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) \tag{1}\\ \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) \tag{2}\\ \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)} \tag{3}\\ \Rightarrow\,\, &U^{'\mu}(x)-U^\mu(x)=\xi^\nu(x)\partial_\nu U^{'\mu}(x)-\partial_\nu\xi^\mu(x)U^\nu(x) \tag{4} \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?


1 Answer 1


Write $U'(x)=U(x)+\delta U(x)$ you see this extra $\delta U(x)$ would be discarded to first order as you already have $\xi^\nu$ contracted in front of its derivative.


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