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Simple question

  1. May 9, 2008 #1

    nrqed

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    A very very dumb question......I am almost embarassed...

    We have the relation

    [tex] 1 = \Omega_{\Lambda} + \Omega_{matter} - \frac{ k c^2}{H_0} [/tex]

    So if we consider a universe with no cosmological constant and [itex] \rho = 0 [/itex], this implies that the curvature is not zero? Am I missing something really obvious?
     
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  3. May 10, 2008 #2

    Wallace

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    No, your not missing anything, that is correct, a completely empty universe (no matter, no cosmological constant) has spatial curvature in the FRW co-ordinates that the equation you mention is expressed in. It seems weird, that an empty universe can be curved, since we know an empty Universe is equivalent to special relativity in which there is no curvature. What this demonstrates is that 'spatial curvature' is a co-ordinate dependent property of a given space-time. The overall 4D curvature is still zero in the empty FRW universe, it's just that the way time and space has been sliced up into co-ordinates, spatial slices of constant time have curvature.
     
  4. May 27, 2008 #3

    nrqed

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    Thanks for the reply. But I am still puzzled by this.
    What you wrote seems to imply that even if k is not zero, there is exist a change of avriable that can bring the FLRW metric in a flat spacetime metric? I don't see this.

    Another way to phrase my question is this: let's look at the Ricci scalar. I might be wrong but I think it's (setting c=1)

    [tex] R = 6 ( \dot{H} + 2 H^2 + \frac{k}{a^2} ) [/tex]

    If there is no spacetime curvature, R must be zero. I don't see how this expression can be zero if k is not zero. According to the euqation I gave in my first post, if there is no matter and no cosmological constant, k must be [tex] - H_0 [/tex]. Replacing this in R does not give zero at all. Unless a(t) has a very specific time evolution. But even then, looking at the FLRW metric I don't see at all how having a nontrivial a(t) and k not zero could lead to a flat spacetime metric even after a change of coordinates. Can you show me how this works out?

    Thanks
     
  5. May 27, 2008 #4

    Wallace

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    I'm sorry, I didn't mean to imply this at all. I will try and be more clear. In general, all FRW metrics are conformally related to the Minkowski metric, which means that there exists co-ordinate transformations that allow you to write the metric down as

    [tex]
    ds^2 = f ds^2_{Minkowski}
    [/tex]

    where f is some function that depends on the cosmology. In the case of an empty universe only, f goes to unity, meaning that you can transform to pure Minkowski spacetime as you would expect since there is no energy hence no gravity. The FRW co-ordinates for an empty universe (sometime known as the Milne Model) have spatial curvature, but as always this just demonstrates that how we slice up space and time depends on the co-ordinates we choose.

    In this case (I haven't checked this equation, I'll assume it is correct), we get for the empty universe in FRW co-ordinates that [tex]a(t)=\frac{t}{t_0}[/tex] hence [tex]H(a) = 1/a[/tex] and [tex]\dot{H}=-\frac{1}{a^2}[/tex], and [tex]k=-1[/tex]. Thus R goes to zero for this model just as we would expect. The total spacetime curvature is zero even though we have spatial curvature given the co-ordinates we have chosen.

    However, this only holds for the empty universe. Any other model will have a different H(a) evolution and hence the H and k terms will not cancel out. This says that for any non-empty universe there is spacetime curvature.

    I hope that clears it up, let me know if you have any more questions.
     
    Last edited: May 27, 2008
  6. May 27, 2008 #5

    nrqed

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    Thanks for the clarification. It is starting to make sense but I still have a few things I'd like to clarify.

    There are two equations to satisfy. R=0 but also (from my first post where there is a typo..H_0 should be squared) we have that

    [tex] 1 = - \frac{k}{H_0^2} \Rightarrow k = - H_0^2 [/tex]

    So we should use that value of k in the expression for R, right? And then we must keep [tex] \dot{H} [/tex] nonzero and it must be such that R comes out to be zero.

    That's correct? It seems highly nontrivial to me that for the a(t) which will come out of this the FRW metric will come out to be equivalent to flat spacetime but I am willing to accept that this is true.

