Is there an integral version of Newton's law of gravity?

[Thadriel]

TL;DR

Maybe something summing up infinitesimal masses or densities?

$$F=G\frac{m_1 m_2}{r^2}$$
is presumably for point masses. If the masses weren't a point masses, then wouldn't you need a version of the formula that sums up the gravity for each infinitesimal portion of the masses? And for my money, "summing up" in physics is integrals, right?

So would it be something like this?

$$F = G\int_0^r \int_0^m \frac{m_1 m_2}{r^2} dmdr\\$$

Actually no, if it's three dimensional, I imagine maybe something like this?

$$F = G\int_0^r \int_0^p \frac{p_1 p_2}{r^2} dpdr\\$$

where p is density?

I feel like this would get very complicated. You have the two masses, which are a distance r apart. But within those masses are other masses, each at a different r, and possibly with a different density that is dependent upon R, the radius of the objects. Now, I'm quite sure these two made up integrals are wrong, but can someone point me in the right direction to where I'd want to look for spatially extended objects and Newton's gravity law? Or is it simply that it doesn't matter because of Gauss' law or something?

Thanks.

 

[Ibix]

Ignore the small mass and just try to calculate $g$ as a function of position. Model the large mass as a set of point masses of mass $\delta m$. What is the gravitational acceleration $\delta g$ at a position $\vec{r}$ from one of those elements?

Then think about how you could write $\delta m$ in terms of $\rho$. Sum up. Then see if the usual $\delta\rightarrow d$ gets you something like an integral.

 

[vanhees71]

It's very important to get the vectors right! Of course, you can formulate Newtonian gravity as a non-relativistic classical field theory (it's the paradigmatic example in fact). In the field picture the gravitational force on a test particle is defined by the gravitational field, $\vec{g}(t,\vec{x})$. The force acting on a point mass $m$ at position $\vec{x}$ is given by $\vec{F}=m \vec{g}(t,\vec{x})$.

The field equation uses the fact that the source of the gravitational field is the mass distribution $\rho(t,\vec{x})$, and it's always attractive,
$$\vec{\nabla} \cdot \vec{g}(t,\vec{x})=-4 \pi \gamma \rho(t,\vec{x}).$$
In addition one knows that for a point mass, $M$, at rest at the origin of the coordinate system you get
$$\vec{g}=-\gamma \frac{\vec{x}}{|\vec{x}|^3},$$
and thus the solution for a charge distribution is
$$\vec{g}(t,\vec{x}) = -\gamma \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t,\vec{x}')(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|}, \qquad (*)$$
and thus you also have
$$\vec{\nabla} \times \vec{g}=0.$$
This implies that $\vec{g}$ has a potential,
$$\vec{g}=-\vec{\nabla} \Phi$$
and thus
$$\Delta \Phi(t,\vec{x}) = +4\pi \gamma \rho(t,\vec{x}).$$
From the Green's function of the Laplace operator you immediately get
$$\Phi(t,\vec{x})=-\gamma \int_{\mathbb{R}^3} \frac{\rho(t,\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Taking the gradient of this wrt. $\vec{x}$ leads back to (*), as it should be.

 

[Thadriel]

Ignore the small mass and just try to calculate $g$ as a function of position. Model the large mass as a set of point masses of mass $\delta m$. What is the gravitational acceleration $\delta g$ at a position $\vec{r}$ from one of those elements?

Then think about how you could write $\delta m$ in terms of $\rho$. Sum up. Then see if the usual $\delta\rightarrow d$ gets you something like an integral.

Thanks. I'll think on this a couple of days. Since you brought up $\vec{r}$, I'm wondering if this will look something like Coulomb's law:

$F = \frac{q}{4πε_0} \sum_{i=1}^N q_i \frac{r - r_i}{|r-r_i|^3}$

which honestly is stretching my math. Am I on the right track there, that I'd need the position of the test mass as well as the major mass in there, since the distance is a factor?But you said ignore the small mass, and honestly I'm a bit confused about the difference between a so-called "test mass" and the mass of the small object. If I just ignore it (and they're the same thing), then maybe there's no need for the $r-r_i$.

Actually no, because the acceleration has to be distance dependent. So here's a question: would I be integrating over density as well as distance, or just density? As you can see, there is a gap for me between writing equations and truly understanding what they mean.

 

[Ibix]

But you said ignore the small mass, and honestly I'm a bit confused about the difference between a so-called "test mass" and the mass of the small object. If I just ignore it (and they're the same thing), then maybe there's no need for the $r-r_i$.

I meant calculate $g$ rather than $F$ - you still need the position vectors - you can't escape them. But vanhees71 has shown an easier way which is to calculate the sum of the potentials, then differentiate to get the force.

