Solving a two-body problem using relative motion and reduced mass

Question

I'm having a hard time trying to understand fully this topic and how reduced mass and relative velocity should be used. Let's say we have some sort of mechanical problem regarding the interaction (or just the motion) of two bodies (I'm trying to be the most general possible). For instance we could have a problem of gravitation, or a problem where the two masses are held together by a massless rod while moving in absence of external forces.

So, for example, if I were to calculate the angular momentum ($L$) of my system, I would do that by summing the angular momentum of the center of mass ($CM$) and the angular momentum in the $CM$ frame. The former is easy to do, the problem is the latter. I know that while putting myself in the $CM$ frame the only thing that gives me an angular momentum is the rotational motion of the two masses. Then I know that $\vec{L}=\mu \vec{v}\times \vec{r}= \mu \vec{r} \times (\dot{r} \hat{e}_r + r\dot{\theta}\hat{e}_{\theta})$, where $\mu$ is the reduced mass and $v$ the relatice velocity. And here it is the problem. First of all, WHY does it work? I mean, not only for the angular momentum, but also for calculating the kinetic energy, the effective potential, also for calculating (if I'm not wrong) the total momentum in the frame of the $CM$. So my question is : why does it work? How can your problem be reduced to another problem where we have just one body with the mass equals to $\mu$ and $v$ speed? Also, in the previous formula, I have used a $\theta$, but I really cannot imagine what his meaning is. When considering two different bodies it's all easier to grasp, but in this situation what angle does that $\theta$ refer to? What's his meaning? Thanks for the help.

Answer 1

1. According to Newton's first principle there exists at least one reference frame, called inertial, in which, by definition, if the sum of all forces acting on a mass-point is zero, the point moves uniformly along a straight line . Let us choose such a coordinate frame. Any other coordinate frame that moves uniformly along a straight line with respect to the given inertial reference frame is also an inertial reference frame.

2. According to Newton's second postulate, in an inertial reference frame the sum of all forces acting on a mass point equals the acceleration of the point times its mass.

3. Newton's third principle states the if a mass point $A$ exerts force $F$ on a mass point $B$ then, in return, $B$ exerts force $-F$ on $A$ (recall that $F$ is a vector!).

We are given two mass points $A_1$ and $A_2$ with masses $m_1$ and $m_2$ respectively. They generate a common gravitational field and we want to study how the two points move with time. Let us fix an inertial frame of reference $Oe_1e_2e_3$, which always exists according to Newton's first principle. Here, $O$ is the origin of the coordinate system, while $e_1, e_2$ and $e_3$ are three unit vectors, perpendicular to each other. Let $r_1=r_1(t)=x_1(t)e_1+y_1(t)e_2+z_1(t)e_3$ be the position vector of point $A_1$ at time $t$. Analogously, let $r_2=r_2(t)=x_2(t)e_1+y_2(t)e_2+z_2(t)e_3$ be the position vector of $A_2$ at time $t$. According to Newton's law of gravity, mass point $A_2$ attracts mass point $A_1$ with force $$F_1(r_1,r_2)= \left(\frac{m_1m_2G}{|r_2-r_1|^2}\right)\frac{r_2-r_1}{|r_2-r_1|} =\frac{m_1m_2G}{|r_2-r_1|^3}(r_2-r_1)$$ where $|r_2-r_1|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ is the distance between $A_1$ and $A_2$.

