Weak Field Approx, algebra geodesic equation

[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 don't 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 !

 

[PeterDonis]

My book

Which book?

I have no idea whatsoever where ii comes from.

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).

The geodesic equation reduces from $\frac{d^{2}x^{a}}{dt^{2}}+\Gamma^{a}_{bc}\frac{dx^{b}}{ds}\frac{dx^{c}}{ds} =0$

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 $t$.

to $\frac{d^{2}x^{i}}{dt^{2}}+\Gamma^{i}_{00}\frac{dx^{0}}{ds}\frac{dx^{0}}{ds}=O(\epsilon^{2})$

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.

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.

Yes.

When we replace $s$ with $t$ for the fist term in the geodesic equation , why don't we do it for the second?

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.

 

[PeterDonis]

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?

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 $t$.

 

[stevendaryl]

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: $g_{00} = 1 + \frac{2}{c^2} \Phi$

 

[binbagsss]

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).

.

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 $\frac{1}{2}$ ?

 

[PeterDonis]

Is this correct and then ignore the $\frac{1}{2}$ ?

Basically, yes.

 

[binbagsss]

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.

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

 

[DrGreg]

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

They are the same.

To understand the notation better, see Big O notation.

 

[binbagsss]

One more question sorry, on http://www.mth.uct.ac.za/omei/gr/chap7/node3.html , 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?

 

[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.

 

[PeterDonis]

why the replacement of $^{i}$ , so the partial derivatives, to $\bigtriangledown ^{i}$ ?

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.

equation (41) $R_{00}=...+O(\epsilon)$ , eq 42 $R_{ab}=...+O(\epsilon)$

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).

 

[binbagsss]

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).

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 that's going to cancel
Thanks.

 

[PeterDonis]

I have $U^{\alpha}=(1,0,0,0)$

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 $dx^i / dt$.

 

[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.

 

[PeterDonis]

So $U^{0} U^{0}=\frac{c^{2}}{c^{2}-v^{2}}=\frac{1}{1-\frac{v^{2}}{c^{2}}}=1+\frac{v^{2}}{c^{2}}+...$

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 $g_{00}$.

 

[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 $v$ ?

 

[PeterDonis]

$U_{0}U_{0}=\frac{1+\frac{2\Phi}{c^{2}}}{1-\frac{v^{2}}{c^{2}}}$

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 $O(\epsilon^2)$.

is this gravitational potential a function of $v$ ?

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

 

[binbagsss]

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

I've seen the derivation but I'm still unsure , there's 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 $h_{00}$

 

[PeterDonis]

theres equation 31, but this looks to me like to solve for $\Phi$ you need to integrate $v^{2}$ wrt space coordinates.

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.

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

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.

 

[binbagsss]

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.

.

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

 

[binbagsss]

$U^0$ is the relativistic $\gamma$ factor so it depends on $dx^i / dt$.

How do we know this? Or could you please ref me to any sources explaining it Thanks.

 

[stevendaryl]

How do we know this? Or could you please ref me to any sources explaining it Thanks.

Have you seen the relationship between proper time $\tau$ and coordinate time $t$?

$d \tau = \sqrt{1-\frac{v^2}{c^2}} dt$

You can turn that around to get

$\frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma$

Since $x^0 = t$, $\frac{dx^0}{d\tau} = \gamma$

 

[binbagsss]

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

Apologies, it's only of the order $O(\epsilon)$ so I still have cross terms in the numerator - gives $4\Phi$- what should I do with these, the $O(\epsilon)$ terms have not cancelled.

(It would be $O(\epsilon^{2})$ if it was $v^{2}$ , but we can't conclude an extra $\epsilon$ from another time derivative, although we have said time derivatives are small?)

 

[PeterDonis]

we can't conclude an extra $\epsilon$ from another time derivative, although we have said time derivatives are small?

I think we can, actually, but I'm not sure.

