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 Joschua_S Feb25-12 08:12 AM

Trace - Integration - Average - Tensor Calculus

Hi

Let $D$ be an anisotropic tensor. This means especially, that $D$ is traceless. $\mathrm{tr}(D) = 0$

Apply the representating matrix of $D$ to a basis vector $S$, get a new vector and multiply this by dot product to your basis vector. Than you got a scalar function.

Now integrate this function over a symmetric region, for example the n-dimensional unit-sphere or a n-dimensional Cube or something other symmetric.

$\int SDS ~ \mathrm{d} \Omega$

This integral vanish!

$\int SDS ~ \mathrm{d} \Omega = 0$

My question:

Trace is for me like an average of something. The symmetric integration vanish also, like the trace.

Is there a link between trace zero and the vanishing integral? What is the math behind this.

If you wish one example. A part of a dipole-dipole coupling Hamiltonian in spectroscopy is given by $H_{DD} = SDS$, with the anisotropic zero field splitting tensor $D$ and spin $S$.

Greetings

Re: Trace - Integration - Average - Tensor Calculus

Quote:
 Quote by Joschua_S (Post 3783411) Is there a link between trace zero and the vanishing integral?
Not in the cases like you have presented. A non-zero scalar function has alway non-zero trace. But the integral can be zero or non zero.

 Joschua_S Feb25-12 12:58 PM

Re: Trace - Integration - Average - Tensor Calculus

Hi

This means that the integral vanish here is pure randomness?

There are no theorems in math about anisotropic tensors, trace and integrals? :-(

Greetings

Re: Trace - Integration - Average - Tensor Calculus

Quote:
 Quote by Joschua_S (Post 3783832) Hi This means that the integral vanish here is pure randomness?
In fact I see no reason for your integral to vanish unless you have some additional assumptions (that you did not list) about the dependence of your D and S on space variables.

 Joschua_S Feb27-12 12:46 PM

Re: Trace - Integration - Average - Tensor Calculus

The dipole-dipole Hamiltonian in ESR is given by

$H_{DD} = \dfrac{\mu_0}{2h} g_j g_k \mu_b^2 \left( \dfrac{\vec{S}_j \cdot \vec{S}_k}{r^3_{jk}} - \dfrac{3(\vec{S}_j \cdot \vec{r}_{jk}) \cdot (\vec{S}_k \cdot \vec{r}_{jk})}{r^5_{jk}} \right)$

One can write it as

$H_{DD} = \vec{S} \underline{\underline D} \vec{S}$

with the traceless symmetric Tensor D that fullfills $\int SDS ~ \mathrm{d} \Omega = 0$

Do you know something about the math behind this?

Greetings

 Joschua_S Mar1-12 12:42 PM

Re: Trace - Integration - Average - Tensor Calculus