# Locally inertial reference frame problem

Hey PF, I am working on a problem set, and one of the problems is proving that the Christoffel symbols vanish at the origin of the coordinates ##y^{\alpha}## given by the coordinate transformation: $$y^\alpha (x) = x^\alpha - x^\alpha_{(0)} + \frac {1} {2} (x^\mu - x^\mu_{(0)} )(x^\nu - x^\nu_{(0)} ) \Gamma^\alpha_{(0) \mu \nu}$$ Where anything with a ##(0)## subscript is evaluated at some particular point.
Figuring out that ##\frac {\partial y^{\alpha}} {\partial x^{\mu}} = \delta^{\alpha}_{\mu} + (x^{\lambda} - x^{\lambda}_{(0)}) \Gamma^{\alpha}_{(0) \mu \lambda}## and that ##\frac {\partial^2 y^{\alpha}} {\partial x^\mu \partial x^\nu} = \Gamma^{\alpha}_{(0) \mu \nu}## was fairly straightforward, and I also know the transformation law for the Christoffel symbols, but I can't figure out how to find the inverse of the Jacobian—the ##\frac {\partial x^\alpha} {\partial y^\mu}##’s. Here’s part of the solution (they use squiggles instead of ##y##):

They’re clearly using a Taylor polynomial, but I don’t understand how to find that the unknown matrix ##A^\alpha_{\mu \beta}## ends up being ##\Gamma^\alpha_{\mu \beta}##. I feel like I must be missing something super obvious. Would someone be so kind as to point out what I’m missing?

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Orodruin
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The chain rule.
$$\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}} \dd{x^\alpha}{y^\beta} \dd{y^\beta}{x^\gamma} = \dd{x^\alpha}{x^\gamma} = \delta^\alpha_\gamma.$$
Insert the Taylor expansion into the above expression along with the coordinate transformation and identify ##A## from the linear term in ##x - x_0##.

Pencilvester
Orodruin
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It should be noted that there is a significantly easier way of showing this. Just consider the geodesic equations in ##x## and show that they imply ##\ddot y^\mu = 0## at ##x_0##.

Pencilvester
The chain rule.
Oh, duh. Yup, that would be the obvious thing I was missing.

It should be noted that there is a significantly easier way of showing this. Just consider the geodesic equations in ##x## and show that they imply ##\ddot y^\mu = 0## at ##x_0##.
That definitely is much simpler and straightforward. Thanks!

Orodruin
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Homework Helper
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Yup, that would be the obvious thing I was missing.
Sometimes you need to be pointed to the trees in the forest...

Pencilvester