# Is there a higher-level reason why $\nabla\cdot(\hat{\bf r}/r^2) = 0$ in three dimensions but not two? [duplicate]

I am working through Griffiths, Introduction to Electrodynamics, and finding the divergence of the electric field generated by a single charge sitting at the origin.

$$\mathbf{E}(\mathbf{r}) = \frac{\hat{\bf r}}{|\mathbf{r}|^2}$$

I compute $$\nabla \cdot \mathbf{E}$$ and get 0. This seems very surprising -- I wouldn't have expected it, but the factors cancel out just so.

Sometimes, the book uses two-dimensional drawings, so I assumed that everything would be mostly the same in two dimensions. But to my surprise, it's not -- if I use only two dimensions and compute $$\nabla \cdot \mathbf{E}$$, I don't get zero.

$$\nabla \cdot \mathbf{E} = \frac{\partial}{\partial x}\left(\frac{x}{(x^2 + y^2)^{3/2}}\right) + \frac{\partial}{\partial y}\left(\frac{y}{(x^2 + y^2)^{3/2}}\right) \ne 0$$

Why is this? It seems like such a coincidence that we live in three dimensions, and three dimensions is also where this divergence works out to be zero.

Possible explanations:

• I know vaguely that the inverse square law is related to three dimensions; if I had to summarize my understanding it would be roughly "if you want conservation of energy to work out in three dimensions, you need the force to go as the inverse of the square rather than e.g. as the inverse or the inverse of the cube". Is this related?

## marked as duplicate by Qmechanic♦Oct 30 '18 at 4:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I don't think this relates to energy conservation, but maybe I'm missing something. Can you explain why you think energy conservation is relevant here, and why you think the inverse square law only works in 3D? Also note that for your 3D case the divergence is technically $\delta(\vec r)$. – Aaron Stevens Oct 30 '18 at 2:56
• Possible duplicates: physics.stackexchange.com/q/93/2451 , physics.stackexchange.com/q/47084/2451 and links therein. – Qmechanic Oct 30 '18 at 3:20

## 1 Answer

Seeing how something fits into a larger pattern can be satisfying, so I'll try to address your question from that angle.

In $$D$$-dimensional space, the Gauss law says $$\nabla\cdot\mathbf{E}\propto\rho$$, where $$\rho$$ is the charge density. This is a straightforward generalization of the $$D=3$$ case, with a $$D$$-component gradient $$\nabla$$ and a $$D$$-component electric field $$\mathbf{E}$$.

(The appropriate $$D$$-dimensional generalization of the whole set of Maxwell's equations is more interesting, but I won't go into that here.)

Even though the Gauss law itself doesn't look any different in $$D$$-dimensional space except for the number of components, the solution of the Gauss law depends on $$D$$ in an interesting way. In the case of a point charge, the Gauss law is $$\nabla\cdot\mathbf{E}(\mathbf{r})\propto\delta(\mathbf{r})$$, and the solution is $$\mathbf{E}(\mathbf{r})\propto\frac{\mathbf{r}}{|\mathbf{r}|^D}.$$ This is the $$D$$-dimensional version of Coulomb's law. When $$D=3$$, this agrees with your first equation. For any $$D\geq 1$$, this satisfies $$\nabla\cdot\mathbf{E}(\mathbf{r})=0 \hskip2cm \text{for all }\mathbf{r}\neq 0.$$ Your calculation in two dimensions confirms that the $$D=3$$ version does not work in two-dimensional space. The $$D=2$$ version does. Here's the calculation: \begin{align*} \nabla\cdot\frac{\mathbf{r}}{|\mathbf{r}|^D} &= \frac{\nabla\cdot\mathbf{r}}{|\mathbf{r}|^D} - D\frac{\mathbf{r}}{|\mathbf{r}|^{D+1}}\cdot\nabla |\mathbf{r}| \\ &= \frac{D}{|\mathbf{r}|^D} - D\frac{\mathbf{r}}{|\mathbf{r}|^{D+1}}\cdot\frac{\mathbf{r}}{|\mathbf{r}|} \\ &= 0 \ \text{ whenever }\mathbf{r}\neq 0. \end{align*}

• Thanks Dan! So it sounds like you're saying that, starting from Gauss's Law, we can derive that we must have an inverse power law where the power is $D - 1$. But my understanding is that the inverse square law is the experimental fact, and Gauss's Law is derived. Isn't it surprising that it works out so we have Gauss's Law? Is there any deeper "why" here? – Eli Rose Oct 31 '18 at 4:33
• @EliRose This gets to the heart of how science works. Our understanding of nature is based on models (aka theories). Models can't be deduced from observations. It's the opposite. Observations do inspire models, but deduction only flows one way: we deduce predictions (what we expect to observe) from models. Once we've found a model that accurately predicts a broader scope of things, we adopt it as the new basis for our understanding. This is why Gauss's law is considered more fundamental than the inverse square law; it is part of Maxwell's eq'ns, which explain much more than static forces. – Chiral Anomaly Oct 31 '18 at 12:43
• @EliRose Here's another way to say it: the route that physics followed historically (or the route that textbooks follow for pedagogical reasons) is often different, even opposite, compared to the route that physics follows logically. Gauss's law may have originally been inferred from the inverse square law, and textbooks may present it that way; but logically a much simpler perspective is to start with Maxwell's equations, including Gauss's law, and deduce everything else from there. The goal of physics is to find nature's underlying simplicity, so I view this as a deeper "why". – Chiral Anomaly Oct 31 '18 at 12:49