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 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.
\(D_\mu = \partial_\mu - i A_\mu\)
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 \(F_{\mu\nu} = 0\)
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