Infinitesimal Coordinate Transformation and Lie Derivative

• A
• Baela
In summary, under an infinitesimal coordinate transformation ##x^{'\mu}=x^\mu-\xi^\mu(x)##, the variation of a vector ##U^\mu(x)## can be expressed as $$\delta U^\mu(x)=U^{'\mu}(x)-U^\mu(x)=\mathcal{L}_\xi U^\mu$$ where ##\mathcal{L}_\xi U^\mu## is the Lie derivative of ##U^\mu## wrt the vector ##\xi^\nu##. There may be a discrepancy between the expression for ##\mathcal{L}_\xi U^\mu## and the final expression on the RHS due to neglecting higher order terms in an
Baela
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?

Last edited:
The difference between $\xi^{\nu}\partial_\nu U^{\mu}$ and $\xi^{\nu}\partial_\nu U'^{\mu}$ is second order in $\xi$; for an infinitesimal change such higher order terms are neglected.

EDIT: Rather than expanding $U'$ about $x$, expand it about $x'$. Then $$\begin{split} U'^\mu(x) &= U'^\mu(x' + \xi) \\ &= U'^\mu(x') + \xi^\nu \partial'_\nu U'^\mu(x') + O(\xi^2)\\ &= U'^\mu(x') + \xi^\nu (\partial'_\nu x^\lambda) \partial_\lambda U'^\mu(x') + O(\xi^2)\\ &= U'^\mu(x') + \xi^\nu \left(\delta_\nu^\lambda + \partial'_\nu \xi^\lambda \right) \partial_\lambda U'^\mu(x') + O(\xi^2) \\ &= U'^\mu(x') + \xi^\nu \partial_\nu U'^\mu(x') + O(\xi^2) \end{split}$$ and now substitute for $U'(x')$ to obtain $$\begin{split} U'^\mu(x) &= U^\mu(x) - U^\nu(x)\partial_\nu\xi^\mu + \xi^\nu \partial_\nu \left( U^\mu(x) - U^\lambda(x)\partial_\lambda\xi^\mu \right) + O(\xi^2) \\ &= U^\mu(x) - U^\nu(x)\partial_\nu\xi^\mu + \xi^\nu \partial_\nu U^\mu(x) + O(\xi^2) \end{split}$$ as required.

Last edited:
malawi_glenn and topsquark

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