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Curvature tensor

  1. Aug 6, 2013 #1
    hi
    In a local inertial frame with [itex]g_{ij}=\eta_{ij}[/itex] and [itex]\Gamma^i_{jk}=0[/itex].
    why in such a frame, curvature tensor isn't zero?
    curvature tensor is made of metric,first and second derivative of metric.
     
  2. jcsd
  3. Aug 6, 2013 #2

    WannabeNewton

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    Because it's only at a point. Just because a function vanishes at a point doesn't mean its derivative has to vanish there; this is basic calculus.
     
  4. Aug 6, 2013 #3
    so why first derivative become zero?
     
  5. Aug 6, 2013 #4

    WannabeNewton

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    By construction of Riemann normal coordinates, the Christoffel symbols (as represented in these coordinates) vanish identically at the point the coordinates are setup at. This does not imply that the Riemann curvature tensor vanishes identically at said point, for the reasons stated above.
     
  6. Aug 6, 2013 #5
    we choose a coordinate system that metric becomes SR, at one point.
    I know what's derivative but first derivative become zero!!! and second isn't!!
    is it regular?
    according your reason, first derivative isn't zero!!
    I can't understand you!!!
     
  7. Aug 6, 2013 #6

    WannabeNewton

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    No, Riemann normal coordinates tell us that the metric becomes Minkowski at a given point and separately that the Christoffel symbols vanish identically at said point. The vanishing of the Christoffel symbols at that point is not a direct consequence of the metric being Minkowski at that point just through differentiation of the Minkowski metric; this would make zero sense mathematically.
     
    Last edited: Aug 6, 2013
  8. Aug 6, 2013 #7

    phyzguy

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    Think of the function y = x^2.

    The function and it's first derivative are both equal to 0 at x=0, but the second derivative (it's curvature) is not equal to zero at x=0. In the same way, if you set up Riemann normal coordinates at a point, the metric is Minkowski, and the Christoffel symbols(first derivative) are zero, but the curvature tensor (second derivative) is not zero.
     
  9. Aug 6, 2013 #8

    Nugatory

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    First derivative equal to zero, means that it flat at that point.

    Second derivative not equal to zero, means that the first derivative will change (no longer be zero) when you move off that point.
     
  10. Aug 6, 2013 #9

    stevendaryl

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    Let's take a particular example: The surface of a sphere of radius 1 meter can be described by coordinates [itex]\theta[/itex] and [itex]\phi[/itex]. (You can think of [itex]\theta[/itex] as latitude and [itex]\phi[/itex] as longitude, although the mathematical convention is to have [itex]\theta[/itex] run from 0 to [itex]\pi[/itex], rather than from -90 to +90, and [itex]\phi[/itex] runs from 0 to 2[itex]\pi[/itex], rather than from -180 to +180)

    The components of the metric tensor in this coordinate system are:

    [itex]g_{\theta \theta} = 1[/itex]
    [itex]g_{\phi \phi} = sin^2(\theta)[/itex]

    Take a first derivative to get:

    [itex]\dfrac{\partial}{\partial \theta} g_{\phi \phi} = 2 sin(\theta) cos(\theta)[/itex]

    Take a second derivative to get:

    [itex]\dfrac{\partial^2}{\partial \theta^2} g_{\phi \phi} = 2 (cos^2(\theta) - sin^2(\theta))[/itex]

    At [itex]\theta = \dfrac{\pi}{2}[/itex], we have

    [itex]g_{\theta \theta} = 1[/itex]
    [itex]g_{\phi \phi} = 1[/itex]

    [itex]\dfrac{\partial}{\partial \gamma} g_{\alpha\beta} = 0[/itex]

    where [itex]\alpha, \beta, \gamma[/itex] are either [itex]\theta[/itex] or [itex]\phi[/itex]

    So, the metric components and their first derivatives look just like flat space. But
    the second derivative is nonzero, which means that the Riemann curvature tensor can be nonzero.
     
  11. Aug 6, 2013 #10

    WannabeNewton

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    Also, you may already know this, but keep in mind that a frame is not a coordinate system and that a coordinate system is not a frame. What you are describing are Riemann normal coordinates, which are unfortunately called "locally inertial frames" in some texts; other texts more appropriately call them "locally inertial coordinates". What one does is use a frame to define a coordinate system, which is how Riemann normal coordinates are constructed (on the physics side anyways).
     
    Last edited: Aug 6, 2013
  12. Aug 7, 2013 #11
    i can't understand your reasons!! 'stevendyal' and 'phyzguy'
    there are another examples that first and second derivative equal zero at point
    for example,y=(x-1)^3 +1 at x=1, this function is simple example!
    it isn't a reason that if some functions like y=x^2 have this properties, another has also!!!
    this is not a prove!!
    I think your prove is based according properties of metric tensor!!
    I don't know that!! but I think like this.

    wannabenewton
    i don't know exactly what's difference between "coordinate system" and "frame"!!
    is it possible to explain it for me?
    thanks.
     
