Covariant classical field theory::worksheet

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This is a worksheet for Covariant classical field theory

1 Notation
2 The action integral
3 Variation of the action integral
4 The Euler-Lagrange equations

[edit] Notation

The notation follows that of introduced in the article on jet bundles. Also, let $\bar{\Gamma}(\pi)$ denote the set of sections of $\pi\,$ with compact support.

[edit] The action integral

A classical field theory is mathematically described by

A fibre bundle $(\mathcal{E},\pi, \mathcal{M})$ , where $\mathcal{M}$ denotes an $n\,$ -dimensional spacetime.
A Lagrangian form $\Lambda:J^{1}\pi \rightarrow \Lambda^{n}M$

Let $\star 1\,$ denote the volume form on $M\,$ , then $\Lambda = L\star 1\,$ where $L:J^{1}\pi \rightarrow \mathbb{R}$ is the Lagrangian function. We choose fibred co-ordinates $\{x^{i},u^{\alpha},u^{\alpha}_{i}\}\,$ on $J^{1}\pi\,$ , such that

$\star 1 = dx^{1} \wedge \ldots \wedge dx^{n}$

The action integral is defined by

$S(\sigma) = \int_{\sigma(\mathcal{M})} (j^{1}\sigma)^{*}\Lambda \,$

where $\sigma \in \bar{\Gamma}(\pi)$ and is defined on an open set $\sigma(\mathcal{M})\,$ , and $j^{1}\sigma\,$ denotes its first jet prolongation.

[edit] Variation of the action integral

The variation of a section $\sigma \in \bar{\Gamma}(\pi)\,$ is provided by a curve $\sigma_{t} = \eta_{t} \circ \sigma\,$ , where $\eta_{t}\,$ is the flow of a $\pi\,$ -vertical vector field $V\,$ on $\mathcal{E}\,$ , which is compactly supported in $\mathcal{M}\,$ . A section $\sigma \in \bar{\Gamma}(\pi)\,$ is then stationary with respect to the variations if

$\left.\frac{d}{dt}\right|_{t=0}\int_{\sigma(\mathcal{M})}(j^{1}\sigma_{t})^{*}\Lambda = 0\,$

This is equivalent to

$\int_{\mathcal{M}} (j^{1}\sigma)^{*}\mathcal{L}_{V^{1}}\Lambda = 0\,$

where $V^{1}\,$ denotes the first prolongation of $V\,$ , by definition of the Lie derivative. Using Cartan's formula, $\mathcal{L}_{X}=i_{X}d + di_{X}\,$ , Stokes' Theorem and the compact support of $\sigma\,$ , we may show that this is equivalent to

$\int_{\mathcal{M}} (j^{1}\sigma)^{*}i_{V^{1}}d\Lambda = 0 \,$

[edit] The Euler-Lagrange equations

Considering a $\pi\,$ -vertical vector field on $\mathcal{E}$

$V = \beta^{\alpha}\frac{\partial}{\partial u^{\alpha}}\,$

where $\beta^{\alpha} = \beta^{\alpha}(x,u)\,$ . Using the contact forms $\theta^{j} = du^{j} - u^{j}_{i}dx^{i}\,$ on $J^{1}\pi\,$ , we may calculate the first prolongation of $V\,$ . We find that

$V^{1} = \beta^{\alpha}\frac{\partial}{\partial u^{\alpha}} + \left(\frac{\partial \beta^{\alpha}}{\partial x^{i}} + \frac{\partial \beta^{\alpha}}{\partial u^{j}}u^{j}_{i}\right)\frac{\partial}{\partial u^{\alpha}_{i}}\,$

where $\gamma^{\alpha}_{i} = \gamma^{\alpha}_{i}(x,u^{\alpha},u^{\alpha}_{i})\,$ . From this, we can show that

$i_{V^{1}}d\Lambda = \left[\beta^{\alpha}\frac{\partial L}{\partial u^{\alpha}} + \left(\frac{\partial \beta^{\alpha}}{\partial x^{i}} + \frac{\partial \beta^{\alpha}}{\partial u^{j}}u^{j}_{i}\right)\frac{\partial L}{\partial u^{\alpha}_{i}}\right]\star 1 \,$

and hence

$(j^{1}\sigma)^{*}i_{V^{1}}d\Lambda = \left[(\beta^{\alpha} \circ \sigma)\frac{\partial L}{\partial u^{\alpha}} \circ j^{1}\sigma + \left(\frac{\partial \beta^{\alpha}}{\partial x^{i}} \circ \sigma + \left(\frac{\partial \beta^{\alpha}}{\partial u^{j}} \circ \sigma \right)\frac{\partial \sigma^{j}}{\partial x^{i}} \right)\frac{\partial L}{\partial u^{\alpha}_{i}} \circ j^{1}\sigma \right]\star 1 \,$

Integrating by parts and taking into account the compact support of $\sigma\,$ , the criticality condition becomes

$\int_{\mathcal{M}} (j^{1}\sigma)^{*}i_{V^{1}}d\Lambda \,$	$= \int_{\mathcal{M}} \left[\frac{\partial L}{\partial u^{\alpha}} \circ j^{1}\sigma - \frac{\partial}{\partial x^{i}} \left(\frac{\partial L}{\partial u^{\alpha}_{i}} \circ j^{1}\sigma \right)\right]( \beta^{\alpha}\circ \sigma )\star 1 \,$
	$= 0 \,$

and since the $\beta^{\alpha}\,$ are arbitrary functions, we obtain

$\frac{\partial L}{\partial u^{\alpha}} \circ j^{1}\sigma - \frac{\partial}{\partial x^{i}} \left(\frac{\partial L}{\partial u^{\alpha}_{i}} \circ j^{1}\sigma \right) = 0\,$