    You made me realize that k is a measure of spatial curvature which can be an artifact of the metric used. It's surprising since the CMBR is used to distinguish between k=0,1,-1. I guess that the case of no matter, no cosmological constant the value of k has less meaning?

    Thanks again
     
  7. May 27, 2008 #6

    Wallace

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    No, k can only be -1,0,1 not anything else. Take a step back, I think there was more typo's in your first equation. We start from the basic Friedmann equation

    [tex]

    H^2 = H_0^2 ( \frac{8 \pi G}{3} \rho + \frac{\Gamma}{3} - \frac{k}{a^2})

    [/tex]

    Now, for the empty universe we get

    [tex]

    H^2 = H_0^2 (-\frac{k}{a^2})

    [/tex]

    and we know that k=-1 which gives us

    [tex]

    H^2 = H_0^2 (1/a^2)

    [/tex]

    in other words
    [tex]

    \frac{H}{H_0} = 1/a

    [/tex]

    This is what I said earlier, that for the empty universe, H(a)~1/a. This is the unique solution to the Friedmann equation given the condition that the universe is empty.

    I'm not sure what you mean by 'highly non-trivial'? The solution you get for a(t) in an empty universe leads to R=0 as we would expect. If it this wasn't the case we would be in big trouble, since we know that GR with no gravity must reduce to SR.

    I don't think k has any less meaning. Be careful, saying that 'k is a measure of spatial curvature which can be an artifact of the metric used' implies that someone we are using the wrong or unnatural metric and hence we have some artificial quantity. Remember that there is no right way to define co-ordinates and we use whichever are the most convenient for a problem. We use the FRW metric in cosmology, and in these co-ordinates k is an important parameter. Using these co-ordinates, we can make measurements of the CMB that tells us something about k. Different physics would lead to a different measure of k, so we are always probing the physics, we just need a framework to describe it in.
     
  8. May 27, 2008 #7

    nrqed

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    Thanks for your patience.
    Before going further I must check if I have the correct equation. I do not have a factor of H_0 in my equation. Checking in a book (say Hartle, equation 18.63) I find
    [tex] \dot{a}^2 - \frac{8 \pi \rho}{3} a^2 = - k [/tex]
    so he does not have a factor of H_0 either. Am I missing something?
     
  9. May 27, 2008 #8

    Wallace

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    Take your equation and re-arrange to get

    [tex] (\frac{\dot{a}}{a})^2 = \frac{8 \pi G}{3} -\frac{k}{a^2} [/tex]

    Now, a is dimensionless, however H has dimensions of inverse time, so we cannot simply make the next step


    [tex] H^2 = \frac{8 \pi G}{3} -\frac{k}{a^2} [/tex]

    Instead, we realise that this equation tells us how H(a) evolves as a fraction of the H_0 we measure today. Since H(a)/H(a_0) is dimensionless we can write

    [tex] H^2 = H_0^2 (\frac{8 \pi G}{3} -\frac{k}{a^2}) [/tex]

    What this really says is the the shorthand defintion that [tex] H(a) = \frac{\dot{a}}{a} [/tex] is really [tex] \frac{H(a)}{H_0} = \frac{\dot{a}}{a} [/tex]. Dimensions are a pain in cosmology, since you usually ignore them by setting pretty much everything to 1 (c, G etc) you get used to not worry about them, but they are still very important to get right.
     
    Last edited: May 27, 2008
  10. May 28, 2008 #9

    nrqed

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    That's something I never realized. But I am still a bit confused.

    Clearly [tex] \frac{H}{H_0} is dimensionless. But [tex]
    \frac{\dot{a}}{a} [/tex] is not. So I don't see how you can write [tex] \frac{H(a)}{H_0} = \frac{\dot{a}}{a} [/tex]. Even allowing factors of G and c I cannot make those two sides of the same dimensions. Can you show me the full relation with all the dimensionful factors reinstated?

    Also, the value of H_0 is in inverse second so I thought that [tex] H(a) = \frac{\dot{a}}{a} [/tex] made perfect sense.

