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Trace - Integration - Average - Tensor Calculus

by Joschua_S
Tags: average, calculus, integration, tensor, trace
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Joschua_S
#1
Feb25-12, 08:12 AM
P: 11
Hi

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

Apply the representating matrix of [itex] D [/itex] to a basis vector [itex] S [/itex], 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.

[itex] \int SDS ~ \mathrm{d} \Omega [/itex]

This integral vanish!

[itex] \int SDS ~ \mathrm{d} \Omega = 0[/itex]

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 [itex] H_{DD} = SDS [/itex], with the anisotropic zero field splitting tensor [itex] D[/itex] and spin [itex] S [/itex].

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arkajad
#2
Feb25-12, 12:43 PM
P: 1,411
Quote Quote by Joschua_S View Post
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
#3
Feb25-12, 12:58 PM
P: 11
Hi

This means that the integral vanish here is pure randomness?

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

Greetings

arkajad
#4
Feb25-12, 01:23 PM
P: 1,411
Trace - Integration - Average - Tensor Calculus

Quote Quote by Joschua_S View Post
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
#5
Feb27-12, 12:46 PM
P: 11
The dipole-dipole Hamiltonian in ESR is given by

[itex]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)[/itex]

One can write it as

[itex]H_{DD} = \vec{S} \underline{\underline D} \vec{S}[/itex]

with the traceless symmetric Tensor D that fullfills [itex]\int SDS ~ \mathrm{d} \Omega = 0[/itex]

Do you know something about the math behind this?

Greetings
Joschua_S
#6
Mar1-12, 12:42 PM
P: 11
arkajad?


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