According to Newton's third principle, the force with which $A_1$ attracts $A_2$ is $$F_2(r_1,r_2)=-F_1(r_1,r_2)=\frac{m_1m_2G}{|r_2-r_1|^3}(r_1-r_2)$$ By definition, for $j=1,2$ the velocity of the point $A_j$ is $r^{\prime}_j = r'_j(t) = \frac{dr_j}{dt}$ and its acceleration is $r''_j = r''_j(t) = \frac{d^2r_j}{dt^2}$. Consequently, Newton's second law produces the following system of ordinary differential equations \begin{align*} m_1 r''_1 &=F_1(r_1,r_2)\\ m_2 r''_2 &=F_2(r_1,r_2) = -F_1(r_1,r_2) \end{align*} or more explicitly \begin{align*} m_1 r''_1 &= \frac{m_1m_2G}{|r_2-r_1|^3}(r_2-r_1)\\ m_2 r''_2 &= \frac{m_1m_2G}{|r_2-r_1|^3}(r_1-r_2) \end{align*} Factor out the masses on both sides of the equations and obtain \begin{align*} r''_1 &= \frac{m_2G}{|r_2-r_1|^3}(r_2-r_1)\\ r''_2 &= \frac{m_1G}{|r_2-r_1|^3}(r_1-r_2) \end{align*} We end up with a system of two ordinary vector differential equations or equivalently, a system of six scalar differential equations. Here is where the magic happens and the system reduces from six to three scalar equations describing the motion of one point only.

Let us form the vector $$R=\frac{m_1r_1+m_2r_2}{m_1+m_2}.$$ This vector $R=R(t)$ is the position of the common center of gravity of the two mass points $A_1$ and $A_2$ at time $t$. After differentiating $R$ twice with respect to time $t$, we get \begin{align*} R^{''} &=\frac{1}{m_1+m_2}\Bigl(m_1r''_1+ m_2r''_2\Bigr)=\frac{1}{m_1+m_2}\Bigl(F_1+F_2\Bigr)\\ &=\frac{1}{m_1+m_2}\Big(F_1-F_1\Big)=0. \end{align*} which means that the velocity of the common center of mass of points $A_1$ and $A_2$ is constant, i.e. $R^{'}(t) \equiv V_R$ which means that $R = R(t) = R_0 + tV_R$ moves uniformly along a straight line. Then if we wish, according to Newton's first principle, we can move the coordinate system to $R(t)$ and still obtain an inertial reference frame. But let us keep working with an arbitrary reference frame.

Now, let us set $r = r_2-r_1$. Then \begin{align*} r'' &= r_2''-r''_1 \\ &=-\frac{m_1G}{|r_2-r_1|^3}(r_2-r_1)-\frac{m_2G}{|r_2-r_1|^3}(r_2-r_1)\\ &=-\frac{(m_1+m_2)G}{|r_2-r_1|^3}(r_2-r_1)\\ &=-\left(\frac{(m_1+m_2)G}{|r|^3}\right)\,r \end{align*} Thus, we are left with a system of three scalar differential equations instead of six. $$\frac{d^2r}{dt^2}=-\left(\frac{(m_1+m_2)G}{|r|^3}\right)\,r$$ which can be written in terms of the reduced mass as $$\left(\frac{m_1m_2}{m_1+m_2}\right)\frac{d^2r}{dt^2} = -\left(\frac{m_1m_2G}{|r|^3}\right)\,r$$ The solution $r=r(t)$ to this vector differential equation gives the solution to the original problem \begin{align*} r_1(t) &=R(t) -\left(\frac{m_2}{m_1+m_2}\right)r(t) =R_0+tV_R -\left(\frac{m_2}{m_1+m_2}\right)r(t)\\ r_2(t)&=R(t)+\left(\frac{m_1}{m_1+m_2}\right)r(t) =R_0+tV_R+\left(\frac{m_1}{m_1+m_2}\right)r(t) \end{align*}

Now, let us move our reference frame to the common center of gravity of the two points $A_1$ and $A_2$. This means that $R_0 = 0$ and $V_R = 0$ and thus $R(t) \equiv 0$. Then the motion of the two mass points becomes \begin{align*} r_1(t) &=-\left(\frac{m_2}{m_1+m_2}\right)r(t)\\ r_2(t)&=\left(\frac{m_1}{m_1+m_2}\right)r(t) \end{align*}

