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Force to resist expansion

  1. Feb 14, 2005 #1

    wolram

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    Difficult to explain this question but i will try, Space is expanding
    and un gravitationally bound matter is carried along with it, so two
    unbound bodies will get further apart, now gravity is a weak" force",
    yet it can resist the expansion," i remember and astroid that has its
    own tiny moon", so the force to resist expansion must be tiny, so
    how do we know that when space expands it carries matter with
    it at exactly the same rate," no slippage". and what "force" holds
    matter to space, this again must be tiny, but non zero as matter
    just would not move with space.
     
  2. jcsd
  3. Feb 14, 2005 #2

    jcsd

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    space if you like is the configuartion of matter so there's no force holding holding matter in postion (for start in one frame matter can be moving, but in it's rest frame it's zero), what is meant by space ex[anding is that the density is decreasing/the distance between objects is tending to increase where the rate is a function of distance.
     
  4. Feb 14, 2005 #3

    wolram

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    sorry JSCD but what you have said is quirky and is no real explanation
    there has to be some tie with space and matter to allow expansion to
    work.
     
  5. Feb 14, 2005 #4

    jcsd

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    I'll try to explain better:

    1) we know expansion is isoptopic because we observe iot to be isotropic and GR predicts taht in oour approcxiamtely isotropic universe expansion will be approximately isotropic

    2) There is no identiifable force that could be accurately described as holding matter to space.

    Given a point in space at a certain time, there is no absolute way of identifying that point in space with a point in space at a different time, so it is hard to make sense of your question and why space is best seen as the relationship between objectsrather than a phsyical object in it's own right.
     
  6. Feb 15, 2005 #5

    Chronos

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    If you translate the hubble flow to a gravitational equivalent force, you will get a rough idea where mass density is insufficient to resist the hubble flow.
     
  7. Feb 15, 2005 #6

    wolram

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    By Chronos
    2) There is no identiifable force that could be accurately described as holding matter to space.
    ----------------------------------------------------------------------------------------------
    You will have to be patient with me on this one, as i can not see
    how matter moves with the expanding space if it is not somehow attracted
    to it, "There must be an action to create a reaction", so why should
    mass move just because the volume it lives in is increased, this
    to me is a purely mechanical problem, ie no attachment no force
    no movement.
     
  8. Feb 15, 2005 #7

    wolram

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    I will be happy if there is no reasonable explanation for this question
    as it shows the nonsensical explanation for expanding of"space time", if gravity can resist expansion then there must be something for it to resists against
     
    Last edited: Feb 15, 2005
  9. Feb 15, 2005 #8

    pervect

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    Basically we are running into terminology / conceptual model diffrerences.

    GR does not view gravity as being due to a force. Your question is phrased entirely in the notion that gravity is a force. The result is a conflict of viewpoints.

    From the GR point of view, two particles in an expanding universe do not experience any mutual force. But neither does the Earth orbiting around the sun experienece a force. The particles (and the Earth) are simply following geodesics in space-time.

    As long as the relative velocities are low, and the geometry of space is not too badly distorted (i.e. space is reasonably flat), you can re-interpret the curvature of space-time as a sort of force. You can do that with the Hubble expansion, too, because space in our universe is flat, as long as the velocities are low enough. Just look at the acceleration that two particles a distance 'd' apart would experience, and say that that acceleration is due to a "force".

    The notion of "friction" between space and matter is probably just going to confuse things even further and is best dropped - realize that from the GR point of view, there is no force at all, and that the conversion into force is just sort of a translation from a GR way of thinking to semi-Newtonian way of thinking.

    Here is (I think this is right) how to get an actual number for acceleration out of the hubble constant.

    We have v=Hr, where v is the relative velocity, H is the Hubble constant, and r is the distance.

    Thus dv/dt = H dr/dt. But dr/dt is just v, which is H*r.

