Field theory with differential forms ed

requires

• Lagrangian mechanics
• Hamiltonian mechanics
• Lagrangian field theory
• covariant Hamiltonian field theory (not really, but helpful)
• derivatives of differential form functions

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

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

\[ 0 = \partial_\epsilon S[\omega + \epsilon \, \psi]_{|\epsilon=0} \]

\[ \quad = \int_M \partial_\epsilon \mathcal L(\omega + \epsilon \, \psi, d\omega + \epsilon \, d\psi)_{|\epsilon=0} \]

\[ \quad = \int_M \partial_\epsilon \left(\mathcal L(\omega, d\omega) + \epsilon \frac{\partial \mathcal L}{\partial \omega} \wedge \psi + \epsilon \frac{\partial \mathcal L}{\partial d\omega} \wedge d\psi + \epsilon^2 \dots \right)_{\epsilon=0} \]

\[ \quad = \int_M \frac{\partial \mathcal L}{\partial \omega} \wedge \psi + \frac{\partial \mathcal L}{\partial d\omega} \wedge d\psi \]

\[ \quad = \int_M \left(\frac{\partial \mathcal L}{\partial \omega} + (-1)^{n-k-1} d\frac{\partial \mathcal L}{\partial d\omega} \right) \wedge \psi \pm \underbrace{\int_{M} d \left(\frac{\partial \mathcal L}{\partial d\omega} \wedge \psi \right)}_{=\int_{\partial M} \frac{\partial \mathcal L}{\partial d\omega} \wedge \psi = 0} \]

\[ \Rightarrow \quad \begin{array}{|c|} \hline \\ \frac{\partial \mathcal L}{\partial \omega} + (-1)^{n-k-1} d\frac{\partial \mathcal L}{\partial d\omega} = 0 \\ \\ \hline \end{array} \]

Hamilton ed

To each value of \[\omega\] and \[d\omega\] we can associate a canonical momentum

\[ \pi(\omega, d\omega) = \frac{\partial \mathcal L}{\partial d\omega}(\omega, d\omega) \]

Define the Hamiltonian as

\[ \mathcal H = \pi \wedge d\omega - \mathcal L \]

\[ \mathcal H(\omega,\pi) = \pi \wedge d\omega(\omega, \pi) - \mathcal L(\omega, d\omega(\omega, \pi)) \]

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

\[ \frac{\partial \mathcal H}{\partial \omega} = 0 + \underbrace{(-1)^{(k+1) k}}_{=+1} \pi \wedge \frac{\partial d\omega}{\partial \omega} - \frac{\partial \mathcal L}{\partial \omega} - \underbrace{\frac{\partial \mathcal L}{\partial d\omega}}_{=\pi} \wedge \frac{\partial d\omega}{\partial \omega} = - \frac{\partial \mathcal L}{\partial \omega} \]

\[ \quad = - (-1)^{n-k} d \frac{\partial \mathcal L}{\partial d\omega} = - (-1)^{n-k} d\pi \]

and

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

So our fields solve the canonical equations

\[ \begin{array}{|c|} \hline \\ d\omega = \frac{\partial \mathcal H}{\partial \pi}, \quad d\pi = (-1)^{n-k-1} \frac{\partial \mathcal H}{\partial \omega} \\ \\ \hline \end{array} \]

(other direction also?)

Poisson brackets ed

Let \[f(\omega, \pi)\] be an s form und \[g(\omega, \pi)\] a t form. Then define

\[ \{f, g\} = \frac{\partial f}{\partial \omega} \wedge \frac{\partial g}{\partial \pi} + (-1)^{n-k-1} \frac{\partial f}{\partial \pi} \wedge \frac{\partial g}{\partial \omega} \]

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:

\[ \{\omega, \omega\} = \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \omega}{\partial \pi} \pm \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \omega}{\partial \omega} = 1 \wedge 0 \pm 0 \wedge 1 = 0 = \{\pi, \pi\}\]

\[ \{\omega, \pi\} = \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \pi}{\partial \pi} \pm \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \pi}{\partial \omega} = 1 \wedge 1 \pm 0 \wedge 0 = 1 \] (0 form)

\[ \{\omega, \mathcal H\} = \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} \pm \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = 1 \wedge d\omega \pm 0 \wedge \dots = d\omega \]

\[ \{\pi, \mathcal H\} = \frac{\partial \pi}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} + (-1)^{n-k-1} \frac{\partial \pi}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = 0 \wedge d\omega + (-1)^{2(n-k-1)} 1 \wedge d\pi = d\pi \]

\[ \{f, \mathcal H\} = \frac{\partial f}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} + (-1)^{n-k-1} \frac{\partial f}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = \frac{\partial f}{\partial \omega} \wedge d\omega + \mu (-1)^{2(n-k-1)} \frac{\partial f}{\partial \pi} \wedge d\pi = df \]

\(^_^)/

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:

\[ d\phi = \frac{\partial \mathcal H}{\partial \pi} = \star \pi, \quad d\pi = - \frac{\partial \mathcal H}{\partial \phi} = - m^2 \star \phi \]

\[ \Rightarrow \quad \left( \star d \star d \phi = - m^2 \phi \right) \] 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:

\[ dA = \frac{\partial \mathcal H}{\partial \pi} = \star \pi, \quad d\pi = + \frac{\partial \mathcal H}{\partial A} = j \]

\[ \Rightarrow \quad d\star F = j \]