0
$\begingroup$

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: $$Q_{ij}=\int_V\rho(\vec{x})(3x_ix_j - r^2\delta_{ij})d^3x$$

$\hskip2in$ enter image description here
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 $$2a^2e[\int_V \delta(x)\delta(y)\delta(z-a)dV-2\int_V \delta(x)\delta(y)\delta(z)dV+\int_V \delta(x)\delta(y)\delta(z+a)dV]$$ $$=2a^2e[-a-2+a]=-4a^2e$$
Which contradicts the solution.
Any help is appreciated.

$\endgroup$
  • 1
    $\begingroup$ $x_1$ is $x$, $x_2$ is $y$, and $x_3$ is $z$. $\endgroup$ – JEB Nov 2 '19 at 22:57
  • $\begingroup$ @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? $\endgroup$ – Alessio Popovic Nov 2 '19 at 23:20
1
$\begingroup$

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

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.