How to calculate the derivative of a Newtonian potential inside a box with uniform source distribution?

Question

I'm working on a potential flow problem. I have a box, centered at the origin (i.e. $V=[-x_b,x_b]\times[-y_b,y_b]\times[-z_b,z_b]$) that has inside of it a uniform distribution of source strength. We can assume this distribution has unit strength. The potential induced by the distribution of source strength in the box at a point $(x,y,z)$ will then be

$$\phi(x,y,z) = \frac{-1}{4\pi}\iiint_V \frac{1}{\sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2}} dV$$

I need to know the second derivatives of potential (i.e. $\frac{\partial^2\phi}{\partial x^2}$, etc.) at the origin. I can easily show that for any point $(x,y,z)$, we have

$$\frac{\partial^2\phi}{\partial x^2}(x,y,z) = \frac{1}{4\pi}\iiint_V \frac{1}{\left[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2\right]^{3/2}} - 3\frac{(x-x_0)^2}{\left[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2\right]^{5/2}} dV$$

Since I want to know $\phi$ at $(x,y,z)=(0,0,0)$, and the box is centered at the origin, I can simplify this to

$$\frac{\partial^2\phi}{\partial x^2}(0,0,0) = \frac{2}{\pi}\int_0^{z_b} \int_0^{y_b} \int_0^{x_b} \frac{1}{\left[x_0^2 + y_0^2 + z_0^2\right]^{3/2}} - 3\frac{x_0^2}{\left[x_0^2 + y_0^2 + z_0^2\right]^{5/2}} dx_0 dy_0 dz_0$$

Integrating in $x$, I obtain

$$\frac{\partial^2\phi}{\partial x^2}(0,0,0) = \frac{2}{\pi}\int_0^{z_b} \int_0^{y_b} \frac{x_b}{(y_0^2+z_0^2)\sqrt{x_b^2 + y_0^2 + z_0^2}} - \frac{x_b^3}{(y_0^2+z_0^2)\left[x_b^2 + y_0^2 + z_0^2\right]^{3/2}} dy_0 dz_0$$

But at this point, I am at a loss as to how to proceed. Can these integrals be evaluated? Is there a shortcut to determining $\frac{\partial^2\phi}{\partial x^2}$ based on potential theory?

Answer 1

So I made some grave mistake here: Still I want to show the post and add the real solution at the end:

With Gauss Law: The integral over the divergence on a volume is equal to the integral of that field on the boundary:

(Also see: Divergence Theorem - Wikipedia )

$ \int_V \nabla \mathbf{F} dV = \oint_S \mathbf{F}\cdot\mathbf{n} dS $

So if you have a symmetric problem then this helps you find a solution quickly. But your box is not symmetric? If you are interested in the field or the potential in the inside: You can still imagine a sphere on the inside since. The above integral is only dependent on the inside of the sphere the block outside is irrelevant.

The potential that is created by a point source at the origin at a distance r is:

$ \Phi(r) = \frac{-1}{4\pi |r|} $

Since your integral sums up over exactly such point sources you have a divergence of 1 everywhere in your box.

Thus within a circle of radius r that completely fits into the box the integral over the volume becomes just a multiplication by the volume of the sphere and the integral over the surface becomes just a multiplication by the surface of the sphere:

$ \frac{4\pi r^3}{3} = F \cdot 4\pi r^2 $

This was the mistake: the field from the outside does NOT vanish as it lacks the necessary symetry:

$ F = \frac{1}{3}\cdot r $

If we only go along the x axis

$ \Phi = \int F\,dx $

$ \frac{ \partial \phi}{\partial x } = F $

$ \frac{ \partial^2 \phi}{\partial x^2 } = \frac{1}{3} $

So here is the correct solution:

The field (derviative of the potential) in the middle is zero. If we move a small $dx$ in the block then the what the point sees is a block that is a $dx$ shorter at the front and $dx$ longer at the back. We could model this by keeping the original position but adding a small $dx$ thick plate at the back of the block and remove a small $dx$ plate from the front. Removing could be done by adding a source with negative sign: A plate that is a sink.

Thus a small movement of $dx$ results in a field that is the same as the field of 2 small plates. One with negative source at distance $+x_b$ and one with positive source at distance $-x_b$

Calculating the field $\mathbf{F}$ from a plate is still hard work but a bit simpler then integrating over the complete block. I found 2 sources that have done this already:

• Youtube: Electric field of a finite rectangular plate
• physics stackexchange: Electric field due to a uniformly charged FINITE rectangular plate

$F_x= \frac{atan \left(\frac{y_b z_b}{r x_b} \right)}{\pi}$ where $r=\sqrt{x_b^2 + y_b^2 + z_b^2} $

For small changes the thickness of the plates is proportional to $dx$ and thus is the field. So this is already the 2nd derivative. We only need to multiply by 2 because we have 2 plates.

Thus:

$ \frac{\partial^2 \Phi}{\partial x^2} = \frac{2}{\pi} \cdot atan \left(\frac{y_b z_b}{r x_b}\right) $

If we set $x_b=y_b=z_b$ then we get

$ \frac{2}{\pi} \cdot atan \frac{1}{\sqrt{3}} = \frac{2}{\pi} \cdot \frac{\pi}{6} = \frac{1}{3} $ which matches the result of the initial naive approach.

This result would also appear at: physics stackexchange: Paradox with Gauss' law when space is uniformly charged everywhere

Answer 2

Denote $(a,b,c):=(x_b,y_b,z_b)$, $(x,y,z):=(x_0,y_0,z_0)$, and $I:=\left.\dfrac{\partial^2\phi}{\partial x^2}\right\rvert_{(0,0,0)}$.

Integrating by parts makes quick work of the $x$-integral,

$$-3 \int_0^a \frac{x^2}{\left(x^2+y^2+z^2\right)^{5/2}} \, dx = \frac a{\left(a^2+y^2+z^2\right)^{3/2}} - \int_0^a \frac{dx}{\left(x^2+y^2+z^2\right)^{3/2}} \\ \implies I = \int_0^c \int_0^b \frac{a}{\left(a^2+y^2+z^2\right)^{3/2}} \, dy \, dz$$

Trigonometric or hyperbolic substitutions are standard in dealing with the $y$-integral. If $y=\sqrt{a^2+z^2}\sinh u$, then

$$\int_0^b \frac{dy}{\left(a^2+z^2+y^2\right)^{3/2}} = \frac1{a^2+z^2} \int_0^{\sinh^{-1}\tfrac b{\sqrt{a^2+z^2}}} \frac{du}{\cosh^2u}$$

Can you continue from here?