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Curvature at the origin of a space as described by a metric

  1. Feb 4, 2016 #1
    1. The problem statement, all variables and given/known data
    This is a problem from A. Zee's book EInstein Gravity in a Nutshell, problem I.5.5

    Consider the metric ##ds^2 = dr^2 + (rh(r))^2dθ^2## with θ and θ + 2π identified. For h(r) = 1, this is flat space. Let h(0) = 1. Show that the curvature at the origin is positive or negative according to whether h(r) starts to turn downward or upward. Calculate the curvature for ##h(r) = \frac{sin(r)}{r}## and for ##h(r) = \frac{sinh(r)}{r}##

    2. Relevant equations


    3. The attempt at a solution
    ##ds^2 = dr^2 + (rh(r))^2dθ^2 = dr^2 + (r^2h(r)^2)dθ^2##

    If I let r → 0 then h(r) → 1 but (r^2h(r)^2)dθ^2 → 0
    How can I tell what would be the behavior of h(r) if it will already be gone if I let r → 0.

    For ##h(r) = \frac{sin(r)}{r}## and ##h(r) = \frac{sinh(r)}{r}## I know that when I let r → 0 from calculus the answer would be 1 and -1 respectively, but I don't know what to do with the "show" part. Can anyone give me hints?
     
  2. jcsd
  3. Feb 4, 2016 #2

    Orodruin

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    You do not compute the curvature by simply looking at one component of the metric. You need to actually compute what the curvature is.
     
  4. Feb 4, 2016 #3
    I get your point but Zee did not show how to compute the curvature and stated in this section that I have to wait until some later chapters, that is why I'm confused why is that question in this chapter. I think maybe he wants me to compute the limit of h(r) as r → 0? In which h(r) are the givens above.
     
  5. Feb 4, 2016 #4

    George Jones

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    Have you looked at the bottom of page 65 and top of page 66, and at appendix 1 for section 1.5?
     
  6. Feb 5, 2016 #5
    I have a question on appendix 1. He stated that the curvature is given by ##R = \lim_{radius\rightarrow 0} {\frac{6}{radius^2}(1-\frac{circumference}{2πradius})}##.

    Now from equation (4) in page 65, he gave the metric ##ds^2 = dρ^2 + \sin^2(ρ)dθ^2##. To know if the space is curved or not. We apply the formula above.

    The radius is ##ρ## so, ##ρ = \int_0^ε\, dρ = ε##. I get that, but for the circumference, in flat space it should be ##(radius)(dθ)##, what does ##\sin(ρ)## stand for here? It seems that it is representing the "radius" because he showed that, ##circumference = \int_0^{2π}\, \sin(ε)dθ = 2π\sin(ε)##

    Also, by applying the formula for curvature, ##R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{2π\sin(ε)}{2πε}) = \frac{6}{ε^2}(1-\frac{\sin(ε)}{ε}) = 1}##. How can it be 1 if ##\lim_{radius\rightarrow 0} {\frac{\sin(ε)}{ε}} = 1##, so it should be 0.
     
  7. Feb 5, 2016 #6

    Orodruin

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    It is not representing the radius, it represents how far you will move if you change ##\theta## by an amoun ##d\theta##. Note that, since your space is curved you no longer have the direct connection between the radius and the circumference.
     
  8. Feb 5, 2016 #7
    Oh, then if I picture it in my mind, ##ρ## is the radius (0 to ε) which is a straight line? And ##\sin(ε)## is somehow the distance from ##0## to ##ε## but curved? So technically ##\sin(ε)## has a "longer" length than ##ρ##?
     
  9. Feb 5, 2016 #8
    Oh, then if I picture it in my mind, ##ρ## is the radius (0 to ε) which is a straight line? And ##\sin(ε)## is somehow the distance from ##0## to ##ε## but curved? So technically ##\sin(ε)## has a "longer" length than ##ρ##?
     
