When a two-body problem consisting of two masses $m_1$ and $m_2$ is reduced to a one-body problem, the equation of motion for the particle of mass $m_2$ w.r.t that of mass $m_1$ is given by $$m_2\ddot{\vec r}=\left(1+\frac{m_2}{m_1}\right)F(r)\hat{r},\tag{1}$$ or written more commonly as $$\mu\ddot{\vec r}=F(r)\hat{r}\tag{2}$$ where $\mu=\frac{m_1m_2}{m_1+m_2}$, called the reduced mass and $\vec r=\vec r_2-\vec r_1$ is the position vector of the particle 2 w.r.t particle 1. Now since the particles are moving under their mutual interaction, both the particles are accelerated. Therefore, using any one of them (particle 1, in my description) as the origin of coordinates, is the same as using a non-inertial frame of reference.

Therefore, instead of the equation of the form (1), don't we expect an equation of the form $$m_2\ddot{\vec{r}}=F(r)\hat{r}+\vec F_{\rm fict}$$ where $\vec F_{\rm fict}$ are fictitious forces due to the acceleration of the particle $1$. If so, can we think of the piece $$\frac{m_2}{m_1} F(r)\hat{r}$$ in (1) as fictitious force $\vec F_{\rm fict}$?

  • $\begingroup$ you edit your equations after my answer $\endgroup$
    – Eli
    Jan 20, 2022 at 15:10
  • $\begingroup$ I started editing before you answer. Didn't see you posted the answer, sorry about that. $\endgroup$ Jan 20, 2022 at 15:24

4 Answers 4


I hope to add a more complete answer to this question since this is a similar question to one which I had a personal problem with. I will also be mostly citing Classical dynamics of particles and systems by Thornton and Marion (5e). As Thornton and Marion alludes to, problems like this can be analogized to planets or stars interacting from a distance, but the theory applies to any central force.

The principle question at hand is whether making the transition from just looking at (for example) distant stars and considering the dynamics of a problem from a distant perspective to considering the system from a local perspective. Consider the graphic below from T&M

Center of Mass

This graphic illustrates the idea I just described. From a distant perspective (see graphic (a)) our position vectors are essentially guaranteed to be measured from an inertial reference. This shall be assumed, and is the same thing as saying Newton's Second Law applies to these bodies relative to our inertial frame.

Relative to some arbitrary inertial reference, the (a) graphic shows the position vectors, the center of mass vector, and includes the relative vector $\vec{\bf r}$. The positions of body 1 and body 2 are given by the position vectors $\vec{\bf r}_1$ and $\vec{\bf r}_2$, respectively in this frame.

The short and simple answer is that making this transition from the general position vectors $\vec{\bf r}_1$, $\vec{\bf r}_2$ to the single relative vector $\vec{\bf r}$ does not create "fictitious" forces so long as the center of mass of the system does not itself accelerate. If we were considering orbital motion, then this is the same as requiring that there is no third object affecting the motion of these bodies. If this was about two bodies on a spring on a train, then the train must be stopped or at constant velocity.

All systems of more than two masses have a center of mass defined by

$$ \vec{\bf R}_{_\text{COM}} =\vec{\bf R} =\frac{1}{\sum_i^N m_i}\sum_i^N m_i \vec{\bf r}_i =\frac{1}{M}(m_1\vec{\bf r}_1+m_2\vec{\bf r}_2) $$

We can describe the positions relative to the center of mass.

$$ \vec{\bf r}_1=\vec{\bf R}+\vec{\bf r}'_1\\ \vec{\bf r}_2=\vec{\bf R}+\vec{\bf r}'_2 $$

Where the primed vectors are the $\vec{\bf r}_1$ and $\vec{\bf r}_2$ in the (b) graphic, respectively. If we substitute back into the definition of the center of mass we can get a new relation:

\begin{align*} \vec{\bf R} &=\frac{1}{M}(m_1\vec{\bf r}_1+m_2\vec{\bf r}_2)\\ &=\frac{1}{M}\big(m_1(\vec{\bf R}+\vec{\bf r}'_1)+m_2(\vec{\bf R}+\vec{\bf r}'_2)\big)\\ &=\frac{1}{M}\big((m_1+m_2)\vec{\bf R}+m_1\vec{r}'_1+m_2\vec{\bf r}'_2\big)\\ &=\vec{\bf R}+\frac{m_1}{M}\vec{\bf r}'_1+\frac{m_2}{M}\vec{\bf r}'_2\\ \Rightarrow\ 0&=\frac{m_1}{M}\vec{\bf r}'_1+\frac{m_2}{M}\vec{\bf r}'_2\\ \vec{\bf r}'_1&=-\frac{m_2}{m_1}\vec{\bf r}'_2 \end{align*}

