We know that electrostatic constant $K=\frac{1}{4\pi\epsilon}$.
This $4\pi$ came from the surface area of the surface in which charge is enclosed.
Then, why don't gravitational constant has the factor of $4\pi$?
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$\begingroup$ This is backwards ish, Gauss law is derived from coulombs law /newtons gravitation. The divergence of the g field has a 4pi in it, canceling it from the surface area $\endgroup$– jensen paullCommented Aug 15, 2022 at 9:05
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$\begingroup$ It is a constant, therefore you can always create a new constant from it if you decide so. I just decided to create a new constant $d=\frac{eG}{3\pi^2}$ so $F_g=\frac{3\pi^2d}{e}\frac{Mm}{r^2}$, where $e$ is Euler’s number. It is obvious correct, the question is: is it useful?, does it bring new insights?, does it make your life easier when writing down several pages of calculations involving gravitational force?. The reason for using $4\pi \epsilon$ in $E$ is because it saves you tons of $4\pi$’s when using the most useful Gauss equation for electrodynamics. $\endgroup$– J. ManuelCommented Aug 15, 2022 at 10:04
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$\begingroup$ Same reason we use $\pi$, the ratio of a circle's circumference to its diameter, instead of $\tau$, the ratio of its circumference to its radius. Tradition! There's a very famous song with that as a title: youtube.com/watch?v=kDtabTufxao $\endgroup$– Sean E. LakeCommented Aug 15, 2022 at 10:12
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3 Answers
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You can define it as so. I'll name $4\pi G$ as $\mathcal{G}$.
In electrostatics:
$$F=\frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2}$$
and so Gauss's law is:
$$\nabla \cdot \vec{E}=\frac{\rho}{\epsilon_0}$$
Notice how the $4\pi$ is gone from the denominator.
Equivalently, Newton's law of gravitation:
$$F=-G \frac{Mm}{r^2}$$
has the poisson's equation:
$$\nabla \cdot \vec{g} = -4\pi G\rho$$
where $\vec{g}$ is the acceleration field, similar to $\vec{E}$.
If you used $\mathcal{G}$, then the equations are:
$$F=-\frac{\mathcal{G}}{4\pi} \frac{Mm}{r^2}$$
and
$$\nabla \cdot \vec{g} = -\mathcal{G} \rho$$
It's just a convention.
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$\begingroup$ Shouldn't it be $\nabla\cdot\vec{g}=-4\pi G\rho$? $\endgroup$– J.G.Commented Aug 15, 2022 at 10:12
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The utility in writing the constant in Coulomb's law as $$\frac{1}{4 \pi \epsilon_0}$$ instead of $$\frac{1}{ \varepsilon_0}$$ is that this convention makes Gauss' law, equivalent to Coulomb's law, look simpler: $$\nabla \cdot \mathbf{E} = \rho/\epsilon_0.$$ With the opposite convention, we would have to insert the $4\pi$ in Gauss's law, and it would disappear in Coulomb's law. My guess is that the convention was chosen so that the $4\pi$ doesn't appear in the version which is used the most, which I'm pretty sure is Gauss's law.
In gravitation, however, the 'Coulomb version' $$F=G\frac{m_1m_2}{r^2}$$ seems to be used the most, so it would make sense to adopt the convention such that this equation appears without the $4\pi$.
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The electrostatic constant in Gaussian units is simply one. It's a choice. It's a weird choice for applied physics because the factors of 4𝜋 show up in rectangular problems but disappear in spherical ones. Hence, for practical problems, we use SI units.
Gaussian units are handier in theoretical physics because spherical problems are more common. In Gaussian units you don't need a lot of 4𝜋's in formulae expressed in spherical coordinates. A similar convention is convenient for gravitational problems since they commonly use spherical coordinates.
