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I Reducing Riemann components to 20

  1. Mar 7, 2016 #1
    I used the expression Rabcd=-Rbacd=-Rabdc=Rcdab to reduce the number of components. I also used if a=b=0 the R=0 and if and c=d=0 then R=0.

    This reduced the number of components to 64. How do I get them down to 21? I know I need another equality to reduce it to 20.

    <<Mentor note: Fixed typesetting>>
    Last edited by a moderator: Mar 7, 2016
  2. jcsd
  3. Mar 7, 2016 #2


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    Please describe your reasoning in more detail. The fact that a = b = 0 gives a zero component is not an independent condition, it is already contained in ##R_{abcd} = -R_{bacd}## ...
  4. Mar 7, 2016 #3
    I'm not sure I understand the index gymnastics well enough to intelligently respond to your question. However, I can try to explain my efforts a little more clearly. I first wrote out all 256 possible combinations of the indices. I immediately eliminated any expressions containing a=b and c=d. I then wrote down the expression

    Rabcd = -Rbacd = -Rabdc= Rcdab for the remaing expressions. This left me with 64 terms. My question is how do I further reduce these 64 to 20?
  5. Mar 7, 2016 #4


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    Let's group the indices [itex]R_{abcd}[/itex] into two groups:
    • [itex]a,b[/itex]
    • [itex]c,d[/itex]
    Concentrating on just the first group, there are apparently [itex]16[/itex] possibilities: [itex]00[/itex], [itex]01[/itex], ... [itex]03[/itex], [itex]10[/itex], ...[itex]33[/itex]. However when [itex]a=b[/itex] the tensor is zero, so that knocks out the cases [itex]00, 11, 22, 33[/itex]. So we're down to just 12 possibilities. But because of the antisymmetry--[itex]R_{abcd} = -R_{bacd}[/itex]--half of those are redundant. So there are only 6 independent values:
    [itex]01, 02, 03, 12, 13, 23[/itex]. Let me just call those cases: [itex]A,B,C,D,E,F[/itex] (where [itex]A[/itex] is shorthand for [itex]01[/itex], [itex]B[/itex] is [itex]02[/itex],etc.)

    There are similarly only 6 independent possibilities for [itex]c,d[/itex]. So you'd think that the total number would be 6 x 6 = 36. But there's another symmetry:

    [itex]R_{abcd} = R_{cdab}[/itex]

    That means that for all 4 indices, we need only consider the following 21 independent cases:
    [itex]AA, AB, AC, AD, AE, AF, BB, BC, BD, BE, BF, CC, CD, CE, CF, DD, DE, DF, EE, EF, FF[/itex]

    or in terms of the original indices:
    [itex]0101, 0102, 0103, 0112, 0113, 0123, 0202, 0203, 0212, 0213, 0223, 0303, 0312, 0313, 0323, 1212, 1213, 1223, 1313, 1323, 2323[/itex]

    There is one more symmetry:
    [itex]R_{abcd} + R_{acdb} + R_{adbc} = 0[/itex]

    This allows us to write [itex]R_{0312}[/itex] in terms of [itex]R_{0231}[/itex] and [itex]R_{0123}[/itex]. So we're down to just 20 independent components.
  6. Mar 7, 2016 #5
    Thanks Steven, geez do I feel dumb. I have another pertinent question. What are the physical significance of these terms? For the stress energy tensor, the components all have physical meaning. For example the mass energy, momentum flux, etc. I know that Reimann is supposed to gives us the difference created by parallel transport around a closed loop on a curved surface. How do these terms relate to physical distance? If I'm not mistaken, isn't R1212 the curvature in two dimensions? Is it Reimann that tells the story or is it the connection coefficients?
  7. Mar 7, 2016 #6


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    This is true only for well chosen coordinate systems. For example, you could use coordinates where, at some point, all the basis vectors are lightlike. Then, your normal rules for the meaning of SET components would be out the window. All invariants computed using SET would still be the same - just that interpretation of components would be meaningless.
  8. Mar 8, 2016 #7


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    Choose a local orthonormal set of basis vectors, and consider the six combinations previously mentioned for (a,b) or (c,d) and divide them into two groups

    Group 1: (tx), (ty), (tz)
    Group 2 (yz), (xz), (xy)

    Here t,x,y,z are unit vectors, t is timelike, x,y,z are spacelike and all the vectors are orthogonal.

    Some terminiology that might be helpful: Group 2 is the "Hodges dual" of group 1.

    You can decompose the Riemann into an "electirc part", which can be physically interpreted as tidal forces. To see the interpretation of this tensor component as a tidal force, look at the geodesic deviation equation. If you have two timelike geodesics pointing in the "t" direction, which are initially separated by a separation vector in the "x" direction, the rate of change of their separation second derivative of the separation vector with respect to time will be proportional to ##R_{xtxt}##. But this is just the relative acceleration of two objects in "free fall" that are separated initially in the "x" direction. So we can regard ##R_{xtxt}## as being basically the component of the tidal gravity in the "x" direction.

    The "electric part" requires both halves of the Riemann to be in group 1. You also have a "magnetic part" where one component is in group 1, and the other in group 2, and a topological part where both components are group 2. This breakdown is sometimes known as the Bel decomposition of the Riemann tensor. There's a short wiki article on this, and a writeup (not under the Bel decomposition name) in MTW's text "Gravitation". There's a few writeups on PF here and there too, I'm not quite sure where.

    Sorry if this is a bit sketchy.
    Last edited: Mar 8, 2016
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