Levi Civita symbol

1. Apr 22, 2015

Telemachus

1. The problem statement, all variables and given/known data
I'm trying to demonstrate the dientity: $\displaystyle \int_v \vec {r} \times ( \nabla \cdot T ) d^3x=\oint_S ( \vec r \times T) \cdot \hat n da$

T is a second order tensor, and r a vector.

So basically I should have: $\vec {r} \times \nabla \cdot T= \nabla \cdot ( \vec r \times T)$

When I use indicial notation, I get a term: $\epsilon_{ijk} T_{ki}$ which should equal zero. I see in principle that if T is a symmetric tensor I get the desired result. But I think that the result should be general, and not hold only for a symmetrical tensor T.

What I did was:

$\epsilon_{ijk} T_{ki}=-\epsilon_{ikj} T_{ki}$ (1)

and by the other side I've used that for dummy indices:

$\epsilon_{ijk} T_{ki}=\epsilon_{ikj} T_{ki}$ (2)

This last step is the one that I'm not sure if its right. I thought that as the indices are dummies, I just could interchange in the Levi Civita symbol the j and the k, and then because of the properties of the levi civita symbol, use the indentity (1) to demonstrate that the term equals zero. But I wanted to know if (2) is fine, because I'm not really sure.

I think that actually if I use the dummy indices properly I should have: $\epsilon_{ijk} T_{ki}=\epsilon_{ikj} T_{ji}$ so what I did is actually wrong.

Last edited: Apr 22, 2015
2. Apr 22, 2015

MarcusAgrippa

What does T represent physically? I suspect that the text you are using assumes that T is symmetric.

3. Apr 22, 2015

Telemachus

Is the Maxwell stress tensor. I think that in general it shouldn't be symmetrical.

4. Apr 22, 2015

MarcusAgrippa

The Maxwell stress tensor is symmetric.

5. Apr 22, 2015

Telemachus

I think that my derivation is probably wrong in a previous step, I don't see that the term should in general be zero as I stated.

Are you sure that the Maxwell stress tensor is symmetric? it represents a force per unit area in the particles and fields, I think that in particular situtations it shouldn't be symmetrical. I think that the vector identity should hold in general, but I'm not totally sure.

Looking at the expression for the Maxwell tensor, it looks symmetrical, at least for linear (and I think that isotropic too) media, linearity is an assumption previously used in the textbook. But I think that the identity should hold in general anyway.

Last edited: Apr 22, 2015
6. Apr 22, 2015

Telemachus

I'll show what I did in the previous step:

$\displaystyle \vec r \times ( \nabla \cdot T )=\epsilon_{ijk}r_j \frac{\partial T_{k \beta}}{\partial x_{\beta}}=\frac{\partial }{\partial x_{\beta}}(\epsilon_{ijk}r_j T_{k \beta})-\delta_{j \beta}\epsilon_{ijk}T_{k\beta}$

I've found a mistake when writing this here :p but it looks that the symmetry of T is needed to hold anyway.

7. Apr 22, 2015

MarcusAgrippa

If it is not symmetrical, a material medium would begin to rotate spontaneously. If you don't believe me, consult Griffiths p 352, eq (8.19). A stress tensor can only support an antisymmetric part if there is a torque per unit volume to counterbalance the twisting motion of the stress tensor. You will probably find a proof of that in Landau and Lifschitz, Fluids.

Also, if the tensor is not symmetric, the symbol
would be ambiguous.

Your calculation is correct (except for the horrid habit of pushing all indices to the lower case position!). Of course,

8. Apr 22, 2015

Telemachus

Thank you verymuch :)

I have the habit to use lower case for the indices, because I think that upper indices are related to covariant and contravariant tensors, and I haven't worked that yet (I think I'll have to when I get to relativity).

9. Apr 22, 2015

MarcusAgrippa

You ought to master the covariant and contravariant indices ASAP. The system is consistent and unambiguous. The Cartesian tensor convention (all indices in lower position) is ambiguous.

10. Apr 22, 2015

Telemachus

You are right, it's an habit I got after working in continuum mechanics. We always worked with cartesian tensors there, so the habit. What does ASAP means?

11. Apr 22, 2015

MarcusAgrippa

As Soon As Possible

12. Apr 22, 2015

Telemachus

Yes, I think that I should start right now. Could you write it in contravariant notation? in this case it would be indifferent to put the indices upward or downward? and If you have some sources to get a quick review for this, it would be great.