    I am also confused by one point. You are using H_0 to restrore the proper units. This is confusing to me because it seems to me that only G and c are set equal to one to simplify the equations, never H_0. Are you saying that when in order to write the equations with all dimensionful factors explicitly reinstated one must put back in factors of G,c AND H_0?

    Thanks for all the help!
     
    Last edited: May 28, 2008
  11. May 28, 2008 #10

    nrqed

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    Just to add a comment.

    My initial equation (from the first post) is (after correcting a typo: I had forgotten to type the square of H_0)
    was

    [tex] = 1= \Omega_\Lambda + \Omega_{matter}- \frac{k c^2}{H_0 ^2 a^2}
    [/tex]
    To me it seemed as if the units were working out ok because that would mean that the units of k are [tex] m^{-2} [/tex] and this makes sense since k appears in the metric through [tex] 1 - k r^2 [/tex].
     
  12. May 28, 2008 #11

    Wallace

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    k is most definitely dimensionless.

    This is about to get a bit more confusing for you, but when I wrote down

    [tex] \frac{H(a)}{H_0} = \frac{\dot{a}}{a} [/tex]

    this implicitly assumes that time is in dimensionless units of [tex] t H_0[/tex] which is the usual way these equations are integrated.

    I know this is all very confusing, it feels like it should be simple but units are always were I make my biggest mistakes in cosmology.
     
  13. May 28, 2008 #12

    nrqed

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    Ah!!

    This is an extra layer of monkeying around with units beyond the usual setting of c and G equal to one!

    At first I thought that cleared up everything but then, you were right, I got more confused.

    The dot in your equation indicates a derivative with respect to the dimensionless time, right? (otherwise the equation would make no sense).
    But then, how do you define Hubble's constant? As [tex] H_0 = (\frac{\dot{a}}{a})_{now} [/tex] with a derivative with respect to the dimensionless time or with respect to time in seconds?

    The first choice does not make sense since H_0 is always quoted in [tex] s^{-1} [/tex]...unless you are telling me that the H_0 appearing in all the equations of cosmology has to be converted before being compared to the measured value. But then, how is the conversion done? I don't see how to use factors of c and G to do the conversion.

    The second choice seems as bad. Because then the equations would involve a mixture of dimensionful time and dimensionless time.

    So yes, I am getting more confused. Sorry for being slow! Thanks for the help!

    Patrick
     
  14. May 29, 2008 #13

    hellfire

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    Consider the Robertson-Walker line element:

    [tex]ds^2 = - c^2 dt^2 + a^2 \frac{dr^2}{1 - kr^2} + \dots[/tex]

    This has units of [tex]m^2[/tex]. Thus, if [tex]k[/tex] is taken to be dimensionless, then [tex]a[/tex] cannot be dimensionless and must have units of [tex]m^2[/tex]. On the other hand, if the [tex]a[/tex] is taken to be dimensionless as a scale factor, then [tex]k[/tex] cannot be dimensionless and must have units of [tex]m^{-2}[/tex].

    With this, the units in the Friedmann equation fit well:

    [tex]H^2 = \frac{8 \pi G}{3 c^2} \, \rho - \frac{k c^2}{a^2} [/tex]

    The gravitational constant:

    [tex][G] = m^3 \, kg^{-1} \, s^{-2}[/tex]

    The speed of light:

    [tex][c] = m \, s^{-1}[/tex]

    The energy density:

    [tex][\rho] = J \, m^{-3} = kg \, s^{-2} \, m^{-1}[/tex]

    Assuming [tex][k] = m^{-2}[/tex], and [tex]a[/tex] dimensionless, at the end you get [tex][H^2] = s^{-2}[/tex] which is consistent with the definition of [tex]H = \dot a / a[/tex] being [tex]a[/tex] a dimensionless scale factor.
     
    Last edited: May 29, 2008
  15. May 29, 2008 #14

    nrqed

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    What you wrote agrees completely with what I have always assumed...k in [tex] m^{-2} [/tex]. a dimensionless, t in seconds.... Thank you, at least I know that I am not alone thinking in those terms. So when people say that k=-1,0 or 1, they really mean -1,0 or 1 [tex] m^{-2} [/tex]?

