# Particle co - moving in flat Roberston - Walker space - time

1. Jun 21, 2011

### WannabeNewton

1. The problem statement, all variables and given/known data
Consider a flat FRW model whose metric is (in polar coordinates) $$ds^{2} = -dt^{2} + a^{2}(t)[dr^{2} + r^{2}d\theta ^{2} + r^{2}sin^{2}\theta d\phi ^{2}]$$ where a(t) is the scale factor. Show that, if a particle is shot from the origin at a time and with velocity $$t_{0},V_{0}$$ respectively, with respect to a co - moving observer then asymptotically it comes to rest with respect to the co - moving frame. Express the co - moving coordinate radius at which it comes to rest as an integral over a(t).

3. The attempt at a solution
First off, since the particle is shot from the origin, $$d\theta = d\phi = 0$$. Doing this then dividing both sides of the metric by dt, I got $$(\frac{ds}{dt})^{2} = -1 + a^{2}(t)(\frac{dr}{dt})^{2}$$. Since the particle is shot with respect to a co - moving frame, the distance with respect to the two never changes so ds / dt = 0. Doing this then getting dr / dt on its own I got, $$\frac{dr}{dt} = \frac{1}{a(t)}$$. Since in an expanding universe, a(t) always increases, as a(t) (or as t) increases dr / dt will asymptotically go to zero. I hope this is right so far? The problem came with the second part (as long as the first part is correct): $$dr = \frac{dt}{a(t)}$$ and, denoting by R the distance at which the particle comes to rest with respect to the co - moving frame, $$R = \int_{t_{0}}^{?} \frac{dt}{a(t)}$$ do I use t = infinity when integrating the right - hand side to denote the time at which dr / dt = 0 or do I just denote any t = final time to represent it since it is asymptotic? Sorry if this is an absurdly stupid question.

2. Jun 22, 2011

### George Jones

Staff Emeritus
You didn't use this information.
This true, but not for the reason you give. It is true because the particle follows a geodesic, not because it starts from the origin.

Good.
No. For the particle in question, what is the physical interpretation of $ds^2$? What is the physical interpretation of $ds^2 = 0$?

I think that you also have to use the fact that the particle moves on a geodesic (at least I used this).

If you want mathematics in line with your text, use itex and /itex instead of tex and /tex. Use tex and /tex when you mathematics to appear on a separate line.

3. Jun 23, 2011

### WannabeNewton

Well for a massive particle $ds^{2} < 0$ or greater depending on the convention. $ds^{2} = 0$ would only be for a particle moving on a null geodesic. I don't get how to use the initial velocity and time given. Do I have to solve the geodesic equation for a time - like geodesic and use the initial conditions they give? I'm just confused as to how to factor in the fact that co - moving observers have constant coordinate distance.

4. Jun 23, 2011

### WannabeNewton

Sorry to bother you again but can we say that since the particle and the observer are co - moving (in flat space) that their proper times are their coordinate times as they move along their respective geodesics?

5. Jun 24, 2011

### George Jones

Staff Emeritus
If I remember correctly, you are self-studying, so let's take some time to work through this.
For the convention in the original post, the proper time of the particle is given by $d\tau^2 = -ds^2$

Right, so we don't want to use this, i.e., we don't want to set $ds/dt = 0$ since we are not dealing with a photon.
Let's leave this until near the end.
Yes. Your question comes from Hartle's book, so use Example 8.2 (but with only two coordinates) as guide to write down geodesic conditions for
$$ds^2 = -dt^2 + a \left(t\right)^2 dr^2 .$$
Yes, observers moving with the Hubble flow (galaxies with no peculiar velocities) have constant co-moving coordinate $r$. Because all galaxies have constant $r$, and because no galaxy is special, we can, WLOG, choose $r = 0$ for galaxy A from which the particle is initially shot. Since the particle moves with respect to this galaxy, it has to have changing co-moving coordinate $r$. As the particle moves farther and farther away from galaxy A, it catches up to galaxies that are "receding" from galaxy A with greater and greater "speeds". As $t \rightarrow \infty$, the expansion of the universe "catches up" with the particle, and $dr/dt \rightarrow 0$.

This question asks for the particle's $r$ value when this happens, i.e., the question asks for the particle's $r$ value as $t \rightarrow \infty$.
No, flat space means flat 3-dimensional space, not flat 4-dimensional spacetime, so flat FRW models are not flat spacetimes.

