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Newtonian and GR Friedman equation

  1. Oct 13, 2014 #1

    ChrisVer

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    Consider an homogeneous spherical universe, with mass density [itex] \mathcal{\rho}_M [/itex]. Then the total energy of some test-mass [itex]m[/itex] at radius [itex]R(t) [/itex] from the center is given by:

    [itex]E_{tot} = E_{kin} + E_{pot} = \frac{1}{2} m \dot{R}^2(t) - \frac{4 \pi}{3} Gm \mathcal{\rho}_M R^2(t) [/itex]

    Or that:

    [itex] \Big( \frac{\dot{R}}{R} \Big)^{2} \equiv H^2= \frac{8 \pi G}{3} \mathcal{\rho}_M + \frac{2E_{tot}}{mR^2}[/itex]

    Comparing with the Friedman equation, I have the feeling this derivation , either using the classical Newtonian mechanics (so not taking into consideration the curvature of spacetime and so GR) or the solution of Einstein Equations (Friedman equation) gives the same results by the identification of:
    [itex]k= - \frac{2E_{tot}}{m}[/itex]
    (connected to the total energy of the system, something to be expected of some "energy" existing in the curvature of space)
    where [itex]k[/itex] the curvature in the [itex]g_{rr}^{FRW} = \frac{1}{\sqrt{1-kr^2}}[/itex]

    Are these same results, something to be expected or not? To me it is something unexpected because of the difference between Newtonian and GR mechanics/ideas.
     
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  3. Oct 13, 2014 #2

    PeterDonis

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    The last term should have an ##R^3(t)## in it, shouldn't it? And that messes up the correspondence with the Friedmann equation, doesn't it?

    But ##k## is not determined by the "total energy of the system" (that term doesn't really have a well-defined meaning in the FRW model anyway). It's either 1, 0, or -1, depending on whether the universe is closed, flat, or open. In other words, it can only take on a few discrete values, whereas something determined by the "total energy of the system" should be able to take on a continuous set of values.

    As a general question, the answer to this is that you can sometimes find apparent correspondences between Newtonian and GR equations, but if you do, it will turn out to be a coincidence and not tell you anything really meaningful about the physics.

    In this particular case, the results aren't the same anyway, so it's a moot point. See above.
     
  4. Oct 14, 2014 #3

    ChrisVer

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    No. The grav. potential is [itex]U= G \frac{M m}{R} [/itex]... because the sphere vollume*density is the mass [itex] M= \mathcal{\rho}_M \frac{4}{3} \pi R^3[/itex], the grav. potential gets an [itex]\frac{R^3}{R}=R^2[/itex] dependence.

    As for [itex]k[/itex], I think sometimes in bibliography the corresponding term in the Friedman equation is the "curvature energy density". So it's some existing energy because of the curvature of the space. In classical mechanical view, the result for [itex]k=0[/itex] (flat) would be that the total energy would be 0, so the test mass would decelerate and stop asymptotically at infinity.
    For [itex]k>0 [/itex] the E<0 and so the universe would be closed - the test mass will be trapped.
    For [itex]k<0[/itex] the E>0 and the universe will expand forever.

    It's a rather funny coincidence :)
     
    Last edited: Oct 14, 2014
  5. Oct 14, 2014 #4

    PeterDonis

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    Ah, of course, you're right. Sorry for the confusion on my part.
     
  6. Oct 15, 2014 #5
    Did you use the shell theorem for potential energy? That wouldn't work. With the energies

    [tex]E_{pot} = - \frac{3}{5} \cdot \frac{{G \cdot m^2 }}{R}[/tex]

    [tex]E_{kin} = \frac{3}{5} \cdot \frac{{m \cdot \dot R^2 }}{2}[/tex]

    for the total mass distribution I get

    [tex]H^2 = \frac{{8 \cdot \pi \cdot G}}{3} \cdot \rho + \frac{{10}}{3} \cdot \frac{{E_{tot} }}{{m \cdot R^2 }}[/tex]

    When I got this result for the first time I was also very surprised and I still have no explanation. Maybe it is just a coincidence (similar to the Schwarzschild radius).
     
  7. Oct 15, 2014 #6

    ChrisVer

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    what is that kinetic and potential energy you've written down?
    Also you've used an [itex]m^2[/itex] which doesn't make "sense" I think for the model I'm considering...
     
  8. Oct 15, 2014 #7
    As mentioned above these are the kinetic and potential energy of the total spherical universe.

    Sorry, my m is the mass of the total universe and v the velocity of it's surface. It would have been better to use other symbols to avoid confusions.
     
