# Classical static equilibrium in atomic physics

Consider a collection of $n+1$ mass weighted points in $\mathbb{R}^3$. Suppose we have one mass located at the point (0,0,0) with mass $m\in\mathbb{N}$ and further suppose we have $n$ masses arranged arbitrarily each with mass -1.

Now we want to assign to each mass weighted coordinate a collection of vectors in the following way:

1. There will be a total of $n$ vectors emanating from each weighted point.
2. The assigned vectors $\vec v$ will be of two varieties: $$\vec v\in\{\frac{\hat r_{ij}}{||r_{ij}||^2},\frac{-m\hat r_{i0}}{||\vec r_{i0}||^2}\}$$ where $||r_{ij}||$ is the norm of the vector between points $i$ and $j$, $\hat r_{ij}$ denotes a direction away from both points $i,j$ and along the line between $i$ and $j$. $||r_{i0}||$ is the norm of the vector between a point $i$ and the origin and $-\hat r_{i0}$ denotes a direction from point $i$ to the origin.

I am interested in whether there exists a 3 dimensional configuration of the points where the sum of all the vectors is zero for each point. If this is not possible I hope someone would give some hints on how to prove it or point me towards a proof in the literature.

In essence what I am seeking is whether there exist three dimensional equilibrium configurations for Coulomb's law of electrostatics as you come across the periodic table of the elements. I suspect not but who was the first to prove it?

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This was proven in the negative in 1842 by Samuel Earnshaw. See Earnshaw's Theorem.

JEM: Correct me if i'm wrong but I think Earnshaw's theorem states that you cannot find a stable equilibrium configuration. I only ask for equilibrium.

An example of an equilibrium (not necessarily stable) is three particles with charges $\{q_1,q_2,q_3\}=\{4,-1,4\}$ with coordinates $\mathbf{r}_k=(x_k,y_k,z_k)^\mathsf{T}$. An equilibrium configuration is $$\mathbf{r}_1^\text{eq}=\pmatrix{-d\\0\\0}\\\mathbf{r}_2^\text{eq}=\pmatrix{0\\0\\0}\\\mathbf{r}_3^\text{eq}=\pmatrix{d\\0\\0}$$ for any $d>0$, as can be readily verified by computing $$\nabla_\mathbf{R}U=\pmatrix{0\\0\\0\\0\\0\\0\\0\\0\\0}$$ where $\mathbf{R}=(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3)^\mathsf{T}$ is the generalized coordinate and $$U=\frac{q_1 q_2}{\sqrt{\left(x_1-x_2\right){}^2+\left(y_1-y_2\right){}^2+\left(z_1-z_2\right){}^2}}+\frac{q_3 q_2}{\sqrt{\left(x_3-x_2\right){}^2+\left(y_3-y_2\right){}^2+\left(z_3-z_2\right){}^2}}+\frac{q_1 q_3}{\sqrt{\left(x_1-x_3\right){}^2+\left(y_1-y_3\right){}^2+\left(z_1-z_3\right){}^2}}$$ is the system potential. Meanwhile, the Hessian of the potential becomes $$\nabla\nabla_\mathbf{R}U=\left( \begin{array}{ccccccccc} -\frac{4}{d^3} & 0 & 0 & \frac{8}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 & 0 \\ 0 & \frac{2}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 & 0 & \frac{2}{d^3} & 0 \\ 0 & 0 & \frac{2}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 & 0 & \frac{2}{d^3} \\ \frac{8}{d^3} & 0 & 0 & -\frac{16}{d^3} & 0 & 0 & \frac{8}{d^3} & 0 & 0 \\ 0 & -\frac{4}{d^3} & 0 & 0 & \frac{8}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 \\ 0 & 0 & -\frac{4}{d^3} & 0 & 0 & \frac{8}{d^3} & 0 & 0 & -\frac{4}{d^3} \\ -\frac{4}{d^3} & 0 & 0 & \frac{8}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 & 0 \\ 0 & \frac{2}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 & 0 & \frac{2}{d^3} & 0 \\ 0 & 0 & \frac{2}{d^3} & 0 & 0 & -\frac{4}{d^3} & 0 & 0 & \frac{2}{d^3} \\ \end{array} \right)$$ which has eigenvalues $$\boldsymbol{\lambda}=\left( \begin{array}{c} -\frac{24}{d^3} \\ \frac{12}{d^3} \\ \frac{12}{d^3} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{array} \right)$$ indicating saddle-point instability.

## Example for $n=6$ using octahedral geometry

I have been curious lately about 3D versions of what you've indicated.

