Field theory with differential forms ed

requires

For the unfinished quantum mechanical version see quantum field theory with differential forms.

Table of Contents

Theory ed

We'll be in n dimensions.

Lagrange ed

Let our field \(\omega(x)\) be a k-form and the Lagrangian \(\mathcal L(\omega, d\omega)\) an n-form (volume form). Then define the action as\[ S[\omega] = \int_M \mathcal L \]

Minimizing
Let \(\psi\) be a "test" k-form, vanishing on \(\partial M\)\begin{align}0 &= \partial_\epsilon S[\omega + \epsilon \, \psi]_{|\epsilon=0} \\&= \int_M \partial_\epsilon \mathcal L(\omega + \epsilon \, \psi, d\omega + \epsilon \, d\psi)_{|\epsilon=0} \\&= \int_M \partial_\epsilon \left(\mathcal L(\omega, d\omega) + \epsilon \frac{\partial \mathcal L}{\partial \omega} \wedge \psi + \epsilon \frac{\partial \mathcal L}{\partial d\omega} \wedge d\psi + \epsilon^2 \dots \right)_{\epsilon=0} \\&= \int_M \frac{\partial \mathcal L}{\partial \omega} \wedge \psi + \frac{\partial \mathcal L}{\partial d\omega} \wedge d\psi\end{align}Using the rule \(d\left( \frac{\partial \mathcal L}{\partial d\omega} \wedge \psi \right) = d\frac{\partial \mathcal L}{\partial d\omega} \wedge \psi + (-1)^{\operatorname{deg}(\frac{\partial \mathcal L}{\partial d\omega}) = n-k-1} \frac{\partial \mathcal L}{\partial d\omega} \wedge d\psi\)\[ \quad = \int_M \left(\frac{\partial \mathcal L}{\partial \omega} + (-1)^{n-k} d\frac{\partial \mathcal L}{\partial d\omega} \right) \wedge \psi \pm \underbrace{\int_{M} d \left(\frac{\partial \mathcal L}{\partial d\omega} \wedge \psi \right)}_{=\int_{\partial M} \frac{\partial \mathcal L}{\partial d\omega} \wedge \psi = 0} \]This leads to the Euler-Lagrange-Equations\[ \Rightarrow \quad \begin{array}{|c|} \hline \\ \frac{\partial \mathcal L}{\partial \omega} + (-1)^{n-k} d\frac{\partial \mathcal L}{\partial d\omega} = 0 \\ \\ \hline \end{array} \]

Hamilton ed

To each value of \(\omega\) and \(d\omega\) we can associate a canonical momentum\[ \pi(\omega, d\omega) = \frac{\partial \mathcal L}{\partial d\omega}(\omega, d\omega) \]which means, we have a differential forms function of degree \(n - (k + 1)\). We also need the inverse \(d\omega(\omega, \pi)\).

Define the Hamiltonian as\[ \mathcal H = \pi \wedge d\omega - \mathcal L \]or more explicit:\[ \mathcal H(\omega,\pi) = \pi \wedge d\omega(\omega, \pi) - \mathcal L(\omega, d\omega(\omega, \pi)) \]

Now, let's assume, \(\omega(x)\) solves the Euler-Lagrange equations and \(\pi(x) = \pi(\omega(x), d\omega(x))\) is it's canonical momentum field.

Using the product and chain rules from derivatives of differential form functions we find\begin{align}\frac{\partial \mathcal H}{\partial \omega} &= (-1)^{(k+1) k} 0 + \pi \wedge \frac{\partial d\omega}{\partial \omega} - \frac{\partial \mathcal L}{\partial \omega} - \underbrace{\frac{\partial \mathcal L}{\partial d\omega}}_{=\pi} \wedge \frac{\partial d\omega}{\partial \omega} = - \frac{\partial \mathcal L}{\partial \omega} \\&= (-1)^{n-k} d \frac{\partial \mathcal L}{\partial d\omega} = (-1)^{n-k} d\pi\end{align}

and\[ \frac{\partial \mathcal H}{\partial \pi} = (-1)^{(n-k-1)(k+1)} d\omega + \pi \wedge \frac{\partial d\omega}{\partial \pi} - \underbrace{\frac{\partial \mathcal L}{\partial d\omega}}_{=\pi} \wedge \frac{\partial d\omega}{\partial \pi} = (-1)^{(n-1)(k+1)} d\omega \]