If we re-express $r(t)$ in terms of $r_1(t)$ and $r_2(t)$ we obtain $$r(t) = - \left(\frac{m_1 + m_2}{m_2}\right)r_1(t) = \left(\frac{m_1+m_2}{m_1}\right)r_2(t)$$ and thus, the system of differential equations in the reference frame with respect to the common center of gravity of the two mass points becomes (by plugging each of these two expression for $r$ into the system for $r$) \begin{align*} r''_1&= - \left(\frac{(m_2)^3G}{(m_1+m_2)^2|r_1|^3}\right)r_1\\ r''_2&=-\left(\frac{(m_1)^3G}{(m_1+m_2)^2|r_2|^3}\right)r_2 \end{align*} which as you can see is already a decoupled system of two vector equations. If the mass $m_2$ is much bigger than $m_1$, so that $m_1$ can be ignored, then $(m_1 + m_2) \sim m_2$, and thus

\begin{align*} m_1 r''_1&= - \left(\frac{m_1m_2G}{|r_1|^3}\right)r_1\\ \end{align*} is the motion of a mass point in a gravitational field created by a much bigger mass. This is what Galileo observed in his experiments -- the motion of a body in Earth's gravity does not depend on its mass. Well, technically, Galileo was wrong, as we can see from our equations, but for practical purposes he was right. He didn't have precise enough measurements to see the difference between $(m_1+m_2)$ and $m_1$.

How do we derive \begin{align*} r_1(t) &=R(t) - \left(\frac{m_2}{m_1+m_2}\right)r(t) =R_0+tV_R+\left(\frac{m_2}{m_1+m_2}\right)r(t)\\ r_2(t)&=R(t) + \left(\frac{m_1}{m_1+m_2}\right)r(t) =R_0+tV_R-\left(\frac{m_1}{m_1+m_2}\right)r(t)? \end{align*} Think of vector $R$ as a three dimensional arrow with origin $O$ -- which is the origin of the coordinate system $Oe_1e_2e_3$ -- and end the center of mass of segment $A_1A_2$. This is called a position vector of the center of mass. The vector $r_j$ is the analogous position vector of point $A_j$ -- starts from $O$ and ends at $A_j$, for $j=1,2$. Now the relative position vector $r = r_2-r_1$ is the vector with origin $A_1$ and end $A_2$. The line determined by the mass points $A_1$ and $A_2$ is $A_1A_2$, which is the line passing through the center of gravity $R$ and with vector $r$ lying on it, and its parametric equation is $p(\lambda) = R + \lambda r$. Therefore $r_j = R + \lambda_j r$ for some number $\lambda_j$ where $j=1,2$. We have to determine $\lambda_1$ and $\lambda_2$. Look at the systems \begin{align} r_1 &= R + \lambda_1 r\\ r_2 &= R + \lambda_2 r \end{align}
and \begin{align} R &= R=\frac{m_1r_1+m_2r_2}{m_1+m_2}\\ r &= r_2 - r_1 \end{align}
Plug the first system into the second system and you find equations for the $\lambda$'s. when you solve them you get the expression above.

