Why does reduced mass help when talking about two body problems?

Question

I hear people say that reduced mass, $\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?

I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.

Derivation:

Step [1] - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$ Step [2] - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$ Step [3] - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$ Step [4] - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step): $$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$

Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help.

Answer 1

Regardless of the central force type (gravity, spring, contact, electromagnetic) it causes an acceleration of each body proportional to $\frac{1}{m_i}$. As a result the relative acceleration is proportional to $\frac{1}{m_1}+\frac{1}{m_2}$.

When this relative acceleration is used to calculate the effective mass this force sees (to produce the said acceleration) the result is

$$ \boxed{ \mu = \left(\frac{1}{m_1}+\frac{1}{m_2}\right)^{-1} }$$

It is a kind of grouping for masses, the same way resistors are grouped with $\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}+\ldots$, or springs are grouped $\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}+\ldots$. It is all a consequence of the fact they share something common. For us it is a central force, for resistors it is voltage, and for springs it is also force.

When displacements are shared (the same with springs) then the effective mass would be $\mu=m_1+m_2+\ldots$, and if currents are shared the effective resistance would be $R=R_1+R_2+\ldots$.

I hope I have shed some light into your question.

Answer 2

I hear people say that reduced mass, μ, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?

The answer to both parts of your second question is yes. Reformulating the problem as a one body problem can simplify the description of the motion. (There's no point in doing so if this doesn't simplify things.) The answer to your first question is simple: It's used on those problems where such a treatment simplifies the description of the motion.

A broad class of two body problems where such a treatment does indeed simplify the description of the motion are those problems in which the force (or potential energy) depends on the displacement between the two bodies, and possibly on derivatives of the displacement: $\vec F_{12} = -\vec F_{21} = \vec f(\vec r, \dot{\vec r}, \dots)$. Two examples are the gravitational force and the spring force.

Since the force is a function of the displacement $\vec r$, it would be handy if the equations of motion can be expressed in terms of this vector and its time derivatives. In a Newtonian setting, this means reducing the problem to the form $$\ddot{\vec r} = \vec f(\vec r, \dot{\vec r})$$

I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration.

Note that $\dot{\vec r} = \dot{\vec r}_1 - \dot{\vec r}_2$ and that $\ddot{\vec r} = \ddot{\vec r}_1 - \ddot{\vec r}_2$. In other words, these are the velocity and acceleration of the two particles with respect to one another. Finding an expression for the relative acceleration in terms of the relative position and relative velocity is exactly what is needed to reduce the problem to the above form.

Answer 3

Yeah, it's mainly a mathematical simplification. The equations of motion for the two-body problem (after simplification) look very much like the equations of motion for a single body orbiting in a fixed potential, and so we can treat the problem this way, although we have to use the reduced mass instead of the mass of either of the original bodies. We know how to solve the one-body problem very easily, and it's also simple to think about, so this method is preferred.