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Tethered galaxy problem reconsidered

  1. May 6, 2008 #1
    As a thought experiment, let's imagine assembling a 300 Mpc long rod in intergalactic space, made of an astounding future material which enables it to be both rigid and nearly weightless, while somewhat "stretchy" and "compressible" longitudinally. The rod is centered at the origin of our proper distance coordinate system, and is assembled progressively from the origin. At the origin, we observe the rod to be locally at rest with respect to the Einstein-de Sitter universe with [tex]\Lambda = 0[/tex] and [tex]\Omega = 1.[/tex]

    Will the ends of the rod be stretched apart longitudinally by the expansion of space? No, because the expansion of space does not act like a "force" or "viscous liquid." The expansion of space is the result, not the cause, of the expansion of the universe. The expansion of the universe is best explained simply as the kinematic (or momentum-like) movement of massive objects away from each other because they previously were moving away from each other.

    Conversely, will the ends of the rod be compressed inward longitudinally by the deceleration of the expansion rate? After all, observers at each end of the rod will observe their rod end having peculiar motion longitudinally towards the origin, with respect to the comoving coordinates of local space. I think the answer is no. I think the locally observed peculiar motion at the ends of the rods is "real" only with respect to the observed kinematic recession of local galaxies away from the origin in comoving coordinates. But the rod ends aren't in proper motion with respect to the coordinate origin, and therefore they do not experience acceleration with respect to the origin either; their proper motion and acceleration both are zero. The rod was constructed ex post facto, and therefore never inherited the original recessionary "momentum" of the decelerating expansion.

    If Lambda (with equation of state w = -1) is added to the universe, will the rod ends be stretched away from the origin? I think the answer depends on whether the cosmological constant can "co-inhabit" the same physical space as the rod matter. If it can, then the rod feels expansionary stretching. If it can't, then the cosmological constant is not directly interposed between the origin and the rod ends, so the rod ends to not experience stretching in proper distance.

    ...

    Now let's imagine two galaxies in the same universe, which are far enough apart not to be gravitationally bound together. The first galaxy is located at the coordinate origin. The second galaxy is tethered to the first by a slack rope (please don't ask how!) A winch at the origin galaxy gently tightens the rope until it just smoothly stops the tethered galaxy from moving away in the Hubble flow (i.e., without jerking or pulling it), and then the rope is released. What will the tethered galaxy do, when observed in proper distance coordinates?

    I submit that it will behave exactly like the long rod in the example above. If Lambda=0, the (formerly) tethered galaxy will exhibit NO proper motion towards or away from the origin galaxy even if the expansion is decelerating. It's initial condition is NOT proper motion towards the origin galaxy; it is specifically released at rest with respect to the origin galaxy. Because it has zero proper velocity towards the origin galaxy, it has no proper velocity to decay. Proper velocity and acceleration remain at zero. Certainly the tethered galaxy experiences peculiar motion with respect to the locally comoving expansion, but this is entirely consistent with zero proper motion with respect to the origin galaxy.

    If there is Lambda with (w = -1), then the tethered galaxy will experience proper motion away from the origin galaxy, as a result of the expansionary "force" of the cosmological constant energy interposed between the origin and the tethered galaxy.

    So, I question whether J. Peacock, A.B. Whiting, Davis & Lineweaver, and Francis, Barnes, et al have explained the deceleration scenario correctly in their respective papers.

    Jon
     
    Last edited: May 6, 2008
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  3. May 6, 2008 #2

    Wallace

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    I think this is where the problem lies, you are ignoring the effects of gravity. If the rod is not kept in shape by internal forces then it will contract, due to the gravitational pull of the material in the Universe. Consider a sphere centred on the middle of the rod, with a radius that just reaches to the end of the rod. We can, by the shell theorem, consider that only the material within this sphere effects the rod (we are assuming the Universe is homogeneous). Clearly then, if there is matter within that sphere it will accelerate the rod towards the centre of the sphere, thus contracting it. The more dense the Universe the more rapid the contraction. If dark energy dominates then the rod is pushed apart.

