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Origins of Scale Factor of FRW Metric and Misc Questions of GR Equations

  1. Apr 16, 2012 #1
    In the context of Friedmann's time, 1922, how did he know to make the metric scale factor, a, a function of time when Hubble's redshifts were not yet published? I understand that he took the assumption that the universe is homogenous and isotropic, but does that naturally imply that the universe is not static? If it doesn't, why did he not assume the scale factor is constant; does the intimacy of space with time naturally make it a function of time? If yes, how so (is there an easily relatable analogy)? Likewise, when Einstein did the calculations for his model years earlier, did he also make a a function of time? If not how were his first results able imply expansion and/or contraction (specifically expansion)? Or did he not know that until Friedmann solved it?

    [STRIKE]Additionally, why does the steady state theory of Bondi and Gold, while they hold the scale factor constant, still have the the universe expanding (as they had to have matter continually created to keep ρ constant)? Never mind, I think they made their model after Hubble IIRC [/STRIKE]

    Does that come from the fact that space and time are linked make expansion inherent (because time is constantly increasing)?

    TL;DR: How does one derive that the constant scale factor of a homogenous isotropic 4-dimensional Euclidean space becomes no longer constant when the metric is extended to the geometry of spacetime before evidence of expansion in the universe? EDIT: From ds2=a2(dx2+k((xdx)2/(1-kx2) how did they determine to make a a function of time when extending it to the geometry of spacetime, as in the FRW metric?

    If you have time, other questions I do have are:

    1. The original expansion comes from inflation after the big bang, right? Is it first from the heat from the first energy fluctuation and then the Higgs field? Why is the higgs field considered a false vacuum? How does it factor into the GR equations?
    2. Why is the cosmological constant proportional to the metric tensor gμv?
    3. Can the LHS of the GR equations describe gravity in terms a quantitative value, or is it only curvature?
    4. I understand that the LHS of the GR equations describe the curvature of spacetime and the RHS the content of mass, but do both sides represent an equal but opposite energy of each other? If so, is that why dark energy is positive energy (as it has to counteract the positive energy of mass) whereas the Einstein Tensor (thus, gravity?) represent the negative energy described by Guth? Are the signs arbitrary?
      • If dark energy creates a repulsive force, is that because of it is positive energy? If so, does the positive energy of mass also have a repulsive force, or does it have an attractive force? Or does energy have nothing to do with attraction/repulsion directly, and, instead, it is how its energy density affects spacetime geometry that creates attraction/repulsion?
      • If so how does negative energy distort space time in the analogy of how mass causes indentation of the spacetime fabric? Does it "raise" it instead? I think I can see how that can cause repulsion but how does it cause accelerating, constant expansion?
    5. Why do I sometimes see the conventions of the GR equations with the cosmological constant as + - - - and other times + - + + (and why does wikipedia call the latter - + + + instead?) Does it make a difference in the end?

    Sorry if I'm talking nonsense and misinterpreting things; I'm trying to write a paper on dark energy with no background in physics for an elective course, and while I understand DE, inflation, vacuum energy, GR, et cetera in popularized terms, I want to understand how those conclusions were derived.

    I'd appreciate any and all help. Thanks.

    EDIT: Sorry guys, found a thread with the same topic; not sure if it'll answer my questions but I'll give it a read through; still have my extra, misc questions though D: (i'll make them green)
    EDIT2: Just finished reading the other thread.
    Question I still have is how you can derive this as I asked below (it'll be red)
    EDIT3: Added more to 1st misc question
    Last edited: Apr 17, 2012
  2. jcsd
  3. Apr 17, 2012 #2
    If the scale factor were constant in time as well, the only solution would be a completely empty universe; certainly not a very interesting subject for a publication.
  4. Apr 17, 2012 #3


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    As I understand it, the Big Bang theory was a result of asking the question, "What is the most general sort of universe that might exist given General Relativity which is also homogeneous and isotropic?" To examine this, we can first take a general metric:

    [tex]ds^2 = a(t,x,y,z)dt^2 - b(t,x,y,z)dx^2 - c(t,x,y,z)dy^2 - d(t,x,y,z)dz^2[/tex]

