Dipole moment of given charge distribution

[PhysicsRock]

Homework Statement

Given the charge distribution $\rho(x,y,z) = \frac{Q}{2\pi R^2} \delta(\sqrt{x^2 + y^2 + z^2}-R) [\Theta(z) - \Theta(-z)]$ calculate the dipole moment using cartesian coordinates.

Relevant Equations

$\vec{p} = \int d^3x \vec{x} \rho(\vec{x})$.

I have come up with a solution, however, I'm not sure whether I'm correct. A fellow student of mine has a different result. I'm gonna show my solution, and hopefully one of you can confirm my result or tell me what I did wrong.

$$
\begin{align}
p_z &= \int d^3x z \rho(\vec{x}) \notag \\
&= \frac{Q}{2\pi R^2} \int d^3x z \delta(\sqrt{x^2 + y^2 + z^2} - R) \left[ \Theta(z) - \Theta(-z) \right] \notag \\
&= \frac{Q}{2\pi R^2} \left[ \int_{z>0} d^3x z \delta(\sqrt{x^2 + y^2 + z^2} - R) - \int_{z<0} d^3x z \delta(\sqrt{x^2 + y^2 + z^2} - R) \right] \notag \\
&= \frac{Q}{\pi R^2} \int_{z>0} d^3x z \delta(\sqrt{x^2 + y^2 + z^2} - R) \notag \\
&= \frac{Q}{\pi R^2} \int_{x,y} \sqrt{R^2 - x^2 - y^2} dx dy \notag \\
&= \frac{Q}{\pi R^2} \int_{y} \int_{-\sqrt{R^2 - y^2}}^{+\sqrt{R^2 - y^2}} \sqrt{R^2 - x^2 - y^2} dx dy \notag \\
\text{Let } u^2 &= R^2 - y^2 \notag \\
\Rightarrow p_z &= \frac{Q}{\pi R^2} \int_y \int_{-\sqrt{R^2 - y^2}}^{+ \sqrt{R^2 - y^2}} \sqrt{u^2 - x^2} dx dy \notag \\
&= \frac{Q}{\pi R^2} \int_y \int_{-\sqrt{R^2 - y^2}}^{+\sqrt{R^2 - y^2}} u \sqrt{1 - \left( \frac{x}{u} \right)^2} dx dy \notag \\
\text{Let } \sin(\alpha) &= \frac{x}{u} \Leftrightarrow dx = u \cos(\alpha) d\alpha \notag \\
&= \frac{Q}{\pi R^2} \int_y u^2 \int_{\sin^{-1}(-\sqrt{R^2 - y^2}/u)}^{+ \sin^{-1}(\sqrt{R^2 - y^2}/u)} \sqrt{1 - \sin^2(\alpha)} \cos(\alpha) d\alpha \notag \\
&= \frac{Q}{\pi R^2} \int_y u^2 \int_{-\pi/2}^{+\pi/2} \cos^2(\alpha) d\alpha \notag \\
&= \frac{Q}{\pi R^2} \int_y \frac{\pi u^2}{2} dy \notag \\
&= \frac{Q}{2R^2} \int_y (R^2 - y^2) dy \notag \\
&= \frac{Q}{2R^2} \int_{-R}^{+R} (R^2 - y^2) dy \notag \\
&= \frac{Q}{2R^2} \left[ R^2 y - \frac{y^3}{3} \right]_{-R}^{+R} \notag \\
&= \frac{Q}{2R^2} \left( R^2 \cdot R - \frac{R^3}{3} - R^2 \cdot (-R) - \frac{R^3}{3} \right) \notag \\
&= \frac{Q}{2R^2} \cdot \frac{4R^3}{3} \notag \\
&= \frac{2QR}{3}. \notag
\end{align}
$$

Between the third and fourth line I used a substitution to make both integrals run within the same boundaries, thus allowing me to only calculate one integral. I'm also only stating the $p_z$-component, because we think the dipole only has non-zero values in the $z$-component, because the shift of charges only occurs along that axis, as seen from the charge distribution stated above. Please correct me if we're wrong here.

