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Curvature Invariants

  1. Dec 28, 2013 #1
    I have a basic question regarding the invariants that can be formed from the Riemann curvature tensor, specifically the Kretschmann scalar. Does this quantity have any physical significance, in the sense that it is connected to anything physically measurable or observable ?

    My current understanding of this invariant is that it provides a scalar measure of total curvature effects at a given point; in exterior Schwarzschild space-times this will depend only on the radial coordinate ( due to the symmetries present ), in other space-times it may be a more complicated expression.

    Thanks in advance for any clarification you may be able to provide on this.
     
  2. jcsd
  3. Dec 28, 2013 #2

    WannabeNewton

    User Avatar
    Science Advisor

    What do you mean by "connected to anything physically measurable or observable"?

    On the one hand, the scalar field ##R^{abcd}R_{abcd}## is built from the Riemann tensor and the components of the Riemann tensor are physically measurable quantities: take a congruence of time-like worldlines and use the equation of geodesic deviation to operationally measure the components of the Riemann tensor. So in that sense it is obviously connected to measurable quantities. Here when I say that the components of the Riemann tensor are measurable I mean that a given observer can use the rods and clock of his rest frame to measure the associated physical quantities and represent them component-wise using the basis vectors of his rest frame (where we assume that the measuring apparatus attached to the rest frame is equivalent to that of a momentarily comoving locally inertial frame-something that can always be arranged by precise calibration of the measuring apparatus).

    If, on the other hand, you are asking about whether ##R^{abcd}R_{abcd}## has direct physical meaning in and of itself, then let me ask you a simpler question:

    Recall that if we have a time-like unit vector field ##\xi^a## then ##E^a = F^{a}{}{}_{b}\xi^b## and ##B^a = -\frac{1}{2}\epsilon^{abcd}\xi_{b}F_{cd}## imply that ##\frac{1}{2}F^{ab}F_{ab} = B^2 - E^2## where ##F_{ab}## is the EM field tensor and ##E^a,B^a## are the electric and magnetic fields relative to ##\xi^a## respectively. Does the electromagnetic invariant ##\frac{1}{2}F^{ab}F_{ab} = B^2 - E^2## have direct physical meaning?

    Are you perhaps asking for a geometric interpretation of ##R^{abcd}R_{abcd}##?
     
  4. Dec 28, 2013 #3
    What I meant was whether there is a direct physical meaning in and of itself to these invariants - the same question applies to the electromagnetic field tensor and its invariants, and I am not clear about the answer there either ( of course one can measure E and B, but that's not what I'm after ). I suppose what I am really curious about is whether the Kretschmann scalar is something that could be directly measured, given a suitable instrument or setup, and what that would be. I suspect that is not the case, but I am not sure.

    Btw, you were alluding to a geometric interpretation - I would be curious as to that as well.
     
  5. Dec 28, 2013 #4

    Dale

    Staff: Mentor

    You can always measure any invariant directly simply by building a device which measures each of the components of the tensor wrt some basis and contracting the tensor.
     
  6. Dec 29, 2013 #5
    Ok, that is pretty much what I thought anyway - just needed confirmation. Thank you everyone for your comments.
     
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