Let us come back to the system that features the reduced mass $$\left(\frac{m_1m_2}{m_1+m_2}\right)\frac{d^2r}{dt^2} = -\left(\frac{m_1m_2G}{|r|^3}\right)\,r$$ which I am going to rewrite as $$\mu \frac{d^2r}{dt^2} = -\kappa(r)\,r$$ where $\kappa(r) = \frac{m_1m_2G}{|r|^3}$ is a scalar function and $\mu = \frac{m_1m_2}{m_1+m_2}$. Let $$K(r') = \frac{1}{2} \mu |r'|^2 \,\,\text{ and }\,\, U(r) = -\frac{m_1m_2G}{|r|}$$ be the kinetic and the potential energies respectively. Now, you can form the total energy $$H(r,r') = K(r') + U(r)$$ as well as the Lagrangian $$\mathcal{L}(r,r') = K(r') - U(r).$$ The momentum of a mass point is usually linked to the Lagrangian by $p = \frac{\partial \mathcal{L}}{\partial\, r'} = \mu \, r'$. So indeed, in our case $p = \mu r'$. A direct computation shows that $H(r,r')$ (technically it should be $H(r,p)$) is a constant of motion, and it is in fact the Hamiltonian of our reduced system. Furthermore, one can form the angular momentum $L = r \times p = r \times (\mu r')$ so if we differentiate it with respect to $t$ we get $$L' = r' \times (\mu r') + r \times (\mu r'') = r \times (\mu r'') = r \times (-\kappa(r) r) = -\kappa(r) (r \times r) = 0.$$ Hence, the angular momentum is also conserved. That basically tells you that the motion is planar. And you have conservation of energy. And if you look at the Runge-Lenz vector $$A = P \times L - \mu \, m_1m_2\, G \, \frac{r}{|r|}$$ this is also conserved, which is basically the fact that the major axis of our motion is also constant with time (for elliptical orbit, this is the major axis of the ellipse. It stays unchanged with time). So everything goes down as you have seen it. You can fix your plane of motion, determined by the constant angular momentum $L$, use the Runge-Lenz vector as an axis for your polar coordinates in that plane, and finally use the conservation of energy to derive your simple equation of motion in polar coordinates, where the mass will be the reduced mass. So, your angle $\theta(t)$ is the angle between the motion vector $r = r_2-r_1$ and the Runge-Lenz vector, all moving in the plane of constant angular momentum. Your radial coordinate $\rho(t) = |r(t)| = |r_2(t)-r_1(t)|$ is the distance between the two mass points. These coordinates are not a direct switch from Cartesian inertial frame of reference to some polar reduction of it. They are very closely related to the inertial frame, but the center of the polar coordinates is not exactly the center of mass. You think of them as slightly abstract coordinates, although, as you can see, they have a clear geometric meaning. If you want a direct interpretation simply take one of the equations
$$r^{''}_1 = - \left(\frac{(m_2)^3G}{(m_1+m_2)^2|r_1|^3}\right)r_1$$ and apply absolutely the same arguments as before (total energy, angular momentum, Runge-Lenz vector), and you get your polar coordinate to be exactly a direct switch from Cartesian inertial frame to a polar restriction on the constant angular momentum plane. Here, the center of the polar system is exactly the center of mass (coinciding with the focus of your orbit, which is a conic section), the Runge-Lenz vector is the zero axis of your polar coordinates. But the reduced mass is mixed up in the constants somewhere. Whichever you prefer.

Answer 2

In Newtonian physics, an isolated body has several conserved quantities: its mass $m > 0$, momentum $𝐩$, energy $H$, angular momentum $𝐉$ and (for lack of a better name) its "moving mass moment" $𝐊$. Denoting the body's center of mass position by $𝐫$ and its velocity by $𝐯 = d𝐫/dt$, they may be decomposed into the following: $$𝐩 = m𝐯, \hspace 1em H = ½mv^2 + U, \hspace 1em 𝐉 = 𝐫×𝐩 + 𝐒, \hspace 1em 𝐊 = m𝐫 - 𝐩t.$$

The total energy consists of the kinetic part $T = ½mv^2$ arising from its center of mass motion, and the internal energy $U$ arising from the motion of its parts. as well as from their respective internal energies.

Its angular momentum consists of the "orbital" part $𝐋 = 𝐫×𝐩 = m𝐫×𝐯$ consisting of its motion around a fixed center at $𝐫 = 𝟎$ and its internal angular momentum $𝐒$ arising from the circulatory motion of its parts around its own center of mass, and also of their internal angular momentum, in turn.

The moving mass moment $𝐊 = m(𝐫 - 𝐯t)$ doesn't have an official name, but despite its explicit dependency on the time $t$ it is conserved for isolated bodies. Its conservation is just the law of inertia, in disguise: namely, the statement that the body movies on a trajectory given by $𝐫 - 𝐯t = 𝐫_0$, where the constant $𝐫_0$ is the position at $t = 0$ for that trajectory. That means it moves at a constant speed in a straight line, in the absence of forces.

