As I don't have access to the solutions manual of this book I would by grateful for any comments regarding my solution to this exercise. Is there a mistake somewhere? Or can I improve / shorten it in any way?
Thanks in advance.
0. Preliminary definitions and notations (pp.24-25 in the book)
- Integrals are understood as reaching over all of $\mathbb R^3$, i.e $$\int = \int_{\mathbb R^3}.$$
- The Hamiltonian on the electric and magnetic fields $\mathbf E$ and $\mathbf B$ (both smooth and $\mathbb R^3 \to \mathbb R^3$), respectively is given as the functional $$H(\mathbf E, \mathbf B) = \frac 1 2 \int \|\mathbf E\|_2^2 + \|\mathbf B\|_2^2 \mathrm d^3 x.$$
- For a functional $F$ within this context the functional derivative $\frac {\delta F} {\delta \mathbf E}$ is given by$$\int \frac {\delta F} {\delta \mathbf E} \cdot \delta \mathbf E \;\mathrm d^3 x = \mathop{\mathrm{lim}}_{\varepsilon \to 0} \frac 1 \varepsilon (F(\mathbf E + \varepsilon \delta \mathbf E, \mathbf B)- F(\mathbf E, \mathbf B)).$$ Analogously for $\frac {\delta F}{\delta \mathbf B}$.
- The Maxwell-Poisson bracket is $$\{F, K\} = \int \frac {\delta F} {\delta \mathbf E} \cdot \mathrm {curl}\left(\frac {\delta K}{\delta \mathbf B}\right) - \frac {\delta K} {\delta \mathbf E} \cdot \mathrm {curl}\left(\frac {\delta F}{\delta \mathbf B}\right) \mathrm d^3 x,$$where $F, K$ are also functionals.
- The speed of light has unit $c = 1$.
1. Problem statement (p.29)
Problem 1.6-1. Verify that one obtains the Maxwell equations from the Maxwell-Poisson bracket.
2. My solution attempt
First I evaluate the functional derivatives of the Hamiltonian: $$\begin{align} \int \frac {\delta H} {\delta \mathbf E} \cdot \delta \mathbf E \;\mathrm d^3 x &=\mathop{\mathrm{lim}}_{\varepsilon \to 0} \frac 1 \varepsilon (H(\mathbf E + \varepsilon \delta \mathbf E, \mathbf B)- H(\mathbf E, \mathbf B))\\ &=\mathop{\mathrm{lim}}_{\varepsilon \to 0} \int \frac 1 {2\varepsilon} (\|\mathbf E + \varepsilon \delta E\|^2 - \|\mathbf E\|^2) \mathrm d^3 x\\ &=\mathop{\mathrm{lim}}_{\varepsilon \to 0} \int \mathbf E \cdot \delta \mathbf E + \frac \varepsilon 2 \|\delta \mathbf E\|^2 \mathrm d^3 x\\ &=\int \mathbf E \cdot \delta \mathbf E \; \mathrm d^3 x +\mathop{\mathrm{lim}}_{\varepsilon \to 0} \frac \varepsilon 2 \int \|\delta \mathbf E\|^2 \mathrm d^3 x\\ &=\int \mathbf E \cdot \delta \mathbf E \; \mathrm d^3 x, \end{align}$$ where I have tacitly assumed that $\mathbf E$ and $\delta \mathbf E$ are taken from appropriate function spaces such that this epression is valid (the technical details are not relevant here). Testing this expression with all possible $\delta \mathbf E$ I get $\frac {\delta H} {\delta \mathbf E} = \mathbf E$ and by the same procedure $\frac {\delta H} {\delta \mathbf B} = \mathbf B$.
To proceed I bring the bracket into a different form by making use of the identity $$ \mathrm{div}(\mathbf A \times \mathbf B) = \mathrm{curl}(\mathbf A) \cdot \mathbf B - \mathbf A \cdot \mathrm{curl}(\mathbf B).$$
Proof: With Einstein-$\Sigma$ and $\frac {\partial A} {\partial_{x^i}} =: A_{,i}$.
$\mathrm{div}(\mathbf A \times \mathbf B) = (\epsilon_{ijk} A_j B_k)_{,i} = \epsilon_{ijk}A_{j,i} B_k + \epsilon_{ijk} A_j B_{k,i}$.
Applied to the second summand in the bracket definition I get $$\int \frac {\delta K} {\delta \mathbf E} \cdot \mathrm {curl}\left(\frac {\delta F}{\delta \mathbf B}\right) \mathrm d^3x = \int \mathrm {curl}\left(\frac {\delta K} {\delta \mathbf E}\right) \cdot \frac {\delta F}{\delta \mathbf B} - \mathrm {div}\left(\frac {\delta K} {\delta \mathbf E} \times \frac {\delta F}{\delta \mathbf B}\right) \mathrm d^3 x = \int \mathrm {curl}\left(\frac {\delta K} {\delta \mathbf E}\right) \cdot \frac {\delta F}{\delta \mathbf B} \mathrm d^3 x,$$ where in the last equality again some additional assumption on the vanishing of the involved vector fields is made (they must vanish as $\|x\| \to \infty$, and then representing the integral of 'div' over the space as the limit of an integral over balls of increasing radii and using Gauss's theorem it follows...).