    The same applies to the test particle in the tethered galaxy experiment. In a matter only Universe it starts with no proper motion with respect to the origin but begins to pick up a velocity towards the origin due to the effects of gravity.
     
    Last edited: May 6, 2008
  4. May 7, 2008 #3
    Wallace, your explanation is logical, but I think the answer to the question I posed lies elsewhere.

    I have hypothesized this universe to be filled homogeneously with dust, other than the insignificant gravitational perturbation caused by the rod itself. Therefore, no arbitrarily selected local region can be any different from another. So, I can also imagine two spheres, one centered at one end of the (uncompressed) rod, the other centered at the other end, with radii equal to 1/2 the original length of the rod. If I use the shell theorem to isolate the effect of these dust spheres, the dust within each of the two spheres will attempt to stretch the local half of the rod away from the origin and away from the far end of the rod. This should cancel out the effect of the dust sphere you defined. I submit that the summation of the infinite number of hypothetical dust spheres that can be defined to include part or all of the rod will have a combined effect of cancelling out all of the competing effects of the dust's gravity on the rod.

    The dust particles are moving a further proper distance away from each other as a function of time. If this expansionary momentum is decelerating, then according to Barnes & Francis it should cause the proper velocity of test particles (such as those comprising the two ends of the rod) to decay over time, or at the very least to not accelerate towards each other. If the rod ends start out with zero proper velocity relative to each other and then accelerate inwards, they seem to be behaving contrary to such expectations.

    The scheme used in the 2006 Francis, Barnes et al paper seems to clearly assume that the two galaxies start out with proper velocities > 0 towards each other. The proper velocity towards each other is highest at the start and decreases over time (until the tethered galaxy passes the origin) as the proper velocity decays. This is illustrated in the bottom panels of Fig 1.

    I think Barnes & Francis have this point wrong. The tethered galaxy (which is non-relativistic) first has a proper velocity outward from the origin equal to the recession velocity of its local dust field, as measured by an observer at the origin. Then the tethered galaxy is accelerated towards the origin until its (inward) proper velocity away from its local dust field is exactly equal to its local dust field's proper velocity away from the origin (as measured by an observer at the origin). The two proper velocities cancel out, meaning that the proper velocity of the tethered galaxy relative to the origin must have decelerated to zero.

    It's as if Barnes & Francis are describing a relativistic test particle rather than a matter test particle. The proper speed, as measured from the origin, of a relativistic test particle emitted from the tethered galaxy is exactly the speed of light (c), regardless of whether the proper velocity of the emitting galaxy relative to origin is positive, negative or zero.

    Jon
     
  5. May 7, 2008 #4

    Wallace

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    Absolutely.

    You are so close here, but just missed the mark. What you need to think about is this; we can't change the physical results just by deciding where we want to place the origin and radius of out arbitrary spheres, you are correct in that assertion. However, the predictions that are invariant upon where we imagine the spheres are what the relative motion of parts of the rod are. In my example of a sphere centred on the middle of the rod, I said that all of the rod will be accelerated towards the centre of the sphere. This is correct, but perhaps a clearer description would have been that all the parts of the rod get compressed. The statement that they do this due to an acceleration towards the centre of the rod is of course a statement only true in the arbitrarily chosen sphere.

    In the case of the two spheres you suggest, we could say that in each case the material is accelerated towards the centre of each sphere, however, again a clearer description might be that the material is compressed in both spheres. If you took any two specks of material in the rod and asked 'how will they move with respect to each other?' we get the same answer regardless of how we imagine the problem in terms of our spheres.

    For the rod then, we get the same results regardless of the one sphere or two sphere way of thinking about it. The total compression of the rod is the same. Think about your two spheres example, in both spheres the centre of the rod is pulled towards the end, the results is clearly therefore that the total length of the rod is decreased.

    The shell theorem works and just like many neat techniques in physics, it allows us to reduce a problem to something easy to understand.

    If this was true then a matter filled Universe would never decelerate, regardless of the amount of matter in it.

    You are misquoting or misunderstanding something about the Barnes paper here? In that paper the tethered galaxy starts with zero proper velocity with respect to our chosen origin (you can't define a proper velocity without choosing an origin). In the case of a decelerating universe, the particle starts moving towards the origin, in an accelerating one it moves away.

