## The electromagnetic field

### 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:

\(j^\mu = \Psi^\dagger \gamma^0 \gamma^\mu \Psi\)

\(j^\mu = \Psi^\dagger \gamma^0 \gamma^\mu \Psi\)

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

\(i\gamma^\mu D_\mu \Psi = m\Psi\)

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.

\(D_\mu = \partial_\mu - i A_\mu\)

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:

\(\partial_\mu A_\nu - \partial_\nu A_\mu = 0\)

\(\partial_\mu A_\nu - \partial_\nu A_\mu = 0\)

### 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.

\(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\)

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:

\(F_{\mu\nu} +F_{\nu\mu} = 0\)

\(F_{\mu\nu} +F_{\nu\mu} = 0\)

### 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:

\(\nabla \cdot \overrightarrow{B} = 0\)

\(\nabla \times \overrightarrow{E} + \frac{\partial}{\partial t} \overrightarrow{B} = 0\)

\(F_{\mu\nu , \lambda} + F_{\lambda\mu , \nu} + F_{\nu\lambda , \mu} = 0\)

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 \(F_{\mu\nu} = 0\)

\(\nabla \times \overrightarrow{B} - \frac{\partial}{\partial t} \overrightarrow{E} = 4\pi \overrightarrow{j}\)

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 \(F_{\mu\nu} = 0\)