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I am currently working on my bachelor's thesis on the anapole / toroidal moment and it seems that I am stuck with a tensor decomposition problem.

I have actually never had a course about tensors, so I am a complete newbie.

I need to expand a localized current density, which is done by expressing the current via delta distribution and expanding the latter:

$$\vec{j}(\vec{r},t) = \int\vec{j}(\vec{\xi},t) \delta(\vec{\xi}-\vec{r}) d^3\xi$$ $$\delta(\vec{\xi}-\vec{r}) = \sum_{l=0}^{\infty} \frac{(-1)^l}{l!} \xi_i ...\xi_k \nabla_i ... \nabla_k \delta(\vec{r}) $$

So I get some result containing the following tensor:

$$B_{ij...k}^{(l)}(t) := \frac{(-1)^{l-1}}{(l-1)!} \int j_i \xi_j ... \xi_k d^3\xi$$

So far, I have understood the math. But now comes the tricky part. In the paper, it says that "we can decompose the tensors $B_{ij...k}^{(l)}$ into irreducible tensors, separating the various multipole moments and radii." and further "...of third rank, $B_{ijk}^{(3)}$ can obviously reduced according to the scheme $1 \times (2+0) = (3+1)+2+1$. It can be seen that the representation of weight $l=1$ is extracted twice from $B_{ijk}^{(3)}$." And then follows what seems like the decomposition and I am hopelessly lost.

$$j_i\xi_j\xi_k = \frac{1}{3} \left[ j_i\xi_j\xi_k + j_k\xi_i\xi_j + j_j\xi_k\xi_i - \frac{1}{5} \left( \delta_{ij}\theta_k + \delta_{ik}\theta_j + \delta_{jk}\theta_i \right) \right] - \frac{1}{3} \left( \epsilon_{ijl} \mu_{kl} + \epsilon_{ikl}\mu_{jl}\right)$$ $$+ \frac{1}{6} \left( \delta_{ij}\lambda_k + \delta_{ik}\lambda_j - 2 \delta_{jk}\lambda_i \right) + \frac{1}{5} \left( \delta_{ij}\theta_k + \delta_{ik}\theta_j + \delta_{jk}\theta_i \right)$$

with

$$\mu_{ik} = \mu_i\xi_k + \mu_k\xi_i \ , \ \mu_i=\frac{1}{2} \epsilon_{ijk}\xi_j j_k$$ $$\theta_i=2\xi_i \vec{\xi}\cdot \vec{j} + \xi^2 j_i$$ $$\lambda_i=\xi_i\vec{\xi}\cdot \vec{j} - \xi^2 j_i$$

This decomposition obviously contains many quantities that later on appear also in the multipole expansion, e.g. the magnetic quadrupole moment $\mu_{ik}$. So on the physics side of things, this makes sense to me.

But not on the mathematical side. On this board I found some questions regarding tensor decomposition and in the answers I learned something about symmetric and antisymmetric tensors and that every tensor can be decomposed in several irreducible ones, which better represent physical properties of the system and symmetries.

But I still, some questions are still there... 1.) What do the numbers $\frac{1}{3}$, $\frac{1}{5}$, etc. mean? Is this some kind of normalization? 2.) How exactly does one decompose the tensor? How can I reconstruct what exactly has been done, which steps one has to follow to decompose it like this?

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  • $\begingroup$ $\uparrow$ Which paper? $\endgroup$ – Qmechanic May 12 '14 at 23:20
  • $\begingroup$ "Toroid Moments in Electrodynamics and Solid-State Physics" by V.M. Dubovik and V.V. Tugushev. Should I add this every time? $\endgroup$ – user46208 May 13 '14 at 11:01
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This appears to be related to the decomposition of a totally symmetric tensor into traceless parts, which is a fairly involved process. The general equation is $$\mathcal{C} Q_{a_1 a_2\cdots a_s} = \sum_{k=0}^{[\frac{s}{2}]} (-1)^s \frac{\binom{s}{k} \binom{s}{2k}}{ \binom{2s}{2k}} \delta_{(a_1 a_2} \cdots \delta_{a_{2k-1} a_{2k}} Q_{a_{2k+1}\cdots a_s)}{}^{c_1} {}_{c_1} {}^{c_2}{}_{c_2} {}^{\cdots c_k}{}_{\cdots c_k},$$ where $[\cdot]$ denotes the integer part, Einstein summation is implied and $Q_{(a_1 a_2 \cdots a_s)} \equiv \frac{1}{s!} \sum_{\sigma\in S_s} Q_{a_{\sigma(1)} a_{\sigma(2)} a_{\sigma(3)} \cdots a_{\sigma(s)}}$. For the quadrupole moment it is $\mathcal{C}Q_{ab} = Q_{ab} - \frac{1}{3} Q^c{}_c \delta_{ab}$, for the octupole $\mathcal{C}Q_{abc} = Q_{abc} - \frac{1}{5} (Q^d{}_{dc}\delta_{ab} + Q^d{}_{da} \delta_{bc}+ Q^d{}_{db}\delta_{ac})$; these yield the factors in your question.

An indication (perhaps proof, although I'm not certain about this at the moment) that the traceless part of a totally symmetric tensor is an irreducible representation is easy to see if one uses the hook formula in dimension 3. A totally symmetric tensor of rank $s$ has $\frac{1}{2}(s+1)(s+2)$ degrees of freedom and the traceless one has the latter minus the number of ways to obtain the traces, $\binom{s}{2}$, which yields $2s+1$. This is the dimension of the irreducible representation of the algebra of SO(3) with spin $s$. A full proof of this statement is in Maggiore Gravitational waves - theory and experiments.

Reference: F.A.E. Pirani Lectures on General Relativity 1965.

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  • $\begingroup$ To add to this answer: see Manheim's appendix in "Brane Localized Gravity" for a full derivation of various decompositions of tensors. $\endgroup$ – JamalS May 10 '14 at 21:16
  • $\begingroup$ Thanks for your answer! This has definitely helped, I have taken a look in the decomposition part of Maggiore's lecture. A general question I still have, though, is how to calculate the trace itself, e.g. in $$Q_{\alpha \beta \gamma} = I_{\alpha\beta\gamma} - a (\delta_{(\alpha\beta}I_{\gamma)})_{\mu\mu}$$ Why does the author then calculate $$Q_{\beta\beta\gamma} = I_{\beta\beta\gamma} - \frac{1}{3} a (\delta_{\beta\beta} I_{\gamma\mu\mu} + \delta_{\beta\gamma} I_{\beta\mu\mu} + \delta_{\gamma\beta} I_{\beta\mu\mu}) = I_{\beta\beta\gamma} - \frac{5}{3} a I_{\beta\beta\gamma}$$ ? $\endgroup$ – user46208 May 16 '14 at 13:35
  • $\begingroup$ To specify my question more, things that are unclear to me are: + what does the notation $(\alpha \beta \quad \gamma)\mu\mu$ mean? + why is the trace calculated via $Q_{\beta\beta\gamma}$ and why is the "weight" $\frac{1}{3}$? $\endgroup$ – user46208 May 16 '14 at 13:53
  • $\begingroup$ @user46208 I didn't answer your question directly, why did you accept my answer? I'll have a look at the article over the weekend. $\endgroup$ – auxsvr May 16 '14 at 15:21
  • $\begingroup$ So sorry, I am absolutely new to this platform! I promise to think twice next time! $\endgroup$ – user46208 May 16 '14 at 20:47

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