    No, the particles start out with zero proper motion. See equation 12 for this in detail. I know it looks like they start with a velocity, but this is simply a matter of the chosen horizontal scales of the plots. The behavior of interest was at late times, so the scales are set to show that. If the scales zoomed in more to the start of the journey it would be more clear that the velocites are indeed zero to being with.

    I'm not sure what you are saying here? Yes, the particle has zero proper velocity initially, as intended. How is that wrong?

    I'm really lost now, that paper didn't deal at all with how any of this would be measured (i.e. you mention 'a test particle emitted from the tethered galaxy', not such speculation occurs in that paper) it just played with co-ordinates. The test particle is a massive particle that initially has zero proper velocity with respect to the origin. That is a very simple straightforward initial condition. I have no idea what that has to do with relativistic particles emitted from that test galaxy?
     
  6. May 7, 2008 #5
    Hi Wallace,

    In the example I gave, the rod would be compressed towards the two ends, but clearly would be stretched (i.e., anti-compressed) away from the center.

    Imagine that the two spheres I defined have their center beyond the ends of the rod, and overlap only the very ends of the rod. Then their only effect is to pull the ends of the rod away from the center, i.e., their only effect is to stretch the rod. Imagine a series of many different spheres which overlap the ends of the rod by only 1 cm, 2cm, 3 cm, etc. Each of those spheres contributes both a stretching force - to the part of the rod they don't overlap - and a compression force - to the part of the rod they do overlap. There is no way that the summation of all of the infinite number of spheres that could be defined would in net contribute more compression than stretching to the rod.

    I don't think so. The expansion rate of the universe would decelerate, but the universe would not necessarily contract. If the universe is expanding at a decelerating rate, it is still expanding, not contracting. Why should an expansion, albeit a decelerating one, cause two distant, gravitationally unbound objects to progressively accelerate towards each other?

    Equation 12 is:

    [tex]\dot{R}\left(t_{0}\right)\chi_{0} + R\left(t_{0}\right) \dot{\chi}\left(t_{0}\right) = 0 [/tex]

    In other words, the initial condition is defined such that the sum of the Hubble recession velocity (of the local dust field) and the test particle's inward peculiar velocity equals zero. So you're correct, equation 12 says the initial proper velocity towards the origin is zero.

    You characterize the initial proper velocity of the test particle shown in Fig. 1 to be an inaccuracy due to the scale factor of the chart. But it makes no sense to me, and I see no mathematical justification in the paper, to suggest that the proper velocity of the test particle starts at 0 relative to the origin, achieves some maximum relative to the origin, and then later decays. What is the explanation for a proper velocity which first accelerates and then later decelerates? Attributing both the early acceleration and later deceleration of the test particle to the gravitation of the background dust field seems self-contradictory to me.

    I don't think that the results shown in Fig 1 are consistent with equation 12 being the initial condition.

    Jon
     
    Last edited: May 7, 2008
  7. May 7, 2008 #6

    Wallace

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    No Jon, as I said, think about any two test bits of the rod, you will get the same answer regardless of how you imagine the problem, as long as you do it correctly. You are suggesting the Gauss's law doesn't work, this is not a complex question of cosmology or relativity but first year physics. If you want that explained to you in more detail ask in an appropriate sub forum. I can do no more than what I've already said.

    The expansion of the background material doesn't do anything to the test particles in question. However that material gravitates, and it is this gravity that pulls these two particles, initially at rest, towards each other.

    It's not an 'inaccuracy' of the plot, it's just that you can't see that region in detail. The plot is perfectly accurate.

    Jon, you are going to have to do better than 'it makes no sense to me'. The problem lies in the fact that you are trying to nut this problem out using words alone. You are deciding what words describe something then finding that those words don't work for some other part of the problem and suggesting the paper is at fault. The paper is written in the language of maths, so you are going to have to use that language to understand it. The equations are simply the solutions to the equations of GR, yet you suggest that they are incorrect because they don't give the same answer as what you expected based on the analogies that someone has explained to you for a different situation.

