Totally symmetric tensor

1. Oct 12, 2007

ehrenfest

1. The problem statement, all variables and given/known data
If t_{ab} are the components of a symmetric tensor and v_a are the components of a vector, show that if:

$$v_{(a}t_{bc)} = 0$$

then either the symmetric tensor or the vector = 0.

Let me know if you are not familiar with the totally symmetric notation.

2. Relevant equations

3. The attempt at a solution

You can write out the 6 terms in the equation above and then reduce it to 3 terms through the symmetry of the tensor but I am not sure where to go from there.

2. Oct 13, 2007

CompuChip

I don't really feel like doing that now... can you post what that got you?

3. Oct 13, 2007

ehrenfest

$$v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0$$

4. Oct 13, 2007

Daverz

I think you can do it if you start with the special cases

$$v_{a}t_{aa} = 0$$

$$2 v_{a}t_{ab} + v_{b}t_{aa}= 0$$

Last edited: Oct 13, 2007
5. Oct 13, 2007

ehrenfest

I don't know--

You're first equation only implies that either v_a or t_aa is 0 for a given a. Using the first one with the second one doesn't really imply anything because both v_a and t_aa could be 0 so then there would be no restrictions on t_ab or v_b.

6. Oct 13, 2007

Daverz

v_a is arbitrary, so if v is not identically zero, t_aa must be zero for all a (think of v_1=1, the rest 0, then the same for a=2, 3, ...). Actually, I think that does it, because there's no way t_aa=0 for all a unless t is identically zero.

Otherwise, for v not identically zero, the second equation then gives v_a t_ab = 0, and, again, since v_a is arbitrary, we must have t_ab = 0.

Last edited: Oct 13, 2007
7. Oct 13, 2007

ehrenfest

First of all, v_a is not arbitrary--it is a vector that is part of the problem statement.
Second, why could you not have something like:

v_a = (1,0,1,0,1,1,1,0,...)
t_aa = (0,1,0,1,0,0,0,1,...)

This is a case where v is not identially zero and t_aa is NOT zero for all a

8. Oct 14, 2007

Daverz

Perhaps it's easier to think of the basis for the vector space as being arbitrary.

But the equation has to hold for any basis for the vector space, not just this one.

Last edited: Oct 14, 2007
9. Oct 18, 2007

ehrenfest

That is not stated in the problem and I think it is not safe to assume that. The problem just gives us the components of a tensor and a vector, presumably in a given basis.

I think we should be able to prove if it holds in just one basis.

10. Oct 20, 2007

ehrenfest

Do other people agree with Daverz? I thought when the problem said that "If t_{ab} are the components of a symmetric tensor", that meant it was symmetric in some given basis?