 

[Charles Link]

One comment or two: It is very clumsy to try to work with a one dimensional delta function, integrating over 3 coordinates. Suggestion is to rewrite the argument of the delta function as $r-R$, and integrate over $dr$, (from zero to infinity), with a $2 \pi r^2 \sin(\theta) \, d \theta$ ,( the $\phi$ gives the $2 \pi$), to complete the volume element, and the $z$ for the dipole is just $r \cos(\theta)$.

 

[PhysicsRock]

One comment or two: It is very clumsy to try to work with a one dimensional delta function, integrating over 3 coordinates. Suggestion is to rewrite the argument of the delta function as $r-R$, and integrate over $dr$, (from zero to infinity), with a $2 \pi r^2 \sin(\theta) \, d \theta$ ,( the $\phi$ gives the $2 \pi$), to complete the volume element, and the $z$ for the dipole is just $r \cos(\theta)$.

I would've certainly done that, however, the assignment clearly requested a solution using cartesian coordinates.

 

[PhDeezNutz]

Correct me if I'm wrong but in this problem there is only uniform charge (equal and opposite) on the lower and top hemispherical surfaces?

So you only need to do a double integral instead of a triple integral.

Given that you are required to evaluate in Cartesian I think the first step is to parameterize the spherical surface in Cartesian

Which I get as

$\vec{S} \left(x,y\right) = \left(x,y, \sqrt{R^2 - x^2 - y^2} \right)$ for the top

$\vec{S} \left(x,y\right) = \left(x,y, -\sqrt{R^2 - x^2 - y^2} \right)$ for the bottom

and $dA = \left|\frac{\partial \vec{S}}{\partial x} \times \frac{\partial \vec{S}}{\partial y} \right| dx dy$

 

[PhysicsRock]

Correct me if I'm wrong but in this problem there is only uniform charge (equal and opposite) on the lower and top hemispherical surfaces?

So you only need to do a double integral instead of a triple integral.

Given that you are required to evaluate in Cartesian I think the first step is to parameterize the spherical surface in Cartesian

Which I get as

$\vec{S} \left(x,y\right) = \left(x,y, \sqrt{R^2 - x^2 - y^2} \right)$ for the top

$\vec{S} \left(x,y\right) = \left(x,y, -\sqrt{R^2 - x^2 - y^2} \right)$ for the bottom

and $dA = \left|\frac{\partial \vec{S}}{\partial x} \times \frac{\partial \vec{S}}{\partial y} \right| dx dy$

I would interprete the charge density in a similar way. I'm gonna try your suggested approach and see where it leads me. Thank you.

 

[PhysicsRock]

Correct me if I'm wrong but in this problem there is only uniform charge (equal and opposite) on the lower and top hemispherical surfaces?

So you only need to do a double integral instead of a triple integral.

Given that you are required to evaluate in Cartesian I think the first step is to parameterize the spherical surface in Cartesian

Which I get as

$\vec{S} \left(x,y\right) = \left(x,y, \sqrt{R^2 - x^2 - y^2} \right)$ for the top

$\vec{S} \left(x,y\right) = \left(x,y, -\sqrt{R^2 - x^2 - y^2} \right)$ for the bottom

and $dA = \left|\frac{\partial \vec{S}}{\partial x} \times \frac{\partial \vec{S}}{\partial y} \right| dx dy$

Assuming I have done everything right, leads to the following:
First, calculate $dA$:

$$
\begin{align*}
\frac{\partial \vec{S_1}}{\partial x} &= \begin{pmatrix}
1 \\ 0 \\ -\frac{x}{\sqrt{R^2 - x^2 - y^2}} \\
\end{pmatrix}, \frac{\partial \vec{S_1}}{\partial y} = \begin{pmatrix}
0 \\ 1 \\ -\frac{y}{\sqrt{R^2 - x^2 - y^2}} \\
\end{pmatrix} \\
\Rightarrow dA &= \sqrt{\frac{x^2 + y^2}{R^2 - x^2 - y^2}} dx dy.
\end{align*} \\
$$

Since we still have the $z$ in the integral, the denominator should cancel out, leaving

$$
\int_{x,y} \sqrt{x^2 + y^2} dx dy
$$

on the upper hemisphere. However, this looks more like hyperbolic functions to me than like trig functions. That seems rather unlikely to me, as we are doing integration on the $S^2$.