    If this paper is in error, then there are two possibilities, the authors made a mathematical error or relativity is wrong. However for you to simply assert that
    without any justification of why this is the case, without showing the supposed error and correcting it is very poor form.

    It is fine to ask for some help in explaining something you don't understand, but to start from the position 'I don't understand it, so it must be wrong' is very unhelpful. You are suggesting not only that the authors got it wrong, but that they referee did as well, not to mention the entire community that has read it and not found a fault. It pays to have a bit more reason for this accusation, or you might look a little foolish.

    I've no interest in getting into another protracted discussion here, I've explained where you have got it right (which is a fair bit) and where you've got it wrong (just a few crucial points) and can't do any more than I've done. If you think this paper is in error then demonstrate the error and I'll listen, otherwise take the time to absorb more deeply the explanations I've already given.
     
  8. May 9, 2008 #7
    OK Wallace,

    I expressed doubt about the correctness of the authors' conclusion because I thought I understood what they were saying. You have pointed out that I didn't correctly interpret what they were saying. In part because the scale of the Barnes & Francis Fig 1 conveys a misleading picture of the early times. So now I need to retract my suggestion that they were wrong, pending a better understanding.

    I agree that "doing the math" is important for a deep understanding of this subject. But "doing the math" also requires an understanding of how to apply the equations to the physics in a meaningful way, and how to interpret the results in a way that is consistent with the physical process being modeled by the math. Applying the math with incorrect physics can lead to all sorts of errors. Which is one reason why each of the authors I cited accuses one or more of the other authors of misapplying or misinterpreting the results of their properly calculated "maths".

    I am confident that if Barnes & Francis were engaged in this Forum, they could readily explain the intended physical interpretation of their model in a way that those of us with mediocre math skills could comprehend. It's too bad we don't have that available to us, and that the authors did not choose to include a more complete explanation in their paper. Based on the eyebrow-raising degree of disagreement among the various papers, this obviously is a topic which is not so straightforward that experts can afford to dispense with complete contextual descriptions when they communicate even at stratospheric levels with each other.

    Jon
     
  9. May 10, 2008 #8
    In my last note I should have quoted from the linked Barnes & Francis paper (emphasis added):

     
  10. May 10, 2008 #9
    Here's how I interpret the linked Francis & Barnes paper, the Davis, Lineweaver & Webb paper (1 May 2003) and the A.B. Whiting paper (5 Apr 2004) regarding the tethered galaxy problem in a [tex]\Omega = 1, \Lambda = 0[/tex] universe.

    Proper velocity between the tethered galaxy and the origin begins at zero when the tethered galaxy is released. The tethered galaxy rapidly accelerates to a constant proper velocity. It passes through the origin and continues onward at nearly the same constant proper velocity. Proper velocity decays very slowly over time, solely as a function of the decreasing density of the background dust field over time.

    Clearly the rapid initial proper acceleration towards the origin must be caused by the gravity of the background dust field, and is directly proportional to its density. This is clearly seen in Francis & Barnes Fig 1, comparing lower panel #2 (density << Omega) and #3 (density = Omega). The constant (post-acceleration) proper velocity is about 4x in panel #3 compared to panel #2. The acceleration of the dust field does NOT appear to be a factor, since the tethered galaxy accelerates to a constant proper velocity in panel #2 even though the acceleration of the dust field is zero.

    None of the papers I've read explains why proper velocity accelerates rapidly and then remains relatively constant over time. I assume that, in Newtonian terminology, the tethered galaxy accelerates up to the escape velocity of the dust field, and then tracks that escape velocity over time (which decelerates in panel #3.)

    I think that the decay in proper velocity in panel #3 is a result of the decline over time in the density of the dust field, not the decline in its proper acceleration. Proper acceleration (actually deceleration) is constant over time in panel #3, but this does not prevent velocity from decaying over time.

    Davis & Lineweaver say that positive or negative peculiar velocity (relative to the origin) results from the acceleration of the dust field. Barnes & Francis say the same, which is strange given that their Fig 1 appears to directly contradict this (panel #3 top and bottom). A.B. Whiting says that peculiar velocity does NOT result from the acceleration of the dust field:

    I agree with Whiting on this point, and extend his interpretation to mean that the decline in the proper velocity over time is a result of declining dust field density, and that both the velocity and acceleration of the dust field are irrelevant.

