^{\alpha}_{\rho\sigma}$ - electromagnetic potential $A_\mu$
Curvature tensor $R^{\alpha}_{\beta\gamma\delta}$ - electromagnetic field $F_{\mu\nu}$
No gravitation $R^{\alpha}_{\beta\gamma\delta} = 0$ - no electromagnetic field $F_{\mu\nu} = 0$