Kepler's 3rd law applied to binary systems: How can the two orbits have different semi-major axes? I suddenly came to the realization that I don't understand something about Kepler's law when applied to binary systems, because I encountered an apparent paradox. There must be an error somewhere in my reasoning, but I can't figure out what.
Consider a binary system of stars with masses $m_1$ and $m_2$. Both stars will be in an elliptic orbit around their common center of mass. If, say, star 1 moves on an ellipse with semi-major axis $a$, then  the period of the orbit of star 1 should be given by Kepler's third law,
\begin{align}
T_1^2 = \frac{4\pi a_1^3}{G(m_1 + m_2)}.
\end{align}
The same holds for star 2:
\begin{align}
T_2^2 = \frac{4\pi a_2^3}{G(m_1 + m_2)}.
\end{align}
But both stars must have the same orbital period (because otherwise the center of mass cannot be at rest), which seems to imply that the semi-major axes of the two stars should also be equal, looking at the above formulas. However, if we consider the Sun-Earth system, for instance (neglecting the other planets, etc) we see that this is clearly not the case.
Where is the error in the above reasoning? 
 A: 
If, say, star 1 moves on an ellipse with semi-major axis $a$, then  the period of the orbit of star 1 should be given by Kepler's third law,
  \begin{align}
{T_1}^2 = \frac{4\pi {a_1}^3}{G(m_1 + m_2)}.
\end{align}
  The same holds for star 2:
  \begin{align}
{T_2}^2 = \frac{4\pi {a_2}^3}{G(m_1 + m_2)}.
\end{align}

Minor issue: You are missing a factor of $\pi$: You should have $T^2 = (2\pi)^2 \frac {a^3}{G(m_1+m_2)}$.
That's a minor issue. The major issue: The semi-major axis in that expression is not the semi-major axis of the orbit about the center of mass. It is instead the semi-major axis of one object about another, with the other object fixed, and it doesn't matter which object you choose to be fixed. Suppose one of the objects is a pea and the other a ten solar mass giant. You can view the pea as orbiting the star, the star as orbiting the pea, or view each as orbiting their common center of mass.
Any way you do it, you'll get an ellipse. The math comes out much easier if you view the star as orbiting the pea (or equivalently, if you view the pea as orbiting the star). The math gets considerably messier (but still quite doable) if you view each as orbiting about the center of mass. You implicitly chose the route that requires the messier mathematics. You aren't doing that messier mathematics right, and hence your apparent dilemma.
A: You simply have your definitions wrong in your application of Kepler's third law. In the case where the masses of both bodies are considered, then $a$ is not the semi-major axis of each orbit, it is the sum of the semi-major axes of the two bodies (and is therefore the same in both equations).
$$ a = a_1 + a_2$$
$$ m_1 a_1 = m_2 a_2$$
$$T^2 =\frac{4\pi^2}{G(m_1+m_2)} a^3$$
A: Kepler's Third law takes a slightly different form when you consider motion around the center of mass. The equations of motion are
$$
\begin{align}
m_1\ddot{\boldsymbol{r}}_1 &= - \frac{Gm_1m_2}{|\boldsymbol{r}_1 - \boldsymbol{r}_2|^3}\left(\boldsymbol{r}_1 - \boldsymbol{r}_2\right),\\
m_2\ddot{\boldsymbol{r}}_2 &= \frac{Gm_1m_2}{|\boldsymbol{r}_1 - \boldsymbol{r}_2|^3}\left(\boldsymbol{r}_1 - \boldsymbol{r}_2\right).\\
\end{align}
$$
If you want to describe the relative motion of one celestial body with respect to the other one, you can combine these equations to obtain
$$
\ddot{\boldsymbol{r}}_1 - \ddot{\boldsymbol{r}}_2 =
- \frac{G(m_1+m_2)}{|\boldsymbol{r}_1 - \boldsymbol{r}_2|^3}\left(\boldsymbol{r}_1 - \boldsymbol{r}_2\right),
$$
or in short
$$
\ddot{\boldsymbol{r}} = - \frac{\mu}{r^3}\boldsymbol{r},
$$
where $\boldsymbol{r} = \boldsymbol{r}_1 - \boldsymbol{r}_2$ and $\mu =G(m_1+m_2)$. This is the familiar Kepler problem, with the corresponding 3rd Kepler law
$$
T^2 = \frac{4\pi^2}{\mu}a^3.
$$
On the other hand, if you wish to describe the motion of both celestial bodies with respect to the center of mass, you need to separate $\boldsymbol{r}_1$ and $\boldsymbol{r}_2$ in the equations of motion. You can do this by using the fact that the position of the center of mass remains constant
$$
m_1\boldsymbol{r}_1 + m_2\boldsymbol{r}_2 = \boldsymbol{0},\tag{1}
$$
so that
$$
\boldsymbol{r}_1 - \boldsymbol{r}_2 = \frac{m_1+m_2}{m_2}\boldsymbol{r}_1 =
-\frac{m_1+m_2}{m_1}\boldsymbol{r}_2.
$$
Therefore,
$$\begin{align}
m_1\ddot{\boldsymbol{r}}_1 &= -Gm_1m_2\left(\frac{m_2^3}{(m_1+m_2)^3r^3_1}\right)\left(\frac{m_1+m_2}{m_2}\boldsymbol{r}_1\right),\\
m_2\ddot{\boldsymbol{r}}_2 &= Gm_1m_2\left(\frac{m_1^3}{(m_1+m_2)^3r^3_2}\right)\left(-\frac{m_1+m_2}{m_1}\boldsymbol{r}_2\right),
\end{align}
$$
or
$$\begin{align}
\ddot{\boldsymbol{r}}_1 = -\frac{\mu_1}{r^3_1}\boldsymbol{r}_1,\qquad\text{and}\qquad
\ddot{\boldsymbol{r}}_2 = -\frac{\mu_2}{r^3_2}\boldsymbol{r}_2,
\end{align}
$$
with 
$$\mu_1 = \frac{Gm_2^3}{(m_1+m_2)^2},\qquad\text{and}\qquad
\mu_2 = \frac{Gm_1^3}{(m_1+m_2)^2}.
$$
So once again we have two Kepler problems, but this time the 3rd laws take the form
$$
T^2 = \frac{4\pi^2}{\mu_1}a_1^3,\qquad\text{and}\qquad
T^2 = \frac{4\pi^2}{\mu_2}a_2^3.
$$
Note that this implies $\mu_2a_1^3 = \mu_1a_2^3$, which simplifies to
$m_1a_1 = m_2a_2$, consistent with Eq. (1). Also, $\mu_1 a = \mu a_1$ and
$\mu_2 a = \mu a_2$ lead to $m_2 a = (m_1+m_2)a_1$ and $m_1 a = (m_1+m_2)a_2$, so that indeed $a = a_1+a_2$.
