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 Exercise [19.17] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [19.17]
Suppose that:

(*) R_{ab}t^at^b = T_{ab}t^at^b for every vector t.

Let i, j be any two particular coordinate indices.
(i.e. these are NOT abstract indices, they're constants, in the range 0..3 when working in four dimensions).

Define the vectors:

p = x^i
q = x^j
r = x^i+x^j

(e.g. for i=1 and j=3, p = [0,1,0,0], q = [0,0,0,1], r = [0,1,0,1]; for i=j=0, p,q = [1,0,0,0], r = [2,0,0,0]; ...etc.)

Then:

R_{ab}p^ap^b = R_{ii}
R_{ab}q^aq^b = R_{jj}
R_{ab}r^ar^b = R_{ii}+R_{jj}+R_{ij}+R_{ji} \;\;\; = \;\;\; R_{ii}+R_{jj}+2R_{(ij)}.

Combining the above gives:

(1) R_{(ij)} = \frac{1}{2}\left(R_{ab}r^ar^b-R_{ab}p^ap^b-R_{ab}q^aq^b\right)

And similarly:

(2) T_{(ij)} = \frac{1}{2}\left(T_{ab}r^ar^b-T_{ab}p^ap^b-T_{ab}q^aq^b\right)


By (*), the RHS of (1) and of (2) must be equal; so we have proven that R_{(ij)}=T_{(ij)} for every choice of i, j.
Hence R_{(ab)}=T_{(ab)}.


If R and T are symmetric tensors, R_{ab}\equiv R_{(ab)} and T_{ab}\equiv T_{(ab)}, and so in that case (*) implies that R_{ab}=T_{ab}.


\blacksquare


07 Apr 2012, 18:59

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [19.17]
Another way of looking at this is to note that any (arbitrary) symmetric \left[{}^2_0\right] tensor S^{ab} can be constructed as a sum of vector (outer-)products:


S^{(ab)} = \pm r^ar^b \pm s^as^b \pm \cdots \pm t^at^b


So (*) implies that R_{ab}S^{(ab)} = T_{ab}S^{(ab)} for any S^{ab}.

This is equivalent to R_{(ab)}S^{ab} = T_{(ab)}S^{ab} for any S^{ab} (even non-symmetric).

...So once again we get R_{(ab)} = T_{(ab)}.


07 Apr 2012, 19:30
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