# The Binet equation

Learning objectives

1. This is a point. This is another point. This is another point. This is another point. 2. This is another point. This is another point. This is another point. This is another point. 3. Final point. This is another point. This is another point. This is another point.

Display equations ?

# The Binet equation

The question of whether the earth revolves around the Sun or the Sun revolves around the earth has been greatly debated over centuries until astronomical observations and calculations became sophisticated enough about 300 years ago and the idea that it is the earth that revolves around the Sun became evident. But if you think carefully, you may realise that it is not so easy to even phrase such a question accurately. What does it mean to say the Sun revolves around the earth? Clearly, everyone sees the Sun rise and cross the sky steadily and finally set. Why would it be wrong to say that the Sun revolves around the earth, especially in the presence of such overwhelming visual evidence? But at the same time if you observed the earth from the position of the sun, you’d see the earth revolve (perhpas even rise and set across the Sun’s sky).

The precise problem is that when it is declared that it is the earth that revolves around the Sun, where are you making this observation from? Floating in empty space? There is no special place in empty space from where you can make such a decisive observation. Perhpas you’d choose to observe the earth and sun from another planet. That’d lead to an even greater confusion since the planet itself would be moving.

The point is, if it is left to a visual observation to judge the relative motions of objects, there’s no end to the confusions. It is always better to resort to precise mathematical calculations. In doing so, we will realise that the answer to the above question is that both the earth and the Sun revolve around one another, it is only that the earth, due to its puny mass (compared to the Sun) that its orbit is large, giving us an illusion that it is only the earth that revolves around the Sun.

What we speak of now applies not just to the earth-sun system but to any two bodies that are tugging at one another as they move under the influence of just the forces applied by each of them on the other. More specifically, to avoid complications, we will assume the force the two bodies exert on one another is conservative and central. By this we mean that the force is such that it does the same amount work no matter what path the body takes in moving between two fixed points in space (what we mean by conservative), and that the force acting on each of them is always along the line joining the two bodies (what we mean by central). To stress the inital description of the forces between the two bodies, we must remember that besides the mutual tugging of the two bodies, there is no other external force acting on them. In the case of the earth-sun system, we neglect the forces that other planets and celestial objects apply on the earth and the sun.

But the question remains, where do we make the measurements from? If obtaining the correct description is our goal, we can make the observations from the position of either bodies or any other point in space. The choice of our location of observations will not affect the final description of the motions. But for convenience, we will choose what we call an \textit{inertial frame of reference} to make all measurements. This is not a special position in space for observations but rather a special condition of the observer in that Newton’s laws hold true: any acceleration observed must be associated with a real, measurable force. Such a frame is also sometimes called an unaccelerated frame (but this terminology becomes tricky when we consider gravitational force and so we shall avoid using it).

Along with a frame of reference comes a coordinate system. We call assume spherical polar coordinates for simplicity. In such a coordinate system, the position of any body can be specified by mentioning how far it is directly from the origin ($r$), and at what polar angle ($\theta$) and azimuthal angle ($\phi$). Now our first job is to locate the positions of the two bodies we are concerned with, using position vectors $\vec{r}_1$ and $\vec{r}_2$. Let the mass of the body at position $\vec{r}_1$ be $m_1$ and that of at position $\vec{r}_2$ be $m_2$. As the bodies move under the influence of mutual forces, these position vectors follow them.

The total kinetic energy of the system will simply be the sum of the kinetic energies of each body. $T={1\over 2}m_1 (\dot{\vec{r}}_1)^2+{1\over 2}m_2 (\dot{\vec{r}}_2)^2$

The total potential energy will depend on the separation between the two bodies $r$. The separation is defined as the magntiude of the vector $\vec{r}=\vec{r}_1-\vec{r}_2$. We can denote the potential energy as $U(r)$. The Lagrangian for the system then, defined as the difference of the kinetic and potential energy, will be,

$L=T-U={1\over 2}m_1 (\dot{\vec{r}}_1)^2+{1\over 2}m_2 (\dot{\vec{r}}_2)^2-U(r)$

If the two bodies are viewed as a single system, we can imagine a point lying somewhere between them, that moves as if the entire mass of the system ($m_1+m_2$) were stuffed into it. We call this point the centre of mass of the system. If say from our frame, the two bodies, as they apply forces on one another move in such a manner so that they appear to drift, so that we can say the entire system is drifting in a direction, we can imagine this drift to be described by the centre of mass. Mathematically, we calculate the centre of mass using the relation $\vec{R}={m_1\vec{r}_1+m_2\vec{r}_2\over m_1+m_2}$