    The physical interpretation of the tethered galaxy passing through the origin is a bit surprising. Any two test particles which are momentarily at zero proper velocity with respect to each other will approach each other, and if they narrowly avoid colliding, will speed past each other indefinitely, with a very slowly decaying relative velocity over time. In my scenario with a long fixed rod, the two end points of the rod will attempt to pass through each other (infinite compression) and then stretch infinitely in the opposite coordinate directions. A sphere of massless test particles will approach the center, and if they narrowly avoid colliding, will speed past each other indefinitely, creating an infinitely expanding sphere which is a mirror image (from a 3-dimensional coordinate perspective) of the original sphere. It is not at all intuitive to me that such effects should result from simple gravitational attraction. But it appears to be the case.

    Jon
     
    Last edited: May 10, 2008
  11. May 11, 2008 #10
    Just to be clear, I am not suggesting that there is NO relationship between the deceleration parameter q and the decay of proper velocity. Barnes & Francis bottom panel #2 shows a dust field with zero acceleration, and obviously density is decreasing smoothly over time. Since there is no decay in proper velocity in this scenario, density decrease during the scenario cannot be the cause of proper velocity decay.

    Apparently the more negative q is in the initial conditions, the more proper velocity seems to decay during the scenario. But I think Whiting is correct in characterizing that the deceleration parameter q is a result of the dust field's density, not the cause of changes in the density.

    In kinematic terminology, it appears that the peculiar velocity of a test particle will decay in any universe in which the kinetic energy of expansion does not exceed the gravitational energy of the dust contents by a large factor. That large factor is such that the deceleration parameter q is essentially zero. In other words, the gravitational energy of the dust field is negligible compared to its expansion velocity.

    The kinetic energy of expansion equals the gravitational energy when the expansion rate is equal to the escape velocity of the dust contents.


    Jon
     
    Last edited: May 11, 2008
  12. May 12, 2008 #11
    The Davis & Lineweaver paper says that the acceleration of the untethered test particle is a function of three parameters:

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

    This reinforces their text and Fig 4 which indicate that the test particle's acceleration is directly related to the value of q.

    For a matter-only universe (w=0), q can be defined as follows:

    [tex]q = \frac{1}{2} \left(1 + K / \left( a H \right)^{2} \right) [/tex]

    where K = 1 (overdense universe), 0 (flat universe) or -1 (underdense universe). Thus q can be defined separately for each density condition of the universe:

    Overdense:

    [tex]q = \left( a H \right)^{-2} \right) [/tex]

    Flat:

    [tex] q = \frac{1}{2} [/tex]

    Underdense:

    [tex]q = - \frac{1}{2} \left( a H \right)^{-2} [/tex]

    Whether a universe is overdense, flat, or underdense depends directly on the relationship between Hubble velocity (H) and density. A matter-only universe is flat if its density is at critical density:

    [tex] \Omega_{m} = \frac{3H^{2}}{8 \pi G} [/tex]

    The relationships between the preceding four equations shows that, at any arbitrarily selected mass density, the value of q can be directly changed by arbitrarily setting a higher or lower value for the Hubble velocity H.

    Therefore, in our untethered test particle problem in a matter-only universe, I derive the following:

    At any selected dust density level, the higher the Hubble expansion velocity of the dust field is, the less proper velocity decay will be experienced by the untethered test particle moving through the dust field.

    I further derive from this that the faster each individual background dust particle (of constant mass) is moving with the Hubble flow, the less gravitational effect it exerts on the passing test particle. In other words, the Newtonian potential of a dust particle's gravity field is inversely related to its kinetic energy.

    This is inconsistent with Newtonian physics. Something is wrong in this model.

