Hi(adsbygoogle = window.adsbygoogle || []).push({});

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].

Greetings

**Physics Forums - The Fusion of Science and Community**

# Trace - Integration - Average - Tensor Calculus

Have something to add?

- Similar discussions for: Trace - Integration - Average - Tensor Calculus

Loading...

**Physics Forums - The Fusion of Science and Community**