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]. Greetings
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
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 [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