Why do most of the physical equations have $\frac{1}{4\pi}$ as constants? I have seen that many equations have $\frac{1}{4\pi}$ as constants like Coulomb, pendulum problems, etc. Can anyone tell me why is like this?
Asked
Active
Viewed 327 times
2
-
1This applies to oscillations. Oscillations are projections of rotations. Rotations are circular. Circles are described by $\pi$. – safesphere May 20 '18 at 17:31
-
2I don't think $1/4\pi$ is more common than $4$, $1/4$, $\pi$, $1/\pi$, or $1/2\pi$. Could you provide more examples besides Coulomb's law, so that people might be able to help you identify a common thread? (In that case the constant out front is dependent on which system of units you use, and the factor of $4\pi$ is related to the area of a sphere being $4\pi R^2$.) – Jess Riedel May 20 '18 at 17:33
-
Related: https://physics.stackexchange.com/q/74254/2451 and links therein. – Qmechanic May 20 '18 at 17:57
-
2Possible duplicate of Why is there a factor of $4\pi$ in certain force equations? – Kyle Kanos May 20 '18 at 19:57
-
1The geometry of many of these problems corresponds to a point source for some type of field. This field "emanates" from the point source in spherical "wave fronts". The strength of the field decreases as the total area covered by the field increases. Since the area of a sphere is 4$\pi r^2$, the 4$\pi$ shows up in the denominator. – David White Jul 26 '18 at 17:52
1 Answers
9
It is--for thing involving Gauss's law, because there are $4\pi$ steradians in total. That is, integrating over every direction gives:
$$ {\rlap{\large{\circ}}} {\rlap{{\hspace{15px} \large{\circ}}}} {\rlap{\raise{-7px}{\hspace{5px} {\color{white}{\rule{15px}{20px}}}}}} {\rlap{\raise{-0px}{\hspace{5px} \rule{15px}{1px} }}} {\rlap{\raise{9px}{\hspace{5px} \rule{15px}{1px} }}} {\rlap{\iint}} {\phantom{\iint}} \, \mathrm{d}\Omega=\int_{\theta=0}^{\pi}{\int_{\phi=0}^{2\pi}{\sin{\theta} \, \mathrm{d}\theta \, \mathrm{d}\phi}}=4\pi $$
