Infinitesimal coordinate transformation of the metric

  • #1
237
11

Homework Statement:

The metric transforms as ##g'_{\mu\nu}(x') = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta}(x)##. Show that under the infinitesimal transformation ##x^\alpha \rightarrow x'^\alpha = x^\alpha + \epsilon^\alpha## (##\epsilon^\alpha## is small), ##g'_{\mu\nu}(x^\mu + \epsilon^\mu) = g_{\mu\nu}(x) + (\partial_{\mu} \epsilon^\mu + \partial_{\nu} \epsilon^\nu) g_{\mu\nu}(x) = g_{\mu\nu}(x) + \partial_{\mu} \epsilon_\nu + \partial_{\nu} \epsilon_\mu##.

Relevant Equations:

Already given above
I kinda know how to do this problem, it is just that I hit a sign problem. If I take the partial derivative of the coordinate transformation with respect to ##x'^\mu##, I get

writing it first in the inverse form, ##x^\alpha = x'^\alpha - \epsilon^\alpha##

##\frac{\partial x^\alpha}{\partial x'^\mu} = \delta^\alpha_\mu - \partial'_\mu \epsilon^\alpha = \delta^\alpha_\mu - \frac{\partial x^\alpha}{\partial x'^\mu} \partial_\alpha \epsilon^\alpha##

transposing terms,

##\frac{\partial x^\alpha}{\partial x'^\mu} (1 + \partial_\alpha \epsilon^\alpha) = \delta^\alpha_\mu##

##\frac{\partial x^\alpha}{\partial x'^\mu} \approx \delta^\alpha_\mu - \delta^\alpha_\mu \partial_\alpha \epsilon^\alpha \quad## (to first order)

Now this is going to give me a minus sign instead of a plus sign in the right hand side of the equation to be shown. Did I do something wrong?
 

Answers and Replies

  • #2
502
261
Mmm... I'm not sure, but maybe you're right and should be ##g'_{\mu\nu}=g_{\mu\nu}-\partial_\mu \varepsilon_\nu - \partial_\nu \varepsilon_\mu##.
But in any case, be very careful in how you label the indices, in more than one expression you give the same symbol to different indices, what can lead to errors or, at least, to confuse the people who are trying to read it.
 

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