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Question about the expansion of space.

  1. Jul 11, 2012 #1
    If space is expanding and I put a ruler inside this expanding space, the ruler should expand with the space. Therefore, inside this expanding space two particles would always measure, to an observer with the ruler inside the same space, the same distance away, even as that space expands since the ruler is expanding as well. So how would those particles ever "rip" apart in accordance with the Big Rip theory?
     
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  3. Jul 11, 2012 #2

    marcus

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    A lot of people are fooled by the words "expansion of space" and how they talk in the media.

    What is meant by it is actually based on the concept of a STATIONARY OBSERVER. Someone who isn't moving relative to the uniform soup of ancient (microwave) light that fills the universe.

    I can try to explain that concept to you, or someone else can. It's basic to any understanding of expansion cosmology. the ancient light is called the CMB (cosmic microwave background) and another name for stationary observer is one at CMB rest. That means not moving relative to the CMB.

    All "expansion of space" means is that distances between stationary observers are growing by a certain very small percentage every year. It's about 1/140 of one percent per million years. That is the same as growing a millionth of a percent every 140 years.

    This rate (according to the current accepted model) is expected to continue declining forever but more and more slowly so it levels out at around 1/160 of one percent per million years.

    Professional researchers don't talk about "big rip" any more. Those scenarios kinda went out of style some years ago.

    The expansion is too gradual to notice except with very large distances, beyond the range of things held together by their own forces. Having to wait a million years and then measure 1/140 of one percent is very slow. Think about it. It only makes sense to talk about that slow increase of very very large distances, where a millionth of a percent in 140 years really means something. (Distances beyond galaxies and clusters of galaxies held together by their own forces.)
     
    Last edited: Jul 11, 2012
  4. Jul 11, 2012 #3
    No, the ruler would not expand. The metric of space expands, not the objects that reside within it.

    The idea behind the 'Big Rip' has nothing to do with the ordinary expansion of the universe. Instead, it is a scenario regarding dark energy (the cosmological constant), the negative pressure that accelerates the expansion of the universe. You can think of it as a type of 'anti-gravity', if you like. Because we believe it has a constant value, the expansion of the universe will continue to accelerate. However, in the Big Rip, the strength of dark energy increases, eventually overcoming the forces that hold together atoms, prying everything apart.

    But, we observe that dark energy matches the characteristics of a cosmological constant. That is, it will never change in value. It will continue to accelerate the expansion of the universe, but will never rip matter apart.

    EDIT: Ah, Marcus beat me to it.
     
  5. Jul 11, 2012 #4

    marcus

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    BTW Weinie, there is an FAQ about CMB rest in our FAQ section.
    Maybe you should read it. Believe me it is basic to getting the idea of what is really meant by the pattern of expanding distances called Hubble law.

    You can't take the words they use in the media seriously.

    Hi Mark M, sorry I beat you to it. You are most welcome to take Weiner's question over! I'd consider it a favor. BTW enjoyed your comments in BtSM forum :biggrin:
     
    Last edited: Jul 11, 2012
  6. Jul 11, 2012 #5
    What if the ruler is very, very long (1 Mpc or longer)? Will it be pulled apart?
     
  7. Jul 12, 2012 #6
    Well, dark energy could pull the atoms apart, no matter what the size is (if it was strong enough). The point I was making is that space expands, not objects in space.

    The force exerted by dark enough could, if strong enough, pull this object apart.
     
  8. Jul 12, 2012 #7

    Chronos

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    How much is 70 km/s divided by a megaparsec?
     
  9. Jul 12, 2012 #8

    marcus

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    It's a frequency (a number per unit time). Think of it as the fractional increase per unit time. You can express it as a number per second. or per year, or per million years. Just put it into the google calculator. WMAP measured 70.4 so let's use that. Paste the blue thing into the google search window:

    70.4 km/s per Mpc

    That will give you the fractional increase PER SECOND. A very small number per second. If you want it as a percent per million years then tell the google that. You can tell the google what UNITS you want the answer expressed in. Paste the red thing into the window:

    70.4 km/s per Mpc in percent per million years

    It will give 0.0072 percent per million years. And 0.0072 = 1/139

    That is why I say 1/139 percent per million years. Or (round number) 1/140 percent per million years.
     