    Note that in general, we might also have off-diagonal terms, but since we can always diagonalize the metric, this is a completely general metric in General Relativity. Furthermore, we can get rid of the function in front of [itex]dt^2[/itex] with no loss in generality by redefining what we mean by time:

    [tex]ds^2 = dt^2 - b(t,x,y,z)dx^2 - c(t,x,y,z)dy^2 - d(t,x,y,z)dz^2[/tex]

    Next, the assumption of isotropy guarantees that the different spatial directions are identical, leaving:

    [tex]ds^2 = dt^2 - b(t,x,y,z)(dx^2 + dy^2 + dz^2)[/tex]

    Finally, the assumption of homogeneity means that [itex]b(t,x,y,z)[/itex] cannot actually depend upon space, leaving:

    [tex]ds^2 = dt^2 - b(t)(dx^2 + dy^2 + dz^2)[/tex]

    So, we can simply identify [itex]b(t) = a^2(t)[/itex], where [itex]a(t)[/itex] is now the usual scale factor.

    P.S. One caveat here is that there actually is a way to have [itex]b(t,x,y,z)[/itex] depend upon space and not violate homogeneity: the situation where you have a constant spatial curvature. But it's not quite as easy to demonstrate that this is true, so I'll leave that for the moment.
  5. Apr 17, 2012 #4


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    The expansion associated with the standard hot big bang can be caused by a variety of energy sources. Inflation is but one special kind of expansion. Modern cosmology does not address what "caused" the initial expansion; mathematically, it is a boundary condition on the Friedmann equations (for example, the early radiation-dominated universe could just have easily have been contracting under the exact same energy conditions.) So it is not correct to think of the big bang as an explosion, or of the expansion as being the result of some great release of energy. Before inflation was dreamed up, the standard big bang model described a universe emerging from the Planck time into a radiation dominated phase, then passing into a matter dominated phase, expanding all along. Inflation is a brief period of exponential expansion that the universe might have passed through before the standard radiation- and matter-dominated periods.

    The big bang and inflation likely have nothing to do with the Higgs field. The Higgs field is not considered a false vacuum, at least not anymore. Before the Higgs spontaneously broke the electroweak symmetry, it was in a metastable false vacuum. However, the symmetry breaking occurs as the Higgs decays to its true vacuum -- so the vacuum of today's Standard Model is the true Higgs vacuum (this is not to be confused with the fact that the Higgs has a nonzero vacuum expectation value, which means that the expectation value of the Higgs field, [itex]\langle \phi \rangle[/itex], is nonzero.)

    The Higgs factors into the Einstein equations just as any other source of stress-energy does: through the stress-energy tensor on the RHS.

    It's not; it's constant! I suppose you are referring to the way in which the CC enters into the Einstein equations. The CC term is proportional to [itex]g_{\mu \nu}[/itex] because, as you'll notice, the stress-energy tensor on the RHS is a 2nd rank tensor. That means that the LHS must also be a 2nd rank tensor. If we just stuck the CC in there by itself we'd have:
    [tex]R_{\mu \nu} -\frac{1}{2}Rg_{\mu \nu} + \Lambda[/tex]
    which doesn't give us a 2nd rank tensor. So, like the Ricci scalar, R, we must multiply it by the metric tensor.
    Curvature can have a quantitative value; e.g. properly normalized, the curvature of the sphere is +1.
    The Einstein equations should not be interpreted this way. The RHS is indeed the source energy, the LHS is the way that the geometry accommodates that source energy. What signs are you talking about? What energy described by Guth do you think is negative? If you are referring to the inflationary energy source, that is positive just like the cosmological constant (or dark energy.)
    You mean the metric signature? There are two conventions: +--- and -+++. I'm not sure where you are getting the +-++, which is not correct (it's not Lorentzian.)
  6. Apr 17, 2012 #5
    Thank you all so much for your responses! I apologize the frustration that will come from my bastardized interpretation of your responses; I've never taken any lessons on differential equations or any in-depth lessons on parametrics and scalar functions.

    What exactly is the redefinition of time? Is it that it has no other coordinates but is a single point that constantly increases/moving forward? But that's probably not the case because you've taken out the term in front of it, right? Or am I going about this wrong?

