Consider a scalar function $\phi(x^\mu)$ of a four-vector $x^\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z)$. What is the Taylor expansion of $\phi(x^\mu+\delta x^\mu)$ for infinitesinal $\delta x^\mu$?

In particular, the Taylor expansion of $f(\textbf{r}+\delta\textbf{r})$ where $\textbf{r}\equiv (x,y,z)$ is the position vector is given by $$f(\textbf{r}+\delta\textbf{r})=(\textbf{r})+(\delta\textbf{r}\cdot\nabla)f(\textbf{r})+\frac{1}{2}(\delta\textbf{r}\cdot\nabla)^2f(\textbf{r})+...$$

How does this change for $\phi(x^\mu+\delta x^\mu)$? In particular, I'm not sure whether the metric $\eta_{\mu\nu}$ will enter the expansion, and the dot products in the expansion of $f(\textbf{r}+\delta\textbf{r})$ be replaced by scalar products in spacetime in the expansion of $\phi(x^\mu+\delta x^\mu)$?

  • $\begingroup$ $f(x+a)=f(x)+(a\cdot\partial)f(x)+\frac12(a\cdot\partial)^2f(x)+\cdots$ $\endgroup$ – AccidentalFourierTransform Nov 27 '16 at 20:44
  • $\begingroup$ @AccidentalFourierTransform Does the notation $a\cdot\partial$ stand for $a^\mu \partial_\mu$ or $\eta_{\mu\nu}a^\mu \partial^\nu$? How does the Taylor expansion know that space and time are different? I mean why not $\delta_{\mu\nu}$ instead of $\eta_{\mu\nu}$? $\endgroup$ – SRS Nov 27 '16 at 20:49
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    $\begingroup$ $a\cdot \partial=a^\mu\partial_\mu=\eta_{\mu\nu}a^\mu\partial^\nu$. Taylor doesn't know about the metric, but $\partial$ does: the derivative is defined as $\partial_\mu=\frac{\partial}{\partial x^\mu}$. If you want to raise the index $\mu$ in $\partial$ you must introduce the metric. But as long as you keep your indices where they belong, you dont need to introduce the metric. In other words, Taylor expansion is about $x^\mu\frac{\partial}{\partial x^\mu}$. As long as you keep this structure unchanged, you dont need to introduce the metric. The latter only appears if you want to move an index $\endgroup$ – AccidentalFourierTransform Nov 27 '16 at 20:53
  • $\begingroup$ By definition, $a^\mu \partial_\mu = \eta_{\mu \nu} a^\mu \partial^\nu$: You can lower and raise indices using your metric. In a natural setting, the derivative is covector, so you will always first compute $\partial_\mu$ and then raise the index (if you required). The taylor expansion just uses a inner product which is induced by your metric, which could be euclidean ($\delta_{\mu \nu}$), minkowskian $\eta$ or something more exotic … $\endgroup$ – Faser Nov 27 '16 at 20:56

Let's sort this out then. Treating $x^\mu$ as a black box, we get

$\phi(x^\mu + \delta x^\mu)= \frac{\partial \phi}{\partial x^\mu} \delta x^\mu + \mathcal{O}(\delta x^\mu)$ in first order. If you take a derivative with respect to a four-vector (with index up), you obtain an index-down object (covector), so implicitly we are contracting with the metric here once. You can convince yourself by checking that in the simple example $\frac{\partial}{\partial x^\mu} (x^\nu x_\nu) = 2 x_\mu$: you are acting with the derivative (index down) on a scalar (since $\nu$ is summed over), yielding a covector (index down).

The same argument works for higher orders, giving you a term $\propto \frac{\partial^2 \phi}{\partial x^\mu \partial x^\nu} \delta x^\mu \delta x^\nu$, and so on. Just be careful to use distinctive indices for higher orders … :-)

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