Simple case of Gauss' law for gravity

In summary, the conversation was about a math problem with Gauss's law for gravity for a point mass M, and whether the divergence should be zero or not. One solution suggested was to consider a small but finite mass at the origin, while another option was to define the divergence of the inverse square function as a three-dimensional Dirac Delta function. However, it was noted that point masses are fictitious and a better approach would be to consider a more mathematically defined model, such as a sphere with a finite radius and constant mass density. This led to an easy solution using spherical coordinates and the known formula for the Laplace operator.
  • #1
snoopies622
846
28
TL;DR Summary
I'm having a math problem with Gauss's law for gravity for a point mass M. Or is my interpretation of the law wrong?
i have little experience with the differential form of Gauss's Law, and I've tried three times now to arrive at it for a point mass M (spherically symmetric classical gravitational field) but instead of getting an answer proportional to the mass density I keep getting zero. Is the divergence supposed to be zero? Intuitively I would say no since all the field lines are pointing away from the origin, but then maybe the fact that they get smaller as r increases makes a difference. Scanned and uploaded my work since LaTeX can be a bit tedious.
 

Attachments

  • grav-final-2.jpg
    grav-final-2.jpg
    36.6 KB · Views: 138
Physics news on Phys.org
  • #2
snoopies622 said:
Summary:: I'm having a math problem with Gauss's law for gravity for a point mass M. Or is my interpretation of the law wrong?

i have little experience with the differential form of Gauss's Law, and I've tried three times now to arrive at it for a point mass M (spherically symmetric classical gravitational field) but instead of getting an answer proportional to the mass density I keep getting zero. Is the divergence supposed to be zero? Intuitively I would say no since all the field lines are pointing away from the origin, but then maybe the fact that they get smaller as r increases makes a difference. Scanned and uploaded my work since LaTeX can be a bit tedious.

why don't you show theGauss's law form that you used? No one can diagnose what the problem here is based on your vague description.

Zz.
 
  • #3
(I hope the image is legible, I kept getting "image too large" so I had to shrink it)
 
  • #4
snoopies622 said:
Summary:: I'm having a math problem with Gauss's law for gravity for a point mass M. Or is my interpretation of the law wrong?

i have little experience with the differential form of Gauss's Law, and I've tried three times now to arrive at it for a point mass M (spherically symmetric classical gravitational field) but instead of getting an answer proportional to the mass density I keep getting zero. Is the divergence supposed to be zero? Intuitively I would say no since all the field lines are pointing away from the origin, but then maybe the fact that they get smaller as r increases makes a difference. Scanned and uploaded my work since LaTeX can be a bit tedious.

Yes, interestingly ##\nabla \cdot \frac{\hat r}{r^2} = 0##.

What are you missing?
 
  • Like
Likes snoopies622
  • #5
I used [itex] \nabla \cdot (\nabla \phi) = -4 \pi G \rho [/itex] with [itex] \phi = -GM/r [/itex] in three dimensional Cartesian coordinates. Okay, I wasn't sure if the answer should be zero or [itex] -4 \pi G \rho [/itex] with [itex] \rho = M/((4/3) \pi r^3) [/itex]. Thanks!
 
  • #6
snoopies622 said:
I used [itex] \nabla \cdot (\nabla \phi) = -4 \pi G \rho [/itex] with [itex] \phi = -GM/r [/itex] in three dimensional Cartesian coordinates. Okay, I wasn't sure if the answer should be zero or [itex] -4 \pi G \rho [/itex] with [itex] \rho = M/((4/3) \pi r^3) [/itex]. Thanks!

Is it ##0## everywhere?
 
  • #7
Yes, it appears to be zero everywhere except at the origin, where [itex] \phi [/itex] is undefined.

I don't have an intuitive sense of what this result means. Using a different formula for the force of gravity (pretending it's proportional to [itex] 1/r^3 [/itex] instead of [itex] 1/r^2 [/itex] for example) produces a non-zero divergence, even though in both cases the mass of the object creating the field is non-zero.
 
  • #8
snoopies622 said:
Yes, it appears to be zero everywhere except at the origin, where [itex] \phi [/itex] is undefined.

I don't have an intuitive sense of what this result means. Using a different formula for the force of gravity (pretending it's proportional to [itex] 1/r^3 [/itex] instead of [itex] 1/r^2 [/itex] for example) produces a non-zero divergence, even though in both cases the mass of the object creating the field is non-zero.