    Instead are the terms a(t,x,y,z) and the b(...), c(...), d(...) functions of its respective, specific coordinates, basically saying for a is the function that describes where t is in spacetime, that is, a=t, and similarly b=x, c=y, d=z, where they all have parametrics t,x,y,z? Thus, if the redefinition of t is saying the position of time is not dependent on spaces x,y,z, meaning they aren't parametrics of the function, we can take it to mean time still has a as its function, and thus is its only parametric, right? But since t is what a describes, does that mean that while we can leave the function a(t) in front of dt^2, we take it out because it would be redundant? (How would one formally say what I've probably bastardized?)?

    Is this a^2(t) the same as the a term in the function a(t,x,y,z)dt^2? If so, why is it now squared when made to be a function of the spatial coordinates? I appreciate what you've taught me thus far so much, Chalnoth, and the origins of the scale factor is becoming more clear.

    Which part of the equations reveal this expansion/contraction? The the spatial coordinates are functions of time?

    If that's the case, why does it not have its own term ⟨ϕ⟩ to be explicit, especially if its considered nonzero, in the equations like the CC?

    That makes much more sense by understanding now they both have to be the same rank. However, what does it mean to be second rank, and why does Stress-Energy tensor considered to be so?

    the following interpretation is probably wrong.EDIT2: I now know that tensor describes the number of dimensions needed to describe a coordinate.[STRIKE] Does this mean if the whole LHS was compacted, would it, that is, the Einstein Tensor, be third rank as it needs 4 dimensions to describe it, as it is spacetime?[/STRIKE] Why is the Einstein Tensor only 2nd order when it has the 4th dimension of time? Does it not need 4 indices to describe it? Or does the time coordinate not count towards the rank of the tensor; if so, why? Or does the curvature of spacetime only need 3 spatial dimensions to describe it? What then would be the difference in the GR equations if it only described the three dimensions of space without time? Would that also be rank 2? If the GR equations were only applicable to 3-D space, not spacetime, would it be able to make predictions of gravitational curvature of the present time instance, and of all time for a static-universe? If so, is spacetime preferable because it is useful for predicting future/past events of a non-static universe? Does this relate to how time is independent of space?

    EDIT 3: Just read that time is scalar and is considered a tensor of rank 0, and that whatever I thought a tensor was, is wrong. What is a tensor or index? A vector has a rank of 1 because it can be described in one dimension? I can't picture that. What does 2nd rank mean then? Does the fact that the FRW metric has time dimension separate from the space dimension, 2nd rank? What are the two indices representing?

    I'm not sure if I've understood what he talked about correctly, but I recall him saying that gravity has a negative energy that nearly perfectly cancels out the positive energy of radiation/matter allowing for a "free-lunch"? That's why I assumed if curvature describes gravity, if gravity is negative energy, the LHS must be equal in energy, but opposite in sign to the RHS.

    I guess what I'm trying to grasp is if regular mass is positive energy and the CC/DE is also positive energy, why does CC/DE create an opposite effect of gravity, repulsion? I understand , if we move the CC to the RHS, since it makes the curvature of space time opposite to gravity, it has to be positive energy as well (since the term is being subtracted from the energy-mass tensor), but doesn't that mean that since it now has a negative sign in front of it as it's on the RHS, that it has negative energy? That's why while I know my sources tell me it's positive energy, I have trouble comprehending it.

    Ah, so it doesn't describe the signs in front of the tensors of the GR equations? My question was based on sometimes seeing Rμν - 1/2Rgμν - Λ = -Tμν, which I took to be associated with what wikipedia said was + - - - (the sign in front of each term), whereas I normally see it as Rμν - 1/2Rgμν + Λ = Tμν, which I took to be + - + + which was different from what wikipedia had as - + + +. I guess my question is, does the difference in the format of the equation matter? Thank you, bapowell, for being willing to go through my miscellaneous questions!

    EDIT: One more question: if the scale factor is made to be a function of time, I can sorta see how that would imply either contraction or expansion of space, but why/how, if we take CC to be zero, does show deceleration of expansion?

    Thank you all so much for your help, but please, if you believe I will not be able to comprehend this well enough with my weak background in math, tell me the truth. I may have to end up writing my paper (which was actually due last Friday) without explaining what each equation or term means (i.e. randomly insert a term from out of nowhere and saying if it equals this or that, it means this or that).
    Last edited: Apr 17, 2012
  7. Apr 17, 2012 #6


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    The specific coordinates we use are just a set of numbers. Those numbers have no physical reality, but are merely labels we place on top of reality. The redefinition of time I mentioned is just reorganizing the labels.

