5
$\begingroup$

I’ve been studying Tremaine's Dynamics of Planetary Systems and have hit a roadblock.

In Chapter 4, titled 'The $N$-body Problem,' there is a derivation of the $N$-body equations of motion, which can be expressed as follows:

$$ \ddot{\vec{r}}_{j} = \sum_{k = 0 \\ k \neq j}^{N-1} \frac{Gm_{k} \left(\vec{r}_{k} - \vec{r}_{j} \right)}{|\vec{r}_{j} - \vec{r}_{k}|^{3}} \, , j = 0, ..., N-1. $$

To rewrite these equations of motion in the astrocentric frame, we introduce $\vec{r}^{\, \star}_{m} = \vec{r}_{m} - \vec{r}_{0}$. Substituting this into the equation above allows us to express it as follows: $$ \ddot{\vec{r}}^{\, \star}_{j} = \sum_{k = 0 \\ k \neq j}^{N-1} \frac{Gm_{k} \left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \sum_{k = 1}^{N-1} \frac{Gm_{k} \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \\ \iff \ddot{\vec{r}}^{\, \star}_{j} = - \frac{Gm_{0} \vec{r}^{\, \star}_{j}}{|\vec{r}^{\, \star}_{j}|^{3}} + \sum_{k = 1 \\ k \neq j}^{N-1} Gm_{k}\left[\frac{\left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \frac{ \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \right] \, , j = 0, ..., N-1. $$

However, Equation (4.5) in Tremaine's Dynamics of Planetary Systems includes an extra factor of $m_{j}$ in the first term. Why is that?

$$ \ddot{\vec{r}}^{\, \star}_{j} = - \frac{G\left(m_{0} + m_{j}\right)\vec{r}^{\, \star}_{j}}{|\vec{r}^{\, \star}_{j}|^{3}} + \sum_{k = 1 \\ k \neq j}^{N-1} Gm_{k}\left[\frac{\left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \frac{ \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \right] \, , j = 0, ..., N-1. $$

$\endgroup$
1
  • $\begingroup$ Note that the sum in equation (4.5) vanishes in the case N=1, resulting in the central body-centric two body equation $$\ddot {\vec r}^{\,\star}_1 = -\frac{G(m_0+m_1)\,\vec r^{\,\star}_1}{|\vec r^{\,\star}_1|^3}$$This is of course the correct equation for the astrocentric two body problem. The $G{m_0+m_j)$ term does indeed belong in Tremaine's equation (4.5). See my answer for details. $\endgroup$ Commented Oct 13 at 13:39

1 Answer 1

2
$\begingroup$

I'll be quoting two of the equations in the question and addressing them in turn.

$$ \ddot{\vec{r}}^{\, \star}_{j} = \sum_{k = 0 \\ k \neq j}^{N-1} \frac{Gm_{k} \left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \sum_{k = 1}^{N-1} \frac{Gm_{k} \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \, , j = 0, ..., N-1\tag{1} $$

That is correct. Note that $\ddot{\vec r}^{\,\star}_0$ is tautologically zero from equation (1), so we might as well exclude $j=0$ as a special case (which is what I will do later on).

$$\iff \ddot{\vec{r}}^{\, \star}_{j} = - \frac{Gm_{0} \vec{r}^{\, \star}_{j}}{|\vec{r}^{\, \star}_{j}|^{3}} + \sum_{k = 1 \\ k \neq j}^{N-1} Gm_{k}\left[\frac{\left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \frac{ \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \right] \, , j = 0, ..., N-1 \tag{2} $$

That is not correct, for two reasons. You have included $j=0$ (your equation (2) results in $\frac00$ for $j=0$), and you have combined sums incorrectly.

Going back to equation (1), note that the second sum in this equation includes $k=j$ for non-zero $j$. We need to single out the $k=j$ element of the second sum, resulting in

$$ \ddot{\vec{r}}^{\, \star}_{j} = - \frac{Gm_{j} \vec{r}^{\, \star}_{j}}{|\vec{r}^{\, \star}_{j}|^{3}} + \sum_{k = 0 \\ k \neq j}^{N-1} \frac{Gm_{k} \left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \sum_{k = 1 \\ k \neq j}^{N-1} \frac{Gm_{k} \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \, , j = 1, ..., N-1 \tag{3} $$

In order to combine the two sums, the $k=0$ element in the first sum in equation (3) also needs to be treated as a special term, resulting in

$$ \ddot{\vec{r}}^{\, \star}_{j} = - \frac{G(m_{0}+m_j)\, \vec{r}^{\, \star}_{j}}{|\vec{r}^{\, \star}_{j}|^{3}} + \sum_{k = 1 \\ k \neq j}^{N-1} \frac{Gm_{k} \left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \sum_{k = 1 \\ k \neq j}^{N-1} \frac{Gm_{k} \vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}} \, , j = 1, ..., N-1\tag{4} $$ Now the two sums can be combined, resulting in $$ \ddot{\vec{r}}^{\, \star}_{j} = - \frac{G(m_{0}+m_j)\, \vec{r}^{\, \star}_{j}}{|\vec{r}^{\, \star}_{j}|^{3}} + \sum_{k = 1 \\ k \neq j}^{N-1} Gm_{k} \left(\frac{\left(\vec{r}^{\, \star}_{k} - \vec{r}^{\, \star}_{j} \right)}{|\vec{r}^{\, \star}_{j} - \vec{r}^{\, \star}_{k}|^{3}} - \frac{\vec{r}^{\, \star}_{k}}{|\vec{r}^{\, \star}_{k}|^{3}}\right) \, , j = 1, ..., N-1\tag{5} $$ This is equation (4.5) in Tremaine's Dynamics of Planetary Systems. Note that equations (3) to (5) exclude $j=0$. That's okay because we already know that $\ddot{\vec r}^{\,\star}_0$ is tautologically zero from equation (1).

$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$
    – RKerr
    Commented Oct 17 at 10:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.