  10. Feb 5, 2016 #9

    Orodruin

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    No, ##\rho## is the radius. From ##\rho =0## to ##\epsilon##, the distance is ##\epsilon##. The quantity ##\sin(\epsilon)d\theta## is the distance travelled along the circle of radius ##\epsilon## when you change the ##\theta## coordinate by ##d\theta##. If you go around a full lap, you will therefore get a distance ##\sin(\epsilon)2\pi##, not ##\epsilon 2\pi##. This is the entire point.
     
  11. Feb 5, 2016 #10
    I know, but how can I picture ##\sin(ε)## only, without the ##dθ##. For example, in flat space the circumference is ##rdθ = r(2π)##, and I know ##r## is the distance from ##0## to ##ε##. For curved space, the circumference is ##\sin(r)dθ##, but where is this ##\sin(r)##?
     
  12. Feb 5, 2016 #11

    Orodruin

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    What do you mean by "where it is"? It is a property of the manifold. It is shorter around the circle than ##2\pi r##. It is a property of the circle.
     
  13. Feb 5, 2016 #12
    What I mean is that, for a flat space the circumference is ##r(2π)##, and ##r## is "the distance from the origin to some arbitrary point"". But for a curved space the circumference is ##\sin(r)(2π)##, and ##\sin(r)## is...???

    I know that this is a property of the manifold, etc. I'm just curious if there is a way to visualize this.

    How about the other question,
    ##R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{2π\sin(ε)}{2πε}) = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{\sin(ε)}{ε})} = 1}##. How can it be 1 if ##
    \lim_{radius\rightarrow 0} {\frac{\sin(ε)}{ε}} = 1##. Shouldn't it be ##0##?
     
  14. Feb 5, 2016 #13

    Orodruin

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    I suggest thinking of the surface of a sphere and how the radius and circumference will be related there.

    No, you are completely neglecting the ##1/\epsilon^2## outside the parenthesis! If you struggle with this type of limits I suggest solidifying your calculus before attempting calculus on manifolds.
     
  15. Feb 5, 2016 #14

    Orodruin

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    Just to clarify, it is not sin(r) for a general manifold. The function will be different for different manifolds. It may even be different for different points on the same manifold.
     
  16. Feb 5, 2016 #15
    Got it, L'hopital's rule, but how can I apply it to h(r) in post#1? h(r) is completely arbitrary.

    ##radius = \int_0^ε\, dr = ε##, ##circumference = \int_0^{2π}\,εh(ε)dθ = 2πεh(ε)##

    ##R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-\frac{2πεh(ε)}{2πε})} = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-h(ε))}##
     
    Last edited: Feb 5, 2016
  17. Feb 5, 2016 #16

    Orodruin

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    You got the circumference wrong.
     
  18. Feb 5, 2016 #17
    I've edited my post. Sorry for the carelessness. But still, h(r) is arbitrary, though it is given that h(0) = 1.
     
  19. Feb 5, 2016 #18

    Orodruin

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    Yes, this is given, so what is the limit?
     
  20. Feb 5, 2016 #19
    By applying l'hospital's rule twice. ##R = \lim_{radius\rightarrow 0} {\frac{6}{ε^2}(1-h(ε))} = -3h''(ε)##

    So depending whether ##h''(ε)## is positive or negative, it will tell me if the curvature is positive or negative.
     
  21. Feb 5, 2016 #20

    Orodruin

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    Almost, a limit in which ##\epsilon \to 0## cannot depend on ##\epsilon## (it was what was approaching zero!). Replace ##h''(\epsilon)## by ##h''(0)##. Also note that ##h'(0)## needs to be zero for this limit to be finite. If not you are not dealing with a smooth manifold.

    Edit: Also note that if ##h''(0) = 0##, then the manifold is flat at r = 0 and this does not necessarily imply that it is flat everywhere. An example of this would be ##h(r) = 1 + r^4##.
     
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