How to Decompose a Tensor into Lambda, F, and V Components?

[latentcorpse]

Show that a tensor T can be written as

T_{ij}=\lambda \delta_{ij} + F_{ij} +\epsilon_{ijk} v_{k}

for the tensor
\[ \left( \begin{array}{ccc}<br /> 1 &amp; 2 &amp; 3 \\<br /> 4 &amp; 5 &amp; 6 \\<br /> 7 &amp; 8 &amp; 9 \end{array} \right)\]

find \lambda, F_{ij}, v_k

i can't get anywhere whatsoever with this question?

 

[ThalorB]

If i and j are the rows/columns, then what is k? Is there something missing in the problem description?

 

[lanedance]

i'm not sure about the problem itself, as I'm not sure of the context or meaning of the symbols (what are vk's & F?), but k is a contracted sum, so i think its ok

 

[latentcorpse]

hi. it's straight out a past exam q so should be ok.

not sure if this is what you mean but i noted that because of the levi civita,

if k is 3 then you can have either i=1,j=2 or j=1,i=2

i.e. we get v_3-v_3 but wouldn't that just make this last term trivial as all the terms would just cancel out similarly for k=1 and k=2?

 

[Ben Niehoff]

I think the tensor F should be symmetric. Basically what the question is asking is to separate the tensor into three components:

1. Its trace.

2. A symmetric, traceless part.

3. An antisymmetric part.

 

[latentcorpse]

yes but \lambda \delta_{ij} puts a \lambda on each diagonal element so in order to get 1,5 and 9 on the diagonals, at least one of the other two terms must have non zero diagonal components. clearly the third (antisymmetric) one cannot. so its down to the F_{ij} and it must be symmetric and traceless.

incidentally - this is the only term i don't understand the origin of - why must it be symmetric and traceless?

then how do i solve for F when all i know is that

F_{11}=1-\lambda.F_{22}=5-\lambda,F_{33}=9-\lambda,F_{11}+F_{22}+F_{33}=0
this gives \lambda=5
so then F_{11}=-4.F_{22}=0,F_{33}=4

how do i find the rest of F using v?

 

[Ben Niehoff]

Try first writing

\epsilon_{ijk}v_k = A_{ij}

where A is an antisymmetric tensor. Now relate the components of A, F, and T. Also use the symmetry of F and the antisymmetry of A. You should get enough equations to obtain the rest of the elements.

 

[Ben Niehoff]

incidentally - this is the only term i don't understand the origin of - why must it be symmetric and traceless?

The purpose of this exercise, as far as I can tell, is for you to separate a general tensor T into parts that transform under different representations of the rotation group.

1. The trace is a scalar, and is invariant under rotation: a spin-0 object.

2. The antisymmetric part has 3 independent components, and transforms like a vector under rotations: a spin-1 object.

3. The symmetric, traceless part has 5 independent components, and transforms like a tensor under rotations: a spin-2 object.

 

[latentcorpse]

hang on,

first of all how did you know to write \epsilon_{ijk}v_k=A_{ij}

secondly i get equaitons like

A_{12} + F_{12}=2, A_{21}+F_{21}=4 \Rightarrow F_{12} + F_{21} = 6

but none of the other equaitons refer to either the 12 or 21 components so I am struggling to get an exact value here

 

[tiny-tim]

Hi latentcorpse!

Hint: what are T_ij + T_ji,

and T_ij - T_ji ?

 

[latentcorpse]

do you want me to apply that to F or A?

for the symmetric F the add one will give twice the original and the subtract one will give 0. vice versa for the antisymmetric tensor A.

how does that help us though?

 

[tiny-tim]

do you want me to apply that to F or A?

No … to T

 

[latentcorpse]

T_{ij}+T_{ji}=2\lambda \delta_{ij} + 2 F_{ij}, T_{ij}-T_{ji}=2 \epsilon_{ijk} v_k

can you explain why \epsilon_{ijk} v_k = A_{ij} i can see that it must be anti symmetric because of the levi civita but how does the k fit into it all?

and why is F symmetric and traceless - how do we know that?