Thank you.

13. Apr 22, 2015

MarcusAgrippa

McConnell, Applied Tensor Analysis.
Synge and Schild, Tensor Analysis.
Barry Spain, Tensors.

$[ r⃗ ×(∇⋅T) ]^{i} = \varepsilon_{ijk} x^j (∂_\ell T^{\ell k } )$
$= ∂_\ell (\varepsilon_{ijk} x^j T^{\ell k } ) - \varepsilon_{ijk} (\partial_\ell x^j) T^{\ell k } = ∂_\ell (\varepsilon_{ijk} x^j T^{\ell k } ) - \varepsilon_{ijk} \delta^j_\ell T^{\ell k }$
$= ∂_\ell (\varepsilon_{ijk} x^j T^{\ell k } ) - \varepsilon_{ijk} T^{j k }$

Last edited: Apr 22, 2015
14. Apr 22, 2015

Telemachus

Some quick and concise reference online? I don't want to get so involved like to take a book for reference.

15. Apr 22, 2015

MarcusAgrippa

I'm afraid there is no "royal road" to tensors. It needs application and hard work. But it pays huge dividends in the end.

16. Apr 22, 2015

Telemachus

Can you explain when you use upper indices and down idices how you choose which to use? in other words, what do upper indices indicate, and what do lower indices indicate.

17. Apr 23, 2015

Fredrik

Staff Emeritus
A notation like $T^i{}_{jk}$ really means $T(e^i,e_j,e_k)$. Here $(e_1,\dots,e_n)$ is an ordered basis for the tangent space at a point p, and $(e^1,\dots,e^n)$ is the ordered basis for the cotangent space at p that's dual $(e_1,\dots,e_n)$ in the sense that $e^i(e_j)=\delta^i_j$ for all i,j. (In this example, T is a multilinear map from $V^*\times V\times V$ into $\mathbb R$, where $V$ is the tangent space at p and $V^*$ is its dual space, i.e. the cotangent space at p).

I don't think there's anything wrong with keeping all the indices downstairs in a problem like this. $T_{ij}$ is a perfectly adequate notation for the number on row i, column j of a matrix T. In problems like this, you can assume that you're dealing with a matrix and forget that you've even heard about tensors. That's what I do when I'm dealing with special relativity.

Chapter 3 in "A first course in general relativity" by Schutz is a nice introduction to tensors. Another option is chapter 8 in "Linear algebra done wrong" by Sergei Treil.

18. Apr 23, 2015

Telemachus

Thank you verymuch Fredrik.

19. Apr 23, 2015

Fredrik

Staff Emeritus
I have to say that I find the notations in $\vec r\times(\nabla\cdot T)=\nabla\cdot(\vec r\times T)$ pretty confusing. $\nabla\cdot(\text{something})$ is usually a scalar, but $\nabla\cdot T$ must be a vector for the left-hand side to make sense. I suppose that we can define vectors $\vec v_i=T_{ij}e_j$ and then define $\nabla\cdot T$ by $\nabla\cdot T=(\nabla\cdot\vec v_i)e_i$. Then we have
$$\nabla\cdot T=\partial_j(\vec v_i)_j e_i =\partial_j T_{ij} e_i.$$ The expression $\nabla\cdot(\vec r\times T)$ is even more difficult to interpret.

20. Apr 23, 2015

MarcusAgrippa

I am not entirely in agreement with this statement. There are problems with pushing all indices to the subscript position. Here are a few:

1. It violates good practice in using the summation convention to sum over indices that are both in the subscript position.

2. The results of such a calculation are valid only if the vector basis in use is orthonormal and the space is Euclidean. The results are incorrect if the basis is not orthonormal and / or if the space is not Euclidean.

3. Crystal physics does not use orthonormal bases.

4. It is confusing to novices to be told that an endomorphism $B_{ij}$, where all indices have been pushed into the lower position, has a different transformation law from the coefficients $B_{ij}$ of a quadratic form. It is impossible to explain the difference if one uses the conventions of Cartesian tensors. The distinction must be stated verbally and is not reflected in the mathematics.

My preference is for being taught the difference between covariant and contravariant ab initio. It makes life very much simpler in the long run.