# Slow motion approx, order of acceleration quick Q?

1. Apr 1, 2015

### binbagsss

I'm looking at equation (41) from equation (40) on http://www.mth.uct.ac.za/omei/gr/chap7/node3.html [Broken] and the second term, it says that these equations are given up to $O(\epsilon)$.

Looking at equation (32) for $g_{00}$ I see it is proportional to $\Phi$, and from looking at (31) I see that $\Phi$ is of the same order as $d^{2}x^{i}/dt^{2}$ .

Looking at the 2nd term in (41) which has only been multiplied by a $-1$ from the $g_{00}$ I conclude that it is only possible to be given up to $O(\epsilon)$ if $\Phi$ , and so, $d^{2}x^{i}/dt^{2}$, is of the order $O(\epsilon^{2})$

By definition of the slow approximation we have $dx^{i}/dt=O(\epsilon)$. And time derivatives are neglected compared to space derivatives.

It seems to me quite possible that $d^{2}x^{i}/dt^{2}$ being of the order $O(\epsilon^{2})$ can be justified.

However I'm not sure how to show this properly/explicitly?

Thanks.

Last edited by a moderator: May 7, 2017
2. Apr 2, 2015

bump.

3. Apr 2, 2015

### stevendaryl

Staff Emeritus
Frankly, I don't understand the basis for their order $\epsilon$ analysis.

Normally, I would think that to get Newtonian physics as an approximation to GR, we need something like this:

1. Assume that $g_{\mu \nu} = \eta_{\mu \nu} + O(\epsilon)$
2. Assume that $\frac{dx^j}{dt} = O(\sqrt{\epsilon})$
3. Assume that $\rho = O(\epsilon)$
4. Assume that $p/c^2 \ll \rho$
5. Assume that $\partial_t g_{\mu \nu}$ is negligible compared with spatial derivatives.
As for 4&5, I'm not sure how pressure and time-derivatives of the metric should rank in powers of $\epsilon$

But the big difference with what you've said is that I don't think that $\Phi$ should be $O(\epsilon^2)$. Since $g_{00} = 1 + \frac{2\Phi}{c^2}$ it must be that $\Phi = O(\epsilon)$