 

[Thadriel]

It's very important to get the vectors right! Of course, you can formulate Newtonian gravity as a non-relativistic classical field theory (it's the paradigmatic example in fact). In the field picture the gravitational force on a test particle is defined by the gravitational field, $\vec{g}(t,\vec{x})$. The force acting on a point mass $m$ at position $\vec{x}$ is given by $\vec{F}=m \vec{g}(t,\vec{x})$.

The field equation uses the fact that the source of the gravitational field is the mass distribution $\rho(t,\vec{x})$, and it's always attractive,
$$\vec{\nabla} \cdot \vec{g}(t,\vec{x})=-4 \pi \gamma \rho(t,\vec{x}).$$
In addition one knows that for a point mass, $M$, at rest at the origin of the coordinate system you get
$$\vec{g}=-\gamma \frac{\vec{x}}{|\vec{x}|^3},$$
and thus the solution for a charge distribution is
$$\vec{g}(t,\vec{x}) = -\gamma \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t,\vec{x}')(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|}, \qquad (*)$$
and thus you also have
$$\vec{\nabla} \times \vec{g}=0.$$
This implies that $\vec{g}$ has a potential,
$$\vec{g}=-\vec{\nabla} \Phi$$
and thus
$$\Delta \Phi(t,\vec{x}) = +4\pi \gamma \rho(t,\vec{x}).$$
From the Green's function of the Laplace operator you immediately get
$$\Phi(t,\vec{x})=-\gamma \int_{\mathbb{R}^3} \frac{\rho(t,\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Taking the gradient of this wrt. $\vec{x}$ leads back to (*), as it should be.

So you are integrating in three dimensions, correct? That is what the $\mathbb{R}^3$ means, right? And will you please explain why you have a gamma there? This isn't the same gamma as the Lorentz factor, is it?

 

[vanhees71]

You are very much on the right track! Instead of the electrostatic field from point charges $q_i$ just consider the gravitational field of point masses $m_i$. Written with vector symbols to make clear, that we deal with vectors, you get the gravitational field (with the point charges $m_i$ at positions $\vec{x}_i$),
$$\vec{g}(\vec{x})=-\gamma \sum_{i} m_i \frac{\vec{x}-\vec{x}_i}{|\vec{x}-\vec{x}_i|^3}.$$
This "microscopic model" is, however quite difficult to use in practice and thus one uses a continuum description for extended bodies consisting of very many mass points, i.e., you describe it by a mass distribution $\rho(\vec{x})$, which describes the mass elements $\mathrm{d} m$ located in a very small volume $\mathrm{d}^3 x$ by $\mathrm{d} m=\mathrm{d}^3 x \rho(\vec{x})$. Then making these volumes infinitesimally small the sum gets into an integral. The result you find in my previous posting (where I assumed that the mass distributions causing gravitation may be moving and thus the mass density might be time-dependent). The $\gamma$ is the gravitational constant. Sorry, I should have used your notation $G$ for it!

There's no relativity taken into account of course. For a relativistic treatment of gravitation you need General Relativity.

 

[Thadriel]

I meant calculate $g$ rather than $F$ - you still need the position vectors - you can't escape them. But vanhees71 has shown an easier way which is to calculate the sum of the potentials, then differentiate to get the force.

So just divide F by m then. In easy math, g is the acceleration, which is $F/m_1$, correct? But why is it $F/m_1$? Is it because $m_1$ is negligible compared to the big mass? I guess if we divided by M, should you not get the same acceleration? I mean you can choose either one as the rest frame. Okay, actually thinking through it, when we talk of F, we're talking of the force on the test mass, right? So the acceleration is symmetrical between m and M but the force is not. Unless I'm mistaken here.

That's a bit of a tangent. $g = GM/r^2$, so I have to find a way to convert that into a spatially extended integral, if I understand it correctly.

 

[Thadriel]

Anyway, I appreciate the responses. These will help me. But I have to spend some time to really think about what the math means.

 

[Ibix]

But I have to spend some time to really think about what the math means.

You imagine cutting a large object up into a lot of small objects, then approximate each small object as a point source of gravity. The total gravitational field is approximately the sum of the gravitational fields of the point sources. Then you consider chopping the object into more smaller pieces, and the approximation is better. Then you take the limit of infinitely many infinitely small objects, and the sum becomes an integral and the error from your approximation goes to zero.

You are explicitly relying on the fact that Newtonian gravity is a linear theory, so if you know the gravitational field of object A and of object B, the combined gravitational field of the two objects is the sum of their separate fields.

 

[vanhees71]

Again: It is utmost important to use VECTORS.