Levi-civita and symmetric tensor

In summary: I hope I've got that the right way around.)Hurkyl, you're right, my book doesn't use upper indicies, but I have to or I get confused and forget which indicies are which. :uhh:Also, thank you to both of you for your help! :!!)Hurkyl, you're right, my book doesn't use upper indicies, but I have to or I get confused and forget which indicies are which. :uhh:Also, thank you to both of you for your help! :!!)In summary, the Levi-Civita tensor can be used to separate a tensor into its symmetric and antisymmetric parts, by contracting with the tensor and using the properties
  • #1
Meggle
16
0

Homework Statement


Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric.

Homework Equations


The Attempt at a Solution


The first bit I think is just like the proof that a symmetric tensor multiplied by an antisymmetric tensor is always equal to zero.
[tex]\epsilon_{ijk} = - \epsilon_{jik}[/tex] As the levi-civita expression is antisymmetric and this isn't a permutation of ijk.
[tex]\epsilon_{ijk}a_{ij} = - \epsilon_{jik}a_{ij}[/tex]
[tex]\epsilon_{ijk}a_{ij} = - \epsilon_{jik}a_{ji}[/tex] As [tex]a_{ji}[/tex] is symmetric.
[tex]\epsilon_{ijk}a_{ij} = - \epsilon_{ijk}a_{ij}[/tex] Swapping dummy indicies.
[tex]2\epsilon_{ijk}a_{ij} = 0[/tex]
[tex]\epsilon_{ijk}a_{ij} = 0[/tex]

I think that proves the "if [tex]a_{ij}[/tex] is symmetric" part, but not the "and only if" part. There's a section in my text that claims a tensor is symmetric in similar based on "contracting with [tex]epsilon_{kqp}[/tex]" and using the relationship between the Levi-civita equation and the Kronecker delta:
[tex]\epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}[/tex]
But I didn't understand what it meant by that.

So, here goes:
[tex]\epsilon_{kqp}(\epsilon_{ijk}a_{ij}) = \epsilon_{kqp}(0) = 0[/tex]

[tex](\epsilon_{kqp}\epsilon_{ijk})a_{ij} = (\delta_{qi}\delta_{pj} - \delta_{qj}\delta_{pi})a_{ij}[/tex]

[tex]\epsilon_{kqp}(\epsilon_{ijk}a_{ij}) = (\delta_{qi}\delta_{pj})a_{ij} - (\delta_{qj}\delta_{pi})a_{ij}[/tex]

[tex]\epsilon_{kqp}(\epsilon_{ijk}a_{ij}) = a_{qp} - a_{pq}[/tex]
Which is zero only where a is symmetric, right?

If someone could please tell me if I'm on the right track or if I've done something completely wrong, that'd be really handy. I've been stuck on this one for hours and only just thought of those last two lines while I was writing out the equations on here, but I really don't know if it's right.
 
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  • #2
All tensors can be separated into a symmetric and antisymmetric part.

Tij = bSij + aAij , sometimes written Tij = bS(ij) + aA[ij].

Can you prove this?

Operate the Levi-Civita tensor on each part.
 
  • #3
Phrak said:
All tensors can be separated into a symmetric and antisymmetric part.

Tij = bSij + aAij , sometimes written Tij = bS(ij) + aA[ij].

Can you prove this?

Operate the Levi-Civita tensor on each part.

I can't prove that. It's stated without proof in my course readings and I haven't managed to figure it out. I don't know how to operate the Levi-Civita tensor either (yes, I'm in trouble, I know).
So is this a way to answer the whole question? And what I had done isn't right at all?
 
  • #4
OK, you don't have to do the proof, but can just use it. That's plenty good enough.

I think you should try the direct approch to get a feel for how tensor multiplication works by substituting 1, 2, and 3 into the Levi-Civita tensor.

Remember, [itex]\epsilon_{ijk}[/itex] = 1 for even permutations of (ijk), [itex]\epsilon_{ijk}[/itex] = -1 for odd permutations of of (ijk), and [itex]\epsilon_{ijk}[/itex] = 0 for every other combination.

All together, there are only 6 non-zero elements in [itex]\epsilon_{ijk}[/itex]. You need to write them out for [itex]\epsilon_{ijk}S_{ij}[/itex] and [itex]\epsilon_{ijk}A_{ijk}[/itex]. I'm using S to represent a symmetric tensor and A for an antisymmetric tensor.

[tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + 4\ more\ tems = 0[/tex]
 
  • #5
Phrak said:
All together, there are only 6 non-zero elements in [itex]\epsilon_{ijk}[/itex]. You need to write them out for [itex]\epsilon_{ijk}S_{ij}[/itex] and [itex]\epsilon_{ijk}A_{ijk}[/itex]. I'm using S to represent a symmetric tensor and A for an antisymmetric tensor.

[tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + 4\ more\ tems = 0[/tex]

Ummmmerrrrrr:
[tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + \epsilon_{231}S_{23} + \epsilon_{321}S_{32} + \epsilon_{213}S_{21}+ \epsilon_{132}S_{13}[/tex]

[tex]\epsilon_{ijk}S_{ij} = (1)S_{12} + (1)S_{31} + (1)S_{23} + (-1)S_{32} + (-1)S_{21} + (-1)S_{13}[/tex]

[tex]\epsilon_{ijk}S_{ij} = S_{12} + S_{31} + S_{23} - S_{23} - S_{12} - S_{31} = 0[/tex] because [tex]S_{ij}=S_{ji}[/tex] ?

Following the same logic for the antisymmetric side:
[tex]\epsilon_{ijk}A_{ij} = \epsilon_{123}A_{12} + \epsilon_{312}A_{31} + \epsilon_{231}A_{23} + \epsilon_{321}A_{32} + \epsilon_{213}A_{21}+ \epsilon_{132}A_{13}[/tex]

[tex]\epsilon_{ijk}A_{ij} = (1)A_{12} + (1)A_{31} + (1)A_{23} + (-1)A_{32} + (-1)A_{21} + (-1)A_{13}[/tex]

[tex]\epsilon_{ijk}A_{ij} = A_{12} + A_{31} + A_{23} + A_{23} + A_{12} + A_{31} = 2A_{12} + 2A_{31} + 2A_{23}[/tex] because [tex]A_{ij} = -A_{ji}[/tex]

So, from here I can say that [tex]\epsilon_{ijk}T{ij} = 0[/tex] whenever [tex]T_{ij}[/tex] is a symmetric tensor, and if [tex]\epsilon_{ijk}T_{ij} = 0[/tex], [tex]T_{ij}[/tex] must be symmetric, as if [tex]T_{ij}[/tex] had an antisymmetric component then [tex]\epsilon_{ijk}T_{ij}[/tex] would not equal zero. (So zero antisymmetric component in [tex]T_{ij} = bS_{ij} + aA_{ij}[/tex] .)
Maybe? That looks (to me) like it's complete. I think. :redface:
 
  • #6
eijkSij should have three values, one for each possible value of k...

(P.S. your book doesn't use upper indices??)
 
  • #7
Well done, Meggle. Especially since you had to encode it all in mathtext :q.

What Hurkyl has pointed out is that εijkSij is a vector. After εijk is contracted with Sij there is still one index left over.

We could better write the equation in question as εijkSij = 0k just to remember this.

You're equations for εijkSij would then look like this.

[tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}S_{12} + \epsilon_{312}S_{31} + \epsilon_{231}S_{23} + \epsilon_{321}S_{32} + \epsilon_{213}S_{21}+ \epsilon_{132}S_{13}[/tex]

[tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}(S_{12} - S_{21}) + \epsilon_{312}(S_{31} - S_{13}) + \epsilon_{231} (S_{23} - S_{32})[/tex]

[tex]\epsilon_{ijk}S_{ij} = \epsilon_{123}0_{12} + \epsilon_{312}(0_{31}) + \epsilon_{231} (0_{23})[/tex]

[tex]\epsilon_{ijk}S_{ij} = 0_3 + 0_2 + 0_1[/tex]

[tex]\epsilon_{ijk}S_{ij} = 0_k[/tex]
 
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Related to Levi-civita and symmetric tensor

1. What is a Levi-Civita tensor?

A Levi-Civita tensor, also known as the permutation tensor, is a mathematical object used in tensor calculus to represent the cross product of two vectors in three-dimensional space. It is defined as a tensor of rank 3 with components that are either 1, -1, or 0, depending on the permutation of its indices.

2. How is a Levi-Civita tensor used in physics?

Levi-Civita tensors are used in physics to define the cross product of two vectors, which is an important operation in electromagnetism, fluid dynamics, and other fields. They are also used in Einstein's field equations of general relativity to represent the curvature of spacetime.

3. What is the relationship between a Levi-Civita tensor and a symmetric tensor?

A Levi-Civita tensor is a special type of symmetric tensor, meaning that it is invariant under a change of indices. However, not all symmetric tensors are Levi-Civita tensors, as they may have components other than 1, -1, or 0.

4. How do you calculate the determinant of a Levi-Civita tensor?

The determinant of a Levi-Civita tensor can be calculated by taking the product of all its components. For example, in three-dimensional space, the determinant is equal to 1 if the indices are in ascending order, -1 if they are in descending order, and 0 if any indices are repeated.

5. Can a Levi-Civita tensor be extended to higher dimensions?

Yes, a Levi-Civita tensor can be extended to higher dimensions, such as four-dimensional spacetime. However, the number of components increases exponentially with the number of dimensions, making it more difficult to work with. In higher dimensions, it is more common to use other types of tensors, such as the generalized permutation tensor, to represent the cross product.

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