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

  1. Jun 21, 2011 #1

    WannabeNewton

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    1. The problem statement, all variables and given/known data
    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).

    3. 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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  3. Jun 22, 2011 #2

    George Jones

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    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).

    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.
     
  4. Jun 23, 2011 #3

    WannabeNewton

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    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.
     
  5. Jun 23, 2011 #4

    WannabeNewton

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    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?
     
  6. Jun 24, 2011 #5

    George Jones

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    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 [itex]d\tau^2 = -ds^2[/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.
    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
    [tex]ds^2 = -dt^2 + a \left(t\right)^2 dr^2 .[/tex]
    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].
    No, flat space means flat 3-dimensional space, not flat 4-dimensional spacetime, so flat FRW models are not flat spacetimes.

    Keep asking questions.
     
  7. Jun 24, 2011 #6

    WannabeNewton

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    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?
     
  8. Jun 24, 2011 #7

    George Jones

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    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.
    Yes.
    Yes.
    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]
     
  9. Jun 24, 2011 #8

    WannabeNewton

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    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]
     
  10. Jun 24, 2011 #9

    George Jones

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    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.
     
  11. Jun 24, 2011 #10

    WannabeNewton

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    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.
     
  12. Jun 25, 2011 #11

    George Jones

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    I'm not sure how you got this.
     
  13. Jun 25, 2011 #12

    WannabeNewton

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    Don't you just use the fact that at [itex]t_{0}[/itex], dr / dt is [itex]V_{0}[/itex]?
     
  14. Jun 25, 2011 #13

    George Jones

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    Two things: 1) Are you sure that you rearranged correctly?
     
  15. Jun 25, 2011 #14

    WannabeNewton

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    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].
     
  16. Jun 25, 2011 #15

    George Jones

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    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]
     
  17. Jun 25, 2011 #16

    WannabeNewton

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    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)?
     
  18. Jun 25, 2011 #17

    George Jones

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    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].
     
  19. Jun 25, 2011 #18

    WannabeNewton

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    Ah, so [itex]k = \frac{V_{0}a(t_{0})}{(1 - V^{2}_{0})^{1/2}} [/itex] it would seem.
     
  20. Jun 25, 2011 #19

    George Jones

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    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.
     
  21. Jun 25, 2011 #20

    WannabeNewton

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