 

[PhDeezNutz]

I'm getting $\left|\vec{S}_x \times \vec{S}_y \right| = \sqrt{\frac{x^2 + y^2}{R^2 - x^2 - y^2} + 1}$

Because for

$\vec{S}_x \times \vec{S}_y$ I'm getting

$\left(\frac{x}{\sqrt{R^2 - x^2 - y^2}},\frac{y}{\sqrt{R^2 - x^2 - y^2}} , 1 \right)$

 

[Charles Link]

To evaluate the last integral, can't you use $r^2=x^2+y^2$, and $dxdy=r \, dr \, d \phi$?

 

[PhysicsRock]

I'm getting $\left|\vec{S}_x \times \vec{S}_y \right| = \sqrt{\frac{x^2 + y^2}{R^2 - x^2 - y^2} + 1}$

Because for

$\vec{S}_x \times \vec{S}_y$ I'm getting

$\left(\frac{x}{\sqrt{R^2 - x^2 - y^2}},\frac{y}{\sqrt{R^2 - x^2 - y^2}} , 1 \right)$

Why is there a 1 in the $z$-component though? The last entry of the cross product should be 0 in my opinion, since it is calculated by $1 \cdot 0 - 0 \cdot 1$, unless, of course, I made a mistake determining the derivatives.

 

[PhDeezNutz]

I also think we can intuit that there will only be a z-component of the dipole moment.

 

[PhysicsRock]

I also think we can intuit that there will only be a z-component of the dipole moment.

That's good to hear.

 

[PhDeezNutz]

Why is there a 1 in the $z$-component though? The last entry of the cross product should be 0 in my opinion, since it is calculated by $1 \cdot 0 - 0 \cdot 1$, unless, of course, I made a mistake determining the derivatives.

I’m getting 1x1 - 0x0 when I cover up the third column to get the third component

 

[PhDeezNutz]

I tried the problem in spherical coordinates as well

I got QR

For both Cartesian and Spherical approach

 

[PhysicsRock]

I’m getting 1x1 - 0x0

But isn't the $z$-component of the cross product of two vectors, say $\vec{a}$ and $\vec{b}$, given my $a_x b_y - a_y b_x$? The derivatives should yield a 0 in the $x$-component for $\partial_y$, and the $y$-component for $\partial_x$, meaning the $z$-component must vanish.

 

[PhDeezNutz]

To evaluate the last integral, can't you use $r^2=x^2+y^2$, and $dxdy=r \, dr \, d \phi$?

Is it even the right integral to evaluate though? I’ll post parts of my full solution soon

In an hour or two.

 

[Orodruin]

But isn't the $z$-component of the cross product of two vectors, say $\vec{a}$ and $\vec{b}$, given my $a_x b_y - a_y b_x$? The derivatives should yield a 0 in the $x$-component for $\partial_y$, and the $y$-component for $\partial_x$, meaning the $z$-component must vanish.

Wrong. As per your expression above, you mix the x and y components in the third component of the cross product. It is $a_x b_y - a_y b_x$, not $a_x b_x - a_y b_y$.

By your argumentation, the area element of a surface at fixed z is equal to zero, not $dx\,dy$.

 

[PhysicsRock]

Wrong. As per your expression above, you mix the x and y components in the third component of the cross product. It is $a_x b_y - a_y b_x$, not $a_x b_x - a_y b_y$.

By your argumentation, the area element of a surface at fixed z is equal to zero, not $dx\,dy$.

Haven't I given the exact equation for the $z$-component as you just did?

 

[Orodruin]

Haven't I given the exact equation for the $z$-component as you just did?

Yes, but you are not applying it as you quoted it.

Yes, when taking $\partial_x$ the y-component vanishes and when taking $\partial_y$ the x-component vanishes. However, both those components appear in the second term of the cross product expression and the components appearing in the first are both one.

 

[PhDeezNutz]

What is the determinant of

$$ \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$$

?

Is this not the same as the third component you seek?

 

[PhysicsRock]

@Orodruin, @PhDeezNutz, you're both correct, pardon me. The current lack of sleep doesn't exactly improve my mental capabilities. Okay, I'll try again now including the extra +1 under the square root.