In the first part, we reviewed a brief history of the mathematics of the Quantum Field Theory (not exhaustively, but enough to get the idea). In this part, we will review how to derive Maxwell equations in spacetime following G.B. Folland’s book (Folland, 2008).

## Exterior derivative of a 4-vector

Consider the four-vector

$A^{\mu}(x) = A^{\mu}(t,\boldsymbol{x}) = (\phi, \boldsymbol{A}), \quad \text{or} \quad A_{\mu} = (\phi, -\boldsymbol{A}),$

where $\phi$ is the electric potential that generates the electric fields, and $\boldsymbol{A}$ is the magnetic potential, whose curl equals the magnetic fields.

The electric and magnetic fields $\boldsymbol{E}$ and $\boldsymbol{B}$ are incorporated into one electromagnetic field tensor of second rank (anti-symmetric):

$F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}, \quad \text{or} \quad F^{\mu\nu} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}.$

This can be viewed as the 4 dimensional version of curl operator $\nabla\times: \boldsymbol{C}^{\infty}(\mathbb{R}^3) \to \boldsymbol{C}^{\infty}(\mathbb{R}^3)$, whereas the cross product of the gradient operator in $\mathbb{R}^3$ is replaced by the gradient wedge product with a 4-vector (exterior derivative of a $1$-form):

$F := \operatorname{curl}_4 A = \nabla \wedge A = \nabla_{\mu} \wedge A_{\mu}.$

Let $\hat{\boldsymbol{e}}_{\mu}$ be the unit vector for the 4-dimensional space, we can write

$\nabla_{\mu} (\cdot) := \partial_{\mu} (\cdot)\hat{\boldsymbol{e}}_{\mu} = \sum_{\mu=0}^3 \partial_{\mu} (\cdot)\hat{\boldsymbol{e}}_{\mu}.$

Taking the wedge product of this operator with $A$ we can write:

\begin{aligned} \nabla_{\mu}\wedge A_{\mu} &= (\partial_t,\nabla) \wedge (\phi,-\boldsymbol{A}) \\ & = \sum_{\nu = 0}^3 \nabla_{\mu} A_{\nu} \wedge \hat{\boldsymbol{e}}_{\nu} \\ & = \sum_{\nu = 0}^3 \sum_{\mu = 0}^3 (\partial_{\mu} A_{\nu})\,\hat{\boldsymbol{e}}_{\mu} \wedge \hat{\boldsymbol{e}}_{\nu} \\ & = \Big(\partial_t \phi\, \hat{\boldsymbol{e}}_{0} + \partial_{1} \phi\, \hat{\boldsymbol{e}}_{1}+ \partial_{2} \phi \,\hat{\boldsymbol{e}}_{2} + \partial_{3} \phi \,\hat{\boldsymbol{e}}_{3}\Big) \wedge \hat{\boldsymbol{e}}_{0} \\ &\quad + \sum_{k=1}^{3}\Big(\partial_t A_k\, \hat{\boldsymbol{e}}_{0} + \partial_{1} A_k \, \hat{\boldsymbol{e}}_{1}+ \partial_{2}A_k \,\hat{\boldsymbol{e}}_{2} + \partial_{3} A_k \,\hat{\boldsymbol{e}}_{3}\Big) \wedge \hat{\boldsymbol{e}}_{1} \\ & = \sum_{\mu<\nu} \left( \frac{\partial A_{\nu}}{\partial x^{\mu}} - \frac{\partial A_{\mu}}{\partial x^{\nu}} \right) \hat{\boldsymbol{e}}_{\mu} \wedge \hat{\boldsymbol{e}}_{\nu}. \end{aligned}

Switch the usual Cartesian unit vector $\hat{\boldsymbol{e}}_{\mu}$ with $d x^{\mu}$ we replicate taking the exterior derivative of a 1-form. Written in the matrix form, the field tensor is:

$F = \begin{pmatrix} 0 & \partial_{0} A_1 - \partial_{1} A_0 & \partial_{0} A_2 - \partial_{2} A_0 & \partial_{0} A_3 - \partial_{3} A_0 \\ & 0 & \partial_{1} A_2 - \partial_{2} A_1 & \partial_{1} A_3 - \partial_{3} A_1 \\ & & 0 & \partial_{2} A_3 - \partial_{3} A_2 \\ &\ast & & 0 \end{pmatrix}.$

Using the definition of the classical potential:

$\boldsymbol{B} = \nabla\times \boldsymbol{A}, \quad \text{and}\quad \boldsymbol{E} = -\nabla \phi - \partial_t \boldsymbol{A}.\tag{1}\label{eq:maxwell-1}$

The field tensor represents the electric field $\boldsymbol{E} = (E^1,E^2,E^3)$ and magnetic field $\boldsymbol{B} = (B^1,B^2,B^3)$ at the same time:

$F = F_{\mu\nu} = \begin{pmatrix} 0 & E^1 & E^2 & E^3 \\ -E^1& 0 & -B^1 & B^2 \\ -E^2& B^1& 0 & -B^3 \\ -E^3& -B^2& B^3 & 0 \end{pmatrix}.$

The upstair electromagnetic tensor is obtained by

$F^{\mu\nu} = g^{\mu\alpha} g^{\nu\beta} F_{\mu\nu} =\begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ E^1& 0 & -B^1 & B^2 \\ E^2& B^1& 0 & -B^3 \\ E^3& -B^2& B^3 & 0 \end{pmatrix}.$

## Lagrangian formalism

For an electromagnetic field, the Lagrangian is defined as

\begin{aligned} \mathcal{L} &= -\frac{1}{4} F^2 = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} \\ &= \frac{1}{2} (\boldsymbol{E}\cdot \boldsymbol{E} - \boldsymbol{B}\cdot \boldsymbol{B}) = \frac{1}{2} |\nabla \phi + \partial_t \boldsymbol{A}|^2 - \frac{1}{2} |\nabla\times \boldsymbol{A}|^2. \end{aligned}

This is the Lagrangian only for the field. To develop a Lagrangian for a field with the presence of charges and currents, we need to add the interaction term with a 4-current $J^{\mu} = (\rho, \boldsymbol{J})$, where the $\rho$ stands for the charge, with $\boldsymbol{J}$ being the current density:

\begin{aligned} \mathcal{L} &= -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} - A_{\mu}J^{\mu} \\ &= \frac{1}{2} |\nabla \phi + \partial_t \boldsymbol{A}|^2 - \frac{1}{2} |\nabla\times \boldsymbol{A}|^2 -\rho\phi + \boldsymbol{A}\cdot \boldsymbol{J}. \end{aligned}\tag{2}\label{eq:maxwell-lg}

Using the formula from the previous post:

$\frac{\partial \mathcal{L}}{\partial\phi} - \sum_{\mu=0}^{3}\partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) = 0,$

on $\mathcal{L} = \mathcal{L}(A_{\mu})$ in \eqref{eq:maxwell-lg} to work out explicitly what the Euler-Lagrange equation yields:

$\frac{\partial \mathcal{L}}{\partial A_{\mu}} - \sum_{\nu=0}^{3}\partial_{\nu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu} )}\right) = 0.$

For the first term, in the Lagrangian the field tensor part does not depend on $A_{\mu}$ but rather partial derivatives of $A_{\mu}$, hence we have:

$\frac{\partial \mathcal{L}}{\partial A_{\mu}} = -J^{\mu}= -(\rho,\boldsymbol{J}).$

For the second term we have (here two more indices are added for the partial derivative of $F_{\mu\nu}$ is a rank 4 tensor):

\begin{aligned} \frac{\partial \mathcal{L}}{\partial (\partial_{\sigma}A_{\lambda})} &= -\frac{1}{4} \frac{\partial }{\partial (\partial_{\sigma}A_{\lambda} )} \bigl[ ( \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}) (\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}) \bigr] \\ &= -\frac{1}{2} \frac{\partial }{\partial (\partial_{\sigma}A_{\lambda} )} \bigl[ (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) - (\partial_{\nu} A_{\mu})(\partial^{\mu} A^{\nu}) \bigr] . \end{aligned}

