# Proof for equal eccentricity in a binary star system

What is the proof that the orbits of two stars orbiting around a common center of mass have equal eccentricities?

You can use: $$m_1r_1 = m_2r_2,$$ then say that $$r_1= a(1+e_1)$$, $$r_2=a(1-e_2)$$ and from the condition of the center of mass: $$m_1a_1=m_2a_2.$$ The problem is that the proof for the condition of the center of mass requires knowing that the eccentricities are equal, so you're kind of spiralling around the same thing. What is the actual, complete proof?

We can get most of the way there without even considering gravity or orbital motion. Since the two masses are the only things exerting forces on each other, we can start from Newton's Third Law: $$\vec{F}_{21} = -\vec{F}_{12}$$ $$m_1 \vec{a}_1(t) = -m_2 \vec{a}_2(t).$$ Here, $$\vec{F}_{12}$$ is the force of object 1 on object 2 and vice versa for $$\vec{F}_{21}.$$ The variables $$m$$ and $$a$$ are mass and acceleration, as usual.
Integrating once by time gives $$m_1\left[\vec{v}_1(t) + \vec{v}_{01}\right] = -m_2\left[\vec{v}_2(t) + \vec{v}_{02}\right]$$ where $$\vec{v}_1(t)$$ and $$\vec{v}_2(t)$$ are the velocities of objects 1 and 2 as a function of time and $$v_{01}$$ and $$v_{02}$$ are the initial starting velocities. I am integrating starting from $$t=0,$$ so $$\vec{v}(0) = \vec{0}$$ for both objects. Integrating again by time gives $$m_1\left[\vec{r}_1(t) + \vec{v}_{01}t + \vec{r}_{01}\right] = -m_2\left[\vec{r}_2(t) + \vec{v}_{02}t + \vec{r}_{02}\right]$$ where $$\vec{r}_1(t)$$ and $$\vec{r}_1(t)$$ are the time-dependent positions and $$r_{01}$$ and $$r_{02}$$ are the initial starting positions. Similarly to velocity $$\vec{r}(0) = 0$$ for both objects.
Now, we move to the center of mass of the system. This means we choose a coordinate system such that $$m_1 \vec{r}_{10} + m_2 \vec{r}_{20} = \vec{0}$$ leaving us with $$m_1\left[\vec{r}_1(t) + \vec{v}_{01}t\right] = -m_2\left[\vec{r}_2(t) + \vec{v}_{02}t\right].$$ We also want the center of mass to not move as time goes on. So, we need to boost into the center of momentum frame where the total momentum is zero. Since momentum is conserved, we only need to match the momentum at $$t = 0:$$ $$m_1 \vec{v}_{01} + m_2 \vec{v}_{02} = \vec{0}$$ resulting in $$m_1 \vec{r}_1(t) = -m_2 \vec{r}_2(t).$$ So, if two objects exert a force on each other and those are the only forces acting on those objects, their motions will have the same shape, just flipped in coordinates and scaled by their masses. So, after using the inverse square force of gravity to derive that both objects move in elliptical orbits, you then know that they both must move in ellipses of the same eccentricity because their motions must have the same shape.
Crudely you can transform this type of problem into the center of mass frame then the total $$r$$ can be written as $$m_1r_1+m_2r_2$$ where $$r_1=\frac{m_2}{\sum m}$$ and similar for $$r_2=\frac{m_1}{\sum m}$$, hence you will get a definite value of $$e$$ in the com frame since total energy and angular momentum is conserved