In the book "Reflections on Relativity" by Kevin Brown, there is a chapter called "Relatively Straight", in which he derives the geodesic equations using the Euler equation. Online version

Just after the second mention of the Euler equation (about 80% down), there is the following text: "Therefore, we can apply Euler's equation to immediately give the equations of geodesic paths on the surface with the specified metric

$$ \frac{\partial F}{\partial x^\sigma} - \frac{d}{d\lambda}\frac{\partial F}{\partial \dot{x}^\sigma}$$

For an n-dimensional space this represents n equations, one for each of the coordinates $x_1, x_2, ..., x_n$. Letting $w = (\frac{ds}{dl})^2 = F^2 = g_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta$ this can be written as

$$\frac{\partial w^{1/2}}{\partial x^\sigma} - \frac{d}{d\lambda}\frac{\partial w^{1/2}}{\partial \dot{x}^\sigma} = \frac{d}{d\lambda}\frac{\partial w}{\partial \dot{x}^\sigma} - \frac{\partial w}{\partial x^\sigma} - \frac{1}{2w}\frac{dw}{d\lambda} \frac{\partial w}{\partial \dot{x}^\sigma} = 0 $$

I get the substitution of sqrt(w) for F on the LHS, but can't see how he obtains the middle expression. I have tried using the product/chain rules as is usual with these things, but just cannot see what he is doing here.

I can usually follow Kevin's work, with a bit of effort, but this one seems a little trickier than I am used to. Can anyone help me to understand the trick?

  • $\begingroup$ Thanks for the edits nivag, I was looking for some similar latex to copy & edit . . . $\endgroup$
    – m4r35n357
    May 23, 2014 at 12:13

1 Answer 1


It seems right. You have

$$ \frac{\partial w^{1/2}}{\partial x^\sigma}=\frac1{2\sqrt{w}}\frac{\partial w}{\partial x^\sigma} $$

Change the order of the derivatives in the second term

$$ \frac{d}{d\lambda}\frac{\partial w^{1/2}}{\partial \dot{x}^\sigma}=\frac{\partial}{\partial \dot{x}^\sigma}\left(\frac{d}{d\lambda} w^{1/2}\right)=\frac{\partial}{\partial \dot{x}^\sigma}\left( \frac1{2\sqrt{w}}\frac{d w}{d\lambda } \right) $$

Product rule

$$ \frac{\partial}{\partial \dot{x}^\sigma}\left( \frac1{2\sqrt{w}}\frac{d w}{d\lambda } \right)=-\frac1{4w\sqrt{w}}\frac{\partial w}{\partial \dot{x}^\sigma}\frac{dw}{d\lambda}+\frac1{2\sqrt{w}}\frac{\partial }{\partial \dot{x}^\sigma}\frac{dw}{d\lambda} $$

Subtract, equate to zero, multiply $2\sqrt{w}$ you get

$$ \frac{\partial w}{\partial x^\sigma}+\frac1{2w}\frac{\partial w}{\partial \dot{x}^\sigma}\frac{d w}{d\lambda}-\frac{d}{d\lambda}\frac{\partial w}{\partial \dot{x}^\sigma}=0 $$

  • 1
    $\begingroup$ Thanks for this MBN! I got the first step OK, but on reflection I wasn't going to get the rest of the way in a hurry ;) $\endgroup$
    – m4r35n357
    May 23, 2014 at 12:44

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