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Trace  Integration  Average  Tensor Calculus 
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#1
Feb2512, 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 ndimensional unitsphere or a ndimensional 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 dipoledipole 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 


#2
Feb2512, 12:43 PM

P: 1,411




#3
Feb2512, 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 


#4
Feb2512, 01:23 PM

P: 1,411

Trace  Integration  Average  Tensor Calculus



#5
Feb2712, 12:46 PM

P: 11

The dipoledipole 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 


#6
Mar112, 12:42 PM

P: 11

arkajad?



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