    Perhaps I should have explained this a bit more carefully. The metric is a measure of distance. The coordinates x, y, z, and t are the numbers we assign to the different locations in space-time, while the functions in front of the coordinate terms like [itex]dx^2[/itex] set the actual distances between points.

    So, consider the FRW metric again:

    [tex]ds^2 = dt^2 - a^2(t)(dx^2 + dy^2 + dz^2)[/tex]

    What this is saying is that objects within our universe are set at fixed x,y,z coordinates, but the distances between those coordinate positions changes as [itex]a(t)[/itex] changes.

    So what I was doing before was imagining we have these coordinates, and then taking a completely general set of possible distances between the coordinate points, and arguing that this function has to take on a specific form given the assumptions of homogeneity and isotropy.

    As a side comment, this is not the only way to write this particular metric. I could, if I wanted, relabel the coordinates so that the metric looked completely different. This is just a simple way of writing it.

    No, sorry for the confusion there. Perhaps I should have used a different set of variables. In this case, [itex]a^2(t) = b(t,x,y,z) = c(t,x,y,z) = d(t,x,y,z)[/itex], while I've by hand set the previous [itex]a(t,x,y,z) = 1[/itex] by simply choosing an appropriate labeling of time.

    Yes. The parameter [itex]a(t)[/itex] impacts the distances between the coordinate positions. That is, the spatial difference between any two points at a particular time is given by:

    [tex]d^2 = a^2(t)\left((x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2\right)[/tex]

    ...where [itex]d^2[/itex] is the square of the distance between the positions 0 and 1 above. So, as you can hopefully see, if [itex]a(t)[/itex] increases with time, then the distances between coordinate points is increasing with time. As long as objects within our universe are nearly fixed with respect to these coordinates (which they are), then that means they're getting further apart as [itex]a(t)[/itex] increases.
  8. Apr 17, 2012 #7
    So redefining time doesn't mean defining that time has no scale factor of its own? If I am understanding correctly, the set of numbers have no physical reality is analogous to how if you draw a coordinate plane on a balloon and inflate it, the actual distances between two labeled set of numbers has now changed, or something like that, right? Does it simply mean every time the balloon expands, the scale of time is redrawn to stay constant, thus has no scale factor function of its own?

    So would that mean, had we chosen another metric, we could've redefined anything such that we make it be something like [itex]ds2=a(t,x,y,z)dt2−dx2−c(t,x,y,z)dy2−d(t,x,y,z)dz2[/itex] which would mean the scale of an x axis is constant as it is redrawn every time the balloon expands?

    Is that setting of [itex]a(t,x,y,z) = 1[/itex] the redefinition? Also, is it because this is a measure of distance that the constant factor function [itex]a[/itex] squared? Does that mean if the universe wasn't isotropic, outside of the metric would the square root of the functions a, b, c, d be the constant factors of each specific coordinate?

    Yes, at least I understood something!

    But regarding Einstein's original nonstatic model, it was because he assumed homogeny and isotropy that the constant factors of the spatial coordinates became functions of time, right, but how does he know to give them constant factors at all? Why couldn't/didn't he make the assumption that the numbers placed upon reality were physical reality?

    EDIT: Why are the constant factor functions factors of each other in the first place? Is it a consequence of all four dimensions in space? For example, if we're trying to find a distance between two points on a line, [itex]d^2=dx^2-dy^2[/itex], right? If you say that it's actually [itex]d^2=g(x,y)(dx^2-dy^2)[/itex] where [itex]g(x,y)[/itex] is the constant factor set to =1, how does that mean it is necessary to assume that all coordinates are linked and affect each other? I can't seem to picture that.

    EDIT2: Just also realized that I've been wrongly taking inflation to be synonymous with the whole expansion of the big bang
    So I guess I should take that to mean we can't assume more than, "If space is expanding, it had to expanded from a dense singularity near the beginning of time"? Is it the heat that makes it expand which can be assumed from a cold CMB? But then these are assumptions derived from observations. Just realized what I was supposed to ask:
    And is this boundary condition the metric I'm having trouble with understanding? Does that mean that this metric and its assumption that time is a parameter of space (am I using the word "parameter" right here in making it equivalent to saying the coordinates of space depend on the some scale factor function which is a function of time?) but not vice versa really has no mathematical derivation, rather, it's just one random situation that was used to apply to the GR equations?

