# Weak Field Approx, algebra geodesic equation

1. Mar 10, 2015

### binbagsss

My book says in the slow motion approx, so $v << c$, $v=\frac{dx^{i}}{dt}=O(\epsilon)$
It then states:

i) $\frac{dx^{i}}{ds}=\frac{dt}{ds}\frac{dx^{i}}{dt}=O(\epsilon)$
ii) $\frac{dx^{0}}{ds}=\frac{dt}{ds}=1+O(\epsilon)$

The geodesic equation reduces from $\frac{d^{2}x^{a}}{dt^{2}}+\Gamma^{a}_{bc}\frac{dx^{b}}{ds}\frac{dx^{c}}{ds} =0$ to $\frac{d^{2}x^{i}}{dt^{2}}+\Gamma^{i}_{00}\frac{dx^{0}}{ds}\frac{dx^{0}}{ds}=O(\epsilon^{2})$

Questions:
1) I have no idea whatsoever where ii comes from.
2) I see that in the slow approx $t$ is approximately equal to $s$, and so we can replace these in the first term, i.e $\frac{dt}{ds}=1$, but isn't $\frac{dx^{i}}{dt}=O(\epsilon)$ also negligible?

3)From ii) I think I see why we neglect $\frac{dx^{i}}{ds}\frac{dx^{i}}{ds}$ compared to $\frac{dx^{0}}{ds}\frac{dx^{0}}{ds}$ so the the former gives $(O(\epsilon))^{2}$ and the latter $(1+O(\epsilon))(1+O(\epsilon))=(1+2O(\epsilon)+(O(\epsilon))^{2})$, so as it has a non $\epsilon$ term we don't neglect it.

4) When we replace $s$ with $t$ for the fist term in the geodesic equation , why dont we do it for the second?

5) Finally probably a stupid question, from a different source:http://www.mth.uct.ac.za/omei/gr/chap7/node3.html (just above eq 28), I'm confused why $\frac{dx^{0}}{dt}=c$, I thought $t$ is coordinate time, and $x^{0}$ is also coordinate time, so this would equal $1$? what is $x^{0}$ then?

Your help is really appreciated thanks !

Last edited: Mar 10, 2015
2. Mar 10, 2015

### Staff: Mentor

Which book?

It comes from the fact that the 4-velocity is a unit 4-vector; its magnitude is 1 (or -1, depending on which metric signature convention you use). So we must have

$$1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds}} = \sqrt{ \left( \frac{dx^0}{ds} \right)^2 - \Sigma \left( \frac{dx^i}{ds} \right)^2 }$$

If you work it out you will see that it shows that i) implies ii).

I think you mean $d^2 x^a / ds^2$ in the first term, correct? The geodesic equation starts out being written in terms of derivatives with respect to $s$, not $t$.

Here I think the second term should just be $\Gamma^i{}_{00}$, correct? That factor is already $O(\epsilon^2)$, so you can set $dx^0 / ds \approx 1$ since the $O(\epsilon)$ term in it can be neglected.

Yes.

Essentially, we do; each factor of $dx^0 / ds$ becomes $dx^0 / dt = dt / dt = 1$ if we are only keeping terms to $O(\epsilon^2)$. See above.

3. Mar 10, 2015

### Staff: Mentor

This is just using conventional units in which $c$ is in meters per second, instead of being just a dimensionless number with magnitude 1. In these units, $x^0 = ct$ instead of just $t$.

4. Mar 11, 2015

### stevendaryl

Staff Emeritus
Sort of an aside, it's not generally known that in the low-velocity, weak-gravity limit, the geodesics expressed in the appropriate coordinates is the same as the solutions of Newtonian gravity.

In Newtonian gravity, you have an action

$A = \int dt (\frac{1}{2} m v^2 - m \Phi)$

where $\Phi$ is the Newtonian gravitational potential. You minimize it to find the paths of a test particle. In GR, in the low-velocity, weak-field limit, you have a proper time:

$\tau = \int dt (1 - \frac{1}{2} \frac{v^2}{c^2} + \frac{\Phi}{c^2}) = t - A/mc^2$

So minimizing $A$ is the same as maximizing $\tau$

The Newtonian potential $\Phi$ is related to the GR metric $g_{\mu \nu}$ through: $g_{00} = 1 + \frac{2}{c^2} \Phi$

5. Mar 11, 2015

### binbagsss

Thank you very much !
So I've done this and I've got $\frac{dx^{0}}{ds}=(1+O(\epsilon))^{\frac{1}{2}}=1+\frac{O(\epsilon)}{2}$
Is this correct and then ignore the $\frac{1}{2}$ ?

6. Mar 11, 2015

### Staff: Mentor

Basically, yes.