    Wallace is using a different system which is what I am trying to understand. He/She is using a dimensionless time and a dimensionless k. I am trying to understand what he/she is using for r. And what the units of H are and how it's related to the measured value (which obviously does not use a dimensionless time). You are saying that his "a" is not dimensionless, then?

    thanks
     
  16. May 29, 2008 #15

    hellfire

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    I made a mistake:

    Actually [tex]a[/tex] must have units of [tex]m[/tex].

    Now to your question:

    I would say this is not the case. If the curvature parameter has to get the values -1, 0 and 1 only, then it was scaled as:

    [tex]k \rightarrow k/ \vert k \vert[/tex]

    Otherwise we would have [tex]k = 1/R^2[/tex] being [tex]R[/tex] the curvature radius of the universe (see for example last chapter of Carroll's Lecture Notes on General Relativity). The substitution above makes [tex]k[/tex] take the values -1, 0, 1 only and, moreover, it makes it dimensionless.

    So I would guess that k = -1, 0, 1 dimensionless requires of a dimensionfull [tex]a[/tex] with units of lenght. My impression is however that one usually changes from dimensionless [tex]a[/tex] to dimensionless [tex]k[/tex] in cosmology. It is practice to talk about both as dimensionless quantities. However both cannot be dimensionless at the same time.

    I find this topic also a bit confusing, but this is my understanding.
     
  17. May 30, 2008 #16

    nrqed

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    You are right about everything.

    I found out that d'Inverno explains this pretty clearly. Not all books are careful, I have seen one who states explicitly that a(t) is dimensionless and that k=0,-1 or 1 !

    let's say we start with a metric

    [tex] c^2 dt^2 - A(t)^2 ( \frac{1}{1- K r^2} dr^2 + r^2 d \Omega^2 )[/tex]
    So far this K is in 1/m^2 and r is dimensionful and A(t) is dimensionless.

    Assuming that K is not zero, one defines [tex] K \equiv k |K| [/tex] with k dimensionless and obviously either +1 or -1. Then, in addition one defines a rescaled radial coordinate
    [tex] \tilde{r} \equiv \sqrt{|K|} r [/tex] which is dimensionless. And then, as a third step, one defines [tex] a(t) \equiv \frac{A(t)}{\sqrt{|K|}} [/tex] with a(t) now dimensionful.
    So one gets



    [tex] c^2 dt^2 - a(t)^2 ( \frac{1}{1- k \tilde{r}^2} d\tilde{r}^2 + \tilde{r}^2 d \Omega^2 )[/tex]



    The two things I had not realized were that a is dimensionful, as you pointed out, and that the radial coordinate is dimensionless.


    Now everything works out ok.

    For example, the equation

    [tex] H^2 = - \frac{k c^2}{a_0^2} + \ldots [/tex]

    makes perfect sense. The a_0 is not always shown because most people set the scale factor now to one but it's actually there. The left side is in 1/s^2. The right side is also in 1/s^2 with k dimensionless and a_0 having the dimensions of a length. So it works out ok.

    Note that teh fact that a has dimensions doe snot affect any of the other equations because a appears always in H. Only in the term with k does a appear by itself.

    For k=0, one has the choice of setting r dimensionless and a dimensionful or not.
    So in the flat space case, one can indeed work with a dimensionless a(t).


    Thanks for the help. It is not trivial to sort out because, as I said, some books do explicitly say that the scale factor is dimensionless an dthen later say that k=0,1 or 1!

    Thanks!!
     
  18. Jun 2, 2008 #17
    P equals Zero?

    P equals zero, k equals zero?

    For us who do not reach such heights, is it possible to just apply common sense to any of these equations?

    Have any of those who spend uncountable hours trying to devise such equations ever just put down the pencils and tried to consider what actual nothingness could be?

    Given the number of physicists and great intellectuals who boast from various sides of this philosophy, where the heck are the REAL thinkers? You know, the ones that actually dare to ask the questions that the intellects have no response for, and hate to be confronted with. and why are they so discounted as religiously fanatic as soon as the suggestion of a creator arises, for the sake of attempting to name the unnamable?

    Simply put, how the heck can something be born from nothing? And I mean NOTHING. For before something there could only be nothing, otherwise we have not looked back far enough to discover the before.
     