    Thanks again for your help.
    Last edited: Apr 18, 2012
  9. Apr 18, 2012 #8


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    The scale factor doesn't have any real existence. It's just a parameter we use to make the equations convenient. You certainly can add a scale factor to time as well, but it isn't very convenient to do so in most cases.

    Well, no, we're just labeling time so that changes in time are equal to changes in space-time distance.

    Yes, this could be done. But it isn't a convenient coordinate choice for a homogeneous, isotropic universe.

    Yes to the first two. Sort of to the third. Properly, the distance is the square root of the integral of the metric [itex](ds^2)[/itex] along a path. If you have flat space or flat space-time, then shortest-distance paths are simply straight lines, and you can just do normal geometry. Things aren't so easy in curved space or curved space-time.

    (Note: while our own universe has a curved space-time, in FRW coordinates our universe is highly flat spatially, so that spatial distances, at least, are relatively easy to calculate. You need to do some integrals to get space-time distances, however)

    Einstein wasn't the one that came up with the expanding universe model. He was actually partial to a static universe, until he saw Hubble's results.

    But it is a fundamental statement within General Relativity that the coordinates used to describe the space-time are not physical. That statement can actually be thought of as one of the foundations of GR itself.

    First, the distance is [itex]d^2 = dx^2 + dy^2[/itex]. The minus sign only comes in for space-time distances, where the time coordinate has the opposite sign of the spatial coordinates.

    But I don't understand what you're asking about this. What do you mean "linked"?
  10. Apr 18, 2012 #9


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    Yes. The FRW metric is [itex]g_{\mu \nu} = diag(-1,a(t),a(t),a(t))[/itex] as I think has been pointed our already by someone. This indicates that physical coordinates scale in proportion to the aptly named scale factor, a(t). The easiest way to see that the Friedmann equation gives both expanding and contracting solutions is to note that [itex]H^2 \propto \rho[/itex] has two solutions: [itex] H>0[/itex] (expansion) and [itex]H<0[/itex] (contraction).
    The term [itex]\langle \phi \rangle[/itex] is an altogether different kind of thing than the CC. It is not a source of vacuum energy like the CC or dark energy. The Higgs is in its true vacuum now, meaning that [itex]V(\langle \phi \rangle) = 0[/itex]. The nonzero vacuum expectation value simply means that this true vacuum is shifted away from the symmetric point [itex]\phi = 0 [/itex].

    That's not so say that the term [itex]\langle \phi \rangle[/itex] isn't physically relevant; it gives masses to the gauge bosons. Therefore, terms involving it do indeed appear in the stress-energy tensor as interaction terms between the Higgs sector and gauge fields.
    As you said, scalars are rank 0 tensors, while vectors are rank 1. Tensor of rank 2 are readily represented as matrices. Higher-rank tensors are more complex mathematical structures (e.g. a 3rd rank tensor could be represented as a 3D matrix in the form of a cube.) A tensor itself is just a mapping from the space of vectors (and dual vectors) to the real numbers (see the wikipedia entry for more details on the definition.)