How far the centre of mass is from each of the bodies depends upon how large the mass of each of them is compared to the other. If $m_1>>m_2$, then the centre of mass lies very close to $m_1$ and vice versa. As a general rule, the distance of the centre of mass from $m_1$ is ${m_2\over M}r$ and from $m_2$ is ${m_1\over M}r$. Where $M=m_1+m_2$ is the total mass of the system. We did not derive this simple result here assuming you’ll work it out yourself. Using this and the law of vector addition, we can write the relation, $\vec{r}_1=\vec{R}+{m_2\over M}\vec{r}$ and $\vec{r}_2=\vec{R}-{m_1\over M}\vec{r}$.

It turns out that there is an additional simplicity if we express the Lagrangian in terms of the variables $\vec{R}$ and $\vec{r}$ instead of $\vec{r}_1$ and $\vec{r}_2$. Making this replacement in the Lagrangian expression, we have,

\begin{align*} L&={1\over 2}m_1 (\dot{\vec{r}}_1)^2+{1\over 2}m_2 (\dot{\vec{r}}_2)^2-U(r)\\ &={1\over 2}m_1 \left(\dot{\vec{R}}+{m_2\over M}\dot{\vec{r}}\right)^2+{1\over 2}m_2 \left(\dot{\vec{R}}-{m_1\over M}\dot{\vec{r}}\right)^2-U(r)\\ &={1\over 2}m_1\left(\dot{\vec{R}}^2+{m_2^2\over M^2}\dot{\vec{r}}^2+2{m_2\over M}\dot{\vec{R}}\cdot \dot{\vec{r}}\right)+{1\over 2}m_2\left(\dot{\vec{R}}^2+{m_1^2\over M^2}\dot{\vec{r}}^2-2{m_1\over M}\dot{\vec{R}}\cdot \dot{\vec{r}}\right)-U(r)\\ &={1\over 2}m_1\dot{\vec{R}}^2+{1\over 2}{m_1m_2^2\over M^2}\dot{\vec{r}}^2+{m_1m_2\over M}\dot{\vec{R}}\cdot \dot{\vec{r}}+{1\over 2}m_2\dot{\vec{R}}^2+{1\over 2}{m_2m_1^2\over M^2}\dot{\vec{r}}^2-{m_2m_1\over M}\dot{\vec{R}}\cdot \dot{\vec{r}}-U(r)\\ &={1\over 2}(m_1+m_2)\dot{\vec{R}}^2+{1\over 2}{m_1m_2\over M^2}(m_1+m_2)\dot{\vec{r}}^2-U(r)\\ &={1\over 2} M\dot{\vec{R}}^2+{1\over 2}{m_1m_2\over M}\dot{\vec{r}}^2-U(r) \end{align*}

We call the factor ${m_1m_2\over M}=\mu$, the reduced mass of the system. And so we have the Lagrangian,

$L={1\over 2} M\dot{\vec{R}}^2+{1\over 2}\mu\dot{\vec{r}}^2-U(r)$

This can simply be read as the Lagrangian of the system is the kinetic energy of the centre of mass plus that about the centre of mass minus the potential energy.

Recall that we have been making all these measurements from a general inertial frame of reference that’s floating in empty space. Realising that the centre of mass $\vec{R}$ is unaccelerated, since we assume no external forces acting on the system, we could imagine a frame of reference attached to the centre of mass that continues to remain inertial. Let’s shift to this frame of reference that we shall call the centre of mass frame. We do this because it leads to further simplification: in this frame, $\dot{\vec{R}}=0$, i.e., the centre of mass is at rest in this frame.

$L={1\over 2}\mu\dot{\vec{r}}^2-U(r)$

Does this not look like the Lagrangian of a single body of mass $\mu$ moving under the potential $U(r)$. We made all the above shifts in frames of references to arrive at this simplicity.

We expect the angular momentum $\vec{L}=\vec{r}\times \vec{p}$ to be constant since no external forces or torques act on the system. This takes care of the fact that the motion of the two bodies is flat, two-dimensional. This is because the result of the product $\vec{r}\times\vec{p}$ is a vector perpendicular to both $\vec{r}$ and $\vec{p}$, or simply perpendicular to the plane containing $\vec{r}$ and $\vec{p}$. If $\vec{L}$ is constant, it is constant in both magnitude and direction. This means the vector $\vec{r}\times \vec{p}$ has a fixed direction. This ensures the motion is confined to the same plane. We can therefore use just two variables $r$ and $\phi$ to describe the motion of the bodies ($\theta$ is constant, we might as well set it to zero, without loss of generality).

