# How to calculate the Quadrupole moment via integration?

I have following problem:

calculate the quadrupole moment of following arrangement, where $$e$$ is the charge and $$a$$ is the distance between the charges.
Hint: The quadrupole moment is defined as: $$Q_{ij}=\int_V\rho(\vec{x})(3x_ix_j - r^2\delta_{ij})d^3x$$

$$\hskip2in$$ I know that the charge density is:

$$\rho(\vec{r})=e\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)]$$ and the solution to this problem is:
$$Q_{11}=Q_{22}=-2a^2e,\quad Q_{33}=4a^2e$$

I am having difficulties solving and understanding this integral
my approach is: $$Q_{11}=\int e\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)(3x_1x_1-r^2\delta_{11})dxdydz$$ my guess is that in this case $$x_1=r=a$$ since it represents the distance from the origin to the first charge, which delivers: $$2a^2e\int\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)]dxdydz$$
what bothers me is that I am not sure how to evaluate this integral, since there is a delta function present my best guess is that we integrate from $$-\infty$$ to $$+\infty$$, and apply the sifting property of the delta function $$2a^2e[\int_V \delta(x)\delta(y)\delta(z-a)dV-2\int_V \delta(x)\delta(y)\delta(z)dV+\int_V \delta(x)\delta(y)\delta(z+a)dV]$$ $$=2a^2e[-a-2+a]=-4a^2e$$
Any help is appreciated.

• $x_1$ is $x$, $x_2$ is $y$, and $x_3$ is $z$. – JEB Nov 2 '19 at 22:57
• @JEB that makes sense, Am I right to assume that $r^2 = x^2 + y^2 + z^2$? And if that would be the case what would the value for $z$ be? – Alessio Popovic Nov 2 '19 at 23:20

$$Q_{ij}=\int_Ve\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)](3x_ix_j - (x^2+y^2+z^2)\delta_{ij})dxdydz$$

is zero for $$i \ne j$$.

Set $$i=x$$:

$$Q_{xx}=\int_Ve\delta(x)\delta(y)[\delta(z-a)-2\delta(z)+\delta(z+a)](2x^2 - (y^2+z^2))dxdydz$$

integrate over $$y$$:

$$Q_{xx}=\int_Ve\delta(x)[\delta(z-a)-2\delta(z)+\delta(z+a)](2x^2 - (0^2+z^2))dxdz$$

integrate over $$x$$:

$$Q_{xx}=\int_Ve[\delta(z-a)-2\delta(z)+\delta(z+a)](0^2 - (0^2+z^2))dz$$

finally, $$z$$:

$$Q_{xx}=-e[a^2-2\cdot0^2+(-a)^2]=-2ea^2$$

And likewise $$yy$$

Of course, since you are dealing with points, you can skip the delta function and integrals and just add up moments of points, so for $$Q_{zz}$$ you want to sum over points with the weighting function:

$$w(x, y, z) = q(3z^2-r^2) = q(2z^2 - x^2 -y^2) \propto qr^2Y^0_2(\theta, \phi)$$

since $$x=y=0$$ we are just summing:

$$\sum_i{2ez_i^2}= 2e[a^2+(-2\cdot 0)^2+(-a)^2] = 4ea^2$$

Now the reason I included the $$\propto = Y_2^0$$ is because rank-2 tensors in Cartesian coordinates are obtuse. There are 9 of them, and one transforms like scalar ($$\delta_{ij}$$), then 3 look like a vector (the antisymmetric parts), but there are 5 that are pure natural-form (i.e. traceless) rank-2:

$$Y_2^0$$: that is how oblate or prolate your distributions is. (Your distribution is pure $$Y_2^0$$).

$$Y_2^{\pm 2}$$: represents a lack of cylindrical symmetry along the $$z$$-axis, specifically with 180 degree rotation symmetry (0 for this problem).

Finally:

$$Y_2^{\pm 1}$$ means you have not diagonalized your axes, and these can be removed by the correct choice of axes.