  13. Aug 7, 2013 #12

    WannabeNewton

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    Sure, I'll explain to you the difference between a frame and a coordinate system after resolving the main issue. First things first, have you seen how Riemann normal coordinates (aka "locally inertial coordinates") are actually constructed?
     
  14. Aug 7, 2013 #13
    what's Riemann normal coordinates?
     
  15. Aug 7, 2013 #14

    phyzguy

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    What exactly is your question? I thought from your first post that the question was, "In a local inertial frame with a Minkowski metric and Christoffel symbols equal to zero at a point, why isn't the curvature tensor zero?" So I provided an example function (y = x^2) whose value and first derivative are zero at a point, but whose second derivative at that same point is non-zero. I thought this might help you understand, but apparently it didn't.

    Please tell us exactly what your question is.
     
  16. Aug 7, 2013 #15

    WannabeNewton

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    Looks like there's a bit of background you have to cover. "Riemann normal coordinates" is the mathematical name for the "locally inertial coordinates" you mentioned in your original post. Can you tell me what textbook you're using for GR? And do you know how "locally inertial coordinates" are actually constructed mathematically? Knowing this will clear up your confusions.
     
  17. Aug 7, 2013 #16

    stevendaryl

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    If both the first and the second derivatives of the metric tensor are zero, then the curvature tensor is zero. At least at that single point. A curved space can have a curvature tensor that has different values at different points.
     
  18. Aug 7, 2013 #17
    i didn't here it.
    yes i know how to construct it.
    i study padmanabhan and caroll

    my question is how to prove second derivative of metric isn't zero, while first is zero.
    hoe we can prove it?
    yes, there are some examples that first is zero but second isn't, we speak about metric function!!!!
     
  19. Aug 7, 2013 #18

    stevendaryl

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    Let me illustrate for the simplest case, of 2D space (no time dimension). Suppose you have some local coordinate system [itex]x, y[/itex], and you have two points

    [itex]P_1[/itex] with coordinates [itex](x,y)[/itex]
    [itex]P_2[/itex] with coordinates [itex](x + \delta x, y + \delta y)[/itex]

    The distance between these points is given approximately (for small [itex]\delta x[/itex] and [itex]\delta y[/itex]) by:

    [itex]\delta s^2 = g_{xx} \delta x^2 + 2 g_{xy} \delta x \delta y + g_{yy} \delta y^2[/itex]

    where [itex]g_{xx}, g_{xy}, g_{yy}[/itex] are three functions of position.

    The coordinates are Riemann normal coordinates for the point [itex]P_1[/itex] provided that:

    [itex]g_{xx} = g_{yy} = 1[/itex] at [itex]P_1[/itex]
    [itex]g_{xy} = 0[/itex] at [itex]P_1[/itex]
    [itex]\frac{\partial}{\partial x} g_{xx} = \frac{\partial}{\partial y} g_{xx} = 0[/itex] at [itex]P_1[/itex]
    [itex]\frac{\partial}{\partial x} g_{xy} = \frac{\partial}{\partial y} g_{xy} = 0[/itex] at [itex]P_1[/itex]
    [itex]\frac{\partial}{\partial x} g_{yy} = \frac{\partial}{\partial y} g_{yy} = 0[/itex] at [itex]P_1[/itex]

    Riemann normal coordinates are the closest you can get to Cartesian coordinates.
     
  20. Aug 7, 2013 #19

    stevendaryl

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    Take the second derivative, and see if it's nonzero. That's how you prove that the second derivative is nonzero. I'm not sure I understand what you are asking for. Sometimes the first derivative is zero, but not the second. Sometimes both the first and second derivatives are zero.

    The point about the curvature tensor is that if it is nonzero in one coordinate system, then it will be nonzero in every coordinate system. If it is zero in one coordinate system, then it will be zero in every coordinate system. So the curvature tensor is independent of your coordinate system in a way that the Christoffel coefficients [itex]\Gamma^i_{jk}[/itex] are not.
     
  21. Aug 7, 2013 #20

    WannabeNewton

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    Steven has already explained it brilliantly. I really have nothing else to add with regards to that issue.

    As far as coordinates and frames go: given an ##n##-manifold ##M##, a frame at an event ##p \in M## is just an orthonormal basis ##\{e_{\mu}\}## for ##T_p M##; a coordinate system is a pair ##(U,x^{\mu})## where ##U \subseteq M## is open and ##x^{\mu}: M \rightarrow \mathbb{R}## are a set of ##n## coordinate functions. In the context of GR, a frame corresponds to the instantaneous rest frame of an ideal observer with ##(e_0) = u##, where ##u## is the 4-velocity of said observer, and with ##(e_i), i = 1,2,3## representing the spatial axes of the instantaneous rest frame. These instantaneous rest frames are also called local inertial frames or local Lorentz frames. One can then use these frames to construct coordinates; a local inertial frame at an event ##p## can be used to construct a set of locally inertial coordinates ##x^{\mu}## in a neighborhood of ##p## using the exponential map (see p.112 of Carroll) in which ##g_{\mu\nu}(p) = \eta_{\mu\nu}## and separately ##\Gamma^{\gamma}_{\mu\nu}(p) = 0##.
     
    Last edited: Aug 7, 2013
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