So our fields solve the canonical equations\[ \begin{array}{|c|} \hline \\ d\omega = (-1)^{(n-1)(k+1)} \frac{\partial \mathcal H}{\partial \pi}, \quad d\pi = (-1)^{n-k} \frac{\partial \mathcal H}{\partial \omega} \\ \\ \hline \end{array} \]

(other direction also?)

Poisson brackets ed

Let \(f(\omega, \pi)\) be an \(s\) form und \(g(\omega, \pi)\) a \(t\) form. Then define\[ \begin{array}{|c|} \hline \\ \{f, g\} = \alpha_{s,t} \left( \frac{\partial f}{\partial \omega} \wedge \frac{\partial g}{\partial \pi} + \beta_t \frac{\partial f}{\partial \pi} \wedge \frac{\partial g}{\partial \omega} \right) \\ \\ \hline \end{array} \]with \(\beta_t = (-1)^{n(k+1) + t(n+1)}\) forced by self-consistency and \(\alpha_{s,t} = (-1)^{s+t+nk+1} \) chosen for convenience.

Since \(\omega\) is a \(k\) form and \(\pi\) is a \((n-k-1)\) form, \(\{f,g\}\) is a \((s-k) + (t-(n-k-1)) = s+t -n +1\) form.

Let' play around:\[\begin{array}{rl} \{\omega, \omega\} &= \pm \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \omega}{\partial \pi} \pm \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \omega}{\partial \omega} = \pm 1 \wedge 0 \pm 0 \wedge 1 = 0 \\ \{\pi, \pi\} &= 0\\ \{\omega, \pi\} &= \alpha_{k,n-k-1} \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \pi}{\partial \pi} \pm \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \pi}{\partial \omega} = (-1)^{n(k+1)} 1 \wedge 1 \pm 0 \wedge 0 = \pm 1 \\ \{\omega, \mathcal H\} &= \alpha_{k,n} \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} \pm \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = \alpha_{k,n} (-1)^{(n+1)(k+1)} 1 \wedge d\omega \pm 0 \wedge \dots = \pm d\omega \\ \{\pi, \mathcal H\} &= \pm \frac{\partial \pi}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} + \alpha_{n-k-1,n} \beta_n \frac{\partial \pi}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = \pm 0 \wedge d\omega + \alpha_{n-k-1,n} (-1)^{n(k+1) + n-k} 1 \wedge d\pi = \pm d\pi \end{array} \]

In general:
\[ \begin{array}{|c|} \hline \\ d(f(\omega,\pi)) = \{f, \mathcal H\} + df \\ \\ \hline \end{array} \]
Proof:
\begin{align} \{f, \mathcal H\} &= \alpha_{s,n} \frac{\partial f}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} + \alpha_{s,n} \beta_n \frac{\partial f}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} \\&= \alpha_{s,n} (-1)^{(k+1)(n+1)} \frac{\partial f}{\partial \omega} \wedge d\omega + \alpha_{s,n} (-1)^{n(k+1)} (-1)^{n-k} \frac{\partial f}{\partial \pi} \wedge d\pi \\&= \underbrace{\alpha_{s,n} (-1)^{(k+1)(n+1)+s+k}}_{=1} \underbrace{\left( (-1)^{s-k} \frac{\partial f}{\partial \omega} \wedge d\omega + (-1)^{s-n-k-1} \frac{\partial f}{\partial \pi} \wedge d\pi \right)}_{d(f(\omega,\pi)) - df}\end{align}

