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I've been doing some exercises in my workbook and every time I came across a problem involving a quadrupole moment it was always assumed that it was traceless without providing any proof (at least none that I saw) in the workbook. I mean it didn't really matter in solving the problem but it still annoys me that I don't know why it's traceless.

I've tried googling the proof to that but I couldn't find one. Each source always assumed that it's tracelss.

Now, am I just dumb and don't see it or what? Why is the quadrupole moment traceless?

Edit: The quadrupole moments I'm working with are defined as $$Q_{ij}=\sum_lq_l(3r_{il}r_{jl}-\|\vec{r_l}\|^2\delta_{ij}).$$ Now, I can't seem to figure out that it's traceless just from looking at that formula.

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    $\begingroup$ Just write down the formula for $Q$. Its tracelessness should be self evident. $\endgroup$ Feb 24, 2016 at 14:45

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From your definition, $$Q_{ij}=\sum_lq_l(3r_{il}r_{jl}-\|\vec{r_l}\|^2\delta_{ij}),$$ the trace is $$ \mathrm{Tr}(Q) =\sum_iQ_{ii} =\sum_{i,l}q_l(3r_{il}^2-\|\vec{r_l}\|^2\delta_{ii}). $$ Here you notice (1) that $\sum_i\delta_{ii}=3$, and (2) that $\sum_ir_{il}^2=\|\vec{r_l}\|^2$ for each $l$. The two individual terms for each fixed $l$ then cancel.

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  • $\begingroup$ Is $\sum_i\delta_{ii}=3$ just because it's a 3x3-matrix? $\endgroup$
    – Michael M
    Feb 24, 2016 at 16:31
  • $\begingroup$ Precisely - you're summing over three spatial dimensions. If you're working in a different dimensionality then all of multipole theory will look different. $\endgroup$ Feb 24, 2016 at 16:34
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the definition (from the Wikipedia article on Quadrupoles):

For a discrete system of point charges (or masses in the case of a gravitational quadrupole), each with charge $q_{l}$ (or mass $m_{l}$) and position $\vec{r_l}=(r_{xl},r_{yl},r_{zl})$ relative to the coordinate system origin, the components of the Q matrix are defined by:

$$Q_{ij}=\sum_l q_l(3r_{il} r_{jl}-\|\vec{r_l}\|^2\delta_{ij})$$

The indices $i,j$ run over the Cartesian coordinates $x,y,z$ and $\delta_{ij}$ is the Kronecker delta.

then calculate the trace = $Q(xx) +q(yy) +Q (zz)$ and it should come out to be zero.

If you feel you know this already -try to visualize the basic reason for a vanishing Trace ; it may be related to symmetry of the defined observable for details you may see

https://en.wikipedia.org/wiki/Trace_(linear_algebra)#traceless

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  • $\begingroup$ I'm sorry, but I still don't see it. Wouldn't that be $Q_{11}=q_1\cdot 2r_1^2$ and so on? How is that zero? Or am I messing something up? $\endgroup$
    – Michael M
    Feb 24, 2016 at 16:09
  • $\begingroup$ actually in your expression you have missed an index l which designates the charges and i,j represents (x,y,z) the position vector r(l,i).... $\endgroup$
    – drvrm
    Feb 24, 2016 at 16:17

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