  10. Jul 12, 2012 #9

    marcus

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    We have no scientific evidence that "dark energy" is an energy or a force. So far the evidence suggests that it is simply a constant curvature in the law of gravity. the effect at 1 Mpc scale is very slight.

    The dominant effect at a fixed 1 Mpc scale would simply be the already existing expansion of distance, 70.4 km/s. This speed is actually DECELERATING, getting smaller. Very slowly tending towards 60 km/s. (No contradiction of accelerated expansion here, subtle point.)

    Our local group of galaxies is held together by its own forces, like a metal ruler is held together by its (metallic crystal bond) forces.
    Our local group of galaxies is on the order of a Mpc wide.

    It is not being torn apart.

    Suppose you had a 1 Mpc cable and put observers A and B at the two ends, and you stabilized A so he was at CMB rest (stationary with respect to the ancient light, no doppler hotspot in his sky).

    then all you would need to do for observer B would be start him off at a speed of 70.4 km/s relative to CMB and let him coast. Everything would be calm and unstressfull. A would be at rest relative to CMB at one end of the cable and B would be coasting along at 70.4 km/s relative to CMB. there would be no extra tension on the cable from pulling him.
    Or you could start each observer moving at 35.2 km/s, "split the difference".

    The solar system is moving at about 370 km/s relative to CMB, so 70.4 km/s doesn't seem like a big deal however you split it.

    Over the long haul (billions of years) there would actually be a tendency for the cable to go slack because H is slated to decline to around 60 km/s per Mpc. So you would provide B with some retrorockets to slow himself down just to keep the cable stretched out straight.

    This does not contradict what we know about expansion accelerating. that is something that affects the scalefactor, not H. The Hubble rate H(t) is slated to decline from around 70 to around 60. That is what affects our cable since it is and remains 1 Mpc long.

    the "dark energy" alias the cosmo constant that we have (and which so far seems constant) is not going to pull things apart at the Mpc scale.
    Our local group of galaxies is expected to stay together. Indeed Andromeda and Milkyway are expected to merge, and the others may join us.
     
  11. Jul 12, 2012 #10

    andrewkirk

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    No because the particles of the ruler are held together by electrostatic forces, which are enormously stronger than the expansionary 'force'.

    If you had a ruler-shaped collection of particles with no charge and no mass (neutrinos?) and not subject to either the strong or weak nuclear forces, initially all stationary with respect to one another [I suspect this set of conditions is impossible], then they would almost expand along with the space in which they were situated. The extent to which they would not would be owing to the fact that if their initial proper velocity relative to one another is nil then some particles will have an initial nonzero spatial velocity relative to the CMB frame. For instance if the centre of the group is stationary relative to the CMB frame, the particles on either end will have equal and opposite nonzero velocities relative to the CMB frame, heading towards the centre of the 'ruler'.

    This is described quite nicely in the 2006 paper by Barnes, Frances, James & Lewis http://arxiv.org/abs/astro-ph/0609271v1
     
  12. Jul 12, 2012 #11

    George Jones

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    Also, see the article "In an expanding universe, what doesn't expand?" by Price and Romano,

    http://arxiv.org/abs/gr-qc/0508052,

    which was published in the May 2012 issue of the American Journal of Physics.
     
  13. Jul 12, 2012 #12

    I skimmed the article and the conclusion seems to be :

     
    Last edited: Jul 12, 2012
  14. Jul 12, 2012 #13

    andrewkirk

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    I skimmed the Price and Romano article too and I agree. It would be easy to misinterpret the conclusion, which includes:

    we show that this bound system either completely follows the cosmological expansion, or — after initial transients — completely ignores it.

    This initially sounded to me like it's saying that in the latter case the state of an atom is exactly the same as it would be with no expansionary forces. But I think it's not, because of the qualification about initial transients.

