# Gravitational potential inside infinite grid of masses

Assume an infinite two-dimensional grid with nodes at $i,j\in \mathbb{Z}$, and each node $(i,j)$ hosting a mass $M_{ij}$. Can we find the total gravitational force $F(x,y)$ in closed form a small mass $m_{xy}$ at point $(x,y)$ feels on the plane, using Newton's formula for the force between two masses: $F=G\frac{m_1\cdot m_2}{r^2}$?

Preliminary notes:

Temporarily reducing the problem to the unit square and assuming unit masses $M_{ij}$, it is fairly obvious that the lines shown below (red) serve as axes of symmetry, so for each grid square, $F(x,y)=0$, iff $(x,y)$ falls on any of these lines.

Are there more lines that exhibit symmetry, different from the 8 red ones for each square? • If $(x,y)$ falls on one of the red lines, how can you say $F(x,y)=0$? Can’t you only say that the direction of $F(x,y)$ must be parallel to the line? It certainly seems possible that a small mass at $(0.0001,0.0001)$ would be drawn towards the origin. (Also, in the statement of the problem, do you need to require that $M_{ij}$ is independent of $i$ and $j$? Otherwise, there goes the symmetry. – Steve Kass Sep 7 '15 at 20:10
• Assuming unit masses $m$ and $M_{ij}$ and infinite grid, the forces cancel at $(1/2 + q, 1/2 + r)$ for $(q,r) \in \mathbb{Z}^2$. – mvw Sep 7 '15 at 21:11
• @SteveKass: Assume all $M_{ij}$ are equal or are unit masses to simplify the problem. I am wondering if this case admits a closed form. I have edited the question on this. Thanks for the observation. – Yiannis Galidakis Sep 8 '15 at 4:03
• This question is ill posed both mathematically and physically. Since the series do not converge at all and such a physical system cannot exists physically. See also physics.stackexchange.com/questions/11054/… – mastrok Sep 8 '15 at 4:31
• @mastock The cited convergence statement was not about a 2D grid in 2D space. Of course the masses would attract without fixing them. There are also problems about infinite resistor networks (I believe a cool one was on this site) even if we might not be able to build such objects for lack of space, materials, time etc usually they serve as a limit case. – mvw Sep 8 '15 at 10:19

## 1 Answer

The force $F_{ij}$ from mass $M_{ij}$ at position $r_{ij} = (i, j)^t$ applied to mass $m$ at position $r = (x,y)^t$ should be \begin{align} F_{ij} &= G \, m \, M_{ij} \frac{r_{ij}-r}{\lVert r_{ij}-r \rVert^3} \\ &= G \, m \, M_{ij} \frac{(i-x,j-y)}{\left(\sqrt{(i-x)^2+(j-y)^2}\right)^3} \end{align} The total force then is \begin{align} F &= \sum_{i,j\in\mathbb{Z}} F_{ij} \\ &= G \, m \, \sum_{i,j\in\mathbb{Z}} M_{ij} \frac{r_{ij}-r}{\lVert r_{ij}-r \rVert^3} \\ &= G \, m \, \sum_{i,j\in\mathbb{Z}} M_{ij} \frac{(i-x,j-y)}{\left(\sqrt{(i-x)^2+(j-y)^2}\right)^3} \end{align}

That sum looks ugly, so I asked my girl Ruby what she thinks about the resulting force for the simplified case $G = M_{ij} = m = 1$. I told her my view of the problem and she replied with this:

n = 98: r = Vector[0.5, 0.5], F = Vector[-0.014357405607155703, -0.01435740560715777]
n = 99: r = Vector[0.5, 0.5], F = Vector[-0.014213111904535183, -0.014213111904536712]
n = 100: r = Vector[0.5, 0.5], F = Vector[-0.014071689664539606, -0.01407168966453928]

n = 98: r = Vector[0.25, 0.25], F = Vector[-4.6817100348218865, -4.6817100348219425]
n = 99: r = Vector[0.25, 0.25], F = Vector[-4.681637888660634, -4.681637888660691]
n = 100: r = Vector[0.25, 0.25], F = Vector[-4.681567178203503, -4.681567178203558]

n = 98: r = Vector[0.25, 0.0], F = Vector[-14.708761955833403, -1.0966027348333074e-16]
n = 99: r = Vector[0.25, 0.0], F = Vector[-14.7086898091545, 1.0235546134620965e-16]
n = 100: r = Vector[0.25, 0.0], F = Vector[-14.708619098200186, 9.540979117872439e-17]

n = 98: r = Vector[0.0, 0.25], F = Vector[2.872627537318234e-15, -14.708761955833696]
n = 99: r = Vector[0.0, 0.25], F = Vector[1.2367358656270588e-16, -14.708689809154844]
n = 100: r = Vector[0.0, 0.25], F = Vector[-4.756347496848862e-15, -14.708619098200511]


So the resulting force at $(0.5,0.5)$ seems to vanish etc. This would make a nice 2D plot but I was too lazy to elaborate further.

• Your numerical evidence suggests convergence(?) Could it be related(in terms of scalars only) to the convergence of the series $\sum\limits_{n=0}^\infty \frac{M}{{r_n}^2}$? – Yiannis Galidakis Sep 8 '15 at 11:37
• That post is only precise until the forumula for the total force. I had no good idea how to analyse that vector sum above, so I wrote the script to get some impression what happens and published it for other folks to be able to play with. Perhaps it tells more about the symmetries than convergence. There are slow converging sums that might bite one. Then I sum over square areas where maybe radial ones might be more wise. And that link posted in the comments raises the question if the sum is absolute convergent, it probably is not and that would mean that reordering affects the limit. – mvw Sep 8 '15 at 13:33
• Note that absolutely convergent and (just) convergent are two different things. It might not be absolutely convergent, but it might be convergent, since (roughly) for each term of positively oriented force, there is a negative one that cancels it, close to the axes of symmetry. Further, the function is continuous on $x$ and $y$, so it is afortiori bounded on small closed intervals around the symmetry points. – Yiannis Galidakis Sep 8 '15 at 14:24