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Totally symmetric tensor 
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#1
Oct1207, 09:47 PM

P: 1,996

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. 


#2
Oct1307, 05:01 AM

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#3
Oct1307, 10:57 AM

P: 1,996

[tex] v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0 [/tex]



#4
Oct1307, 01:21 PM

P: 910

Totally symmetric tensor
[tex] v_{a}t_{aa} = 0 [/tex] [tex] 2 v_{a}t_{ab} + v_{b}t_{aa}= 0 [/tex] 


#5
Oct1307, 02:06 PM

P: 1,996

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
Oct1307, 05:53 PM

P: 910

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. 


#7
Oct1307, 08:02 PM

P: 1,996

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
Oct1407, 04:38 PM

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#9
Oct1807, 09:39 PM

P: 1,996

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


#10
Oct2007, 01:45 PM

P: 1,996

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?



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