With this, beginning with the expression $\vec{r}=r\hat{r}$, we obtain the expression for the velocity, $\dot{\vec{r}}$, by differentiating,

\begin{align*}

\vec{r}&=r\hat{r}\\

\dot{\vec{r}}&=\dot{r}\hat{r}+r\dot{\hat{r}}\\

&=\dot{r}\hat{r}+r\dot{\phi}\hat{\phi}

\end{align*}

We have made use of the fact that $\dot{\hat{r}}=\dot{\phi}\hat{\phi}$. This can be arrived at if we realise $\hat{r}=\cos\phi \hat{x}+\sin\phi\hat{y}$ and $\hat{\phi}=-\sin\phi\hat{x}+\cos\phi\hat{y}$.

The Lagrangian then may be written as,

$L={1\over 2}\mu(\dot{r}\hat{r}+r\dot{\phi}\hat{\phi})^2-U(r)$

Now, let us do what the Lagrangian was built for, use it in the Euler-Lagrange equations.

${\partial L\over \partial q}={\mathrm{d}\over\mathrm{d}t}\left({\partial L\over\partial \dot{q}}\right)$

As we saw above, we need just two coordinates to describe the motion, i.e., we shall use $q\equiv r$ and $q\equiv\phi$. First, the radial part reads,

\begin{align*}

{\partial L\over \partial r}&={\mathrm{d}\over\mathrm{d}t}\left({\partial L\over\partial \dot{r}}\right)\\

{\partial \over \partial r}\left({1\over 2}\mu(\dot{r}\hat{r}+r\dot{\phi}\hat{\phi})^2-U(r)\right)

&={\mathrm{d}\over\mathrm{d}t}\left[{\partial \over\partial \dot{r}}\left({1\over 2}\mu(\dot{r}\hat{r}+r\dot{\phi}\hat{\phi})^2-U(r)\right)\right]\\

{\partial \over \partial r}\left({1\over 2}\mu(\dot{r}^2\hat{r}^2+r^2\dot{\phi}^2\hat{\phi}^2+2r\dot{r}\dot{\phi}\hat{r}\cdot\hat{\phi})-U(r)\right)

&={\mathrm{d}\over\mathrm{d}t}\left[{\partial \over\partial \dot{r}}\left({1\over 2}\mu(\dot{r}^2\hat{r}^2+r^2\dot{\phi}^2\hat{\phi}^2+2r\dot{r}\dot{\phi}\hat{r}\cdot\hat{\phi})-U(r)\right)\right]\\

\mu r\dot{\phi}^2-{\partial U\over\partial r}&={\mathrm{d}\over\mathrm{d}t}[\mu\dot{r}]\\

\mu r\dot{\phi}^2-{\mathrm{d} U\over\mathrm{d} r}&=\mu\ddot{r}\\

\end{align*}

We made use of the fact that $\hat{r}\cdot\hat{\phi}=0$ and $\hat{\phi}^2=\hat{\phi}\cdot\hat{\phi}=1$ and $\hat{r}^2=\hat{r}\cdot\hat{r}=1$, i.e., $\hat{r}$ and $\hat{\phi}$ make an orthonormal basis, and the fact that since $U(r)$ is a function of $r$ alone, we can write its derivative as an ordinary derivative. We call this equation

$-{\mathrm{d} U\over\mathrm{d} r}=\mu\ddot{r}-\mu r\dot{\phi}^2$

Doing the same for $q\equiv\phi$, we get,

\begin{align*}

{\partial L\over \partial \phi}&={\mathrm{d}\over\mathrm{d}t}\left({\partial L\over\partial \dot{\phi}}\right)\\

{\partial \over \partial \phi}\left({1\over 2}\mu(\dot{r}\hat{r}+r\dot{\phi}\hat{\phi})^2-U(r)\right)

&={\mathrm{d}\over\mathrm{d}t}\left[{\partial \over\partial \dot{\phi}}\left({1\over 2}\mu(\dot{r}\hat{r}+r\dot{\phi}\hat{\phi})^2-U(r)\right)\right]\\

0&={\mathrm{d}\over\mathrm{d}t}[\mu r^2\dot{\phi}]

\end{align*}

Recognise this? $\mu r^2\dot{\phi}$ is just the angular momentum $\vec{L}$. And all this equation does is reinforce the law of conservation of angular momentum that we’ve already discussed. Perhaps a sign that we are heading in the right direction.