    Thus dv/dt = H^2 r

    The exact value of H is currentl being debated, I am using for this example H = 70 km/ megaparasec

    Using google to do the conversion of units

    google unit conversion

    I get 5*10^-33 m/s^2 for r = distance = 1 km for this value of H
     
  10. Feb 15, 2005 #9

    hellfire

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    Note that the Hubble parameter depends on time. Therefore:

    [tex]\dot v = \dot H r + \dot r H[/tex]
    [tex]a = H\dot r + H^2 r[/tex]

    You would get the same result with the definition of H in terms of the scale factor R

    [tex]H = \frac{\dot R}{R}[/tex]

    But usualy the deceleration parameter is used to define the acceleration of expansion:

    [tex]q = - \frac{R \ddot R}{\dot R^2} = - \frac{\dot H}{H^2} - 1[/tex]

    The accepted value is around q ~ -0.6. This is an adimensional quantity and is usually the referred one when talking about acceleration of expansion.

    You could measure also a "kind of acceleration" with [itex]\dot H[/itex]. However, the meaning of the first time derivative of the Hubble parameter is very different than the meaning of the deceleration parameter.
     
    Last edited: Feb 15, 2005
  11. Feb 16, 2005 #10

    pervect

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    Thanks for the correction.

    I'm not quite sure what the 'deacceleration' parameter you refer to is, I haven't run across it in my textbooks :-(.

    I think that what I'm actually looking for is

    [tex]\frac{\ddot{a}}{a}[/tex]

    where a(t) is defined by it's role in the flat FRW metric

    ds^2 = dt^2 + a(t)^2(dx^2+dy^2+dz^2)

    I think you are calling what I call a(t) R(t) (?).

    The above expression should give the ratio of the acceleration of two nearby geodesics to their distance. (It's the value of [tex]\mbox{R^x{}_{txt}}[/tex]).

    However, upon review, I'm not really confident in how to calculate this expression from the Hubble constant.

    A numerical example for the relative acceleration of two geodesics 1km apart in m/s^2 would be helpful, if it's not too much work.
     
  12. Feb 16, 2005 #11
    Expansion can be looked upon as creating an effective spatial acceleration - if the Hubble sphere of radius R is considered to be uniformly expanding at velocity c, then the volumetric acceleration is 8(pi)R(c^2). Applying the divergence theorem, the effective acceleration acting outwardly by expanding space is 2(c^2)/R. For a central mass M exerting a gravitaional force on a mass wafted along by expansion, The balance point r would be in the neighborhood of (RGM/2c^2)^1/2
     
  13. Feb 16, 2005 #12

    wolram

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    Thank you all for replies, but they don't really address the problem,
    ie why do two ungravitationly bound bodies get further apart, when
    no force has acted on them and as is said ,space is not exerting any
    influence on them.
    In a purely mechanical way a reaction has to start with an action, and
    increasing a volume would have no effect on the distance between
    two objects.
    so if mass is not somehow tied to space, how does mass keep in step
    with spacial expansion
    It could be that space is expanding faster than we think it is, how
    could we tell if the distance change between two objects is a % of
    spacial expansion?
     
    Last edited: Feb 16, 2005
  14. Feb 16, 2005 #13

    hellfire

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    OK, the value you are talking about is the 00 component of the Ricci tensor:

    [tex]R_{00} = -3 \frac{\ddot a}{a}[/tex]

    (In my previous post I was using "R" instead of "a" for the scale factor, because I did use "a" before for the acceleration. But now I will use "a" for the scale factor). In terms of the Hubble parameter you can compute:

    [tex]\frac{\ddot a}{a} = H^2 + \dot H[/tex]

    This is the same as:

    [tex]\frac{\ddot a}{a} = - H^2 q[/tex]

    With q = -0.6 and H = 71 Km / s Mpc: 3.12 × 10-36 s-2. This is 3.12 × 10-33 m / s-2 for every km.