The last relation we need is the plain ol' relative r vector defined by $\vec{\bf r}=\vec{\bf r}_1-\vec{\bf r}_2$. If we substitute for the prime vectors we get:

\begin{align*} \vec{\bf r} &=\vec{\bf r}_1-\vec{\bf r}_2\\ &=(\vec{\bf R}+\vec{\bf r}'_1)-(\vec{\bf R}+\vec{r}'_2)\\ &=\vec{\bf r}'_1-\vec{\bf r}'_2\\ \end{align*}

The last part we need is to express both position vectors solely in terms of $\vec{\bf R}$ and $\vec{\bf r}$. All we need to do is substitute.

\begin{align*} \vec{\bf r} &=\vec{\bf r}'_1-\vec{\bf r}'_2 \\ &=(1+\frac{m_1}{m_2})\vec{\bf r}'_1\ \Rightarrow \vec{\bf r}'_1=\frac{1}{(1+\frac{m_1}{m_2})}\vec{\bf r}=\frac{m_2}{m_1+m_2}\vec{\bf r} \\ &=-(\frac{m_2}{m_1}+1)\vec{\bf r}'_2\ \Rightarrow \vec{\bf r}'_2=-\frac{1}{(\frac{m_2}{m_1}+1)}\vec{\bf r}=-\frac{m_1}{m_1+m_2}\vec{\bf r} \\ \end{align*}

And so

$$ \vec{\bf r}_1=\vec{\bf R}+\frac{m_2}{m_1+m_2}\vec{\bf r}\\ \vec{\bf r}_2=\vec{\bf R}-\frac{m_1}{m_1+m_2}\vec{\bf r} $$

I should note here that for a system to not be inertial here would entail that the center of mass of the system is accelerating. That is to say, it's under an external force. This would be similar to a star and planet system orbiting a black hole, or for springs on a train example, the train would be coming to a stop or leaving the train stop. Regardless of what system we are talking about, in order for the system to be experiencing only behavior associated with the central force, there must be no external forces acting on the system. This is the same as saying that the acceleration of the center of mass of our system is zero. Mathematically we would say $\ddot{\bf R}=0$.

Ultimately these fictitious forces do not appear in our equations unless we are attempting to "shift" our equations from the external perspective to a local perspective attached to the center of mass. If we see new so-called "fictitious forces" ($\vec{\bf F}_\text{Fict}$) acting on our system that appear to come from nowhere and are caused by nothing, then we call these inertial forces and we know that the system must be experiencing an external force that is not caused by it's constituent members.

If the system as a whole is experiencing inertial motion, we can write the acceleration portion of newtons second law:

$$ m_1\ddot{\vec{\bf r}}_1=\frac{m_1m_2}{m_1+m_2}\ddot{\vec{\bf r}}=\mu\ddot{\vec{\bf r}}\\ m_2\ddot{\vec{\bf r}}_2=-\frac{m_1m_2}{m_1+m_2}\ddot{\vec{\bf r}}=-\mu\ddot{\vec{\bf r}}\\ $$

What this tells us is that for central forces acting on a two-body system we can move from thinking about individual bodies moving with relation to each other, to thinking about a reduced mass body orbiting around a central point, the center of mass. The dynamics are the same and there are no inertial forces present.

  • $\begingroup$ Made slight changes to make the vector notations uniform. Hope you don't mind. $\endgroup$ Jan 24, 2022 at 14:34
  • 1
    $\begingroup$ The concept of reduced mass $\mu$ arises when one body is viewed from the other body, not from the center of mass. It appears to an observer on $m_1$, that instead of a particle of mass $m_2$, there is a particle of mass $\mu$ which moves under the force exerted by $m_1$. $\endgroup$ Jan 24, 2022 at 14:58
  • $\begingroup$ A system is considered inertial if the origin of the coordinate system you are using to define your position vectors is not accelerating with respect to an inertial external observer. If the origin is accelerating, such as if we define the relative position vector $\vec{\bf r}$ with respect to the center of mass and the system is a binary stellar system, then when a larger third star causes the system to be flung away, a future human that is approximately inertial with the center of mass of the binary system will see the third star's effects as a "ficticious force" acting on everything. $\endgroup$
    – Gerald
    Jan 25, 2022 at 1:31
  • $\begingroup$ As far as the edits go, I don't mind. It was very intentional insofar as I was trying to convey the book results with bold vectors, and what I was adding with non-bold vectors, but I understand the potential for confusion and it probably adds more clarity to modify it. $\endgroup$
    – Gerald
    Jan 25, 2022 at 1:35