Using this result and the functional derivates I obtain the following bracket expression $$ \{F, H\} = \int \frac {\delta F} {\delta \mathbf E} \cdot \mathrm {curl}(\mathbf B) - \mathrm {curl} (\mathbf E) \cdot \frac {\delta F}{\delta \mathbf B} \mathrm d^3 x. $$ In order to obtain the i-th component of $\mathbf E, \mathbf B$, $\dot{\mathbf E}$ and $\dot{\mathbf B}$ at $x$, I introduce the evaluation functionals $$E^i(x)(\mathbf E, \mathbf B) := \int \delta(y - x) e^i \cdot \mathbf E \; \mathrm d^3 y,\\ B^i(x)(\mathbf E, \mathbf B) := \int \delta(y - x) e^i \cdot \mathbf B \; \mathrm d^3 y,$$ where $e^i$ denotes the $i$-basis vector in the standard basis of $\mathbb R^3$. Their time derivative counterparts are obviously $$\dot{E}^i(x)(\mathbf E, \mathbf B) := \int \delta(y - x) e^i \cdot \dot {\mathbf E} \; \mathrm d^3 y,\\ \dot{B}^i(x)(\mathbf E, \mathbf B) := \int \delta(y - x) e^i \cdot \dot {\mathbf B} \; \mathrm d^3 y.$$ In order to not confuse the evaluation functionals from the actual field values I'm going to denote the actual field values by bold symbols, i.e. $E^i(x)(\mathbf E, \mathbf B) = \mathbf E^i(x)$, and analogously for all other fields.
The Maxwell equations can now be retrieved by evaluating $$\dot{E}^i(x)(\mathbf E, \mathbf B) = \{E^i(x), H\}(\mathbf E, \mathbf B),\\ \dot{B}^i(x)(\mathbf E, \mathbf B) = \{B^i(x), H\}(\mathbf E, \mathbf B), $$ for all $x$ and $i$.
In order to do so the functional derivatives $\frac{\partial E^i(x)}{\partial \mathbf E}$, $\frac{\partial E^i(x)}{\partial \mathbf B}$, $\frac{\partial B^i(x)}{\partial \mathbf E}$ and $\frac{\partial B^i(x)}{\partial \mathbf B}$ need to be computed:
$$ \begin{align} \int \frac {\delta E^i(x)} {\delta \mathbf E} \cdot \delta \mathbf E \;\mathrm d^3 y &=\mathop{\mathrm{lim}}_{\varepsilon \to 0} \frac 1 {\varepsilon} (E^i(x)(\mathbf E + \varepsilon \delta \mathbf E, \mathbf B)- E^i(x)(\mathbf E, \mathbf B))\\ &=\mathop{\mathrm{lim}}_{\varepsilon \to 0} \frac 1 {\varepsilon} \int \delta(y - x) e^i \cdot (\mathbf E + \varepsilon \delta \mathbf E) - \delta(y - x) e^i \cdot \mathbf E \; \mathrm d^3 y\\ &= \int \delta(y - x) e^i \cdot \delta \mathbf E \; \mathrm d^3 y, \end{align}$$ and hence $\frac {\delta E^i(x)} {\delta \mathbf E} = \delta(y - x) e^i$. Repeating the same procedure I also obtain $\frac {\delta E^i(x)} {\delta \mathbf B} = 0$, $\frac {\delta B^i(x)} {\delta \mathbf E} = 0$ and $\frac {\delta B^i(x)} {\delta \mathbf B} = \delta(y - x) e^i$.
Now I can evaluate the brackets: $$ \begin{align} \{E^i(x), H\}(\mathbf E, \mathbf B) &= \int \frac {\delta E^i(x)} {\delta \mathbf E} \cdot \mathrm {curl}(\mathbf B) - \mathrm {curl} (\mathbf E) \cdot \frac {\delta E^i(x)}{\delta \mathbf B} \mathrm d^3 y\\ &= \int \delta(y - x) e^i \cdot \mathrm {curl}(\mathbf B)\mathrm d^3 y\\ &= \mathrm {curl}(\mathbf B)^i(x), \end{align} $$ and $$ \begin{align} \{B^i(x), H\}(\mathbf E, \mathbf B) &= \int \frac {\delta B^i(x)} {\delta \mathbf E} \cdot \mathrm {curl}(\mathbf B) - \mathrm {curl} (\mathbf E) \cdot \frac {\delta B^i(x)}{\delta \mathbf B} \mathrm d^3 y\\ &= -\int \mathrm {curl} (\mathbf E) \cdot (\delta(y - x) e^i) \mathrm d^3 y\\ &= -\mathrm {curl} (\mathbf E)^i(x). \end{align} $$ Since $\dot{E}^i(x)(\mathbf E, \mathbf B) = \dot{\mathbf E}^i(x), \; \dot{B}^i(x)(\mathbf E, \mathbf B) = \dot{\mathbf B}^i(x)$ the poisson equations become $$ \begin{align} \dot{\mathbf E} &= \mathrm{curl} (\mathbf B)\\ \dot{\mathbf B} &= -\mathrm{curl} (\mathbf E). \end{align} $$ $\square$