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 $F_{\mu\nu} = 0$

The gravitational field

Today we will start implementing the 7 point roadmap in the case of the gravitational field. Technically gravity does not form a gauge theory but since it was the starting point of Weyl's insight, I will start with this as well and next time I will show how the program works in case of the electromagnetic field.

1. The gauge group

The "gauge group" in this case is the group of general coordinate transformations in a real four-dimensional Riemannian manifold M. Now the argument against Diff M as a gauge group comes from locality. An active diffeomorphism can move a state localized near the observer to one far away which can be different. However, for the sake of argument I will abuse this today and considered Diff M as a "gauge group" because of the deep similarities (which will explore in subsequent posts) between this and proper gauge theories like electromagnetism and Yang-Mills.

2. The covariant derivative giving rise to the gauge group

For a vector field $f^\alpha$ the covariant derivative is defined as follow:

$D_\rho f^\alpha = \partial_\rho f^\alpha +{\Gamma}^{\alpha}_{\rho\sigma} f^\alpha$

where ${\Gamma}^{\alpha}_{\rho\sigma}$ is called an affine connection. If we demand that the metric tensor is a covariant constant under D we can find that the connection is:

${\Gamma}^{\sigma}_{\mu\nu} = \frac{1}{2}[g_{\rho\mu,\nu} + g_{\rho\nu,\mu} - g_{\mu\nu,\rho}]$

where $f_{\rho,\sigma}  = \partial_\sigma f_\rho$ 

3. The integrability condition

We define this condition as the commutativity of the covariant derivative. If we define the notation: $D_\mu D_\nu f_\sigma = f_{\sigma;\nu\mu}$ we can write this condition as:

$f_{\rho;\mu\nu} - f_{\rho;\nu\mu} = 0$

Computing the expression above yields:

$f_{\rho;\mu\nu} - f_{\rho;\nu\mu} = f_\sigma {R}^{\sigma}_{\rho\mu\nu}$

where

${R}^{\sigma}_{\rho\mu\nu} = {\Gamma}^{\tau}_{\rho\mu}{\Gamma}^{\sigma}_{\tau\nu} - {\Gamma}^{\tau}_{\rho\nu}{\Gamma}^{\sigma}_{\tau\mu} + {\Gamma}^{\sigma}_{\rho\mu,\nu} - {\Gamma}^{\sigma}_{\rho\nu,\mu}$

4. The curvature

From above the integrability condition is ${R}^{\sigma}_{\rho\mu\nu} = 0$ and R is called the Riemann curvature tensor.

5. The algebraic identities

The algebraic identities come from the symmetry properties of the curvature tensor which reduces the 256 components to only 20 independent ones. I am too tired to type the proof of the reduction to 20, but you can easily find the proof online.

6. The homogeneous differential equations

If we take the derivative of the Riemann tensor we obtain a differential identity known as the Bianchi identity:

${R}^{\sigma}_{\rho\mu\nu;\tau} + {R}^{\sigma}_{\rho\tau\mu;\nu} + {R}^{\sigma}_{\rho\nu\tau;\mu} = 0$

7. The inhomogeneous differential equations

This equation is of the form:

geometric concept = physical concept

And in this case we use the stress energy tensor $T_{\mu\nu}$ and we find a geometric object with the same mathematical properties: symmetric and divergenless build out of curvature tensor. The left-hand side is the Einstein tensor:

$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$

The constant of proportionality comes from recovering Newton's gravitational equation in the nonrelativistic limit. In the end one obtains Einstein's equation:

$G_{\mu\nu} = 8\pi G T_{\mu\nu}$

Next time I will go through the same process for the electromagnetic field and map the similarities between the two cases. Please stay tuned.