Trace - Integration - Average - Tensor Calculus

  1. Feb 25, 2012 #1
    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
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2
    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.
     
  4. Feb 25, 2012 #3
    Hi

    This means that the integral vanish here is pure randomness?

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

    Greetings
     
  5. Feb 25, 2012 #4
    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.
     
  6. Feb 27, 2012 #5
    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
     
  7. Mar 1, 2012 #6
    arkajad? :blushing:
     
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