6. Jun 24, 2011

### WannabeNewton

Yeah I'm self - studying this from Hartle's book atm =D. So is that a no for proper time being equal to coordinate time for co - moving objects? I ask because if the particle is moving on a geodesic then using the euler - lagrange equations and for r I get $\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\partial L}{\partial \dot{r}}) = 0$ so $\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}$ where k = const. So I can't replace proper time with coordinate time in this situation?

7. Jun 24, 2011

### George Jones

Staff Emeritus
Yes, for co-moving objects, coordinate time is proper times, since for co-moving objects, since $dr = d \theta = d \phi = 0$ gives that $ds^2 = - dt^2$.

No, for the particle, coordinate time is not proper time, since $dr \ne 0$ for the particle.
Yes.
Yes.
I am not sure what you mean. For the particle, proper is not coordinate time, but you can use $\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}$ to eliminate proper time in the equations for the particle. Solve for $d \tau$, and substitute this into
$$ds^2 = -d \tau^2 = -dt^2 + a \left(t\right)^2 dr^2 .$$

8. Jun 24, 2011

### WannabeNewton

I wrongly assumed that the particle would not have a changing co - moving radial coordinate so I equated the proper time and coordinate time. I kept thinking of it in terms of how cosmological time is defined. When I substituted in for tau I got a not so elegant expression for dr / dt so I fear that I might have gone wrong in the algebra: $$\frac{\mathrm{d} r}{\mathrm{d} t} = \frac{k}{a(t)[a^{2}(t) + k^{2}]^{1/2}}$$

9. Jun 24, 2011

### George Jones

Staff Emeritus
If the particle had an unchanging co-moving coordinate, then it wouldn't be "shot" anywhere. Note that if the universe expands forever, then $\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}$ gives that the particle asymptotically comes to rest with respect to a co-moving frame.
This seems right to me. The constant $k$ is determined from initial conditions.

10. Jun 24, 2011

### WannabeNewton

Ok well then that came out to $k^{2} = V_{0}^{2}a^{4}(t_{0}) - a^{2}(t_{0})$. I took the darn co - moving coordinates part as being much more integral to the problem than it really was. Thanks a million for clearing that up. You know the problems on the gravitational waves chapter were much easier...much less extrapolation =p. Thanks again.

11. Jun 25, 2011

### George Jones

Staff Emeritus
I'm not sure how you got this.

12. Jun 25, 2011

### WannabeNewton

Don't you just use the fact that at $t_{0}$, dr / dt is $V_{0}$?

13. Jun 25, 2011

### George Jones

Staff Emeritus
Two things: 1) Are you sure that you rearranged correctly?

14. Jun 25, 2011

### WannabeNewton

Probably not heh. I did it again and got $$k^{2} = \frac{V_{0}^{2}a^{4}(t_{0})}{1 - V_{0}^{2}a^{2}(t_{0})}$$.

15. Jun 25, 2011

### George Jones

Staff Emeritus
Yes. Now, the other thing: 2) although it seems reasonable, unfortunately, $V_0$ is not $dr/dt$ evaluated at $t_0$. For a co=moving observer, proper distance is given by $ar$ (see equation (18.3)) and proper time is $t$, so speed measured by a co-moving observer is
$$\frac{d}{dt} \left( ar \right).$$

16. Jun 25, 2011

### WannabeNewton

So since the speed measured by the comoving - observer is actually $\dot{a(t)}r + a(t)\frac{\mathrm{d} r}{\mathrm{d} t}$, for the initial velocity would the r in $\dot{a(t)}r$ just be the initial distance from the origin (in this case 0)?

17. Jun 25, 2011

### George Jones

Staff Emeritus
Right. You are given that at time $t_0$, the particle is at the origin and moving with speed $V_0$.

18. Jun 25, 2011

### WannabeNewton

Ah, so $k = \frac{V_{0}a(t_{0})}{(1 - V^{2}_{0})^{1/2}}$ it would seem.

19. Jun 25, 2011

### George Jones

Staff Emeritus
Yes. This makes sense physically because
$$k = a^2 \frac{dr}{d \tau} = a \frac{dt}{d \tau} a \frac{dr}{d t},$$

and $dt/d\tau = \gamma$, just as it does in special relativity.

20. Jun 25, 2011

### WannabeNewton

Oh cool didn't see that. Thanks a bunch sir. This was the main problem that was bugging me from the chapter.