    Don't confuse the rank of the tensor with the dimensions of the space! For example, a rank 1 tensor (a vector) can be defined for any number of dimensions (3-vector, 4-vector,...n-vector...) All of these are still rank 1 tensors. The rank tells you how many degrees of freedom are described at each point in space. A scalar (rank 0) assigns a single value to each point in space. A vector (rank 1) assigns a set of d values at each point (where d is the dimension of the space). Think velocity vector here. The vector is rank 1 so it has a single index, where the index keeps track of which of the d-values we might wish to refer to. The components of a 3-vector can be written as [itex]v_j[/itex], where j = x, y, z, or 1, 2, 3, or what have you. Lastly, a 2nd rank tensor can be represented as a 2D matrix at each point -- that's dxd values at each point!
    Yes, I'm familiar with this explanation, but I've never been comfortable with it. It's true that the potential energy of the gravitational field is negative, while gravitational sources have positive energy. But where is the potential gravitational energy in a homogeneous and isotropic universe? I've never been able to find it ;) The simpler and I think more correct (though less satisfying) answer is that energy simply isn't conserved in GR.
    Aha! Because it has a negative pressure! This is why vacuum energy is such odd stuff. It satisfies the equation of state: [itex]p = -\rho[/itex], whereas ordinary stuff like matter and radiation obey [itex]p=0[/itex] and [itex]p=\rho/3[/itex], respectively. These things exert positive pressure on their surroundings, whereas vacuum energy exerts a negative pressure. What happens to the space in this case? Look up Guth for a nice mechanical description of what happens in a piston filled with a negative-pressure fluid.
    Ah, I see. No, the metric signature has nothing to do with the signs in front of the terms in the equations.
    You have to look at the behavior of the scale factor to see it mathematically. But, intuitively, we should expect the expansion to slow under the influence of ordinary matter. It has to do with that non-negative pressure -- the matter wants to condense under the influence of gravity.
  11. Apr 18, 2012 #10
    So the scale factor [itex]a(t)[/itex] describing the expansion between space isn't real? Without it then, how would we know that at some future point in time, the distance between two points at rest in an expanding spacetime has increased? Or is its purpose merely to change the metric from showing the distance between the two static points in an expanding space are the "same" coordinate distance away in all times, to show instead the distance between the two static points in relation to some past coordinate scale?

    So if I stick to my balloon analogy, time is merely labelled to be a function of its inflation?

    Right, so does that mean there can be some hypothetical universe where time has a scale factor that is a function of space and not vice versa?

    So the redefinition of time does mean, rather than not having a scale factor, that it is not affected by space or itself?

    But I thought he introduced the cosmological constant in 1917 to counter expansion/contraction, before Hubble published his results or Friedmann came up with his solutions?

    Is this similar to my above interpretation of the purpose of scale factors?

    *Face palm* I'm so embarrassed. Why does time have an opposite sign though?

    I guess what i'm trying to see is if there are two dimensions, is there is a possibility their coordinates will both be a function of each other? Maybe I should've done d^2=g(x,y)dx^2+h(x,y)dy^2?

    Gonna post my EDIT2 which I submitted after your post:
    EDIT2: Just also realized that I've been wrongly taking inflation to be synonymous with the whole expansion of the big bang

    So I guess I should take that to mean we can't assume more than, "If space is expanding, it had to expanded from a dense singularity near the beginning of time"? Is it the heat that makes it expand which can be assumed from a cold CMB? But then these are assumptions derived from observations. Just realized what I was supposed to ask:

    And is this boundary condition the metric I'm having trouble with understanding? Does that mean that this metric and its assumption that time is a parameter of space (am I using the word "parameter" right here in making it equivalent to saying the coordinates of space depend on the some scale factor function which is a function of time?) but not vice versa really has no mathematical derivation, rather, it's just one random situation that was used to apply to the GR equations?

    Thanks for putting up with me. Gah, bapowell, I see you've responded as well. Thanks, but unfortunately I have to get to class.

    I apologize for my increasing incoherence; I've not slept a wink.
  12. Apr 18, 2012 #11
    Sorry for the double post.

    So a rank 1 vector can be manipulated to be represented on a 1D plane as a dot (i.e. coming straight out of the plane)? If a 1st order tensor has the three dimensions to be its one index, what is the other index of a 2nd rank tensors of the GR equations (is it time?)?

    When the universe expands, does the gravitational field, that is the curvature of space around a mass as well as the curvature from its pressure, also expand? Could that be where the potential energy increases come from?

    Ah, so all these derivations I've seen for an energy-momentum conservation does ≠ energy conservation. If that's so where does the momentum come into play in context of dark energy?

    I couldn't find a negative-pressure "fluid" unless a vacuum can also be taken to meant to be fluid. In the piston examples it seemed to describe negative pressure as a sucking sort of force. How does a force that sucks cause the universe to expand? I'd understand if there was a vacuum outside our universe sucking the universe towards it to cause expansion, but then there is nothing outside our universe because nothing doesn't exist. So then how does this work? I also read that curvature of spacetime that causes gravity is also responsible for causing repulsion. I can see if a positive mass was pushing down on the fabric of spacetime to make a gravitational well, so I can see how a negative uniform mass can cause the space time fabric to expand all over by pulling it up, but the problem is dark energy is positive energy, hence positive mass, right?