The partial derivatives for the field tensor can be computed by

$\frac{\partial F_{\mu\nu}}{\partial (\partial_{\sigma}A_{\lambda})} = \frac{\partial (\partial_{\mu} A_{\nu})}{\partial (\partial_{\sigma}A_{\lambda} )} - \frac{\partial (\partial_{\nu} A_{\mu})}{\partial (\partial_{\sigma}A_{\lambda} )} = \delta^{\sigma}_{\mu} \delta^{\lambda}_{\nu} - \delta^{\sigma}_{\nu} \delta^{\lambda}_{\mu}.$

Notice this is a rank 4 tensor, wherein $\delta_{\mu}^{\sigma}$ is a rank 2 tensor. If the Kronecker delta is written as $\delta^{\sigma\nu}$ or $\delta_{\sigma\nu}$, it is a scalar function being either 1 or 0. Plugging into above equation while the repeated indices get summed up, and use the metric tensor $g^{\alpha\beta}$ to raise the index

\begin{aligned} \frac{\partial }{\partial (\partial_{\sigma} A_{\lambda})}\Bigl[ (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \Bigr] &= (\partial^{\mu} A^{\nu}) \frac{\partial (\partial_{\mu} A_{\nu})}{\partial (\partial_{\sigma} A_{\lambda})} + (\partial_{\mu} A_{\nu}) \frac{\partial (\partial^{\mu} A^{\nu})}{\partial (\partial_{\sigma} A_{\lambda})} \\ &= (\partial^{\mu} A^{\nu}) \delta^{\sigma}_{\mu} \delta^{\lambda}_{\nu} + (g^{\mu\alpha} g^{\nu\beta} \partial_{\mu}A_{\nu}) \frac{\partial (\partial_{\alpha} A_{\beta})}{\partial (\partial_{\sigma} A_{\lambda})} \\ &= 2\partial^{\sigma} A^{\lambda}. \end{aligned}

Similarly, it can be computed:

$\frac{\partial }{\partial (\partial_{\sigma} A_{\lambda})}\Bigl[ (\partial_{\nu} A_{\mu})(\partial^{\mu} A^{\nu})\Bigr] = 2\partial^{\lambda} A^{\sigma}.$

Eventually we have:

$\frac{\partial \mathcal{L}}{\partial (\partial_{\nu}A_{\mu})} = - (\partial^{\nu} A^{\mu}-\partial^{\mu} A^{\nu} ) = -F^{\nu\mu}.$

Therefore, the Euler-Lagrange equation based on the Lagrangian \eqref{eq:maxwell-lg} is

$-J^{\mu} + \sum_{\nu=0}^{3}\partial_{\nu} F^{\nu\mu} = 0.$

When $\mu = 0$, $\partial_{\nu} F^{\nu 0} = J^0$ is equivalent to

$\sum\limits_{\nu=1}^{3}\partial_{\nu} F^{\nu 0} = J^0$

which is

$\nabla \cdot \boldsymbol{E} = \rho. \tag{3}\label{eq:maxwell-gauss}$

Letting $\mu = 1,2,3$, we have

\begin{aligned} \partial_{0} F^{0 1} + \partial_{2} F^{2 1} + \partial_{3} F^{3 1}= J^1 \implies -\partial_t E_1 + \partial_2 B_3 -\partial_3 B_2 = J^1, \\ \partial_{0} F^{0 2} + \partial_{1} F^{1 2} + \partial_{3} F^{3 2}= J^2 \implies -\partial_t E_2 - \partial_1 B_3 + \partial_3 B_1 = J^2, \\ \text{ and }\; \partial_{0} F^{0 3} + \partial_{1} F^{1 3} + \partial_{2} F^{2 3}= J^3 \implies -\partial_t E_3 + \partial_1 B_2 - \partial_2 B_1 = J^3. \end{aligned}