    If you are not happy with the deceleration parameter q, you could use the fact that the 00 component of the Ricci tensor leads to the first Friedmann equation:

    [tex] -3 \frac{\ddot a}{a} = \frac{4 \pi G}{c^2} (\rho + 3p) - \frac{\Lambda}{c^2}[/tex]

    If one includes the cosmological constant as a dark energy with equation of state [itex]p = - \rho[/itex], this can be written as:

    [tex] -3 \frac{\ddot a}{a} = \frac{4 \pi G}{c^2}(\rho + 3p) [/tex]

    This should allow to calculate R00, considering 0.7 of matter and 0.3 of dark energy component (as fractions of the critial density).

    If you assume only a dark energy content equal to the critical density you will see (using the equation of state and inserting the definition of critical density in the previous formula) that:

    [tex]\frac{\ddot a}{a} = H^2[/tex]

    Comparing with the second formula in this post, this means [itex]\dot H = 0[/itex] (the Hubble parameter does not change with time) and then your formula applies.

    It is not clear to me is that this quantity R00 is the acceleration between two nearby geodesics. Could you elaborate on that?
     
    Last edited: Feb 16, 2005
  15. Feb 16, 2005 #14

    hellfire

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    I realize now you were not talking about R00, but about R1010, and R00 = R1010 + R2020 + R3030; all of them are equal in an homogeneous and isotropic space which explains the factor 3. Right? Anyway, could you explain how R1010 describes the acceleration for the geodesic deviation in this case?
     
  16. Feb 16, 2005 #15

    pervect

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    That much I can do right away, I have yet to go over your previous responses. The basic idea is that the Riemann tensor can be physically interpreted in terms of tidal forces via the geodesic deviation equation.

    The geodesic deviation equation

    [tex]
    \frac{d^2 x^a}{dt^2} = R^a{}_{bcd} u^b \xi^c u^d
    [/tex]

    gives the acceleration between a pair of points on neighboring geodesics. We need to specify each geodesic, which can be done by specifying a point in space, and a four-velocity, which defines the geodesic. The relative acceleration, though, depends only on the relative position. So we need to specify two four velocities, and one relative position.

    Here u^b and u^d are the four-velocities (one four velocity for each geodesic), and [tex]\mbox{\xi^c}[/tex] is the spatial separation of the pair of geodesics.


    [tex]
    R^a{}_{bcd} u^b u^d
    [/tex]

    maps a vector [tex]\xi^c[/tex] that describes an spatial offset into a relative acceleration.


    Now we assume that u^b and u^d both represent zero velocity (which means they are unit vectors in the time directon). This means we are interested in the tidal force on an object where both ends are stationary

    Next, in this case, we are interested in the acceleration in the x direction introduced by a displacement between geodesics in the x direction. This mean that [tex]\mbox{\xi}[/tex] is a unit vector in the x direction.

    The end result is that with these values for [tex]u^b, \xi^c, u^d[/tex], we in essence "pick out" one component of the Riemann

    This is [tex]R^x{}_{txt}[/tex]

    It gives the tidal force in the x direction due to an x offset - it's the x component of the "stretching" tidal force. By isotropy, the y and z stretching components will be equal.

    Conveniently, we don't have to worry too much about scale factors with these combinations of superscripts and subscripts. If we scale everything in the x direction by a factor of .1, both the separation vector and the acceleration vector get scaled by the same amount, so the ratio remains constant. The only thing we need to worry about is that the time vector is scaled properly. Whle four velocities are always one in minkowski space they might not be one in a general metric.

    In this case g_00 is -1, so there isn't a problem.

    [add]
    Note that [tex]R^y{}_{txt}[/tex] is zero in this case. If it existed, it would represent a tidal torque, rather than a tidal stretching force. The diagonal elements of [tex]R^a{}_{bcd} u^b u^d[/tex] are all stretching or compressing tidal forces, the off-diagonal components are all tidal torques.
     
    Last edited: Feb 16, 2005
  17. Feb 16, 2005 #16

    wolram

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    Please feel free to debate theory, which has no bounds on reality,
    im sure one day it will be realised that your ricci tensors are in a twist.
     