Mathematically, the divergence theorem does not apply in this case because of the function at ##r=0##.

One solution is to consider a small but finite mass at the origin. In which case, all the divergence is in that small region, and zero outside it, and thew divergence theorem holds.

The other option is to define the divergence of the inverse square function as a three-dimensional Dirac Delta function. In this case, you use Gauss's law (the value of surface integral) and insist that the volume integral must have the same value, hence the divergence must be a Delta function.
 
  • #9
The problem is that "point masses" are fictitions that make our lives often simpler but only as long as we don't look at the place where the "point mass" sits.

To solve your problem rather consider a mathematically better defined model for the source of your gravitational field first. The most simple you can think of is a sphere with a finite radius ##a## located at the origin of the coordinate system with a constant mass density ##\rho##, i.e.,
$$\rho(\vec{x})=\begin{cases} \rho_0 & \text{for} \quad |\vec{x}| \leq a \\0 & \text{for} \quad |\vec{x}|>a. \end{cases}.$$
Now you can apply the usual math of vector calculus. You know that the gravitational field ##\vec{g}(\vec{x})## is a conservative field, i.e., that there is a scalar potential ##\Phi## such that
$$\vec{g}=-\vec{\nabla} \phi$$
and then there's "Gauss's Law",
$$\vec{\nabla} \cdot \vec{g}=-\Delta \Phi=-4 \pi G \rho,$$
where ##G## is Newton's gravitational constant (the factor ##4 \pi## is conventional).

Now you can solve the problem very easily by first thinking about the symmetries of the problem. Obviously with our choice of a spherically symmetric mass distribution it makes sense to introduce spherical coordinates. Then ##\rho(\vec{x})=\rho(r)## and obviously the ansatz ##\Phi(\vec{x})=\Phi(r)## can be expected to work. Indeed using the known formula for the Laplace operator in spherical coordinates leads to
$$\Delta \Phi(r)=\frac{1}{r} \partial_r^2 (r \Phi)=4 \pi \rho.$$
That's easy to solve for our problem. Let's start with the region ##r>a##. Then the right-hand side vanishes, and we have
$$(r \Phi)''=0 \; \Rightarrow \; r \Phi=A +B r \; \Rightarrow \; \Phi(r)=\frac{A}{r} + B.$$
The constant ##B## is irrelevant and can be set to 0, because adding a constant to the potential doesn't change the field ##\vec{g}=-\vec{\nabla} \Phi##. We can't determine ##A## yet. For that we need to also solve the equation for ##r<a##, where
$$\frac{1}{r} (r \Phi)''=4 \pi \rho_0.$$
Now
$$(r \Phi)''=-4 \pi G \rho_0 r \; \Rightarrow \; (r \Phi)'=2 \pi G \rho_0 r^2 + B' \; \Rightarrow \; r \Phi=\frac{2}{3} \pi \rho_0 r^3 + B' r +A' \; \Rightarrow \; \Phi=-\frac{2}{3} G \pi \rho_0 r^2 + B'+\frac{A'}{r}.$$
Now since there are no singularities in ##\rho## at ##r=0##, there shouldn't be singularities there for ##\Phi## either. Thus we have to set ##A'=0##. Further the potential as well as the field
$$\vec{g}=-\vec{\nabla} \Phi=-\Phi'(r) \vec{e}_r$$
should be continuous at ##r=a##, from which
$$\frac{2}{3} \pi G \rho_0 a^2 +B'=\frac{A}{a}, \qquad (*)\\
\frac{4}{3} \pi G \rho_0 a =-\frac{A}{a^2}.$$
From the latter equation we find
$$A=-\frac{4}{3} \pi G a^3 \rho_0=-M G,$$
where ##M=\rho_0 V=4 \pi a^3 \rho/3## is the total mass of our spherical body.
Finally we get from (*)
$$\frac{M G}{2 a}+B'=-\frac{M G}{a} \; \Rightarrow \; B'=-\frac{3 M G}{2a}.$$
Now let's see what happens, when we make ##a \rightarrow 0## but keeping ##M=\text{const}##. Since ##\rho_0=M/V=3 M/(4 \pi a^3)## we have ##\rho_0 \rightarrow \infty##.