Above three equations are combined into the following

$\nabla\times \boldsymbol{B} = \partial_t \boldsymbol{E} + \boldsymbol{J}. \tag{4}\label{eq:maxwell-amp}$

For people with the training of calculus of variations, the index notation is not intuitive enough to see the process. We can use the vector notation in \eqref{eq:maxwell-lg} to compute the first variation of $A_{\mu} = (\phi,-\boldsymbol{A})$ directly, thus to obtain the Euler-Lagrange equation. The action is

$S\big[(\phi,\boldsymbol{A})\big]:= \int^{\tau}_0\int_{\Omega} \mathcal{L} \,d\boldsymbol{x} dt.$

Then the Euler-Lagrange equation can be obtained by setting the Gâteaux derivative to be 0:

$\lim_{\epsilon\to 0}\frac{d}{d\epsilon} S\big[(\phi,\boldsymbol{A}) + \epsilon(\psi,\boldsymbol{v})\big] = 0, \quad\text{for any test function pair } (\psi,\boldsymbol{v}), \tag{5}\label{eq:maxwell-gat}$

where the test function pair $(\psi,\boldsymbol{v})$ does not change the boundary/initial value of $(\phi,\boldsymbol{A})$ in both space and time.

By the arbitrariness of the test pair $(\psi,\boldsymbol{v})$, letting $\psi = 0$ firstly, and plugging the expression of Lagrangian into \eqref{eq:maxwell-gat} we have:

$0 = \int^{t}_0\int_{\Omega} \Big( (\partial_t \boldsymbol{A} + \nabla\phi) \cdot\partial_t \boldsymbol{v} -\nabla \times \boldsymbol{A} \cdot \nabla \times \boldsymbol{v} +\boldsymbol{J} \cdot \boldsymbol{v} \Big) \,d\boldsymbol{x} dt.$

Plugging the electric and magnetic potentials \eqref{eq:maxwell-1} in above equation leads to:

$0 = \int^{\tau}_0\int_{\Omega} \Big( -\boldsymbol{E}\cdot\partial_t \boldsymbol{v} -\boldsymbol{B} \cdot \nabla\times \boldsymbol{v} +\boldsymbol{J} \cdot \boldsymbol{v} \Big) \,d\boldsymbol{x} dt.$

Integrating by parts in both the spacial and temporal variables, together with the fact that the test function should not change the boundary value of the electric and magnetic fields, we have for any $\boldsymbol{v}$:

$0 = \int^{\tau}_0\int_{\Omega} \Big( \partial_t \boldsymbol{E}\cdot\boldsymbol{v} -\nabla\times \boldsymbol{B} \cdot \boldsymbol{v} + \boldsymbol{J}\cdot \boldsymbol{v} \Big) \,d\boldsymbol{x} dt.$

This replicates the derivation of Amp`{e}re’s law \eqref{eq:maxwell-amp}:

$\nabla\times \boldsymbol{B} = \partial_t \boldsymbol{E} + \boldsymbol{J}.$

Letting $\boldsymbol{v} =\boldsymbol{0}$ in the test pair, integrating by parts one more time and using the fact $\psi$ does not change the boundary value of the electric potential, we reach the following

\begin{aligned} 0 &= \int^{\tau}_0\int_{\Omega} \Big((\partial_t \boldsymbol{A} + \nabla\phi) \cdot \nabla \psi -\rho \psi\Big) \,d\boldsymbol{x} dt \\ &= \int^{\tau}_0\int_{\Omega} \Big(-\boldsymbol{E} \cdot \nabla \psi -\rho \psi\Big) \,d\boldsymbol{x} dt \\ &= \int^{\tau}_0\int_{\Omega} \Big(\nabla \cdot \boldsymbol{E} \, \psi -\rho \psi\Big) \,d\boldsymbol{x} dt. \end{aligned}

The replicates the derivation of Gauss’s law \eqref{eq:maxwell-gauss} for electric field:

$\nabla \cdot \boldsymbol{E} = \rho.$

## References

1. Folland, G. B. (2008). Quantum Field Theory: A tourist guide for mathematicians (Number 149). American Mathematical Soc.

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