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I'm trying to understand the following problem:

We are looking at an infinitesimal coordinate transformation $$ x^\mu \rightarrow x^\mu + \epsilon u^\mu(x), \space \epsilon \rightarrow 0 $$ and we are interested in it's effect on the line element from $x$ to $x+dx$ $$ ds^2 = g_{\mu\nu} dx^\mu dx^\nu $$ where $g_{\mu\nu}$ is the Minkowski metric.

The book I'm reading (Local Quantum Physics, Fields, Particles, Algebras by R. Haag) gives the following formula: $$ \delta ds^2 = \epsilon g_{\mu\nu} ( \frac{\partial u^\mu}{dx^l} dx^l dx^\nu + \frac{\partial u^\nu}{dx^l} dx^l dx^\mu). $$

However when I try and compute $\delta ds^2$ I don't get the $\epsilon$ constant:

\begin{split} \delta ds^2(x) &=& \frac{d}{d\epsilon} ds^2(x+ \epsilon u(x))|_{\epsilon=0} \\ &=& \frac{d}{d\epsilon} g_{\mu\nu} (dx^\mu + \epsilon( u^\mu(x+dx) - u^\mu(x))) (dx^\nu + \epsilon( u^\nu(x+dx) - u^\nu(x)))|_{\epsilon=0} \\ &=& g_{\mu\nu}((u^\mu(x+dx) - u^\mu(x)) (dx^\nu + \epsilon( u^\nu(x+dx) - u^\nu(x))) \\ &&+ (dx^\mu + \epsilon( u^\mu(x+dx) - u^\mu(x))) ( u^\nu(x+dx) - u^\nu(x))) |_{\epsilon=0} \\ &=& g_{\mu\nu} ((u^\mu(x+dx) - u^\mu(x)) dx^\nu + dx^\mu( u^\nu(x+dx) - u^\nu(x))) |_{\epsilon=0} \\ &=& g_{\mu\nu} (\frac{\partial u^\mu}{\partial x^l} dx^l dx^\nu + \frac{\partial u^\nu}{\partial x^l} dx^l dx^\mu) \\ \end{split}.

Any ideas what I'm doing wrong here? (Also let me know f this would be more appropriate for the physics stack exchange, I just thought the topic was more the computation than the underlying theory.)

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  • $\begingroup$ It seems Haag is writing the first term in the Taylor expansion, while you computed the coefficient of this term. So you just have to multiply by $\epsilon$. $\endgroup$ Commented Jan 20, 2022 at 11:49
  • $\begingroup$ Thank you! So this just depends on the convention you use for what $\delta ds^2$ is exactly? I feel like I never saw it with the $\epsilon$ multiplied. It doesn't really make a difference and looking at it like a Taylor expansion makes sense but I just wanted to practice this kind of computation so I wanted to be sure. $\endgroup$
    – diesmond
    Commented Jan 20, 2022 at 13:05

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