  18. Feb 16, 2005 #17

    pervect

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    Sorry to get technical on you, but before I can give you a correct numerical answer, I have to go through the equations to be sure that I'm computing it correctly myself. Besides, it is a good exercise (you've lucked out and raised an interesting question - sorry if you don't feel lucky).

    If you are not interested in the numbers, but just the concepts, think of a baloon being inflated with little dots on it. (That's the standard anology).

    As the baloon inflates, the dots get further apart. It might be possible to attribute the dots getting further apart to "forces", but it's better not to at this point.

    The fact that the dots are moving further apart means that they have some relative velocity. The rate of change of this velocity will be their relative acceleration.

    It turns out that the velocity and the acceleration will be proportioanl to distance, at least for two points which are reasonably close together. I.e. points which are twice as far apart will be moving away from each other twice as fast, and will be accelerating away from each other twice as hard as well.

    It remains only to work out the exact value of the constant (the ratio of the acceleration to the distance). Dimensionally, it will be (meters/sec^2) / meters, or 1/sec^2. Roughly speaking, it should be something like 1 / (age of the universe)^2, but maybe there are some additional factors of 2, or 8 pi, or something-or-other involved. Hence the discussion.

    Switching viewpoints back to the "force" point of view, you can interpret this physically as a "tidal force", othe same as the tidal force that the moon (and sun, for that matter) exerts on the Earth. Like the tidal force the moon exerts on the Earth, you can explain it in terms of matter if you like (in this case, you need to consider contributions from all the matter in the uinverse). You can also consider the tidal force to be due to the curvature of space-time (just as you can say that the curvature of space-time due to the moon and/or sun is what causes tides on the Earth).

    I call it a tidal force because it is a stretching force (acceleration) that is directly proportional to distance (as long as the distance is small). This is exactly what a tidal force is. This is exactly what causes the Earth's tides.

    It doesn't make a lot of sense (IMO) to attribute these tidal forces to friction. Friction is not really involved.
     
  19. Feb 16, 2005 #18

    pervect

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    Look on the bright side - whatever replaces General Relativity will probably be in complexity and ease of calculation to GR as GR is to Newtonian gravity. We'll all be working out "simple" problems in GR wherever we can, saving the more complex calculations in the new theory for cases where the additonal complexity is needed.

    Heck, we might even be in the unfortunate position of M-theory, where we can't actually calculate any numbers at all with the theory....
     
  20. Feb 16, 2005 #19

    pervect

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    Yes, the factor of 3 comes from [tex]R^0{}_0 = R^0{}_{000} + R^1{}_{010} + R^2{}_{020}+R^3{}_{030}[/tex] the first of which is zero, and the last three of which are equal. (The ricci just contracts slots 1 and 3 of the Riemann).

    The bottom line is that my first answer was off by a factor of -q, I should multiply it by .6, right?
     
    Last edited: Feb 16, 2005
  21. Feb 16, 2005 #20

    pervect

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    OK, I'm totally convinced. H = [tex]\frac{\dot{a}}{a}[/tex] as you said

    http://scienceworld.wolfram.com/physics/ExpansionParameter.html

    and q, which I was not previously familair with is indeed, as you also said

    [tex]-\frac{\ddot{a} a}{\dot{a}^2} = -\frac{\ddot{a}}{H^2 a}[/tex]

    http://scienceworld.wolfram.com/physics/DecelerationParameter.html

    So obviously [tex]\frac{\ddot{a}}{a} = -qH^2[/tex].

    We can use either an argument similar to the one given in the first link above ("expansion parameter"), or direct calculation of the Riemann to find that the acceleration / unit distance is indeed [tex]\frac{\ddot{a}}{a}[/tex]

    Actually measuring q is apparently difficult, though.

    http://www.astro.psu.edu/users/rbc/a480/lec13n.pdf

    Thank you, Hellfire, for an interesting and enlightening post.
     
    Last edited: Feb 16, 2005
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