- #1

Baela

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- TL;DR Summary
- Problem in attempt to express the variation of a vector ##U^\mu(x)## under infinitesimal coordinate transformation, as a Lie derivative ##\mathcal{L}_\xi U^\mu##

I need to prove that under an infinitesimal coordinate transformation ##x^{'\mu}=x^\mu-\xi^\mu(x)##, the variation of a vector ##U^\mu(x)## is $$\delta U^\mu(x)=U^{'\mu}(x)-U^\mu(x)=\mathcal{L}_\xi U^\mu$$ where ##\mathcal{L}_\xi U^\mu## is the Lie derivative of ##U^\mu## wrt the vector ##\xi^\nu##.

I have performed the following steps and have a question in the final result.

By general coordinate transformation rule we know that

\begin{align}

& U^{'\mu}(x')=\frac{\partial x^{'\mu}}{\partial x^\nu} U^\nu(x) \\

\Rightarrow\,\, &U^{'\mu}(x^\nu-\xi^\nu(x))=\Big[\frac{\partial x^\mu}{\partial x^\nu}-\frac{\partial \xi^\mu(x)}{\partial x^\nu}\Big]U^\nu(x)\\

\Rightarrow\,\, &U^{'\mu}(x^\nu)-\xi^\nu(x)\frac{\partial U^{'\mu}(x)}{\partial x^\nu}=\delta^\mu_\nu U^\nu(x)-\partial_\nu\xi^\mu(x)U^\nu(x)\quad \text{(Taylor expansion upto first order in}\,\, \xi^\nu\, \text{on LHS)} \\

\Rightarrow\,\, &U^{'\mu}(x)-U^\mu(x)=\xi^\nu(x)\partial_\nu U^{'\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x)

\end{align}

The final expression on the RHS is $$\xi^\nu(x)\partial_\nu U^{'\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x)$$ but the expression for ##\mathcal{L}_\xi U^\mu## is $$\mathcal{L}_\xi U^\mu=\xi^\nu(x)\partial_\nu U^{\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x).$$ Why this discrepancy? What's the resolution?

I have performed the following steps and have a question in the final result.

By general coordinate transformation rule we know that

\begin{align}

& U^{'\mu}(x')=\frac{\partial x^{'\mu}}{\partial x^\nu} U^\nu(x) \\

\Rightarrow\,\, &U^{'\mu}(x^\nu-\xi^\nu(x))=\Big[\frac{\partial x^\mu}{\partial x^\nu}-\frac{\partial \xi^\mu(x)}{\partial x^\nu}\Big]U^\nu(x)\\

\Rightarrow\,\, &U^{'\mu}(x^\nu)-\xi^\nu(x)\frac{\partial U^{'\mu}(x)}{\partial x^\nu}=\delta^\mu_\nu U^\nu(x)-\partial_\nu\xi^\mu(x)U^\nu(x)\quad \text{(Taylor expansion upto first order in}\,\, \xi^\nu\, \text{on LHS)} \\

\Rightarrow\,\, &U^{'\mu}(x)-U^\mu(x)=\xi^\nu(x)\partial_\nu U^{'\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x)

\end{align}

The final expression on the RHS is $$\xi^\nu(x)\partial_\nu U^{'\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x)$$ but the expression for ##\mathcal{L}_\xi U^\mu## is $$\mathcal{L}_\xi U^\mu=\xi^\nu(x)\partial_\nu U^{\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x).$$ Why this discrepancy? What's the resolution?

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