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## Homework Statement

Consider the infinitesimal form of the Lorentz tranformation: ##x^{\mu} \rightarrow x^{\mu}+{\omega^{\mu}}_{\nu}x^{\nu}##.

Show that a scalar field transforms as ##\phi(x) \rightarrow \phi'(x) = \phi(x)-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\phi(x)## and hence show that the variation of the Lagrangian density is a total derivative ##\delta \mathcal{L}=-\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L})##.

Using Noether's theorem deduce the existence of the conserved current ##j^{\mu}=-{\omega^{\rho}}_{\nu}[{T^{\mu}}_{\rho}x^{\nu}]##.

The three conserved charges arising from spatial rotational invariance define the

*total angular momentum*of the field. Show that these charges are given by ##Q_{i}=\epsilon_{ijk}\int d^{3}x (x^{j}T^{0k}-x^{k}T^{0j})##.

Derive the conserved charges arising from invariance under Lorentz boosts. Show that they imply ##\frac{d}{dt}\int d^{3}x (x^{i}T^{00})= \text{constant}## and interpret this equation.

## Homework Equations

## The Attempt at a Solution

Under the infinitesimal form of the Lorentz tranformation given by

##x^{\mu} \rightarrow x^{\mu}+{\omega^{\mu}}_{\nu}x^{\nu}={\delta^{\mu}}_{\nu}x^{\nu}+{\omega^{\mu}}_{\nu}x^{\nu}=({\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu})x^{\nu}={(\Lambda)^{\mu}}_{\nu}x^{\nu}##, so that ##{(\Lambda^{-1})^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}-{\omega^{\mu}}_{\nu},##

a scalar field transforms as

##\phi(x) \rightarrow \phi'(x) = \phi(\Lambda^{-1}x) = \phi({(\Lambda^{-1})^{\mu}}_{\nu}x^{\nu}) = \phi(({\delta^{\mu}}_{\nu}-{\omega^{\mu}}_{\nu})x^{\nu}) = \phi(x^{\mu}-{\omega^{\mu}}_{\nu}x^{\nu}) = \phi(x)-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\phi(x),##

where in the last step, I used the following Taylor expansion of ##f(x+\delta x)## (where ##f## is a scalar and ##x## is a ##4##-vector):

##f(x+\delta x) = f(x) + x^{\nu}\frac{\partial f}{\partial x^{\nu}}+\dots##

and we ignore all the higher order terms since the transformation ##{\omega^{\mu}}_{\nu}## is infinitesimal.

Similarly,

##\mathcal{L}(x) \rightarrow \mathcal{L}'(x) = \mathcal{L}(\Lambda^{-1}x) = \mathcal{L}({(\Lambda^{-1})^{\mu}}_{\nu}x^{\nu}) = \mathcal{L}(({\delta^{\mu}}_{\nu}-{\omega^{\mu}}_{\nu})x^{\nu}) = \mathcal{L}(x^{\mu}-{\omega^{\mu}}_{\nu}x^{\nu}) = \mathcal{L}(x)-{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\mathcal{L}(x)##

so that

##\delta \mathcal{L} = -{\omega^{\mu}}_{\nu}x^{\nu}\partial_{\mu}\mathcal{L}(x)##

##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))+{\omega^{\mu}}_{\nu}\mathcal{L}(x)\partial_{\mu}x^{\nu}##

##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))+{\omega^{\mu}}_{\nu}\mathcal{L}(x)\delta^{\nu}_{\mu}##

##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))+{\omega^{\mu}}_{\mu}\mathcal{L}(x)##

##\implies \delta \mathcal{L} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu}\mathcal{L}(x))##,

where the last line follows from the antisymmetry ##{\omega^{\mu}}_{\nu}=-{\omega^{\nu}}_{\mu}## of ##{\omega^{\mu}}_{\nu}## so that the diagonal elements, and by extension the sum of the diagonal elements ##{\omega^{\mu}}_{\mu}##, is ##0##.

Am I correct so far?