    The argument seems to be about potential wells and whether the expansionary 'force' will cause the electron to break out of the electrostatic potential well around the atom and escape completely. They show that in some (probably almost all) cases this doesn't happen. But that doesn't mean the electron doesn't follow an ever so slightly different 'orbit' within the potential well than it would if there were no expansionary 'force'. That slight difference would be covered by what they call 'initial transients', although I think the words are poorly chosen because they could be interpreted as suggesting the tiny effect disappears after a while and the body of their paper doesn't seem to indicate that it does.

    It seems very odd to be discussing orbits of electrons as though the Rutherford model of the atom were still considered valid, but perhaps it's close enough for the purposes of this issue.
     
  15. Jul 12, 2012 #14

    marcus

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    The "initial transients" also occur in my example of the 1 Mpc cable.There would be an initial tug on the cable until observer B is up to speed (70.4 km/s)Assuming you hold observer A at CMB rest.
    But I imagined equipping B with rockets to boost him to that speed before joining him to the cable, that boost is the "initial transient."

    After that there is essentially no tension on the cable. Expansion does not need to pull things apart. And in this case the slight measured "acceleration" has no such effect either.
    Not sure why one should bother with the Rutherford atom. Expansion and acceleration would not even strain a system that is 1 Megaparsec (3.26 million light years) long. So one would not expect strain on a smaller system bound by its own forces (metal bonds, organic molecular bonds, electrostatic attraction, gravity, whatever)
     
    Last edited: Jul 12, 2012
  16. Jul 13, 2012 #15

    andrewkirk

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    Marcus that sounds rather like the 'tethered galaxy' thought experiment.
    It seemed to me that there would remain a tension in the cable, at least based on SR-type reasoning as set out below (I haven't tried to work out what the geodesic for B would be under GR):

    Take A as the origin of an inertial coordinate system. Then for a CMB-stationary, free-falling object C near B at t=0 let x(t) denote the proper distance from A to C. Assume the Hubble parameter is constant at 60.

    Then by Hubble's law we have x' = Hx, where a prime ' indicates differentiation with respect to t. This DE has solution x(t) = x(0)eHt. Hence C is subject to an acceleration x'' = H2x in a direction away from A. We can consider this as a force per unit mass of H2x, caused by dark energy.

    If we assume the force of dark energy is independent of an object's velocity then B is subject to the same force as C, so there should always be a tension in the cable of mBH2d(A,B) where m is B's mass and d(A,B) is the distance from A to B.

    We can consider another free-falling object D that starts off next to B but with the same initial velocity as B, ie heading towards A at velocity Hd(A,B) relative to the CMB frame. Because of the force mDH2d(A,B) on D, it will accelerate away from A.

    Does that make sense? Is the assumption that the dark energy forces on B and C would be identical reasonable? Would it be different if we tried to calculate the trajectories via GR geodesics rather than by SR? What about the fact that H is expected to reduce over time from currently around 70 to asymptotically approach 60?
     
    Last edited: Jul 13, 2012
  17. Jul 13, 2012 #16
    That's what I'm thinking as well. The cable should be in the class of things not affected by cosmological expansion since the mechanical/electromagnetic forces in the tether are presumably unaffected. In order for the endpoints to stay at rest relative to their local CMB, the cable's length must increase over time. No doubt there are complications here with different frames at the endpoints, but surely the distance between its constituent atoms must grow and increase the mechanical tension, eventually breaking the cable.

    Is this wrong?
     
  18. Jul 13, 2012 #17

    marcus

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    Andrew and Allo, H2x is not an acceleration ("force per unit mass") caused by dark energy.
    It would be there even if the cosmological constant were zero, and H(t) still taken to be approximately constant. IOW it's not caused by "dark energy", it's just a feature of ordinary Hubble law expansion in the special case where H(t) is imagined to be non-decreasing.

    We know that H(t) is decreasing so that the equation x'' = H2x is not exactly correct. For much of history H(t) has decreased rapidly so that x" has actually been declining. A distance between two stationary observers would be increasing but with a decreasing slope. Now the curve has inflected and slope is increasing but it's still not exponential growth.

