11.1 Binary Dynamics

The next day, our three friends get together again. Alice looks a bit sleepy, obviously has stayed up late the previous night.

Carol:

Fortunately, we have a blackboard in this room.

Bob:

And more important, we have Alice in the room!

Alice:

But only a half-awake Alice. Yesterday evening it took me longer than I thought to figure everything out. What I found looks easy, it always does in the end, but I didn't realize how much I had forgotten about the classical dynamics course that I had followed in my freshman year.

Bob:

Some day I'd like to know how you derived whatever you derived, but for now, I'm really eager to fish for binaries, so let's not worry too much about derivations. Perhaps you can just show us what you found.

Carol:

I second that. Let's see whether it works first, and if it turns out to be useful, I'd love to learn more about the details.

Alice:

Fine with me, and just as well, since I don't think I'm quite together enough to give a complete pedagogical derivation.

Alice walks to the blackboard, and writes down the equations for the energy $E$ and angular momentum $L$ for a two-body system. Here the masses of the two bodies are $M_1$ and $M_2$. The relative position vector, pointing from $M_1$ to $M_2$, is given by ${\bf r}$, and its time derivative is the relative velocity vector ${\bf v}$. As before, $r$ is the absolute value of ${\bf r}$, a scalar, equal to the length of the vector ${\bf r}$, and similar $v$ is the length of ${\bf v}$. The energy is given by the sum of the negative potential energy and the positive kinetic energy. The angular momentum points along the outer product, also called the cross product, of the relative position and velocity vectors. This implies that it points in a direction perpendicular to the orbit of the two bodies.

$$E = -\, G\:\frac{M_1 M_2}{r} \;+\, {\ifmmode{{1 \over 2}}\else{${1 \over 2}$}\fi}\,\frac{M_1 M_2}{M_1 + M_2}\:v^2$$ (11.1)

$$L = \frac{M_1 M_2}{M_1 + M_2} \; {\bf r}\times {\bf v}$$ (11.2)

Note that we could have defined ${\bf r}$ equally well as pointing from $M_2$ to $M_1$, as long as ${\bf v}$ would still be the time derivative of ${\bf r}$. In the definition of the energy, the lengths of $r$ and $v$ would not change, and in the definition of the angular momentum, the two minus signs that would thus be introduced at the right would cancel each other.

These equations simplify considerably when we introducing the symbol $\mu$ for the reduced mass, defined as follows:

$$\mu = \frac{M_1 M_2}{M_1 + M_2}$$ (11.3)

we can now write the energy per unit of reduced mass as

$$\tilde E \equiv \frac{E}{\mu} = -\, G\:\frac{M_1 + M_2}{r} \;+\, {\ifmmode{{1 \over 2}}\else{${1 \over 2}$}\fi}\,\:v^2$$ (11.4)

and similarly the angular momentum per unit of reduced mass as

$$\tilde L \equiv \frac{L}{\mu} = {\bf r}\times {\bf v}$$ (11.5)

Alice:

The shape of a binary orbit is given by the values of the semi-major axis $a$ and the eccentricity $e$. We can find those values in two steps. First we invert the expression that gives $\tilde E$ in terms of $a$, to obtain an expression for $a$:

$$\tilde E = -\, G\:\frac{M_1 + M_2}{2 a}$$ (11.6)

$$a = -\, G\:\frac{M_1 + M_2}{2 \tilde E}$$ (11.7)

Alice:

And similarly, we invert the expression that gives $\tilde L$ in terms of $a$ and $e$, to obtain an expression for $e$, or to keep it simple, for $e^2$ (we can always take the square root later):

$$\tilde L^2 = G\: (M_1 + M_2)a(1-e^2)$$ (11.8)

$$e^2 = 1 - \frac{\tilde L^2}{G(M_1 + M_2)a}$$ (11.9)

Alice:

The interpretation is as follows: as the name suggests, $a$ is half the length of the longest axis of the ellipse that describes the relative orbit of the two bodies; in this way $2a$ is size of the orbit. The eccentricity $e$ indicates the relative displacement of the focus of the orbit from the center of the orbit. The closest approach between the two bodies occurs at a distance of $ae$, also called the pericenter distance, while the largest separation occurs at a distance of $a(1-e)$, the apocenter distance.