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

In summary, a particle is shot from the origin with a given velocity and time, and comes to rest at a given radius asymptotically.
  • #1
WannabeNewton
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Homework Statement


Consider a flat FRW model whose metric is (in polar coordinates) [tex]ds^{2} = -dt^{2} + a^{2}(t)[dr^{2} + r^{2}d\theta ^{2} + r^{2}sin^{2}\theta d\phi ^{2}] [/tex] where a(t) is the scale factor. Show that, if a particle is shot from the origin at a time and with velocity [tex]t_{0},V_{0}[/tex] 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).

The Attempt at a Solution


First off, since the particle is shot from the origin, [tex]d\theta = d\phi = 0[/tex]. Doing this then dividing both sides of the metric by dt, I got [tex](\frac{ds}{dt})^{2} = -1 + a^{2}(t)(\frac{dr}{dt})^{2} [/tex]. 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, [tex]\frac{dr}{dt} = \frac{1}{a(t)}[/tex]. 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): [tex]dr = \frac{dt}{a(t)}[/tex] and, denoting by R the distance at which the particle comes to rest with respect to the co - moving frame, [tex]R = \int_{t_{0}}^{?} \frac{dt}{a(t)}[/tex] 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.
 
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  • #2
WannabeNewton said:
if a particle is shot from the origin at a time and with velocity [itex]t_{0},V_{0}[/itex] respectively, with respect to a co - moving observer

You didn't use this information.
WannabeNewton said:
First off, since the particle is shot from the origin, [itex]d\theta = d\phi = 0[/itex].
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.

WannabeNewton said:
Doing this then dividing both sides of the metric by dt, I got [tex](\frac{ds}{dt})^{2} = -1 + a^{2}(t)(\frac{dr}{dt})^{2} .[/tex]

Good.
WannabeNewton said:
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.

No. For the particle in question, what is the physical interpretation of [itex]ds^2[/itex]? What is the physical interpretation of [itex]ds^2 = 0[/itex]?

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
George Jones said:
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 [itex]ds^2[/itex]? What is the physical interpretation of [itex]ds^2 = 0[/itex]?

I think that you also have to use the fact that the particle moves on a geodesic (at least I used this).
Well for a massive particle [itex]ds^{2} < 0[/itex] or greater depending on the convention. [itex]ds^{2} = 0[/itex] 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
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
If I remember correctly, you are self-studying, so let's take some time to work through this.
WannabeNewton said:
Well for a massive particle [itex]ds^{2} < 0[/itex] or greater depending on the convention.
For the convention in the original post, the proper time of the particle is given by [itex]d\tau^2 = -ds^2[/itex]

WannabeNewton said:
[itex]ds^{2} = 0[/itex] would only be for a particle moving on a null geodesic.[/itex]

Right, so we don't want to use this, i.e., we don't want to set [itex]ds/dt = 0[/itex] since we are not dealing with a photon.
WannabeNewton said:
I don't get how to use the initial velocity and time given.

Let's leave this until near the end.
WannabeNewton said:
Do I have to solve the geodesic equation for a time - like geodesic and use the initial conditions they give?

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
[tex]ds^2 = -dt^2 + a \left(t\right)^2 dr^2 .[/tex]
WannabeNewton said:
I'm just confused as to how to factor in the fact that co - moving observers have constant coordinate distance.

Yes, observers moving with the Hubble flow (galaxies with no peculiar velocities) have constant co-moving coordinate [itex]r[/itex]. Because all galaxies have constant [itex]r[/itex], and because no galaxy is special, we can, WLOG, choose [itex]r = 0[/itex] 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 [itex]r[/itex]. 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 [itex]t \rightarrow \infty[/itex], the expansion of the universe "catches up" with the particle, and [itex]dr/dt \rightarrow 0[/itex].

This question asks for the particle's [itex]r[/itex] value when this happens, i.e., the question asks for the particle's [itex]r[/itex] value as [itex]t \rightarrow \infty[/itex].
WannabeNewton said:
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?

No, flat space means flat 3-dimensional space, not flat 4-dimensional spacetime, so flat FRW models are not flat spacetimes.

Keep asking questions.
 
  • #6
George Jones said:
No, flat space means flat 3-dimensional space, not flat 4-dimensional spacetime, so flat FRW models are not flat spacetimes.
Keep asking questions.
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 [itex]\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\partial L}{\partial \dot{r}}) = 0[/itex] so [itex]\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}[/itex] where k = const. So I can't replace proper time with coordinate time in this situation?
 
  • #7
WannabeNewton said:
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?

Yes, for co-moving objects, coordinate time is proper times, since for co-moving objects, since [itex]dr = d \theta = d \phi = 0[/itex] gives that [itex]ds^2 = - dt^2[/itex].

No, for the particle, coordinate time is not proper time, since [itex]dr \ne 0[/itex] for the particle.
WannabeNewton said:
I ask because if the particle is moving on a geodesic then using the euler - lagrange equations and for r I get [itex]\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\partial L}{\partial \dot{r}}) = 0[/itex]

Yes.
WannabeNewton said:
so [itex]\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}[/itex] where k = const.

Yes.
WannabeNewton said:
So I can't replace proper time with coordinate time in this situation?

