## The gravitational energy of an isotropic sphere.

In this post we find an expression for the gravitational energy of an isotropic sphere. An isotropic sphere is a sphere such that the mass density can be written as $\sigma(r)$.

The gravitational energy of two point masses $m_p$ and $m_q$ is given by

$\displaystyle U_{pq} = - \frac{G m_p m_q}{|\vec{r}_p-\vec{r}_q|}$

The gravitational potential of a point mass $m_p$ is given by

$\displaystyle \Phi_p(\vec{r}) = -\frac{Gm_p}{|\vec{r}-\vec{r}_p|}$

Thus we can write

$\displaystyle U_{pq} = m_p \left( - \frac{G m_q}{|\vec{r}_p-\vec{r}_q|}\right) = m_p \Phi(\vec{r}_p)$.

When we use a mass density $\sigma(\vec{r})$ we can write

$\displaystyle U = -\frac{1}{2} G \int dm' \int dm \frac{1}{|\vec{r}_{dm'} - \vec{r}_{dm}|}$

where $dm = \sigma dx dy dz$. Let $\ell = |\vec{r}_{dm'} - \vec{r}_{dm}|$ then it is clear that

$\ell^2 = \left(r_{dm}\right)^2 + \left(r_{dm'}\right)^2 - 2 r_{dm} r_{dm'} \cos(\theta)$

Therefore

$\ell d\ell = r_{dm} r_{dm'} \sin(\theta)$

where $r=r_{dm}$ and $r'=r_{dm'}$ Now

$\displaystyle \int dm \frac{1}{\ell} = \int_0^R dr \int_0^\pi d\theta \int_0^{2\pi} d\phi \frac{\sigma(r) r^2 \sin(\theta)}{\ell}$

So

$\displaystyle \int dm \frac{1}{\ell} = 2 \pi \int_0^R dr \int_{|r-r'|}^{|r+r'|} d\ell \frac{r \sigma(r)}{r'}$

Thus

$\displaystyle \int dm \frac{1}{\ell} = 2 \pi \int_0^R dr \frac{r \sigma(r)}{r'} \left(|r+r'| - |r-r'|\right)$

Then it is clear that

$\displaystyle \int dm \frac{1}{\ell} = 2 \pi \left\{ \int_0^{r'} dr \frac{r \sigma(r)}{r'} \left(|r+r'| - |r-r'|\right) + \int_{r'}^R dr \frac{r \sigma(r)}{r'} \left(|r+r'| - |r-r'|\right) \right\}$

So

$\displaystyle \int dm \frac{1}{\ell} = 4 \pi \left\{ \frac{1}{r'} \int_0^{r'} dr r^2 \sigma(r) + \int_{r'}^R dr r \sigma(r) \right\}$

For the total energy we have

$\displaystyle U = -8 \pi^2 G \int_0^R dr' r'^2 \sigma(r') \left\{ \frac{1}{r'} \int_0^{r'} dr r^2 \sigma(r) + \int_{r'}^R dr r \sigma(r) \right\}$

In case $\sigma$ is constant we obtain

$\displaystyle U = -8 \pi^2 G \sigma^2 \int_0^R dr' r'^2 \left\{ \frac{1}{r'} \int_0^{r'} dr r^2 + \int_{r'}^R dr r \right\}$

Then

$\displaystyle U = -8 \pi^2 G \sigma^2 \int_0^R dr' r'^2 \left\{ \frac{1}{2} R^2 - \frac{1}{6} r'^2 \right\} = -8 \pi^2 G \sigma^2 \left\{ \frac{1}{6} - \frac{1}{30}\right\} R^4 = -\frac{16}{15} \pi^2 G \sigma^2 R^5$

as expected and well known.

Let us define $\displaystyle M(r) = \int dr r^2 \sigma(r)$ and $\displaystyle Q(r) = \int dr r \sigma(r)$. Then we can write

$\displaystyle U = -8 \pi^2 G \left\{ \int_0^R dr' r' \sigma(r') M(r') - \int_0^R dr' r' \sigma(r') M(0) + \int_0^R dr' r'^2 \sigma(r') Q(R) - \int_0^R dr' r'^2 \sigma(r')Q(r') \right\}$

So

$\displaystyle U = -8 \pi^2 G \left\{ 2\int_0^R dr' r' \sigma(r') M(r') - Q(R) M(0) - Q(0) M(0) + M(R) Q(R) - M(0) Q(R) - M(R)Q(R) + M(0)Q(0) \right\}$

Thus

$\displaystyle U = -16 \pi^2 G \left\{ \int_0^R dr' r' \sigma(r') M(r') - Q(R) M(0) \right\}$

But $M(0)=0$ so

$\displaystyle U = -16 \pi^2 G \int_0^R dr' r' \sigma(r') M(r')$

and using the definition of $M(r)$ we can write

$\displaystyle U = -16 \pi^2 G \int_0^R dr' r' \sigma(r') \int_0^{r'} dr r^2 \sigma(r)$

Again the case of a constant $\sigma$ then we get

$\displaystyle U = -16 \pi^2 G \sigma^2 \int_0^R dr' r' \int_0^{r'} dr r^2 = -16 \pi^2 G \sigma^2 \int_0^R dr' \frac{1}{3} r'^4 = -\frac{16}{15}\pi^2 G \sigma^2 R^5$

The gravitational potential of an isotropic sphere is given by

$\displaystyle U = -16 \pi^2 G \int_0^R dr' r' \sigma(r') \int_0^{r'} dr r^2 \sigma(r)$

where $\sigma(r)$ is the radial mass density.