How to Correctly Differentiate the Acceleration Equation in N-Body Simulations?

[cc94]

Homework Statement

I was reading about the Hermite integration scheme for N-body simulations, as seen here: http://www.artcompsci.org/kali/vol/two_body_problem_2/ch11.html#rdocsect76

This scheme uses jerk, the time rate of change of acceleration. The problem is that I don't know how to correctly evaluate the derivative of the acceleration equation shown there.

Homework Equations

\vec{a} = \frac{GM}{r^2}\vec{\hat{r}}
\vec{j} = GM\left(\frac{\vec{v}}{r^3} - 3 \frac{(\vec{r} \cdot \vec{v})\vec{r}}{r^5}\right)

The Attempt at a Solution

First I rewrote the unit vector in a to get:
\vec{a}= \frac{GM\vec{r} }{r^3}

To differentiate, I applied the quotient rule:
\frac{d\vec{a}}{dt} = GM\frac{r^3\frac{d\vec{r}}{dt} - \vec{r}\frac{dr^3}{dt}}{r^6}

The first term is obviously v/r^3, but the second term is where I'm confused. I can get the right answer by using the chain rule:

\frac{\frac{dr^3}{dt}}{r^6} = \frac{3r^2\frac{d\vec{r}}{dt}}{r^6} = \frac{3\vec{r}\frac{d\vec{r}}{dt}}{r^5} = \frac{3(\vec{r}\cdot\vec{v})}{r^5}

but I'm just arbitrarily changing the scalar r into the vector r. What's the right way to do this step?

 

[andrewkirk]

The answer is correct, but the steps are all wrong: the first because it equates a scalar to a vector, the second because it contains a meaningless multiplication of two vectors, and the third because it then somehow gets back to a scalar quantity.
Instead, keep the amounts as scalars all the way along. Start with:
$$\frac{d(r^3)}{dt}=3r^2\frac{dr}{dt}$$
Then draw a diagram and try to express $\frac{dr}{dt}$ in terms of $\vec v$ and $\vec r$. You may find it convenient to use $\hat r$, the unit vector in the direction of $\vec r$, along the way.

 

[cc94]

Ok so, if we draw \vec{r} = r\hat{r} on the plane, for instance, then the change in length will depend on which direction v is pointing. In the limit of a small time step dt, the change of length r is equal to v*dt*cos(θ), where θ is the angle between v and r. I.e.
\frac{dr}{dt} = vcos\theta

So we pull one of the 'r's from in front to get r*v*cos(θ), which is indeed equal to the dot product r·v. The remaining r cancels one in the bottom to get the r^5 term.

 

[andrewkirk]

Yes, that's it!