This shows, what's going on for the "point-particle limit". You get
$$\rho(\vec{x})=\begin{cases} 0 & \text{for} \quad \vec{x} \neq 0, \\ \rightarrow \infty & \text{for} \quad \vec{x} \rightarrow 0,
\end{cases}
$$
but
$$M=\int_{V} \mathrm{d}^3 x \rho(\vec{x})$$
for any volume ##V## which contains the origin. Thus the correct limit for the mass density needs an extension of our application of vector calculus to socalled "generalized functions" or "distributions". In this case we get the famous Dirac ##\delta## distribution,
$$\rho(\vec{x})=M \delta^{(3)}(\vec{x}),$$
and
$$\Phi(\vec{x})=-\frac{M G}{r},$$
which diverges for ##r \rightarrow 0##. So you cannot apply the usual notions of vector calculus, where curl, div, grad are given as derivatives with respect to the coordinates for ##r=0##, but everything gets consistent by defining
$$\Delta \Phi(\vec{x})=4 \pi G \rho(\vec{x})=4 \pi G M \delta^{(3)}(\vec{x}),$$
from which you get the extended meaning of the Laplace operator in the sense of generalized functions:
$$\Delta \frac{1}{r} = -4 \pi \delta^{(3)}(\vec{x}).$$
As you see, the apparently "simple" case of a "point mass" needs the introduction of an entire new kind of mathematics, and indeed the Dirac ##\delta## distribution, introduced in a handwaving way by Dirac to make calculations in quantum theory simpler, triggered the development of the entire new mathematical discipline of functional analysis.
 
  • Like
Likes PeroK
  • #10
Wow, thanks PeroK and vanhees71 ~ so much detail!

I was actually more curious about the fact that [itex] \nabla ^2 \phi =0 [/itex] outside of the sphere even though the gravitational field itself is non-zero. Rather like how the Schwarzschild metric represents a non-zero gravitational field and is itself a "vacuum solution" to Einstein's equation. In fact what prompted this subject for me was the beginning of chapter 8 of Schutz's "A First Course in General Relativity" in which he arrives at Einstein's field equation using the Gauss law as a kind of analogy. If I remember correctly Einstein himself does too in his original paper.
 
  • Like
Likes vanhees71
  • #11
snoopies622 said:
I was actually more curious about the fact that [itex] \nabla ^2 \phi =0 [/itex] outside of the sphere even though the gravitational field itself is non-zero.

Divergence identifies sources of the field. In regions where there is no matter (or electric charges in EM field theory), you have Laplace's equation: $$\nabla^2 \phi = 0$$.
 
  • Like
Likes vanhees71 and snoopies622

Related to Simple case of Gauss' law for gravity

1. What is Gauss' law for gravity?

Gauss' law for gravity is a fundamental principle in physics that describes the relationship between mass and gravitational force. It states that the gravitational flux through a closed surface is proportional to the enclosed mass, and is independent of the shape or size of the surface.

2. How is Gauss' law for gravity related to Newton's law of gravitation?

Gauss' law for gravity is essentially a more general form of Newton's law of gravitation. While Newton's law only applies to point masses, Gauss' law can be applied to any distribution of mass, making it a more versatile and powerful tool for calculating gravitational forces.

3. What are the applications of Gauss' law for gravity?

Gauss' law for gravity has many practical applications in various fields of science and engineering. It is commonly used in celestial mechanics to calculate the gravitational forces between planets and other celestial bodies. It is also used in geophysics to study the Earth's gravitational field and in engineering to design structures that can withstand gravitational forces.

4. How is Gauss' law for gravity derived?

Gauss' law for gravity can be derived from the more general Gauss' law of electromagnetism. By assuming that the gravitational field behaves similarly to the electric field, and applying the divergence theorem, we can arrive at the equation for Gauss' law for gravity.

5. Is Gauss' law for gravity valid in all situations?

While Gauss' law for gravity is a very useful and accurate principle, it is not valid in all situations. It assumes a static and spherically symmetric distribution of mass, and does not take into account factors such as rotation, tidal forces, or the effects of general relativity. In these cases, more complex equations and models must be used.

Similar threads

Replies
10
Views
1K
Replies
1
Views
700
  • Classical Physics
Replies
5
Views
1K
  • Classical Physics
Replies
24
Views
791
  • Introductory Physics Homework Help
Replies
26
Views
683
  • Classical Physics
Replies
16
Views
876
  • Classical Physics
Replies
9
Views
1K
  • Classical Physics
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
803
Replies
6
Views
1K
Back
Top