Well, as an example, there are stable octahedral arrangements for $n=6$, although they have irrational charges. To see this, let $$\{q_1,q_2,q_3,q_4,q_5,q_6,q_7\}=\{q_p,q_m,q_m,q_m,q_m,q_m,q_m\}$$ where $q_p>0$ is the "nuclear" charge and $q_m<0$ is the "electron" charge. We have $$U=\sum_{j=2}^7\sum_{k=1}^j\frac{q_jq_k}{|\mathbf{r}_j-\mathbf{r}_k|}$$ and with the octahedral ansatz $$\mathbf{R}^\text{eq}=(\mathbf{r}_1^\text{eq},\mathbf{r}_2^\text{eq},\mathbf{r}_3^\text{eq},\mathbf{r}_4^\text{eq},\mathbf{r}_5^\text{eq},\mathbf{r}_6^\text{eq},\mathbf{r}_7^\text{eq})=\begin{pmatrix} 0&d&-d&0&0&0&0\\ 0&0&0&d&-d&0&0\\ 0&0&0&0&0&d&-d \end{pmatrix}$$ we obtain the reduced potential $$U_\text{red}=\frac{3 q_m \left(\left(1+4 \sqrt{2}\right) q_m+4 q_p\right)}{2 d}$$ and $$\frac{\partial U_\text{red}}{\partial d}=0\Longrightarrow q_m=-\frac{4 q_p}{1+4 \sqrt{2}}.$$ We then verify the ansatz by computing the gradient of the full potential, which becomes $$\nabla_\mathbf{R}U(\mathbf{R}^\text{eq})=\mathbf{0}$$ meaning that this choice of geometry and charge yields an equilibrium configuration.

As before, the Hessian becomes (using MatrixPlot in Mathematica, because the full matrix is really big) $$\nabla\nabla_\mathbf{R}U(\mathbf{R}^\text{eq})=$$

which for $q_p=1,d=1$ has eigenvalues

$$\boldsymbol{\lambda}=\left( \begin{array}{c} \frac{2 \left(-9-5 \sqrt{2}-\sqrt{68283+29658 \sqrt{2}}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{2 \left(-9-5 \sqrt{2}-\sqrt{68283+29658 \sqrt{2}}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{2 \left(-9-5 \sqrt{2}-\sqrt{68283+29658 \sqrt{2}}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{2 \left(-9-5 \sqrt{2}+\sqrt{68283+29658 \sqrt{2}}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{2 \left(-9-5 \sqrt{2}+\sqrt{68283+29658 \sqrt{2}}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{2 \left(-9-5 \sqrt{2}+\sqrt{68283+29658 \sqrt{2}}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{36 \left(-8-\sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{36 \left(-8-\sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{24 \left(8+\sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{24 \left(8+\sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{24 \left(8+\sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{4 \left(9+5 \sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{4 \left(9+5 \sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ \frac{4 \left(9+5 \sqrt{2}\right)}{\left(1+4 \sqrt{2}\right) \left(33+8 \sqrt{2}\right)} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{array} \right)$$ which is saddle-point unstable.

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Correct me if i'm wrong but I think Earnshaw's theorem states that you cannot find a stable equilibrium configuration. I only ask for equilibrium. –  JEM Apr 23 '14 at 20:48
@JEM: There are many examples of equilibrium configurations, I'll edit my answer to include one. However, as far as explaining periodic trends go, this is still not useful, as just one tiny bump will cause the system to self-destruct. –  DumpsterDoofus Apr 23 '14 at 21:25
Thx for the discussion DD. I have been curious lately about 3D versions of what you've indicated. I suppose we will need to have $n+1\geq4$ in order to get a spatial investigation underway. Anything less will define a plane or in your example a line. If we consider Lithium as our intuition, is there a spatial static equilibrium? Stable or otherwise? Maybe we could invoke some more symmetry with Carbon: 4 on a square and 2 orthogonal to the plane of the square? Ive been playing with this for a while and cant seem to build one. –  JEM Apr 23 '14 at 23:04
@JEM: I'm pretty sure there are entire classes of nonplanar equilibrium configurations. Still, I don't really see the point, because they'll always be unstable and fall apart, and quantum mechanics is, in many cases, consistent with observed spectral data on the elements to absurd levels of accuracy. –  DumpsterDoofus Apr 23 '14 at 23:32
@JEM: I'll post an example of a nonplanar system in a moment. –  DumpsterDoofus Apr 23 '14 at 23:41