    Why does it still decelerate at a point of expansion where the gravity they feel from another is insignificant? Or does gravity constantly take away the energy of expansion as it expands slowing it down? Is it synonymous to escape velocity? Thanks for your help, bapowell!
  13. Apr 19, 2012 #12


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    Unfortunately, I don't have enough time to respond to all of this, so I'll pare it down a bit.

    It's not "real" in the sense that there are other ways to describe the expansion. The parameter [itex]a(t)[/itex] is a description of a real phenomenon, though there are a great many equivalent ways of describing the exact same phenomenon.

    If one coordinate was a function of the other, then you wouldn't have two coordinates any longer. You'd just have one coordinate.

    The most you can do in this regard is to have coordinates that don't meet at right angles. Then, instead of just having the diagonal terms in the metric, you also have off-diagonal terms. For example:

    [tex]ds^2 = f(x,y)dx^2 + g(x,y)dxdy + h(x,y)dy^2[/tex]

    You can always find a coordinate transformation which gets rid of the off-diagonal terms, however, so we usually just use orthogonal coordinates.
  14. Apr 19, 2012 #13
    No worries, you've given me more than enough help! In fact, if you feel you have to abandon this thread for good, please have no hesitation (my questions seem to be increasing exponentially).

    Ah so I can take the scale factor to be the function of the proper real distance between any two points in relation to time? **OOOH, so is the constant factor the function of expansion?** Now that I know why the scale factor is a function of time, and that it's because of this is the cause of why the universe is either expanding or contracting, I guess I should change my original question to: what is the function of a(t) such that its second derivative shows an eventual deceleration in a universe without dark energy and where is this derived from?

    How would deceleration affect recessional velocities of galaxies? Would the deceleration mean the recession velocity still increase as it increases its proper distance from us, only slower until it reaches a constant?

    Ah, so the scale factors a(...),b(...), c(...), etc in front of the coordinates in the metrics are terms just to describe the coordinate (comoving?) distance in proper distance, and that that doesn't mean the coordinates are functions of each other?

    I think I just lost all the progress I made (since I thought the scale factors had to be placed in front of the coordinates because I thought the scale factors were from taking the coordinates to be functions of each other): so how do we know there has to be scale factors in the first place?

    Thanks for your help!
  15. Apr 19, 2012 #14


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    To figure out how [itex]a(t)[/itex] changes in time, we have to use the Friedmann equations, which are derived from Einstein's equations which relate the metric to matter.

    Right, if [itex]a(t)[/itex] is increasing but at a decelerating rate, then the distances between galaxies are increasing at a decelerating rate.

    Well, I think you're terminology in distance is a bit off here, but yes, the functions in front of the coordinate terms determine to how changes in the coordinates relate to changes in distances.

    The scale factor is a convenient way of talking about the expansion.
  16. Apr 19, 2012 #15


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    I'm not sure what you mean by the first sentence. I don't know what you mean by a rank 1 vector or what a 1D plane is. Consider a 2D plane (like a sheet of paper.) You could define a scalar at each point of the plane by simply assigning a number to each point. Looking at a temperature map of the Mercator-projected Earth is an example. You could also define a vector at each point, by assigning 2 numbers at each point -- one for each basis direction (like x,y) -- and taking the resultant. A map showing the instantaneous direction of ocean currents is a good example . A second rank tensor would be a more complex object that we could not readily visualize in this way.

    In 3D space, our 1st rank tensor (vector) can have three components, say (x,y,z). But in GR we deal with 4D space-time; and so we deal with so-called 4-vectors (4-component vectors). The 4-vectors can also be written [itex]v_i[/itex], where i stands for (x,y,z,t), where t is time. So, space-time is really just a 4D geometry with time as an additional coordinate to the three familiar spatial directions. So, what about the second index in a 2nd rank tensor?

