<p>This is <a href="https://en.wikipedia.org/wiki/Poisson%27s_equation" rel="nofollow noreferrer">Poisson's equation</a> and it derives from <a href="https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity" rel="nofollow noreferrer">Gauss's law</a>:</p>
<blockquote>
<p>Since the gravitational field has zero curl (equivalently, gravity is a conservative force) as mentioned above, it can be written as the gradient of a scalar potential, called the gravitational potential:</p>
<p><span class="math-container">$$g =-\nabla \phi$$</span></p>
<p>Then the differential form of Gauss's law for gravity becomes Poisson's equation:</p>
<p><span class="math-container">$$\nabla ^{2}\phi =4\pi G\rho$$</span></p>
<p>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.</p>
</blockquote>

CC BY-SA 4.0
2022-09-25T03:15:59.743Z
2022-09-26T13:53:13.640Z
2022-09-26T13:53:13.640Z
DialFrost
<p>Poisson's law is ultimately based on Newton's hypothesis for a law of universal gravitation. Mathematically, the acceleration <span class="math-container">$\vec{g}(\vec{r})$</span> caused by some distribution of mass, characterised by a density <span class="math-container">$\rho(\vec{r}')$</span> is written
<span class="math-container">$$ \vec{g}(\vec{r}) = -G\int \rho \frac{\left(\vec{r}'-\vec{r}\right)}{|\vec{r}'-\vec{r}|^3}\ dr'^{3}, $$</span>
where <span class="math-container">$\vec{r}'$</span> is a position in the source density distribution and the integral is over the entire volume of the source.</p>
<p>The above equation is a hypothesis - it cannot be proved (and turns out to be wrong in detail, which is why is must be replaced with Einstein's field equations).</p>
<p>Taking the divergence of both sides of this equation, and using the <a href="https://peppyhare.github.io/r/notes/griffiths/ch1-5" rel="nofollow noreferrer">vector calculus identity</a> that
<span class="math-container">$$\nabla \cdot \frac{\vec{r}'-\vec{r}}{|\vec{r}'-\vec{r}|^3} = 4\pi \delta(\vec{r}'-\vec{r}),$$</span>
where <span class="math-container">$\delta$</span> is the 3D <a href="https://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow noreferrer">Dirac delta function</a>, we have
<span class="math-container">$$\nabla \cdot \vec{g}(\vec{r}) = -\int 4\pi G \rho \delta(\vec{r}'-\vec{r})\ dr'^3 = -4\pi G \rho$$</span></p>
<p>If we then define a scalar potential such that <span class="math-container">$\vec{g}=-\nabla \Phi$</span> and insert this into the equation above, we get
<span class="math-container">$$\nabla \cdot \nabla \Phi = \nabla^2\Phi = 4\pi G \rho\ .$$</span></p>

CC BY-SA 4.0
2022-09-27T06:30:30.877Z
2022-09-28T08:52:21.680Z
2022-09-28T08:52:21.680Z
ProfRob
<p>How can it be shown that <span class="math-container">$$\nabla^2\phi = 4\pi G \rho(\vec{r})$$</span> ?</p>

CC BY-SA 4.0
2022-09-25T03:17:54.380Z
Maybe you can explain what the $r$ in your equation means? It doesnt appear in the possion eq
DialFrost
CC BY-SA 4.0
2022-09-25T08:36:57.147Z
I've been waiting 2.5 years for this to happen again! :-) [How to best format the question mark that immediately follows an equation in a question sentence?](https://math.meta.stackexchange.com/q/31283/284619)
uhoh
<galactic-dynamics><potential-energy>
How is the potential-density pair relation derived?
realanswers
# How is the potential-density pair relation derived?

(Asked by: username_0 on 2022-09-25)

How can it be shown that $$\nabla^2\phi = 4\pi G \rho(\vec{r})$$ ?

**Tags:** galactic-dynamics, potential-energy

> **Comments:**
> **username_1**: Maybe you can explain what the $r$ in your equation means? It doesnt appear in the possion eq  
> **username_2**: I've been waiting 2.5 years for this to happen again! :-) [How to best format the question mark that immediately follows an equation in a question sentence?](https://math.meta.stackexchange.com/q/31283/284619)  

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## 2 answers

### Answer by username_1 on 2022-09-25. Score: 4

This is [Poisson's equation](https://en.wikipedia.org/wiki/Poisson%27s_equation) and it derives from [Gauss's law](https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity):

> Since the gravitational field has zero curl (equivalently, gravity is a conservative force) as mentioned above, it can be written as the gradient of a scalar potential, called the gravitational potential:
>
> $$g =-\nabla \phi$$
>
> Then the differential form of Gauss's law for gravity becomes Poisson's equation:
>
> $$\nabla ^{2}\phi =4\pi G\rho$$
>
> 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.


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### Answer by username_3 on 2022-09-27. Score: 3

Poisson's law is ultimately based on Newton's hypothesis for a law of universal gravitation. Mathematically, the acceleration $\vec{g}(\vec{r})$ caused by some distribution of mass, characterised by a density $\rho(\vec{r}')$ is written
$$ \vec{g}(\vec{r}) = -G\int \rho \frac{\left(\vec{r}'-\vec{r}\right)}{|\vec{r}'-\vec{r}|^3}\ dr'^{3}, $$
where $\vec{r}'$ is a position in the source density distribution and the integral is over the entire volume of the source.

The above equation is a hypothesis - it cannot be proved (and turns out to be wrong in detail, which is why is must be replaced with Einstein's field equations).

Taking the divergence of both sides of this equation, and using the [vector calculus identity](https://peppyhare.github.io/r/notes/griffiths/ch1-5) that
$$\nabla \cdot \frac{\vec{r}'-\vec{r}}{|\vec{r}'-\vec{r}|^3} = 4\pi \delta(\vec{r}'-\vec{r}),$$
where $\delta$ is the 3D [Dirac delta function](https://en.wikipedia.org/wiki/Dirac_delta_function), we have
$$\nabla \cdot \vec{g}(\vec{r}) = -\int 4\pi G \rho \delta(\vec{r}'-\vec{r})\ dr'^3 = -4\pi G \rho$$

If we then define a scalar potential such that $\vec{g}=-\nabla \Phi$ and insert this into the equation above, we get
$$\nabla \cdot \nabla \Phi = \nabla^2\Phi = 4\pi G \rho\ .$$


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