Gauss' law for gravity

In physics, Gauss' law for gravity, also known as Gauss' flux theorem for gravity, is a law of physics which is essentially equivalent to Newton's law of universal gravitation. Its form is mathematically similar to Gauss' law for electricity; in particular, Gauss' law for gravity bears the same mathematical relation to Newton's law that Gauss' law for electricity bears to Coulomb's law.

The law is expressed in terms of the gravitational field. The law has two forms, a "differential form" and an "integral form", which are equivalent to each other due to the divergence theorem.

Although Gauss' law for gravity is physically equivalent to Newton's law, there are many situations where Gauss' law for gravity offers a more convenient and simple way to do a calculation than Newton's law.

Definition of the gravitational field

The gravitational field g (also called gravitational acceleration) is a vector field – a vector at each point of space (and time). It is defined so that the gravitational force experienced by a particle is:$mathbf\{F\}\_g\; =\; mmathbf\{g\}(mathbf\{r\})$where:"m" is the mass of a particle,:r is the position vector of the particle.

Integral form

The integral form of Gauss' law for gravity states::$oint\_\{part\; V\}mathbf\{g\}cdot\; dmathbf\{A\}\; =\; -4\; pi\; GM$where:∂"V" is any closed surface,:"d"A is a vector, whose magnitude is the area of an infinitesimal piece of the surface ∂"V", and whose direction is the outward-pointing surface normal (see surface integral for more details),:"M" is the total mass enclosed within the surface ∂"V".

The left-hand side of this equation is called the flux of the gravitational field. Note that it is always negative (or zero), and never positive. This can be contrasted with Gauss' law for electricity, where the flux can be either positive or negative. The difference is because "charge" can be either positive or negative, while "mass" can only be positive.

Differential form

The differential form of Gauss' law for gravity states::$ablacdot\; mathbf\{g\}\; =\; -4pi\; G\; ho$where:$ablacdot$ denotes divergence,:"G" is the gravitational constant of the universe,:"ρ" is the mass density at each point.

Relation to the integral form

The two forms of Gauss' law for gravity are mathematically equivalent. The divergence theorem states::$oint\_\{part\; V\}mathbf\{g\}cdot\; d\; mathbf\{A\}\; =\; int\_V\; ablacdotmathbf\{g\}\; dV$where:"V" is a closed region bounded by a simple closed oriented surface ∂"V",:g is a continuously differentiable vector field defined on a neighborhood of "V",:"dV" is an infinitesimal piece of the volume "V" (see volume integral for more details).

Given also that:$M\; =\; int\_\{V\}\; ho\; dV$we can apply the divergence theorem to the integral form of Gauss' law for gravity, which becomes::$int\_V\; ablacdotmathbf\{g\}\; dV\; =\; -4\; pi\; Gint\_\{V\}\; ho\; dV$which can be rewritten::$int\_V(\; ablacdotmathbf\{g\})\; dV\; =\; int\_\{V\}\; (-4\; pi\; G\; ho)\; dV$This has to hold simultaneously for every possible volume "V"; the only way this can happen is if the integrands are equal. Hence we arrive at:$ablacdotmathbf\{g\}\; =\; -4pi\; G\; ho$which is the differential form of Gauss' law for gravity.

It is possible to derive the integral form from the differential form using the reverse of this method.

Although the two forms are equivalent, one or the other might be more convenient to use in a particular computation.

Relation to Newton's law

Deriving Gauss' law from Newton's law

Gauss' law for gravity can be derived from Newton's law of universal gravitation, which states that the gravitational field due to a point mass is::$mathbf\{g\}(mathbf\{r\})\; =\; -GMfrac\{mathbf\{e\_r\{r^2\}$where:e_r is the radial unit vector,:"r" is the radius, |r|.:"M" is the mass of the particle, which is assumed to be a point mass located at the origin.

