totally symmetric tensor


by ehrenfest
Tags: symmetric, tensor, totally
ehrenfest
ehrenfest is offline
#1
Oct12-07, 09:47 PM
P: 1,998
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:

[tex] v_{(a}t_{bc)} = 0 [/tex]

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.
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
CompuChip
CompuChip is offline
#2
Oct13-07, 05:01 AM
Sci Advisor
HW Helper
P: 4,301
Quote Quote by ehrenfest View Post
You can write out the 6 terms in the equation above and then reduce it to 3 terms through the symmetry of the tensor
I don't really feel like doing that now... can you post what that got you?
ehrenfest
ehrenfest is offline
#3
Oct13-07, 10:57 AM
P: 1,998
[tex] v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0 [/tex]

Daverz
Daverz is offline
#4
Oct13-07, 01:21 PM
P: 883

totally symmetric tensor


Quote Quote by ehrenfest View Post
[tex] v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0 [/tex]
I think you can do it if you start with the special cases

[tex] v_{a}t_{aa} = 0 [/tex]

[tex] 2 v_{a}t_{ab} + v_{b}t_{aa}= 0 [/tex]
ehrenfest
ehrenfest is offline
#5
Oct13-07, 02:06 PM
P: 1,998
Quote Quote by Daverz View Post
I think you can do it if you start with the special cases

[tex] v_{a}t_{aa} = 0 [/tex]

[tex] 2 v_{a}t_{ab} + v_{b}t_{aa}= 0 [/tex]
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.
Daverz
Daverz is offline
#6
Oct13-07, 05:53 PM
P: 883
Quote Quote by ehrenfest View Post
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.
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.
ehrenfest
ehrenfest is offline
#7
Oct13-07, 08:02 PM
P: 1,998
Quote Quote by Daverz View Post
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.
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
Daverz
Daverz is offline
#8
Oct14-07, 04:38 PM
P: 883
Quote Quote by ehrenfest View Post
First of all, v_a is not arbitrary--
Perhaps it's easier to think of the basis for the vector space as being 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
But the equation has to hold for any basis for the vector space, not just this one.
ehrenfest
ehrenfest is offline
#9
Oct18-07, 09:39 PM
P: 1,998
Quote Quote by Daverz View Post
But the equation has to hold for any basis for the vector space, not just this one.
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.
ehrenfest
ehrenfest is offline
#10
Oct20-07, 01:45 PM
P: 1,998
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?


Register to reply

Related Discussions
totally antisymmetric tensor Advanced Physics Homework 4
Query about non-symmetric metric tensor General Physics 6
dimension of symmetric and skew symmetric bilinear forms Linear & Abstract Algebra 2
Permutation rules of the completely anti symmetric tensor Quantum Physics 0
totally lost totally insane General Discussion 13