 

[tiny-tim]

T_{ij}+T_{ji}=2\lambda \delta_{ij} + 2 F_{ij}, T_{ij}-T_{ji}=2 \epsilon_{ijk} v_k

Yes, half the sum (this is for anything, not just tensors) is the symmetric part, and half the difference is the anti-symmetric part.

can you explain why \epsilon_{ijk} v_k = A_{ij} i can see that it must be anti symmetric because of the levi civita but how does the k fit into it all?

k is a dummy index …

just draw the matrix, and you'll see where the v coordinates fit in

and why is F symmetric and traceless - how do we know that?

We don't … we choose F and lambda so that trace(F) = 0, by subtracting trace(T_ij + T_ji) from T_ij + T_ji

 

[latentcorpse]

so the sums involving T_ij and T_ji give me the extra equations i was needing to find all the constants but I am still having trouble seeing what's going on:

i agree that k is a dummy index but i can't really see how it works here:
if i pick k=1 that restricts i and j to 1 or 2
so the matrix has a 1 in the (12) component and a -1 in the (21) component - is this along the right lines?

and i still don't follow how/why we choose F and lambda so that trace(F)=0 or why we'd even want trace(F) to be 0 in the first place?

 

[lanedance]

so i think it means
\epsilon_{ijk} v_{k} = <br /> \[ \left( \begin{array}{ccc}\epsilon_{11k} v_{k} &amp; \epsilon_{12k} v_{k} &amp; \epsilon_{13} v_{k} \\\epsilon_{21k} v_{k} &amp; .. &amp; .. \\.. &amp; .. &amp; .. \end{array} \right)\]

if i remember correctly about the levi civita
\epsilon_{ijk} = 0, if i=j or j=k or k=i
\epsilon_{ijk} = 1, for even permutation of i,j,k
\epsilon_{ijk} = -1, for odd permutation of i,j,k

which gives
\epsilon_{ijk} v_{k} = <br /> \[ \left( \begin{array}{ccc} 0 &amp; \epsilon_{123} v_{3} &amp; \epsilon_{132} v_{2} \\\epsilon_{213} v_{3} &amp; .. &amp; .. \\.. &amp; .. &amp; .. \end{array} \right)\]

which gives
\epsilon_{ijk} v_{k} = <br /> \[ \left( \begin{array}{ccc} 0 &amp; +v_{3} &amp; -v_{2} \\ -v_{3} &amp; .. &amp; .. \\.. &amp; .. &amp; .. \end{array} \right)\]

which is antisymmetric with zero trace

 

[tiny-tim]

if i pick k=1 that restricts i and j to 1 or 2
so the matrix has a 1 in the (12) component and a -1 in the (21) component - is this along the right lines?

No, you're missing the point (and it's not 1, it's 3 ) …

the point of the ε_ijk is to pick up bits of v and plonk them down into a matrix …

if you pick k=3 that restricts i and j to 1 or 2
so the matrix has v₃ in the (12) component and -v₃ in the (21) component …

see lanedance's last matrix!

 

[latentcorpse]

kl. can i ask a quick unrelated question while your at it:

let V=span(1,1)^T

that is the set of all linear combinations, i.e. V=\{\lambda \vec{i} + \mu \vec{j} \} if we're in \mathbb{R^2}. so would it be correct to plot this in \mathbb{R^2} as a series of infinite lines parallel to the line y=x. or is it just the line y=x?

 

[tiny-tim]

that is the set of all linear combinations, i.e. V=\{\lambda \vec{i} + \mu \vec{j} \} if we're in \mathbb{R^2}. so would it be correct to plot this in \mathbb{R^2} as a series of infinite lines parallel to the line y=x. or is it just the line y=x?

uhh? \{\lambda \vec{i} + \mu \vec{j} \} is R²

 

[latentcorpse]

so what does the span of (1,1)^T look like then is it just the line y=x?

 

[tiny-tim]

so what does the span of (1,1)^T look like then is it just the line y=x?

Yes, it's \{\lambda \vec{i} + \lambda \vec{j} \} which is y = x.