If the body is made up of parts, these quantities are totaled from those of its constituents and this partly accounts for how $U$ and $𝐒$ arise in composite bodies. But, then, the natural question - going in the upwards direction - is: how do you combine the contributions from separate parts of a body to arrive at a total for the body made out of those parts, and what is the total? In particular: how do you combine the totals from two constituents? You can always think of a composite body as being made of two constituents, each of them possibly, themselves, being composite, so this is actually reflective of the general situation.

Let the two bodies have respective invariants $$\left(m_0, 𝐩_0, H_0, 𝐉_0, 𝐊_0\right), \hspace 1em \left(m_1, 𝐩_1, H_1, 𝐉_1, 𝐊_1\right),$$ and let their respective center of mass positions and velocities be: $$𝐫_0, \hspace 1em 𝐯_0 = \frac{d𝐫_0}{dt}, \hspace 1em 𝐫_1, \hspace 1em 𝐯_1 = \frac{d𝐫_1}{dt}.$$ In particular, we assume: $$𝐩_i = m_i𝐯_i, \hspace 1em H_i = ½m_iv_i^2 + U_i, \hspace 1em 𝐉_i = 𝐫_i×𝐩_i + 𝐒_i, \hspace 1em 𝐊_i = m_i𝐫_i - 𝐩_it_i \hspace 1em (i = 0, 1).$$

Because of the interaction of the two parts of the body with each other, their "invariants" need not any longer be preserved, but we'd expect the total for the entire body to be, if the entire body is isolated. We'll suppose the two parts interact with each other through some kind of potential energy $V$, so that the total energy of the body should include this as a contribution. We assume that all the other quantities are additive; thereby leading to the following totals: $$m = m_0 + m_1, \hspace 1em 𝐩 = 𝐩_0 + 𝐩_1, \hspace 1em H = H_0 + H_1 + V, \hspace 1em 𝐉 = 𝐉_0 + 𝐉_1, \hspace 1em 𝐊 = 𝐊_0 + 𝐊_1.$$

Now the question arises: can they be made to satisfy a similar deconstruction? If so, then what are $U$ and $𝐒$? It's here that you'll find the reduced mass arises, as well as reduced versions of the position vector and velocity for a fictitious body. Call them $μ$, $𝞀$ and $𝞄$, respectively.

In reference to your question: the angular coordinates are those obtained from writing the position $𝞀$ of the fictitious body in spherical coordinates.

When dealing with two-body dynamics, the total quantities - being conserved - recede to the background and this reduces the focus on just these constituents. To find what they are, first set up the totals for the momentum and moving mass moment. $$m𝐯 = m_0𝐯_0 + m_1𝐯_1, \hspace 1em m𝐫 - m𝐯t = m_0𝐫_0 - m_0𝐯_0t + m_1𝐫_1 - m_1𝐯_1t.$$ From this, we can almost directly read off the solutions for $𝐫$ and $𝐯$: $$𝐫 = \frac{m_0𝐫_0 + m_1𝐫_1}{m_0 + m_1}, \hspace 1em 𝐯 = \frac{m_0𝐯_0 + m_1𝐯_1}{m_0 + m_1},$$ and confirm that $𝐯 = d𝐫/dt$.

That's the center of mass position and velocity for the body and is actually where it comes from: the additivity of the momentum and moving mass moment.

