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