 

[binbagsss]

Sorry to re-bump, I'm looking at the trace expression, eq 38, I believe it is given to the order $\epsilon$, is this correct?
$U_{a}$ is the four velocity of the fluid, am I correct in thinking that $U_{i}=0$, $i=1,2,3$ the spatial dimensions?
( I have another q to follow, I'd just like to confirm that $U_{i}=0$ before I get ahead of myself...thanks.)

 

[binbagsss]

$U_{a}$ is the four velocity of the fluid, am I correct in thinking that $U_{i}=0$, $i=1,2,3$ the spatial dimensions?

Since in the energy-momentum tensor $p$ and $\rho$ are measured in the rest frame of the fluid

 

[stevendaryl]

Sorry to re-bump, I'm looking at the trace expression, eq 38, I believe it is given to the order $\epsilon$, is this correct?
$U_{a}$ is the four velocity of the fluid, am I correct in thinking that $U_{i}=0$, $i=1,2,3$ the spatial dimensions?
( I have another q to follow, I'd just like to confirm that $U_{i}=0$ before I get ahead of myself...thanks.)

I know it's a pain to do so, but it would be helpful to try to reproduce the equations you're talking about, rather than have people have to go through old messages and URLs to find them.

Equation 37 is:

$T_{\alpha \beta} = (\rho + \frac{p}{c^2})U_\alpha U_\beta + p g_{\alpha \beta}$

There is no approximation involved in going to equation 38. Taking the trace gives:

$T^\alpha_{\alpha} = (\rho + \frac{p}{c^2})U^\alpha U_\alpha + p g^\alpha_{\alpha}$ (where $\alpha$ is summed over. It's always true that:

$U^\alpha U_\alpha = -c^2$
$g^\alpha_\alpha = 4$

To see the first one, remember that $\tau$ is defined by:

$\delta \tau^2= - \frac{1}{c^2} (g_{\alpha \beta} \delta x^\alpha \delta x^\beta)$

Divide both sides by $\delta \tau^2$ to get

$1= - \frac{1}{c^2} (g_{\alpha \beta} \frac{\delta x^\alpha}{\delta \tau} \frac{\delta x^\beta}{\delta \tau})$
$= - \frac{1}{c^2} g_{\alpha \beta}U^\alpha U^\beta$
$= - \frac{1}{c^2}U^\alpha U_\alpha$

The second equation is $g^\alpha_\alpha = 4$.

To see this, start with
al
$g_{\beta \alpha}$

Now, "raise" the index $\beta$ using the tensor $g^{\beta \lambda}$:

$g^\lambda_\alpha = g^{\beta \lambda} g_{\beta \alpha}$

But the definition of $g^{\beta \lambda}$ is that it is the inverse of $g_{\beta \lambda}$. That is,

$g^{\beta \lambda} g_{\beta \alpha} = \delta^\lambda_\alpha$

where $\delta^\lambda_\alpha$ is 1 when $\lambda = \alpha$ and 0 otherwise. So

$g^0_0 = g^1_1 = g^2_2 = g^3_3 = 1$

So $\sum_\alpha g^\alpha_\alpha = 4$

 

[binbagsss]

A light cone is the path that a flash of light, emanating from a single event (localized to a single point in space and a single moment in time) and traveling in all directions, would take through spacetime. I

To see the first one, remember that $\tau$ is defined by:

$\delta \tau^2= - \frac{1}{c^2} (g_{\alpha \beta} \delta x^\alpha \delta x^\beta)$

Divide both sides by $\delta \tau^2$ to get

$1= - \frac{1}{c^2} (g_{\alpha \beta} \frac{\delta x^\alpha}{\delta \tau} \frac{\delta x^\beta}{\delta \tau})$
$= - \frac{1}{c^2} g_{\alpha \beta}U^\alpha U^\beta$
$= - \frac{1}{c^2}U^\alpha U_\alpha$

Thanks for that proof.

And it is wrong to assume $U_{i}=0$?

 

[stevendaryl]

A light cone is the path that a flash of light, emanating from a single event (localized to a single point in space and a single moment in time) and traveling in all directions, would take through spacetime. I


Thanks for that proof.

And it is wrong to assume $U_{i}=0$?

Well that assumption isn't necessary to get equation 38 from equation 37