    Well, let me discuss briefly what the metric tensor is all about, and that should shed some light on this issue. The metric tensor is basically a mathematical device (function) that tells us how to measure distances in a given geometry. For example, in 2D Euclidean space, the distance between the origin and the point (x,y) is given by [itex]s^2 = x^2+y^2[/itex]. For reasons that I won't go into here, it is necessary to consider instead from now on the infinitesimal distance, [itex]ds^2 = dx^2 + dy^2[/itex], where the terms are now differentials. Allow me to rewrite this quantity more compactly:
    [tex]ds^s = g_{ij}dv^i dv^j = g_{00}dv^0dv^0 + g_{01}dv^0dv^1 + g_{10}dv^1 dv^0 + g_{11}dv^1 dv^1[/tex]
    This might look complicated but it's really not. We get the second inequality by summing over the repeated indices (this is called Einstein summation is just a compact notation for writing sums). Now, let's define our vector: [itex]v^i=(x,y)[/itex], so that [itex]v^0=x[/itex] and [itex]v^1=y[/itex]. This gives us
    [tex]ds^2 = g_{00}dx^2 + g_{01}dxdy + g_{10}dy dx + g_{11}dy^2[/tex]
    This is looking close to what we wrote down for the Euclidean distance measures. In fact, we recover it if we observe that [itex]g_{00} = g_{11} = 1[/itex] and [itex]g_{01}=g_{10}=0[/itex]. This [itex]g_{ij}[/itex] is our metric tensor. It tells us how distances are computed in our geometry. We can write it as a matrix:
    [tex]g_{ij} = \left( \begin{array}{cc}
    1 & 0\\
    0 & 1\end{array} \right)[/tex]
    where the indices label the component of the matrix: i refers to the row and j the column. Euclidean geometry is simple, it's just the identity matrix. But more generalized geometries have more interesting distance measures, and the metric tensor encodes all of this information. So, the two indices really indicate that the metric tensor deals with two vectors -- it finds the distance between two points (each one can be represented by a vector in the space by drawing an arrow from the origin to the point). Some notation that you might see that makes this relationship more apparent is
    [tex]g_{ij} = g(v_i,v_j)[/tex]
    The two indices of the 2nd rank tensor correspond to the indices of its arguments: you can think of an nth-rank tensor as a function that acts on n different vectors.
  17. Apr 20, 2012 #16
    Ah, so basically metrics are tensors which describe the distance between two different points from an origin, and since they're from an origin, they can both be considered vectors? Everything makes much more sense now (I was wrongly conflating the coordinate elements with indices), thanks!

    So how does one come to the conclusion that there is expansion? In Lemaitre's paper, "A Homogeneous Universe of Constant Mass and INcreasing Radius accounting for the Radial Velocity of Extra-galactic Nebula", he says
    But what does he mean, in layman's term, by all events are perfectly equivalent in space-time (i.e. "events"); "the partition of space-time" (what does partition mean in this sense, and how does that disturb homogeneity?); how can a particle at rest at the centre describe a geodesic of the universe and one elsewhere not describe a geodesic (how does this destroy homogeneity and what is the so-called "horizon" and what sort of "paradoxical results" are produced there?); what does he mean by a "corresponding partition of space and time" (to preserve the homogeneity of the universe)--these are the scale factor constants, right? From there I understand from what Chalnoth has taught me:
    On another tangent: regarding the negative pressure I've found one of your old posts, Chalnoth:

    So can the box be taken to be analogous to the fabric of space time? Hence if photons are exerting positive pressure on it, space-time exerts a negative work back, causing gravitational attraction, whereas if dark energy is thought of as exerting negative pressure on space-time, space-time exerts back a positive expanding force/repulsion? Does this opposite pressure that the box exerts on the space in it exceed whatever is inside exerting (why is it not "every action has an opposite and equal reaction" ∴ no change in volume)? If so, where does this extra energy in both contracting & expanding energy come from?

    Thanks again, bapowell & Chalnoth--just tell me if you tire of my bottomless abyss of questions.
  18. Apr 20, 2012 #17


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    Unfortunately, I'm not entirely sure what he means here. The problem is that these statements don't make sense when we consider the modern coordinates that we use in cosmology today, as geodesics are stationary with respect to the spatial coordinates we use today. I'm sure what he's saying there is perfectly sensible with the coordinate system he's chosen to use, but it just doesn't make sense with more modern choices of coordinates.

    It doesn't really come from anywhere. Energy just isn't conserved in an expanding universe in general. You can make it conserved by defining a potential energy in some specific cases, but not all.
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