  19. Jun 3, 2008 #18

    nrqed

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    Thanks Wallace for your help.

    Now that I have cleared up the issue of the units, I went back to my original question.
    What was puzzling me was this: if we consider a totally empty universe (no cosmological constant, no matter) the equation at the present epoch

    [tex] 1 = - \frac{k c^2}{H_0^2 a_{now}^2} [/tex]

    seemed to indicate that k was not zero (and had to be minus one as you pointed out). This, in turn, implied that there had to be a nontrivial time dependence for the scale factor a(t). This puzzled me because going back to the FRW metric, I did not see how any change fo coordinate could bring it into a flat metric!

    But now that I have thought about all this, I think that the above equation is not valid in the case of an empty universe.

    The time-time part of Einstein's equation is (setting pressure, density and cosmolgical constant equal to zero)

    [tex] H(t)^2 = - \frac{k c^2}{a(t)^2} [/tex]

    If I set t= t_now in the above and divide out buy H_0^2 I get the equation quoted above.

    The summed space-space equation is

    [tex] \dot{H} + H^2 = 0 [/tex]

    and the Ricci scalar is

    [tex] R = 6/c^2 ( \dot{H} + H^2 + \frac{k c^2}{a^2} ) [/tex]

    Now, the only way I have found to set R=0 and solve the above two equations is to set

    [tex] k=0, a= constant \Rightarrow H = \dot{H} = 0 [/tex]

    in which case the space time is trivially flat.


    So then the question is: what went wrong with the very first argument (using the first equation above) that seemed to indicate that k had to be minus 1? Now it seems to me that this equation is simply not valid because it i sobtained from the time-time equation by dividing out by [tex] H_0^2 [/tex] which is not allowed here since H_0 is zero! So this invalidates that equation and solves the problem.

    Does that make sense?


    Patrick
     
  20. Jun 3, 2008 #19

    Wallace

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    I'm not sure I follow exactly what you are asking? For an underdense Universe k=-1, and the extreme case of an underdense universe is an empty one, but k is still -1. This lead us to the Friedmann equation for the empty universe of

    [tex] H^2 = H^2_0(\Omega_k a^{-2})[/tex]

    But since in this case

    [tex]\Omega_k = \Omega_{total} [/tex]

    this reduces to

    [tex] H = \frac{H_0}{a}[/tex]

    Note that this time (as opposed to last time I demonstrated this derivation) we are explicitly putting in the H_0, but it was there all along in the quote above implicitly.

    If you follow this through as I did before you find that the Ricci scalar is zero.

    Note that it is not true to say the H_0 = 0 in an empty universe if you describe it in FRW co-ordinates. Again, this is known as the Milne model. Setting H_0=0 means that a(t)=constant and hence the FRW metric just become Minkowski, which as you say is trivially flat.
     
  21. Jun 4, 2008 #20

    hellfire

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    The equation

    [tex]1 = - \frac{k c^2}{H^2 a^2}[/tex]

    is actually valid for an empty universe since it follows directly from the Friedmann equation. May be it is useful for you to see this as follows. That equation is equivalent to

    [tex]1 = - \frac{k c^2}{\left(\displaystyle \frac{\dot a}{a} \right)^2 a^2}[/tex]

    [tex]1 = - \frac{k c^2}{\dot a^2}[/tex]

    This means

    [tex]\dot a = c \, \sqrt{- k}[/tex]

    and

    [tex]a = c \, \sqrt{- k} \, t + a_0 = A \, t + B[/tex]

    Now, consider this form of the line element

    [tex]ds^2 = a \, (- c^2 dt^2 + dr^2 + \dots)[/tex]

    that you get from the standard one by choosing [tex]t = a t[/tex]. Choose

    [tex]t^{\prime} = \frac{1}{2} (a + B) \, t[/tex]

    etc., and then you will verify that

    [tex]dt^{\prime} = a \, dt[/tex]

    and finally you get a flat space-time

    [tex]ds^2 = - c^2 dt^{\prime}^2 + dr^{\prime}^2 + \dots[/tex]
     
    Last edited: Jun 4, 2008
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