Kevin McHugh said:
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>>
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.