  9. Oct 15, 2014 #8

    ChrisVer

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    The potential energy of a test mass [itex]m[/itex] at some position [itex]R(t)[/itex] depends on the mass existing in the shaded "volume" in the attached picture. http://prntscr.com/4whjko
    The shaded area will contain a mass: [itex]M(R)= \rho_M \times \frac{4}{3} \pi R^3[/itex] (it's the interior sphere after applying Gauss's law)

    So the [itex]E_{pot} = - G \frac{ m M(R) }{R} [/itex] that's why I don't understand your factor [itex]\frac{3}{5}[/itex] and the existence of the same masses [itex]m[/itex] in the numerator. In your case you are setting [itex]m=M(R)[/itex] and I don't understand the reason. So it's not what they denote, but how you used them.

    The kinetic energy again makes some sense apart from the [itex]3/5[/itex] again, and the way you define [itex]m[/itex] to be the total mass of the universe. That's because the universe is not moving wrt to itself (??) so there would be no velocity. So indeed your [itex]m[/itex] has to denote a test mass (maybe a galaxy) as in my image
     
  10. Oct 15, 2014 #9
    ...and on the mass outside the shaded volume. The potential energy of a test mass inside a sphere (not a ball) is equal to its potential energy on the surface of the sphere. To get the total potential energy of the the test mass you need to add the integral of its potential energy over all spheres from R to the surface of the entire mass distribution.

    It's better to use Gauss's law only. For a homogeneous and isotropic mass distribution it gives the tidal acceleration

    [tex]\ddot r = - \frac{4}{3} \cdot \pi \cdot G \cdot \rho \cdot r = - \frac{{4 \cdot \pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{3 \cdot r^2 }}[/tex]

    for two test masses with the distance r. Integration results in

    [tex]H^2 = \frac{{8 \cdot \pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{3 \cdot r}} + k = \frac{8}{3}\pi \cdot G \cdot \rho + k[/tex]

    This method does not allow to specify the integration constant k but it is not limited to finite spherical mass distributions. It also works for an infinite mass distribution.


    [edit]To get the potential and kinetic energy of the total mass distribution you need to integrate the potential energy and the kinetic energy of any particle over the volume of the total mass distribution. That's where the factor 3/5 and the square of the total mass comes from. But I have to admit that am not sure if I solved all integrals correctly.[/edit]
     
    Last edited: Oct 15, 2014
  11. Oct 15, 2014 #10

    PeterDonis

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    Can you give an example, please? I'm not sure the integral involved for an infinite mass distribution is convergent.
     
  12. Oct 16, 2014 #11
    I'm not sure, what you mean. The formulas in my last post apply to an infinite homogeneous mass distribution and the integral is convergent. I just forgot some r². The correct result should be

    [tex]H^2 = \frac{{8 \cdot \pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{3 \cdot r^3 }} + \frac{k}{{r^2 }} = \frac{8}{3} \cdot \pi \cdot G \cdot \rho + \frac{k}{{r^2 }}[/tex]

    You are right that the integral of the potential energy is divergent for infinite mass distributions. That's why ChrisVer's approach works for finite mass distributions only (btw: I double checked my corresponding calculations and did not found an error). But that doesn't affect my second approach. It don't uses potential energy or even forces but tidal accelerations only and the tidal accelerations within an infinite mass distribution are always finite and well defined.
     
  13. Oct 16, 2014 #12

    PeterDonis

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    I think I'm not understanding your notation. What are ##\rho_0## and ##r_0##?
     
  14. Oct 16, 2014 #13
    ##\rho_0## is the density of the mass distribution when the distance between two test particles is ##r_0##.
    ##\rho## is the density of the mass distribution when the distance between the same two test particles is ##r##
     
  15. Oct 16, 2014 #14

    PeterDonis

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    Ok, so you're basically adopting coordinates where one test particle is at rest at the origin, and ##r## is the radial coordinate of an arbitrary test particle in this chart. (Angular coordinates can be ignored since this universe is homogeneous and isotropic.) Then ##r## for a given test particle is a function of coordinate time ##t##, and so is the density ##\rho## (but ##\rho## is not a function of ##r## since the universe is homogeneous). ##\rho_0## and ##r_0## are just initial conditions.

    That all makes sense, but now I'm not sure how you derived your equation for ##\ddot{r}## (more precisely, the second equality, with ##\rho_0## and ##r_0## in it).
     
    Last edited: Oct 16, 2014
  16. Oct 17, 2014 #15

    It is not necessary that one particle is at rest at the origin. r is just the difference between the radial coordinates of two arbitrary test particles. Actually it is not even possible to determine if the particles are accelerated or not because the rest system of all particles are free falling.