The solution to the radial equation is what we’re after in the most ultimate sense. It tells us the trajectory or the orbit of the bodies around each other. But it does so in terms of time, i.e., the solution would be, $r(t)$. Using this you could locate the position of the bodies at any instant of time. But it’d be much more enlightening if we obtained a solution of the form $r(\phi)$, so that we can locate the bodies for different azimuthal angles $\phi$, giving us a direct glimpse at the nature of the orbit.

To do this, let us first rewrite the radial equation, realising that angular momentum $L=\mu r^2\dot{\phi}$ or $\dot{\phi}={L\over \mu r^2}$. Therefore,

\begin{align*}

-{\mathrm{d} U\over\mathrm{d} r}&=\mu\ddot{r}-\mu r\left({L\over \mu r^2}\right)^2\\

F(r)&=\mu\ddot{r}-{L^2\over \mu r^3}

\end{align*}

Or simply,

$\mu\ddot{r}=F(r)+{L^2\over \mu r^3}$

We’ve made use of the fact that in a conservative force field, the force can be expressed as a negative gradient of the potential function, $F=-{\mathrm{d}U\over\mathrm{d}r}$.

If we want the equation to yield solutions of the form $r(\phi)$, we better rewrite it in terms of some function $u(\phi)$ that is related to the radial coordinate by $r={1\over u}$. We then will also have to convert the derivatives into ones with respect to $\phi$, rather than $r$. To do this, consider,

\begin{align*}

\dot{r}={\mathrm{d}r\over\mathrm{d}t}&={\mathrm{d}r\over\mathrm{d}\phi}{\mathrm{d}\phi\over\mathrm{d}t}\,\,\,\,\mathrm{(using \,\,chain\,\, rule)}\\

&={\mathrm{d}r\over\mathrm{d}\phi}\,\,{L\over \mu r^2}

\end{align*}

Using $r={1\over u}$,

\begin{align*}

\dot{r}&={\mathrm{d}\left({1\over u}\right)\over\mathrm{d}\phi}{Lu^2\over \mu}\\

&=\left(-1\over u^2\right){\mathrm{d}u\over\mathrm{d}\phi}{Lu^2\over \mu}\\

&={-L\over \mu}{\mathrm{d}u\over\mathrm{d}\phi}

\end{align*}

Differentiating this again,

\begin{align*}

\ddot{r}={\mathrm{d}\dot{r}\over\mathrm{d}t}&={\mathrm{d}\over\mathrm{d}t}\left({-L\over\mu}{\mathrm{d}u\over\mathrm{d}\phi}\right)\\

&={-L\over \mu}{\mathrm{d}\over\mathrm{d}\phi}\left({\mathrm{d}u\over\mathrm{d}\phi}\right){\mathrm{d}\phi\over\mathrm{d}t}\\

&={-L\over \mu}{\mathrm{d}^2u\over\mathrm{d}\phi^2}{L\over\mu r^2}\\

&={-L^2u^2\over \mu^2}{\mathrm{d}^2u\over\mathrm{d}\phi^2}

\end{align*}

Substituting these results back into the radial equation,

\begin{align*}

\mu\left({-L^2u^2\over \mu^2}{\mathrm{d}^2u\over\mathrm{d}\phi^2}\right)&=F+{L^2u^3\over\mu}\\

{-L^2u^2\over \mu}{\mathrm{d}^2u\over\mathrm{d}\phi^2}&=F+{L^2u^3\over\mu}

\end{align*}

Rearranging,

$F=-{L^2u^2\over\mu}\left(u+{\mathrm{d}^2u\over\mathrm{d}\phi^2}\right)$

This equation has central importance in studying the two-body problem and is known as the \textbf{Binnet equation}, in honour of Jacques Philippe Marie Binet, a French physicist who first derived it during the early 19th C. The solution of this equation essentially tells us what the orbit of any object must be when subjected to a given central force $F$, or even what the force is if the orbit is known.

Keywords

## Don’t we all love to watch something?

We’ll have something for you to watch soon on Physics Capsule TV. If you want us to inform you when we’re live and send you updates, please subscribe to our newsletter.