Next, substitute into the additive relation for $𝐉$, ignoring the internal angular momenta for now, to find out what addition contribution is being made to $𝐒$ from the circulatory motions of the two bodies about their common center of mass: : $$\begin{align} 𝐋_0 + 𝐋_1 - 𝐋 &= 𝐫_0×𝐩_0 + 𝐫_1×𝐩_1 - 𝐫×𝐩 \\ &= m_0𝐫_0×𝐯_0 + m_1 𝐫_1×𝐯_1 - \frac{m_0𝐫_0 + m_1𝐫_1}{m_0 + m_1}×\left(m_0𝐯_0 + m_1𝐯_1\right) \\ &= \frac{m_0m_1\left(𝐫_0×𝐯_0 - 𝐫_0×𝐯_1 - 𝐫_1×𝐯_0 + 𝐫_1×𝐯_1\right)}{m_0 + m_1} \\ &= \frac{m_0m_1}{m_0 + m_1} \left(𝐫_0 - 𝐫_1\right)×\left(𝐯_0 - 𝐯_1\right). \end{align}$$ From this, you can almost read off the candidates for $μ$, $𝞀$ and $𝞄$: $$μ = \frac{m_0m_1}{m_0 + m_1}, \hspace 1em 𝞀 = 𝐫_0 - 𝐫_1, \hspace 1em 𝞄 = 𝐯_0 - 𝐯_1,$$ and confirm that $d𝞀/dt = 𝞄$. Therefore, we can write: $$𝐋_0 + 𝐋_1 = 𝐋 + μ𝞀×𝞄.$$

Thus, putting back in the contributions from the internal angular momenta, we find: $$\begin{align} 𝟎 &= 𝐉_0 + 𝐉_1 - 𝐉 \\ &= 𝐋_0 + 𝐋_1 - 𝐋 + 𝐒_0 + 𝐒_1 - 𝐒 \\ &= μ𝞀×𝞄 + 𝐒_0 + 𝐒_1 - 𝐒. \end{align}$$ Therefore $$𝐒 = 𝐒_0 + 𝐒_1 + μ𝞀×𝞄.$$ So there is an additional contribution to the internal angular momentum for the body, beyond those that arise from the two component bodies that is equal to the "orbital" angular momentum of a fictitious body with reduced mass $μ$, position $𝞀$ and velocity $𝞄$. The position is not really relative to anything, per se, but could be thought of as the position of either body as seen by the other. To get the vantage point from the opposite body, you could just as well flip the signs on both $𝞀$ and $𝞄$, and, instead, write $𝞀 = 𝐫_1 - 𝐫_0$ and $𝞄 = 𝐯_1 - 𝐯_0$, without any material change, since all the contributions arising from them will be quadratic.

Finally, for the total energy, again ignoring the internal contributions, we find the following reduction: $$\begin{align} T_0 + T_1 - T &= ½m_0{v_0}^2 + ½m_1{v_1}^2 - ½mv^2 \\ &= ½m_0{v_0}^2 + ½m_1{v_1}^2 - ½\frac{\left|m_0𝐯_0 + m_1𝐯_1\right|^2}{m_0 + m_1} \\ &= ½\frac{m_0m_1{v_0}^2 + m_0m_1{v_1}^2 - 2m_0m_1𝐯_0·𝐯_1}{m_0 + m_1} \\ &= ½\frac{m_0m_1}{m_0 + m_1}{\left|𝐯_0 - 𝐯_1\right|}^2 \\ &= ½μ|𝞄|^2. \end{align}$$ Therefore, $$T_0 + T_1 = T + ½μ|𝞄|^2.$$

Thus, in a similar way, we find: $$\begin{align} 0 &= H_0 + H_1 + V - H \\ &= T_0 + T_1 - T + U_0 + U_1 + V - U \\ &= ½μ|𝞄|^2 + U_0 + U_1 + V - U. \end{align}$$ Thus $$U = U_0 + U_1 + ½μ|𝞄|^2 + V.$$

So, the total of the conserved quantities consists of the sum of the contributions from two component bodies, plus a contribution to the internal angular momentum and internal energy from a fictitious body. If the composite body is free from external interactions, then the focus of all the dynamics falls on the fictitious body and its attributes: its mass $μ$, position $𝞀$ and velocity $𝞄$, acting under the influence of a potential energy $V$. This account also spells out, explicitly, the possible contributions that the internal angular momenta and internal energies of the component bodies may make to the dynamics of the fictitious body. It's possible to have energy and angular momentum transfers along these channels.

It may be possible to do this deconstruction in tandem with three or more component bodies, instead of breaking down components two at a time.