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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.

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  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.

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  • $\begingroup$ Ok I think I have understood the "trick" there, I have just a few doubts. $\endgroup$ – tommy1996q Aug 23 '16 at 9:30
  • $\begingroup$ First, how do you find $r_1=R+r$? Is it just algebraic operations? Secondly, you managed to find that $F(r)=\mu \ddot{r}$, so this means that (being the Newton formula the most basic one), even all other quantities could be written in the same way using relative position and reduced mass? So for instance in the $CM$ frame you would have that $p=\mu v$, with $\mu$ reduced mass and $v$ relative velocity? Last thing: as you can see in the previous question, there is that $\theta$ bothering me. When considering relative motion, what does it mean? $\endgroup$ – tommy1996q Aug 23 '16 at 9:40
  • $\begingroup$ See my edits. I have added some new content. $\endgroup$ – Futurologist Aug 23 '16 at 13:22
  • $\begingroup$ Forty two equations to answer "Why does this work?" Is it possible to add a tl;dr with the main points? This is a heroic effort to reproduce the entire derivation here, but is there also a more concise, or "Aha!" way to express this besides because these forty two equations are all true? $\endgroup$ – uhoh Sep 23 '17 at 4:05

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