    In any case it's probably not a good idea to think of x" as corresponding to a force on objects moving thru space. The Hubble law is about distances between stationary observers, x(t) is a distance, x'(t) is the slope of the x(t) curve and x"(t) tells how the slope changes. It is a feature of how distances have been and are behaving. Observers at CMB rest don't feel any acceleration, they just get farther apart from each other without going anywhere. They don't experience anything we would recognize as force.

    I think the practical problem would be to KEEP enough tension in the cable (after "initial transients" of the startup period are over.) It would tend to buckle under its own gravity so one might have to provide a little bit of extra tension artificially to keep it stretched out. Or replace it by a stiff rod of some wonderfully strong light material. But I said cable---so be it.:biggrin:

    Remember that the cable is exactly one Megaparsec long.
    So the two observers are always exactly 1 Mpc apart.
    One is at CMB rest.
    The other is moving 70.4 km/s relative to CMB towards his buddy. So he always stays 1 Mpc from him.
    Notice that stationary observers at exactly that distance would find the distance increasing at exactly 70.4 km/s. So the guy's speed (relative to CMB) towards his buddy exactly cancels the increase in distance.

    He doesn't need the wire to keep him always at the same distance of 1 Mpc. (I'm neglecting two small corrections to the picture, law of conservation of momentum does not hold exactly in the CMB rest frame, there is a very very gentle deceleration. But then that is partly balanced by the fact that the number 70.4 is also very slowly declining.) Essentially he doesn't need the cable.
     
    Last edited: Jul 13, 2012
  19. Jul 14, 2012 #18

    andrewkirk

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    Are you saying that even if [itex]\Lambda[/itex] were zero, H(t) would still be bounded below by some positive number?
    Looking at the second Friedmann equation [itex] \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}[/itex] I would have thought that in that case H(t) would asymptotically approach zero, as both p and ρ would approach zero as the universe continued to expand, despite the absence of acceleration.

    Is there a feasible model in which H(t) would maintain an almost constant positive value despite the absence of a nonzero [itex]\Lambda[/itex] or other forms of dark energy like quintessence or moduli?

    Thanks.
     
  20. Jul 14, 2012 #19

    marcus

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    No. Not saying that.
     
  21. Jul 14, 2012 #20

    andrewkirk

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    Sorry to keep picking away at this Marcus, and feel free to ignore this if it is irritating, but I feel like I'm just starting to understand this expansion business so this thought experiment is quite important to me in that the conclusion you reach challenges that understanding.

    When you say the CMB-stationary objects don't feel any acceleration I don't see that as implying there is no acceleration there. Those objects are free-falling, just following geodesics, so by definition they do not feel their acceleration. This is the same as how an orbiting satellite or an object falling towards Earth in a vacuum does not feel gravity, despite being subject to it.

    My understanding is that an object will feel an acceleration iff it is not following a geodesic. If that is correct then the question of whether there is tension in the cable becomes one of whether B is following a geodesic. My crude SR/Newtonian calc suggests it isn't. Unfortunately I think the maths of a proper GR analysis of the situation is beyond me.

    My guess, based on the SR/Newtonian calc, is that, if we assume that the mass of A, B and the connecting cable are nil, and the universe is a homogeneous, isotropic FLRW universe with nonzero [itex]\Lambda[/itex], the geodesic of an object with B's initial velocity (70.4 km/s towards A relative to the CMB frame) will diverge from the geodesic of A, meaning that the proper distance from A to an untethered version of B increases with cosmic time. If that is the case, I would expect the cable to have a nonzero tension, in order to pull B away from its geodesic.

    If the two geodesics converged rather than diverged then the cable would go slack and start to bunch up as A and B came closer to one another.

    Only if the geodesics remained a constant proper distance apart could the cable remain straight with no tension.

    How can we resolve which of these three would occur, with more confidence than comes from my simplistic calc?

    Thanks.
     
    Last edited: Jul 14, 2012
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