In this section, two alternative proofs of this fact are presented. The first proof is more visual and intuitive, while the second proof is more mathematical.

pecial case: Spherical surface centered at a point mass

Consider a spherical surface of radius "r" centered at a point-mass of mass "M". The total flux of the gravitational field g over a closed surface ∂"V", according to Newton's law, is::$oint\_\{part\; V\}mathbf\{g\}cdot\; dmathbf\{A\}\; =\; oint\_\{part\; V\}-frac\{GM\}\{r^2\}\; mathbf\{e\_r\}cdot\; dmathbf\{A\}$However, the magnitude of the infinitesimal area element "d"A is just the area of the infinitesimal solid angle "dΩ", given by:$dmathbf\{A\}\; =\; r^2mathbf\{e\_r\}\; dOmega$which gives us:$oint\_\{part\; V\}mathbf\{g\}cdot\; dmathbf\{A\}\; =\; -GMoint\_\{part\; V\}frac\{1\}\{r^2\}\; mathbf\{e\_r\}cdot\; r^2mathbf\{e\_r\}\; dOmega$By observing that e_r·e_r = 1, and that the integral of unity over a closed surface with respect to the solid angle is the surface area of a unit sphere, 4π, we arrive at:$oint\_\{part\; V\}mathbf\{g\}cdot\; dmathbf\{A\}\; =\; -4\; pi\; GM$which is the integral form of Gauss' law for gravity, for this special case.

General case: Field lines (visual proof)

To move to the general case, we use the method of field lines. The gravitational field can be depicted via field lines, a set of lines or curves that follow the direction of the gravitational field. The magnitude of the field is required to be proportional to the density of field lines. Moreover, it can be shown that the flux of the field through a surface is proportional to the net number of field lines that pass through the surface (the term "net" means, specifically, the number that pass outward minus the number that pass inward).

Newton's law implies that the field lines will extend directly, radially inward towards the point mass in every direction. Moreover, the special case above shows that if we imagine a series of concentric spheres centered at the point mass, the same number of field lines will pass through each one. So in other words, the field lines all begin at infinity, and go directly inward towards the point mass, ending at the point mass, and coming in uniformly from all directions.

For "any" finite closed surface (not necessarily spherical) that encloses the point mass, each of the field lines will start at infinity outside the surface, pass through the surface at some point, and end at the point mass inside the surface. Therefore, the flux through the surface is a constant -4π"GM", regardless of the shape of the surface, as long as the point mass is inside.

Likewise, for any finite closed surface that does "not" enclose the point mass, some of the field lines will pass into and then back out of the surface, and some field lines will not touch the surface at all. Regardless, the "net" flux through the surface is zero.

In every case, this is consistent with Gauss' law. To finish off the proof, we need to consider the case where there is more than one mass (or even infinitely many masses comprising a continuous distribution). The simplest way to handle this case is to say that both Newton's law and Gauss' law obey the superposition principle, so if Gauss' law is a consequence of Newton's law for a single mass, then it's a consequence of Newton's law for any number of masses. Alternatively, one can note that the net number of field lines that enter a surface equals the number of field lines that end on a mass inside the surface, which is proportional to the "total" mass inside the surface.

General case: Mathematical proof

The differential form of Gauss' law for gravity can also be derived from Newton's law of universal gravitation. Using the expression from Newton's law, we get the total field at r by using an integral to add up the field at r due to the mass at each other point in space with respect to an s coordinate system, to give:$mathbf\{g\}(mathbf\{r\})\; =\; -Gint\_V\; frac\{\; ho(mathbf\{s\})(mathbf\{r\}-mathbf\{s\})\}\{|mathbf\{r\}-mathbf\{s\}|^3\}\; dV(mathbf\{s\})$If we take the divergence of both sides of this equation with respect to r, and use the known theorem [See, for example, cite book | author=Griffiths, David J. | title=Introduction to Electrodynamics (3rd ed.) | publisher=Prentice Hall | year=1998 | id=ISBN 0-13-805326-X | page=50] :$abla\; cdot\; (frac\{mathbf\{s\{|mathbf\{s\}|^3\})\; =\; 4pi\; delta(mathbf\{s\})$where δ(s) is the Dirac delta function, the result is:$ablacdotmathbf\{g\}(mathbf\{r\})\; =\; -4pi\; Gint\_V\; ho(mathbf\{s\})\; delta(mathbf\{r\}-mathbf\{s\})\; dV(mathbf\{s\})$Using the "sifting property" of the Dirac delta function, we arrive at:$ablacdotmathbf\{g\}\; =\; -4pi\; G\; ho$which is the differential form of Gauss' law for gravity, as desired.