    I started from the differential form of the gauss law for Newtonian gravity:


    [tex]\nabla \cdot \vec g = - 4 \cdot \pi \cdot G \cdot \rho[/tex]


    Due to isotropy the derivation of g should be identical in any direction:


    [tex]\frac{{dg_r }}{{dr}} = \frac{{dg_x }}{{dx}} = \frac{{dg_y }}{{dy}} = \frac{{dg_x }}{{dx}}[/tex]


    This results in


    [tex]\frac{{dg_r }}{{dr}} = - \frac{4}{3} \cdot \pi \cdot G \cdot \rho[/tex]


    and after integration finally in


    [tex]g_r = \ddot r = - \frac{4}{3} \cdot \pi \cdot G \cdot \rho \cdot r[/tex]


    Now I just need to include the variation of the density. Imagine a spherical volume with initial radius ##r_0## and initial density ##\rho_0## within the mass distribution. The mass within this sphere remains constant even when its radius changes:


    [tex]m = \frac{4}{3} \cdot \pi \cdot \rho \cdot r^3 = \frac{4}{3} \cdot \pi \cdot \rho _0 \cdot r_0^3[/tex]


    This gives


    [tex]\ddot r = - \frac{4}{3} \frac{{\pi \cdot G \cdot \rho _0 \cdot r_0^3 }}{{r^2 }}[/tex]


    This works because during collapse or expansion of the mass distribution all distances scale with the same factor. Replacing the radius by the scale factor


    [tex]a = \frac{r}{{r_0 }}[/tex]


    results in an equation of the same structure:


    [tex]\ddot a = - \frac{4}{3} \frac{{\pi \cdot G \cdot \rho _0 }}{{a^2 }}[/tex]
     
  17. Oct 17, 2014 #16

    PeterDonis

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    Ok, this makes the derivation clearer; and it makes the issue with divergence clearer as well:

    Over what range of integration? What this integral is saying is that the magnitude of ##\vec{g}## is linear in ##r##, so for an infinite universe, ##\vec{g}## can increase without bound as ##r \rightarrow \infty##. The integral does not converge.

    Reframing the integral in terms of ##m = (4/3) \pi G \rho_0 r_0^3## doesn't help either, because in an infinite universe, ##m## also increases without bound as ##r_0## increases, i.e., as you consider particles with larger and larger initial separations.

    Also, physically, even if we leave out the above issues, the picture given by these equations isn't really consistent. For example, suppose I have particle A which starts out halfway between particles B and C. If I compute particle A's ##\ddot{r}## based on its separation from particle B, I get an acceleration in one direction; but if I compute it based on A's separation from particle C, I get an acceleration in the opposite direction. Both can't be right, since they contradict each other.

    More generally, if the universe is homogeneous and isotropic, then a given particle moving freely at any point in the universe should not be accelerated in any particular direction, because that would pick out a particular direction and violate isotropy. But your equations seem to be saying that a given particle will be accelerated in some particular direction.

    If your response is that the actual acceleration of a given particle is obtained by summing the results for all possible values of ##r## and all possible directions, then we're right back to the divergence issue. And even if we leave that out again and say that the sum must be zero (by isotropy), that invalidates any useful concept of "potential energy", since ##\vec{g}## is supposed to be the gradient of the potential energy, so if it's zero, the "potential energy" must be the same everywhere, which amounts to saying it's physically meaningless.
     
  18. Oct 17, 2014 #17
    Over the distance between two test particles.

    No, you don't get an acceleration of a single particle. You get the second derivation of the distance between two particles with respect to time.

    Homogeneity and isotropy do not require constant escape velocities.

    The actual acceleration of a given particle depends on the frame of reference. Each particle can be assumed to be at rest. Than the acceleration of all other particles result from their distance from this particle according to the equations above. But there is no absolute acceleration.

    If you have problems with the equations for the distance between specific test particles, just use the equation with the scale factor instead.
     
  19. Oct 17, 2014 #18

    PeterDonis

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    Same problem: any given particle can be paired with multiple other particles, and the results obtained in the individual cases will not be consistent. The fact that you are not using a well-defined coordinate chart (or, if you like, you're switching coordinate charts depending on which pair of particles you are looking at) may be preventing you from seeing this.

    I would say that homogeneity and isotropy are inconsistent with the concept of escape velocity, just as they are inconsistent with the concept of gravitational potential energy. Perhaps you are using a non-standard definition of these terms: can you state more precisely how you are defining them?
     
  20. Oct 17, 2014 #19
    I don't know what you mean. Can you please demonstrate this with a specific example?

    That means all modern cosmological models are inconsistent with the cosmological principle?
     
  21. Oct 17, 2014 #20

    PeterDonis

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    How so? The cosmological principle doesn't say anything about escape velocity.
     
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