6.3. The Local Group timing argument¶
One of the first indications that the Local Group—consisting of the Milky Way, M31, and many smaller galaxies—is much more massive than the total stellar masses of all of its constituents came from the Local Group timing argument. This argument was first made by Kahn & Woltjer (1959). The crux of the argument is the following: Because the Milky Way and M31 are by far the largest galaxies in the Local Group, it makes sense to approximate the mass of the Local Group as being all contained in the Milky Way and M31. A simple way to model their dynamics then is to assume that both large galaxies are point masses that have been orbiting each other since the Big Bang. At the Big Bang, both (proto)-galaxies were at the same position \(r=0\) and were launched on a zero-angular momentum orbit to where we see them today. From the age of the Universe and the present-day relative velocity and distance between the Milky Way and M31, we can solve for the orbit and the total mass of the system. We can solve for the mass from these three observables, because the orbit is characterized by two parameters only: the initial radial velocity (because the initial separation and the angular momentum are zero) and the current orbital phase. Because a collision between the galaxies would cause them to merge, we can assume that the orbit is still in its first radial period and \(T_r > t_\mathrm{H}\), with \(t_\mathrm{H} = 13.7\,\mathrm{Gyr}\) the age of the Universe.
The orbit of two point masses can be solved by going to the center-of-mass reference frame. If the Milky Way’s mass is \(M_\mathrm{MW}\) and M31’s mass is \(M_\mathrm{M31}\), then the total mass is \(M = M_\mathrm{MW} + M_\mathrm{M31}\), the reduced mass is \(\mu = M_\mathrm{MW}\,M_\mathrm{M31}/M\); the Milky Way’s position is \(\vec{x}_\mathrm{MW}\) and M31’s position is \(\vec{x}_\mathrm{M31}\) and the center of mass is therefore at a vector \(\vec{X} = (M_\mathrm{MW}\,\vec{x}_\mathrm{MW} + M_\mathrm{M31}\,\vec{x}_\mathrm{M31})/M\). Because of the conservation of momentum in this system, the center of mass is fixed. It is then straightforward to show that the time dependence of the separation vector (or displacement vector) \(\vec{r} = \vec{x}_\mathrm{M31}-\vec{x}_\mathrm{MW}\) follows the same orbit as a particle with a mass equal to \(\mu\) in a fixed point-mass potential with mass \(M\), with the center of mass \(\vec{X}\) as the origin. This is the orbit that we solved in Chapter 4.2.2.
Rather than directly solving Kepler’s equation for the mass \(M\) implied by the observations, let’s use the conservation of energy and angular momentum to make a simple estimate. The energy in a point-mass potential is given in terms of the semi-major axis \(a\) by Equation (4.32), and energy conservation during the orbit therefore means that \begin{equation} -\frac{G\,M}{2\,a} = \frac{v^2}{2}-\frac{G\,M}{r}\,, \end{equation} where \(r = |\vec{r}|\), the magnitude of the separation vector and \(v = |\vec{v}|\) is the magnitude of the relative velocity. Re-writing this as follows gives the relative velocity in terms of the mass and displacement \begin{equation} v^2 = G\,M\,\left[-\frac{1}{a}+\frac{2}{r}\right]\,. \end{equation} We can use Kepler’s third law (Equation 4.36)—a statement of the conservation of angular momentum—to substitute the period \(T_r\) for the semi-major axis \(a\) in this equation and derive a relation between the relative velocity, the period, and the separation; writing this as a cubic equation for \(x = G\,M\), we find \begin{equation}\label{eq-lgtiming-deriv-2} -\frac{8\,x^3}{r^3}+\left(\frac{12\,v^2}{r^2}+\frac{4\pi^2}{T_r^2}\right)\,x^2-\frac{6\,v^4}{r}\,x+v^6 = 0\,. \end{equation}
The current separation between the Milky Way and M31 is \(r \approx 740\,\mathrm{kpc}\) and the current relative velocity is \(v \approx -125\,\mathrm{km\,s}^{-1}\) (where we’ve included the sign to indicate that M31 is currently approaching the Milky Way). To apply Equation (6.10), we need \(T_r\). Let’s estimate \(T_r\) by approximating the time until the next collision as \(t_c = -r/v\) and adding this to the age of the Universe to estimate \(T_r\): \(T_r \approx t_\mathrm{H}-r/v\). Then we can solve the cubic equation. Here's the code that does this:
[6]:
import numpy
import astropy.constants as apyconst
import astropy.units as u
# Observables
r= 740.*u.kpc
v= -125.*u.km/u.s
tH= 13.7*u.Gyr
# Estimate T_r
Tr= tH-r/v
# We will rewrite the cubic equation in terms of M12 = M/(10^12 Msun)
# so we need a tG = G x 10^12 Msun
tG= apyconst.G*10.**12.*u.Msun
sol= numpy.roots([-8.*(tG**3./r**3.).to((u.km/u.s)**6).value,
((12.*v**2./r**2.+4*numpy.pi**2./Tr**2.)*tG**2.)\
.to((u.km/u.s)**6).value,
-6.*(v**4./r*tG).to((u.km/u.s)**6).value,
(v**6.).to((u.km/u.s)**6).value])
sol= sol[numpy.isreal(sol)]
print(f"The approximate LG timing mass is {sol.real[0]:.2f} x 10^12 Msun")
The approximate LG timing mass is 4.00 x 10^12 Msun
The mass for the Local Group implied is, therefore, approximately \(M = M_\mathrm{MW}+M_\mathrm{M31} = 4\times10^{12}\,M_\odot\). The radial period \(T_r\) will in reality be somewhat smaller than the rough estimate that we used and it is easy to check (by running the above code for different \(T_r\)) that a smaller \(T_r\) leads to a larger mass. The approximate timing mass that we have derived here is therefore a lower limit. An upper limit can be obtained by setting \(T_r = t_\mathrm{H}\), which gives \(M = 5.5\times 10^{12}\,M_\odot\), so without doing any complicated Keplerian orbit modeling, we find that
\begin{equation} 4.0 \times 10^{12}\,M_\odot < M_\mathrm{MW} + M_\mathrm{M31} < 5.5 \times 10^{12}\,M_\odot\,.\end{equation}Li & White (2008) demonstrated, by applying the timing argument to groups with two large galaxies in the Millenium simulation, that the timing argument has an uncertainty of a factor of two, larger than the range that we get from our simple derivation above.
The full, exact solution that solves for \(T_r\) by directly solving for the orbit is left as an exercise. That solution gives \(T_r = 16.6\,\mathrm{Gyr}\). If you plug this into the code above, you find that this implies \(M = 4.6\times10^{12}\,M_\odot\).
The mass for the Local Group implied by the timing argument is at least an order of magnitude larger than the total mass in stars in all Local Group galaxies. The timing argument therefore provided an early indication that galaxies like the Milky Way are surrounded by extended, massive halos of dark matter.
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