    Jon

    EDIT: I want to substitute the 3 equations for q (overdense, flat, underdense) into Davis & Lineweaver's equation at the top of this post this yields:

    Overdense:

    [tex]\ddot{D} = \frac{D}{a_{2}} [/tex]

    Flat:

    [tex]\ddot{D} = - \frac{1}{2} H^{2} D [/tex]

    Underdense:

    [tex]\ddot{D} = \frac{1}{2} \frac{D}{a_{2}} [/tex]

    Jon
     
    Last edited: May 12, 2008
  13. May 12, 2008 #12
    If D and [tex]a^{-2}[/tex] each are arbitrarily fixed at unity (1) at a point in time, then the three equations compare to each other as follows:

    Overdense:

    [tex]\ddot{D} \propto 1 [/tex]

    Flat:

    [tex]\ddot{D} \propto - \frac{1}{2} \dot{a}^{2} [/tex]

    Underdense:

    [tex]\ddot{D} \propto \frac{1}{2} [/tex]

    In order to change density from overdense to flat, or from flat to underdense, while holding [tex]a^{2} [/tex] constant, we make an arbitrary change in the value of [tex]\dot{a}^{2} [/tex] within a certain range. Clearly that change in [tex]\dot{a}^{2} [/tex] will change the test particle's proper acceleration rate at that instant in time.

    Jon
     
  14. May 18, 2008 #13
    OK now I think I have the explanation for why the proper velocity of the particle decays after the particle passes the origin. Although the origin coordinate is arbitrarily selected, it creates the perspective point against which the particle's proper motion is judged. This is important to the outcome.

    At first, the test particle will accelerate towards origin. Its proper acceleration rate will flatten out rather quickly as the test particle reaches the Newtonian escape velocity of the dust field. However, its acceleration towards the origin will be slightly above zero, because the density of the dust field causes the test particle to behave as if it is in a contracting universe, causing the proper distance between the particle and the origin to accelerate together slightly.

    After the test particle passes through the origin, it will continue moving away at a nearly flat proper velocity. The proper velocity will now decay over time because the contractive gravity of the dust field causes a slightly acceleration of the dust particle backwards towards the origin. And because the dust field is actually expanding (not contracting), its density decreases over time, causing the backwards acceleration (velocity decay) to decrease over time. But this decrease in backwards acceleration slows over time since the expansion rate is decelerating.

    A very complex movement behavior. And one that is primarily dictated by the absolute density of the dust field, with a 2nd order effect caused by the decrease in density over time, and a 3rd order effect caused by the deceleration of the expansion.

    Jon
     
  15. May 22, 2008 #14
    My last post seems to explain the shapes of the curves in the Barnes & Francis and Davis & Lineweaver papers well enough, but I'm still not satisfied that the gravitational effect of the dust field as I described it is self-consistent and reasonable. I hope that someone will jump in here and help me analyze whether my dissatisfaction is justified.

    Imagine the background dust field as a set of individual dust particles, each with a tiny rocket engine attached. The coordinate origin point is at the center of the dust field. These engines initially all fire continuously at a rate that exactly prevents the dust particles from collapsing gravitationally towards the origin; in other words, their proper distances from the origin remain constant.

    If the engine of one particle (the "test particle") located at proper distance "D" from the origin is then shut off, the gravity of the dust field will accelerate the test particle towards the origin. Since the density of the dust field is homogeneous, the gravitational force on the test particle does not increase as the test particle approaches the origin. In fact, the shell theorem tells us that the gravitational force declines as distance to the origin decreases. The gravitational force is determined by the total mass of the sphere with its center at the origin and its radius at the test particle. As the test particle approaches the origin, the radius of the sphere shrinks and the total mass in the sphere decreases in proportion to its volume.

    So, I think my previous reference to the "escape velocity" of the dust field was incorrect. What's really happening is that as the test particle approaches the origin, its proper velocity increases but its proper acceleration decreases. So the proper velocity of the test particles increases rapidly at first and then flattens out to nearly zero, and exactly to zero as the test particle crosses the origin.

    What happens after the test particle passes through the origin? At that point, the relationship between proper velocity and acceleration should switch to the reverse of what I just described. At first, the velocity away from the origin is high, and the acceleration "backwards" towards the origin is low. But as proper distance increases, the backwards acceleration increases and therefore the rate of velocity decline should increase. If the density of the dust cloud remained constant, by the time the test particle reaches -D its proper velocity should decrease to zero. Exactly symmetrical acceleration and deceleration.