\(^_^)/

Properties of the Poisson bracket ed

Theorem
\[ \begin{array}{|c|} \hline \\ \{f \wedge g, h\} = (-1)^{(\deg f + 1) \deg g} f \wedge \{g, h\} + (-1)^{\deg f} g \wedge \{f, h\} \\ \\ \hline \end{array} \]
Proof (\(\deg f = r, \deg g=s, \deg h = t\))
\begin{align}\{f \wedge g, h\} &= \alpha_{r+s,t} \left( \frac{\partial (f \wedge g)}{\partial \omega} \wedge \frac{\partial h}{\partial \pi} + \beta_t \frac{\partial (f \wedge g)}{\partial \pi} \wedge \frac{\partial h}{\partial \omega} \right) \\&= \alpha_{r+s,t} \left( (-1)^{ks} \frac{\partial f}{\partial \omega} \wedge g \wedge \frac{\partial h}{\partial \pi} + f \wedge \frac{\partial g}{\partial \omega} \wedge \frac{\partial h}{\partial \pi} \right. \\&\qquad \qquad + \left. \beta_t \left( (-1)^{(n-k-1)s} \frac{\partial f}{\partial \pi} \wedge g \wedge \frac{\partial h}{\partial \omega} + f \wedge \frac{\partial g}{\partial \pi} \wedge \frac{\partial h}{\partial \omega} \right) \right) \\&= \alpha_{r+s,t} \left( (-1)^{rs} g \wedge \frac{\partial f}{\partial \omega} \wedge \frac{\partial h}{\partial \pi} + \dots + \beta_t (-1)^{rs} g \wedge \frac{\partial f}{\partial \pi} \wedge \frac{\partial h}{\partial \omega} + \dots \right) \\&= \alpha_{r+s,t} \big( \alpha_{r,t} (-1)^{rs} g \wedge \{f,h\} + \alpha_{s,t} f \wedge \{g,h\} \big) \\&= (-1)^{r} g \wedge \{f,h\} + (-1)^{s(r+1)} f \wedge \{g,h\}\end{align}

Examples ed

Klein-Gordon ed

Using a scalar real field \( \phi \).

Lagrangian:
\[ \mathcal{L}(\phi, d\phi) = \frac{1}{2} \star \langle d\phi, d\phi \rangle - \frac{1}{2} m^2 \star \phi^2 \]

momentum
\[ \pi = \frac{\partial \mathcal L}{\partial d\phi} = \star d\phi \]

Hamiltonian
\[ \mathcal H = \pi \wedge d\phi - \mathcal L = \frac{1}{2} \star \langle \pi, \pi \rangle + \frac{1}{2} m^2 \star \phi^2 \]

canonical equations:
\[ d\phi = \frac{\partial \mathcal H}{\partial \pi} = \star \pi, \quad d\pi = - \frac{\partial \mathcal H}{\partial \phi} = - m^2 \star \phi \]\[ \Rightarrow \quad \left( \star d \star d \phi = - m^2 \phi \right) \text{ or } \left( \square + m^2 \phi \right) \phi = 0 \]

Maxwell ed

1-form potential \(A\), 2-form \(F=dA\), 3-form current density \(j\).

Lagrangian
\[ \mathcal L(A, dA) = \frac 12 \star \langle dA, dA \rangle - j \wedge A \]

momentum
\[ \pi = \frac{\partial \mathcal L}{\partial dA} = \star dA = \star F \]

Hamiltonian
\[ \mathcal H = \pi \wedge dA - \mathcal L = \frac 12 \star \langle \pi, \pi \rangle + j \wedge A \]

canonical equations:
\[ dA = \frac{\partial \mathcal H}{\partial \pi} = \star \pi, \quad d\pi = + \frac{\partial \mathcal H}{\partial A} = j \]\[ \Rightarrow \quad d\star F = j \]

Categories: Physik, Gedanken zur Physik