Deriving Newton's law from Gauss' law

Strictly speaking, Newton's law cannot be derived from Gauss' law alone, since Gauss' law does not give any information regarding the curl of g (see Helmholtz decomposition). However, Newton's law "can" be proven from Gauss' law if, in addition, an extra assumption is made. There are a number of such extra assumptions possible (e.g., that gravity is a conservative force). Here, we will choose to use the "obvious" assumption that the gravitational field from a point mass is spherically-symmetric. Taking ∂"V" in the integral form of Gauss' law to be a spherical surface of radius "r", centered at the point mass "M", we have: $oint\_\{part\; V\}mathbf\{g\}cdot\; dmathbf\{A\}\; =\; -4\; pi\; GM$By the assumption of spherical symmetry, the integrand is a constant which can be taken out of the integral. The result is: $4pi\; r^2mathbf\{e\_r\}cdotmathbf\{g\}(mathbf\{r\})\; =\; -4pi\; GM$Again by spherical symmetry, g points in the radial direction, and so we get: $mathbf\{g\}(mathbf\{r\})\; =\; -GMfrac\{mathbf\{e\_r\{r^2\}$which is Newton's law.

Relation to gravitational potential and Poisson's equation

Since the gravitational field has zero curl (equivalently, gravity is a conservative force), it can be written as the gradient of a scalar potential, called the gravitational potential::$mathbf\{g\}=-\; ablaphi$, Then the differential form of Gauss' law for gravity becomes Poisson's equation::$\{\; abla\}^2phi\; =\; 4pi\; G\; ho$,This provides an alternate means of calculating the gravitational potential and gravitational field. Although computing g via Poisson's equation is mathematically equivalent to computing g directly from Gauss's law, one or the other approach may be an easier computation in a given situation.

In radially symmetric systems, the gravitational potential is a function of only one variable (namely, $r=|mathbf\{r\}|$), and Poisson's equation becomes (see Del in cylindrical and spherical coordinates)::$frac\{1\}\{r^2\}frac\{partial\}\{partial\; r\}(r^2frac\{partial\; phi\}\{partial\; r\})\; =\; 4pi\; G\; ho(r)$while the gravitational field is::$mathbf\{g\}(mathbf\{r\})\; =\; -mathbf\{e\_r\}frac\{partial\; phi\}\{partial\; r\}$

Applications

Gauss' law can be used to easily derive the gravitational field in certain cases where a direct application of Newton's law would be more difficult (but not impossible). See the article Gaussian surface for more details on how these derivations are done. Three such applications are as follows:

Bouguer plate

We can conclude (by using a "Gaussian pillbox") that for an infinite, flat plate (Bouguer plate) of any finite thickness, the gravitational field outside the plate is perpendicular to the plate, towards it, with magnitude 2πG times the mass per unit area, independent of the distance to the plate (see also gravity anomalies).

More generally, for a mass distribution with the density depending on one Cartesian coordinate "z" only, gravity for any "z" is 2πG times the difference in mass per unit area on either side of this "z" value.

In particular, a combination of two equal parallel infinite plates does not produce any gravity inside.

Cylindrically symmetric mass distribution

In the case of an infinite cylindrically symmetric mass distribution we can conclude (by using a cylindrical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of 2G/r times the total mass per unit length at a smaller distance (from the axis), regardless of any masses at a larger distance.

For example, inside an infinite hollow cylinder, the field is zero.

pherically symmetric mass distribution

In the case of a spherically symmetric mass distribution we can conclude (by using a spherical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of G/r² times only the total mass within a smaller distance than r. All the mass at a greater distance than r from the center can be ignored.

For example, a hollow sphere does not produce any net gravity inside. The gravitational field inside is the same as if the hollow sphere were not there (i.e. the resultant field is that of any masses inside and outside the sphere only).

Although this follows in one or two lines of algebra from Gauss' law for gravity, it took Isaac Newton several pages of cumbersome calculus to derive it directly using his law of gravity; see the article Shell theorem for this direct derivation.

ee also

*Carl Friedrich Gauss
*Divergence theorem
*Flux
*Gaussian surface
*Gauss' law for electricity
*Gauss' law for magnetism

References

*For usage of the term "Gauss' law for gravity" see, for example, [http://link.aps.org/doi/10.1103/PhysRevLett.70.1195 this article] .