7. Mar 14, 2015

### binbagsss

Sorry just a side question, how does neglecting $(O(\epsilon))^{2}$ compare to neglecting $O( \epsilon^{2} )$?
Thanks.

8. Mar 14, 2015

### DrGreg

They are the same.

To understand the notation better, see Big O notation.

9. Mar 14, 2015

### binbagsss

One more question sorry, on http://www.mth.uct.ac.za/omei/gr/chap7/node3.html [Broken] , going from (29) to (30), why the replacement of $,^{i}$, so the partial derivatives, to $\bigtriangledown ^{i}$? , so this is the gradient function ? It has no significance right?

Last edited by a moderator: May 7, 2017
10. Mar 14, 2015

### binbagsss

Sorry another, this may be a stupid question but, equation (41) $R_{00}=...+O(\epsilon)$, eq 42 $R_{ab}=...+O(\epsilon)$ and then we sub (29) into (45), and obviusly don't neglect $O(\epsilon)$ in this substitution, but how is this consistent with neglectng it as done in (41) and (42)? Thanks.

11. Mar 14, 2015

### Staff: Mentor

He's using them both to mean the same thing. That's not really standard notation; in standard notation, $\nabla$ means the covariant derivative, not the partial derivative.

No. What he means when he says those equations are correct to first order in $\epsilon$ is that any additional terms are $O(\epsilon^2)$. The equations as he writes them contain all terms up to $O(\epsilon)$; they are not being neglected. The reason no terms in $\epsilon$ appear in equation (41) is that the $O(\epsilon)$ terms cancel when you expand the 00 component of equation (40).

12. Mar 27, 2015

### binbagsss

So in eq (40) I have $U^{\alpha}=(1,0,0,0)$ and $g_{00}=\eta_{00}+\epsilon h_{00}$, I can't see anything thats going to cancel
Thanks.

13. Mar 28, 2015

### Staff: Mentor

No, you don't; the matter is slowly moving but you can't assume it's stationary. The article says (earlier on) that $dx^i / dt$ is $O(\epsilon)$, and $U^0$ is the relativistic $\gamma$ factor so it depends on $dx^i / dt$.

14. Mar 29, 2015

### binbagsss

So $U^{0} U^{0}=\frac{c^{2}}{c^{2}-v^{2}}=\frac{1}{1-\frac{v^{2}}{c^{2}}}=1+\frac{v^{2}}{c^{2}}+...$
I would conclude to neglect the 2nd term onward, I'm not seeing where the $c^{2}$ comes from, thanks.

15. Mar 29, 2015

### Staff: Mentor

Yes, but you need $U_0 U_0$ to expand equation (40). Lower the index on $U^0$ with $g_{00}$ and see what happens. Note that equation (32) gives you an explicit expression for $g_{00}$.

16. Mar 29, 2015

### binbagsss

$U_{0}U_{0}=\frac{1+\frac{2\Phi}{c^{2}}}{1-\frac{v^{2}}{c^{2}}}$, $\Phi$ the gravitational potential, unsure what to do next, is this gravitational potential a function of $v$ ?

17. Mar 29, 2015

### Staff: Mentor

Actually, you need two factors of $g_{00}$, not one, because you're lowering two indexes (one for each $U$).

Also, I'm no longer sure there will be any terms of $O(\epsilon)$ in the expansion. $v^2 / c^2$ is $O(\epsilon^2)$.

Look at the equation for it earlier in the article and how that equation is derived.

18. Mar 30, 2015

### binbagsss

I've seen the derivation but I'm still unsure , theres equation 31, but this looks to me like to solve for $\Phi$ you need to integrate $v^{2}$ wrt space coordinates.

We also have $g_{00}=\eta_{00}+\epsilon h_{00}$, but we don't have an explicit expression for $h_{00}$

19. Mar 30, 2015

### Staff: Mentor

The LHS of equation (31) is the acceleration $dv / dt$, not $v^2$. So it's telling you that the space derivative of $\Phi$ is equal to the time derivative of $v$.

However, you don't necessarily need to find an actual solution for $\Phi$; you just need to determine whether it's $O(\epsilon)$, $O(\epsilon^2)$, etc.

Yes, you do; comparing equations (30) and (31) gives you the relationship between $h_{00}$ and $\Phi$. That relationship is expressed in equation (32); since $g_{00} = \eta_{00} + \epsilon h_{00}$, you can read off $h_{00}$ from that equation.

20. Mar 31, 2015

### binbagsss

• Ah thanks, so $\Phi$ is of the order $O(\epsilon^{2})$ and so $\frac{(1+2\Phi)^{2}}{1-\frac{v^{2}}{c^{2}}}=\frac{c^{4}}{c^{2}}+O(\epsilon^{2})=c^{2}+O(\epsilon^{2})$