Trace - Integration - Average - Tensor Calculus

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Discussion Overview

The discussion revolves around the relationship between the trace of an anisotropic tensor and the vanishing of a specific integral involving that tensor and a basis vector. Participants explore the implications of these mathematical properties in the context of tensor calculus and their applications in spectroscopy.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the trace of an anisotropic tensor \( D \) is zero and proposes that this might relate to the vanishing integral of the form \( \int SDS \, \mathrm{d} \Omega = 0 \) over symmetric regions.
  • Another participant argues that there is no direct link between a zero trace and a vanishing integral, stating that a non-zero scalar function typically has a non-zero trace, while the integral can be either zero or non-zero.
  • A participant questions whether the vanishing integral indicates randomness and expresses a desire for theorems regarding anisotropic tensors, traces, and integrals.
  • Another participant emphasizes that the integral's vanishing may require additional assumptions about the spatial dependence of \( D \) and \( S \).
  • A further contribution presents the dipole-dipole Hamiltonian in electron spin resonance (ESR) and reiterates the relationship \( H_{DD} = \vec{S} \underline{\underline D} \vec{S} \), noting that it fulfills the condition \( \int SDS \, \mathrm{d} \Omega = 0 \) and asks for insights into the underlying mathematics.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the zero trace of the tensor and the vanishing integral, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Some participants note the potential need for additional assumptions regarding the dependence of the tensor and vector on spatial variables to understand the conditions under which the integral vanishes.

Joschua_S
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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
 
Last edited:
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Joschua_S said:
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.
 
Hi

This means that the integral vanish here is pure randomness?

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

Greetings
 
Joschua_S said:
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.
 
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
 
arkajad? :blushing:
 

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