Field theory with differential forms ed

requires

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

Hamilton ed

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

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

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

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

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

(other direction also?)

Beispiel: Klein-Gordon ed

(reelles Feld...)

Differentialformen ed

Lagrange-Dichte:
\[ \mathcal{L}(\phi, d\phi) = \frac{1}{2} d\phi \wedge \star d\phi - \frac{1}{2} m^2 \star \phi^2 \]

Minimiere Wirkung (modulo Vorzeichen):

\[ 0 = \partial_\epsilon S[\phi + \epsilon \Psi] = \int_M d\Psi \wedge \star d\phi + m^2 \star \phi \Psi = \int_M \Psi \cdot d \star d\phi + m^2 \star \phi \Psi - \int_{\partial M} \dots \]
\[ \Rightarrow \Delta \psi = m^2 \psi \]

Hamilton
\[ \pi = \frac{\partial \mathcal L}{\partial d\phi} = \star d\phi \]
\[ \mathcal H = \pi \wedge d\phi - \mathcal L = \frac{1}{2} \pi \wedge \star \pi + \frac{1}{2} m^2 \star \phi^2 \]

Kanonische Gleichungen:

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

Beispiel: Maxwell ed

Tensoren ed

Die Lagrange-Dichte des elektromagnetischen Feldes ist

\( \mathcal L ( A^\mu , \partial_\mu A^\nu ) = - \frac 14 F^{\mu \nu} F_{\mu \nu} - j_\mu A^\mu \)

mit \( F^{\mu \nu} := \partial^\mu A^\nu - \partial^\nu A^\mu \) und dem Potential \( A^\mu \) . Daraus folgt

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

Problem: \( \partial_\mu A^\nu \) lässt sich nicht durch \( \pi^{\mu \nu} \) ausdrücken. m(-_-)m

Differentialformen ed

1-Form \[A\], \[F=dA\], 3-Form \[j\]

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

Minimiere Wirkung:

\[ \partial_\epsilon S[A + \epsilon \Psi] = \int_M d\Psi \wedge \star dA - j \wedge \Psi = \int_M \Psi \wedge d \star dA - j \wedge \Psi - \int_{\partial M} \dots \]
\[ \Rightarrow d\star F = j \]

Hamilton
\[ \pi = \frac{\partial \mathcal L}{\partial dA} = \star dA \]
\[ \mathcal H = \pi \wedge dA - \mathcal L = \frac 12 \pi \wedge \star \pi + j \wedge A \]

Kanonische Gleichungen:

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

Poisson-Klammern ed

Seien \[f(\omega, \pi)\] eine s-Form und \[g(\omega, \pi)\] eine t-Form. Dann definiere

\[ \{f, g\} = \frac{\partial f}{\partial \omega} \wedge \frac{\partial g}{\partial \pi} - \frac{\partial f}{\partial \pi} \wedge \frac{\partial g}{\partial \omega} \]

Mit \[\omega\] einer k-Form und \[\pi\] einer \[(n-k-1)\]-Form ist \[\{f,g\}\] eine \[(s-k) + (t-(n-k-1)) = s+t -n +1\] Form.

Es gilt:

\[ \{\omega, \mathcal H\} = \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} - \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = 1 \wedge d\omega - 0 \wedge \dots = d\omega \]
\[ \{\pi, \mathcal H\} = \frac{\partial \pi}{\partial \omega} \wedge \frac{\partial \mathcal H}{\partial \pi} - \frac{\partial \pi}{\partial \pi} \wedge \frac{\partial \mathcal H}{\partial \omega} = 0 \wedge \dots - 1 \wedge (- d\pi) = d\pi \]
\[ \{\omega, \omega\} = \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \omega}{\partial \pi} - \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \omega}{\partial \omega} = 1 \wedge 0 - 0 \wedge 1 = 0 = \{\pi, \pi\} \]
\[ \{\omega, \pi\} = \frac{\partial \omega}{\partial \omega} \wedge \frac{\partial \pi}{\partial \pi} - \frac{\partial \omega}{\partial \pi} \wedge \frac{\partial \pi}{\partial \omega} = 1 \wedge 1 - 0 \wedge 0 = 1 \] (0-Form)

\(^_^)/