I am not sure what you mean. For the particle, proper is not coordinate time, but you can use [itex]\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}[/itex] to eliminate proper time in the equations for the particle. Solve for [itex]d \tau[/itex], and substitute this into
[tex]ds^2 = -d \tau^2 = -dt^2 + a \left(t\right)^2 dr^2 .[/tex]
 
  • #8
George Jones said:
I am not sure what you mean. For the particle, proper is not coordinate time, but you can use [itex]\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}[/itex] to eliminate proper time in the equations for the particle. Solve for [itex]d \tau[/itex], and substitute this into
[tex]ds^2 = -d \tau^2 = -dt^2 + a \left(t\right)^2 dr^2 .[/tex]

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: [tex]\frac{\mathrm{d} r}{\mathrm{d} t} = \frac{k}{a(t)[a^{2}(t) + k^{2}]^{1/2}}[/tex]
 
  • #9
WannabeNewton said:
I wrongly assumed that the particle would not have a changing co - moving radial coordinate so I equated the proper time and coordinate time.

If the particle had an unchanging co-moving coordinate, then it wouldn't be "shot" anywhere. Note that if the universe expands forever, then [itex]\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}[/itex] gives that the particle asymptotically comes to rest with respect to a co-moving frame.
WannabeNewton said:
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: [tex]\frac{\mathrm{d} r}{\mathrm{d} t} = \frac{k}{a(t)[a^{2}(t) + k^{2}]^{1/2}}[/tex]

This seems right to me. The constant [itex]k[/itex] is determined from initial conditions.
 
  • #10
George Jones said:
If the particle had an unchanging co-moving coordinate, then it wouldn't be "shot" anywhere. Note that if the universe expands forever, then [itex]\frac{\mathrm{d} r}{\mathrm{d} \tau } = \frac{k}{a^{2}(t)}[/itex] gives that the particle asymptotically comes to rest with respect to a co-moving frame.


This seems right to me. The constant [itex]k[/itex] is determined from initial conditions.

Ok well then that came out to [itex]k^{2} = V_{0}^{2}a^{4}(t_{0}) - a^{2}(t_{0})[/itex]. 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
WannabeNewton said:
Ok well then that came out to [itex]k^{2} = V_{0}^{2}a^{4}(t_{0}) - a^{2}(t_{0})[/itex].

I'm not sure how you got this.
 
  • #12
George Jones said:
I'm not sure how you got this.

Don't you just use the fact that at [itex]t_{0}[/itex], dr / dt is [itex]V_{0}[/itex]?
 
  • #13
Two things: 1) Are you sure that you rearranged correctly?
 
  • #14
Probably not heh. I did it again and got [tex]k^{2} = \frac{V_{0}^{2}a^{4}(t_{0})}{1 - V_{0}^{2}a^{2}(t_{0})}[/tex].
 
  • #15
Yes. Now, the other thing: 2) although it seems reasonable, unfortunately, [itex]V_0[/itex] is not [itex]dr/dt[/itex] evaluated at [itex]t_0[/itex]. For a co=moving observer, proper distance is given by [itex]ar[/itex] (see equation (18.3)) and proper time is [itex]t[/itex], so speed measured by a co-moving observer is
[tex]\frac{d}{dt} \left( ar \right).[/tex]
 
  • #16
So since the speed measured by the comoving - observer is actually [itex]\dot{a(t)}r + a(t)\frac{\mathrm{d} r}{\mathrm{d} t}[/itex], for the initial velocity would the r in [itex]\dot{a(t)}r[/itex] just be the initial distance from the origin (in this case 0)?
 
  • #17
Right. You are given that at time [itex]t_0[/itex], the particle is at the origin and moving with speed [itex]V_0[/itex].
 
  • #18
Ah, so [itex]k = \frac{V_{0}a(t_{0})}{(1 - V^{2}_{0})^{1/2}} [/itex] it would seem.
 
  • #19
Yes. This makes sense physically because
[tex]k = a^2 \frac{dr}{d \tau} = a \frac{dt}{d \tau} a \frac{dr}{d t},[/tex]

and [itex]dt/d\tau = \gamma[/itex], just as it does in special relativity.
 
  • #20
Oh cool didn't see that. Thanks a bunch sir. This was the main problem that was bugging me from the chapter.
 

Related to Particle co - moving in flat Roberston - Walker space - time

What is particle co-moving in flat Robertson-Walker space-time?

Particle co-moving refers to the motion of a particle in a given space-time that is in sync with the expansion of the universe. In flat Robertson-Walker space-time, the universe is considered to be spatially flat and expanding at a constant rate.

How does particle co-moving affect the behavior of particles?

Particle co-moving allows for a simplified understanding of the behavior of particles in an expanding universe. It eliminates the need to account for the expansion of the universe in calculations and allows for a more straightforward interpretation of results.

What are the advantages of using particle co-moving in flat Robertson-Walker space-time?

The use of particle co-moving in flat Robertson-Walker space-time simplifies calculations and allows for a clearer understanding of the behavior of particles in an expanding universe. It also allows for a more accurate prediction of how particles will behave over time.

What are the limitations of using particle co-moving in flat Robertson-Walker space-time?

Particle co-moving assumes a constant rate of expansion of the universe, which may not always be the case. It also does not take into account the effects of gravity or other forces on the behavior of particles.

How does particle co-moving in flat Robertson-Walker space-time relate to the expanding universe?

Particle co-moving in flat Robertson-Walker space-time is a way to simplify the understanding and calculation of particle behavior in an expanding universe. It allows for a more accurate prediction of how particles will move and behave as the universe continues to expand.

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