# Angular Momentum of Two Non-interacting Particles

I'm reading a book (An Introduction to Mechanics by Kleppner) where they calculate the angular momentum $l$ of a system of two non-interacting particles, but I don't understant what are they doing.

Consider two non-interacting particles with $m_1$ and $m_2$ moving toward each other with constant velocities $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$. Their paths are offset by distance $b$, as shown in the sketch.

In general the energy of a system of two particles, relative to the center of mass, can be written

$$E=\frac{1}{2}\mu v^{2}+U(r), \qquad (1)$$

being $\mu$ the reduced mass. Or, using $v^2=\dot r^2+r^2 \dot \theta ^2$ and $l=\mu r^{2}\dot{\theta}$,

$$E=\frac{1}{2}\mu\dot{r}^{2}+\frac{l^{2}}{2\mu r^{2}}+U(r), \qquad(2)$$

So the book calculates the angular momentum $l$ of the system using (1) and (2) and the fact that for a system of two non-interacting particles $U(r)=0$.

The book says:

The relative velocity is $$\mathbf{v}_{0}=\dot{\mathbf{r}}:=\dot{\mathbf{r}}_{1}-\dot{\mathbf{r}}_{2}=\mathbf{v}_{2}-\mathbf{v}_{1}$$ with $\mathbf{v}_{0}$ is constant since $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are constant (which I think it's right).

The energy of the system relative to the center of mass is (see equation 1)

$$E=\frac{1}{2}\mu v_{0}^{2}$$ or (from equation 2)

$$E=\frac{1}{2}\mu\dot{r}^{2}+\frac{l^{2}}{2\mu r^{2}}$$

Now here comes the argument I really don't understand:

Argument a): When $m_1$ and $m_2$ pass each other, $r=b$ and $\dot r = 0$ (But the book said earlier that $\mathbf{v}_{0}=\dot{\mathbf{r}}$, a constant vector in time!). Hence

$$\frac{l^{2}}{2\mu b^{2}}=\frac{1}{2}\mu v_{0}^{2} \qquad (3)$$

I'm confused because I could apply the same condition in an arbitrary point, say $r=2b$, then $\dot r=0$, and (3) wouldn't be valid.

So, my question is: Is the argument a) wrong? If not, please help me to clarify the ideas behind this.

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You are mixing $\dot{\vec r}$ and $\dot r$. Begin writing the coordinates of $\vec r$, then get $\dot{\vec r}$ and $\dot r$, and you will see the difference between the two. –  Trimok Oct 28 '13 at 19:33

Elaborating on Trimok's comment, you are mixing up $\dot{\mathbf r}$ and $\dot r$. $\dot{\mathbf r}$ is the rate of change of the separation vector, while $\dot r$ is the rate of change of the separation distance.
We can quickly see they are different. By definition, we know that $\mathbf r = r \,\hat{\mathbf r}$. Take the time derivative of this: $$\dot{\mathbf r} = \dot r\, \hat{\mathbf r} + r \, \dot{\hat{\mathbf r}}$$
In this case, the direction between the two is changing, i.e. $\dot{\hat{\mathbf r}}\neq 0$, so both terms on the right-hand side of this equation contribute to the value of $\dot{\mathbf r}$. In particular, $\dot{\mathbf r}$ can remain constant while $\dot r$ changes, so long as the above equation is obeyed.
OK. I understand that. But why is $\dot r =0$ when $r=b$? [Argument a)] –  Anuar Oct 28 '13 at 20:51
As the particles get closer to each other, $\dot r<0$, and after they pass and get further away, $\dot r>0$. When $r$ is minimized (which is the same as saying its first derivative $\dot r$ is zero) the particles are at their closest distance. Looking at the diagram above, the closest they can get given those constant velocities is $r=b$. –  Flavin Oct 29 '13 at 0:58
You are right. I arrived to the same conclusion while traveling in the subway xD As you say $\dot r <0$ while they are getting closer and $\dot r >0$ after passing the closest distance $b$. So there is a change of sign in $\dot r$, i.e. it must be a zero of $\dot r$ somewhere. This point is precisely when $r=b$. Thanks! –  Anuar Oct 29 '13 at 4:02