Why is there a factor of $4\pi$ in certain force equations?

Question

I mean to ask why there is $4\pi$ present in force equations governing electricity? Though all objects in universe are not spherical and circular, the constant of proportionality in both equations contain $4\pi$. Why?

Answer 1

For example, if you mean $k_e=\frac1{4\pi\epsilon_0}$, it comes from a natural "Gauss's law" understanding of Coulomb's law, where the electric field is distributed over the surface of a sphere of area $4\pi r^2$.

$$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2 $$

This is also the explanation for the $1/r^2$ term (and in other classical fields), and there are also attempts to extend this sort of reasoning to gravitational fields.

Answer 2

If you want to avoid factors of $\pi$ in the more fundamental equations like $\nabla . E = \rho / \epsilon_0$, you have to accept them where they belong, for instance in: $E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}$.

As remarked by others, Newton failed to put a factor $4 \pi$ into his gravitation equation (he stipulated $g = G \frac{M}{r^2}$, instead of $g = G \frac{M}{4 \pi r^2}$) and as a result we have to live with factors of $\pi$ in the more fundamental Gauss' law for gravity, and more importantly, also in Einstein's theory of gravity.

Answer 3

The physical reason for the appearance of a $4\pi$ somewhere in the theory is the spherical symmetry of the problem and is discussed more in other answers . Here I want to quote an interesting argument from Arnold Sommerfeld's Lectures on Theoretical Physics Vol III, which has a section dedicated to this issue.

If you remove the $4\pi$ from the force law you will have it in more fundamental Maxwell's equation:

$$\nabla.\mathbf{D}=4\pi\rho$$ and also it will distorts the energy density into: $$W=\frac{1}{8\pi}\mathbf{E}.\mathbf{D}$$

But Heaviside, as is said in the mentioned book, who fought a life-long battle for the rational units $(4\pi$ present in the force law$)$, has another interesting argument about advantage of this system to others. He points to the capacity of a capacitor:

A plate capacitor (with area $A$ , plate separation $d$) in this rational units $(4\pi$ present in the force law$)$ and in other units $($with $ 4\pi$ present in the Maxwell's equations$)$ has a capacity of
$$C=\frac{\epsilon A}{d}\tag{rational}$$ $$C=\frac{\epsilon A}{4\pi d}\tag{others}$$ and for a spherical capacitor (radius $R$, outer sphere imagined at infinity): $$C=4\pi \epsilon R\tag{rational}$$ $$C=\epsilon R\tag{others}$$ We see that with rational units the factor $4\pi$ appears for the sphere. For other units it is missing for the sphere and appears for the plane capacitor.

Heaviside then makes the following striking comparison: In passing from the measurement of distance to the measurement of area one might define as unit of area the area of a circle of radius $1$. This would be logically possible. It would however lead to the strange result that a square with the side $1$ would have the area $\dfrac{1}{\pi}$. Everyone would then say that $\pi$ was at the wrong place. We said the same of the factor $4\pi$ in the above formulas for capacities.

Answer 4

Any differential equation of the form $\nabla \cdot A = \alpha$ and $\nabla \wedge A = 0$ in $n$-dimensions has as its Green's function (that is, the solution for a point source, for $\alpha = \delta$, the Dirac delta function) a field $G$ of the form

$$G(r) = \frac{1}{S_{n-1}} \frac{\hat r}{|r|^{n-1}}$$

where $S_{n-1}$ is the surface area of a unit $n$-ball, which we know to be $4\pi$ in 3d. The factor of $4\pi$ that appears in many such force equations is inherently geometrical, and as has been stated, it is confounding to lump it into a constant, especially if one works outside of 3d (see, for example, the electric field of a line charge, which is a 2d problem and thus has $2\pi$ appear in it, as circumference of a unit circle).

Answer 5

The main reason is that it makes the calculation easier and the results look nicer. For example, suppose the field (or force) is given by $$ E = \frac{1}{4\pi} f(\mathbf{r})$$ for some function $f(\mathbf{r})$.

If the system process rotational symmetry, after sum over the density, you will get a factor of $2\pi$. If the system process spherical symmetry, you will get a factor of $4\pi$. In both case, the resulting equation contains no $\pi$ factor. It is particular useful in EM as we usually consider a density distribution process some symmetry.

So, why there is no $\pi$ in the Newtonian gravity $F=GMm/r^2$ ?

Because there is no such need. You cannot craft a planet like macroscopic object into a infinitely long rod, a disk shape, or some funny shape. Gravity only have significant effect when it accumulates a large mass, at the same time, it become a sphere by its own gravity. So, basically, it is already a good point like particle when we look at it some radius away. The extra $\pi$ won't make any calculation easier.

Answer 6

Some people (myself included) would regard field equations like Gauss' Law as more "fundamental" than force equations. The most obvious reason for this is that Coulomb's Force Law only works when the charges in question are held static - it has to be modified once they are allowed to move.

As the others have stated, the $4 \pi$ has to do with the surface area of a sphere. Gauss' Law (I'll use the integral form to make the surface area connection more apparent) for the electric field tells us:

$$\int \mathbf{E} \cdot \mathrm{d} \mathbf{A}= \frac{q}{\epsilon_0}$$

In words, this means that if you enclose some charges in an imaginary surface, then the sum of the electric field sticking out of that surface multiplied by the area of the surface is equal to the total charge enclosed by it, multiplied by some constant factor. If you take a single point charge and enclose it with a spherical surface of radius $r$, then it reduces to:

$$4 \pi r^2 E= \frac{q}{\epsilon_0}$$

Do some rearranging and it's easy enough to see how it relates to Coulomb's Law.

There's also a form of Gauss' Law for (Newtonian) gravity:

$$\int \mathbf{g} \cdot \mathrm{d} \mathbf{A}= 4 \pi GM$$

This field equation actually contains the factor $4 \pi$ already, so when you enclose a mass with a spherical surface the factor cancels on both sides. This is simply because when Newton wrote down his force law for gravity he didn't know about things like Gauss' Law, and so neglected to include the $4 \pi$ in the force equation. And since then the convention has stuck around, so we're left with a slightly confusing hodgepodge of some field equations needing factors of $\pi$ and some not. In general, if you see a factor of $\pi$ in a field equation then you're probably looking at something gravity-related.

Answer 7

It's just a result of Gauss theorem applied in symmetry in this case surface area of sphere with charge placed at it's center.

Which is 4 Pi R^2