Second derivative of tensor

1. Jun 16, 2014

electricspit

Hello,

I have two problems.

I'm going through the Classical Theory of Fields by Landau/Lifshitz and in Section 32 they're deriving the energy-momentum tensor for a general field. We started with a generalized action (in 4 dimensions) and ended up with the definition of a tensor:

$T^{k}_i =q_{,i} \frac{\partial \Lambda}{\partial q_{,k}}-\delta^{k}_i \Lambda$

Where $q_{,i} \equiv \frac{\partial q}{\partial x^i}$ and $\Lambda$ is the Lagrangian density of the field. This led to the conclusion that:

$\frac{\partial T^{k}_i}{\partial x^k}=0$

Which is the first thing I'm confused about.

Second, using previous results about four divergences:

$\frac{\partial A^k}{\partial x^k} = 0$

If this is true, then it is equivalent to saying $\int A^k dS_k$ is conserved. This led to:

$P^i = const. \int T^{ik}dS_k$

The constant was determined to be $\frac{1}{c}$ but that is unimportant to my question for now. They say the defintion of $T^{ik}$ is not unique since we can add a 2nd rank tensor to this and still retrieve the same result:

$T^{ik}+\frac{\partial \psi^{ik\ell}}{\partial x^{\ell}}$

Where $\psi^{ik\ell}=-\psi^{i\ell k}$. This apparently still yields:

$\frac{\partial T^{ik}}{\partial x^k}=0$

(for now let's ignore the switching between mixed/contravariant). In other words, the symmetric operator $\frac{\partial^2}{\partial x^k \partial x^{\ell}}$ applied to the antisymmetric (in $k$ and $\ell$):

$\frac{\partial^2 \psi^{ik\ell}}{\partial x^k \partial x^{\ell}} = 0$

This is my second question. Why is this zero? Can anyone show me the math behind this? I'm having trouble sorting it out.

Thank you!

2. Jun 16, 2014

Orodruin

Staff Emeritus
First: $T^k_i$ is the kth component of the conserved current related with translations in the ith direction (I assume your Lagrangian density is not explicitly dependent on the coordinates) which means the divergence equal to zero follows directly from Noether's theorem.

Second: This is just like any contraction of symmetric vs anti-symmetric indices. Swapping the order of the indices and then renaming them gives back minus the original expression so it must be zero.

3. Jun 16, 2014

electricspit

Okay so the second question definitely makes a lot of sense now and actually was quite simple.

The first question, I don't think I have strong enough understanding of Lagrangian mechanics or tensors to understand. From what I know, Noether's theorem states that if a system has certain symmetries, there will be conserved quantities associated with them. What does it mean in this specific context? You were right in assuming the Lagrangian density is not explicitly dependent on the coordinates. What do you mean by conserved current?

Thank you again.

4. Jun 16, 2014

Orodruin

Staff Emeritus
Noether's theorem states that if a transformation of the fields and the variables is a symmetry of the Lagrangian, then there is a corresponding conserved current $J^\mu$ for which $\partial_\mu J^\mu =0$. If you make the transformation x → x+a and your lagrangian density is not explicitly coordinate dependent, the transformation is a symmetry. The expression you have for T is the corresponding conserved current and therefore has zero divergence. If I do not misremember the wikipedia page on Noether's theorem is quite informative.

Often in classical mechanics you will see Noether's theorem as giving a conserved charge Q, i.e., dQ/dt = 0. This however assumes that the parameter is time only. When you go to field theory you, the fields also depend on the spatial coordinates and you get the generalization $\partial_\mu J^\mu =0$.

5. Jun 16, 2014

electricspit

Thank you for the information!