    Of course the density of the dust field does not remain constant; the dust field expands, so its density decreases over time. This effect causes the deceleration of the dust particle to decrease, while simultaneously the increasing proper distance causes the acceleration to increase. What surprises me is that the two competing effects nearly cancel each other out. Even in the "overdense" universes in Barnes & Francis Fig. 1 bottom panels 4 and 5, the test particle does not come to a stop at -D, nor at any greater negative distance on the chart. The charts suggest that it never comes to a stop. Given how quickly the test particle originally accelerated from a dead stop, I find it quite puzzling that the deceleration effect after passing through the origin is so attenuated. Particularly since the expansion rate of the dust field is itself decreasing over time. Intuitively I expected the test particle to stop and reverse course at some point, passing back through the origin, and continuing to occillate in that manner indefinitely.

    Comments would be appreciated.

    Jon
     
  16. May 22, 2008 #15

    Wallace

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    My only comment is that you aren't going to get any further with this unless you stop trying to understand this with words alone and play around with the equations. Physics is the application of physical principles described mathematically. It's not simply playing with abstract equations, that's pure maths, you need to take the conceptual approach you have been using but apply that to the equations.

    For instance to answer the main question in the last paragraph above, you need only look at the equation for the particles acceleration to find the answer. You need to think conceptually in order to gain physical information from the maths, but without the maths you will forever go in circles.
     
  17. May 22, 2008 #16
    Hi Wallace,

    The equations I showed in this thread tell me that in a flat universe, the test particle's proper velocity will decline after passing through the origin if [tex] \dot{H}^{2} [/tex] is greater than [tex] \dot{D} [/tex]. But selecting any particular value of H and D seems arbitrary to me, and suggests that a variety of different curves can be generated using different values. That doesn't seem very enlightening.

    I also note that in an underdense or overdense universe (and regardless of how much over- or under- they are), the equations seem to say that the test particle's acceleration is perfectly constant as a function of time. With no relationship to H or D. That's a head-scratcher too.

    I think I've made more progress thinking about the physical model than thinking about the math.

    Jon
     
    Last edited: May 22, 2008
  18. May 22, 2008 #17

    Wallace

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    Yes they are arbitrary, but you only pick the arbitrary value once. I.e. set H=1 when the particle is released and then H(t)/H(t_0) tells you the comparative value at some later time.

    Which acceleration? (proper or co-moving) What equation are you referring to?

    It is a fool's errand to think you can separate 'the physical model' from 'the maths', given that the physical model is maths! I cannot help you if you insist on trying to learn science by analogy.
     
  19. May 22, 2008 #18
    Hi Wallace.

    The equations are in my post of 05.13.08, 00:07. As I look at it, I see that I referred in my last post to the wrong equations for underdense and overdense universes. In those cases, the acceleration depends on the ratio between D and [tex]a^{2}[/tex], so it is not a constant over time.

    Everything in this thread is about proper, not comoving, acceleration, velocity and distance.

    Jon
     
    Last edited: May 22, 2008
  20. May 22, 2008 #19
    I think I got the sign wrong on the overdense equation. If I make that change, and replace H with its own equation, the equations should be:

    Overdense:

    [tex]\ddot{D} = - \frac{D}{a_{2}} [/tex]

    Flat:

    [tex]\ddot{D} = - \frac{1}{2} H^{2} D [/tex]

    Underdense:

    [tex]\ddot{D} = \frac{1}{2} \frac{D}{a_{2}} [/tex]
     
  21. May 22, 2008 #20
    I meant to replace H in the flat equation:

    Overdense:
    [tex]\ddot{D} = - \frac{D}{a_{2}} [/tex]

    Flat:
    [tex]\ddot{D} = - \frac{1}{2} \dot{a}^{2} \frac{D}{a_{2}} [/tex]

    Underdense:
    [tex]\ddot{D} = \frac{1}{2} \frac{D}{a_{2}} [/tex]
     
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