The motion using the reduced mass is typically derived using an inertial frame assuming no external forces, or external forces such that ${\vec F_1^{\,e} \over m_1} = {\vec F_2^{\,e} \over m_2}$. See Symon, Mechanics. $\vec F_1^{\,e}$ and $\vec F_2^{\,e}$ are the total external forces on the two masses $m_1$ and $m_2$, respectively. $\vec F_1^{\,i} = -\vec F_2^{\,i}$ where $\vec F_1^{\,i}$ and
$\vec F_2^{\,i}$ are the internal forces exerted by each mass on the other. $M = m_1 + m_2$ is the total mass.

Using an inertial frame, it can be shown- see Symon, Mechanics- $$(1) \enspace M \ddot {\vec R} = \vec F$$ $$ (2) \enspace \mu \ddot {\vec r} = \vec F_1^{\,i}$$ where $\vec R $ is the position of the center of mass (CM) in the inertial frame, $\vec F = \vec F_1^{\,e} + \vec F_2^{\,e}$, $\vec r$ is the relative position of $m_1$ relative to $m_2$, and $\mu = {m_1m_2 \over (m_1 + m_2)}$ is the reduced mass.

Relationship (1) is the typical relationship for motion of the center of mass (CM) for a system of particles (two particles here). The CM can be accelerating in the inertial frame if $\vec F$ is not zero.

Relationship (2) is the motion of a particle of mass $\mu$ acted on by the internal force body 2 exerts on body 1; the motion of body 1 as viewed from body 2 is the same as if body 2 is fixed and body 1 has mass $\mu$.

There are no fictitious forces involved since relationships (1) and (2) are derived in an inertial frame. But, these relationships assume the condition on external forces as specified earlier.


On the right side of your first formula (and later), you show an undefined vector product (no dot or cross). If you are going to indicate direction with the unit radius vector, you do not need the arrow over the F(r). In the more general case of a mass in (any) accelerated coordinate frame, the vector (F) can represent the sum of external forces and you can add the pseudo-force to that.

  • $\begingroup$ Thanks, that's a typo. I have corrected it. I am asking whether I can interpret the second term on the RHS of (1) as a fictitious force. If so, how can we relate it to the standard forms of the fictitious forces (Coriolis, centrifugal or Euler, or a combination of all)? $\endgroup$ Jan 20, 2022 at 15:27
  • $\begingroup$ In an accelerating frame, the fictitious (or pseudo) force on a mass is generally (-ma), where the vector (a) is the acceleration of the frame. $\endgroup$
    – R.W. Bird
    Jan 20, 2022 at 16:11
  • $\begingroup$ Even that is not apparent here. $\endgroup$ Jan 20, 2022 at 16:15

lets look at the equations

$$m_1\mathbf{\ddot{r}}_1=-F(r)\mathbf{\hat{r}}\\ m_2\mathbf{\ddot{r}}_2=F(r)\mathbf{\hat{r}}$$

from here $$\mathbf{\ddot{r}}_1=-\frac{1}{m_1}\,F(r)\mathbf{\hat{r}}\tag 1$$ $$\mathbf{\ddot{r}}_2=\frac{1}{m_2}\,F(r)\mathbf{\hat{r}}\tag 2$$


$$ \mathbf{\ddot{r}}_2-\mathbf{\ddot{r}}_1=\mathbf{\ddot{r}}=\left(\frac{1}{m_2}+ \frac{1}{m_1}\right)\,F(r)\mathbf{\hat{r}}\tag 3$$ $$\mu\,\mathbf{\ddot{r}}=F(r)\mathbf{\hat{r}}$$

now your approach multiply equation (3) with $~m_2~$

$$ m_2\mathbf{\ddot{r}}=\left(1+ \frac{m_2}{m_1}\right)\,F(r)\mathbf{\hat{r}}=F(r)\mathbf{\hat{r}}+ \underbrace{\frac{m_2}{m_1}\,F(r)\mathbf{\hat{r}}}_{\ne F_{\text{fict}}}$$

so your approach is not correct

  • $\begingroup$ The question is since particle 1 is a non-inertial frame, we expect the extra term must be a fictitious force. Of course, it doesn't look like any standard fictitious force. Hence, the question. $\endgroup$ Jan 20, 2022 at 15:26
  • $\begingroup$ so the equation is correct, but nothing to do with the fictitious force $\endgroup$
    – Eli
    Jan 20, 2022 at 17:54

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