IPhO 2017 · Theory problem 1

Dark Matter

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The first formal inference for the existence of dark matter was provided by Fritz Zwicky based on his observation on the dynamics of the Coma galaxy cluster, a cluster of galaxies that consist of about a thousand of galaxies. Zwicky used the Virial theorem to estimate the mass of the galaxy cluster. In a simple sun-planet system, where the planet revolves around the sun in a circular orbit, the Virial theorem states that the kinetic energy of the planet is exactly related to its gravitational potential energy. While in a general case for a system of many particles bounded by some interaction, the Virial theorem, relates the time average total kinetic energy with its time average total potential energy.

In 1933, based on his observation on the velocity of the galaxies near the edge of the Coma galaxy cluster, Zwicky estimated that the cluster has more mass than what was visually observed (i.e. the galaxies). The gravitational attraction from the observable matter (the galaxies) was too small to account for the velocities of the galaxies. Thus there must be some hidden masses that account for such a large velocity. That hidden mass is the dark matter mass. In what follows, assume that the mass of each galaxy is the sum of its visible mass and the mass of the dark matter which moves together with that galaxy, and dark matter interacts with visible matter only gravitationally.

A. Cluster of Galaxies

Consider a cluster of galaxies consist of a large number \(N\) of galaxies and dark matter that distributed homogeneously in a sphere of radius \(R\) with the total mass (galaxies and dark matters) of the cluster \(M\). Assume that the average total mass (visible and dark matter) of a galaxy is \(m\).

A.1 Assuming a continuous distribution of matter in the cluster, find the total gravitational potential energy of the cluster, in terms of \(M\) and \(R\). 1.0 pt.

Due to cosmological expansion, any distance object will be moving away from an observer on Earth with a speed that depends on the distance from the observer to the object. A certain Lyman (a hydrogen emission spectral line) frequency from a type IA supernova on the \(i\)-th galaxy in the galaxy cluster is observed to be \(f_i\), with \(i=1,\ldots,N\), while the same corresponding Lyman frequency on Earth is \(f_0\).

A.2 Determine the average speed \(V_{cr}\) of the whole galaxy cluster moving away from the Earth in terms of \(f_i\) (with \(i=1,\ldots,N.\)), \(f_0\) and \(N\). Note that a galaxy's speed is very small compared to the speed of light \(c\). 0.5 pt.
A.3 By assuming that the galaxies velocities with respect to the center of the cluster are isotropic (the same in all direction), determine the root-mean square speed \(v_{rms}\) of the galaxies with respect to the center of the cluster in terms of \(N\), \(f_i\) (with \(i=1,\ldots,N\)), and \(f_0\). From this result determine the mean kinetic energy of a galaxy with respect to the center of the cluster in terms of \(v_{rms}\) and \(m\). 1.5 pt.

To find the total mass of the cluster, one can use Virial theorem. The theorem stated that for a system of particles bounded by their conservative force,

\[ \langle K\rangle_t=-\gamma\langle U\rangle_t, \]

where \(\langle K\rangle_t\) is the time average total kinetic energy, \(\langle U\rangle_t\) is the time average total potential energy, and \(\gamma\) is a constant. This theorem can be derived by assuming that for a system of particles bounded by its own interaction, the magnitudes of the position and momentum of each particle are finite, and thus the following quantity

\[ \Gamma=\sum_i\vec p_i\cdot\vec r_i \]

is finite.

A.4 Using the fact that the time average over a long period of time of \(d\Gamma/dt\) vanishes, \(\left\langle d\Gamma/dt\right\rangle_t=0\), determine \(\gamma\) in the Virial theorem above for the case of gravitational interaction. (Hint: Try to do the problem with the summation in \(\Gamma\) for a small finite number of galaxies). 1.7 pt.
A.5 From the previous results determine the total dark matter mass of the cluster in terms of \(N\), \(m_g\), \(R\), and \(v_{rms}\), where \(m_g\) is the average total visible mass of a galaxy. Note that the root-mean-square speed of the dark matter is the same as that of the galaxies. 0.5 pt.

B. Dark Matter in a Galaxy

Dark matter also exists inside and around a galaxy. Consider a spherical galaxy with a visible edge radius \(R_g\) (an approximate outermost distance where a large number of stars still visible, but note that a very small number of stars may still be distributed in the region beyond \(R_g\)). Assume that the stars in the galaxy are point particles with an average mass \(m_s\). The stars in the galaxy, distributed homogeneously with a number density \(n\), are assumed to move in circular orbits.

B.1 If the galaxy consists only of stars, find the speed \(v(r)\) of a star as a function of its distance to the center of the galaxy and sketch \(v(r)\) for \(r<R_g\) and \(r\ge R_g\). 0.8 pt.

The existence of dark matter can be inferred from the galaxy rotation curve, which is a plot of \(v(r)\) obtained from observations. The figure below shows a common pattern of the galaxy rotation curve. You may assume simplifyingly that \(v(r)\) is a linear function for \(r\le R_g\) and a constant \(v_0\) for \(r>R_g\).

Figure 1

Fig. 1 Plot of galaxy rotation curve in a galaxy.

B.2 Find the total mass \(m_R\) of that part of the galaxy which lies within the sphere of radius \(R_g\) in terms of \(v_0\) and \(R_g\). 0.5 pt.

The discrepancy between the figure in B.2. and the plot obtained in B.1 indicates the existence of dark matter.

B.3 Determine the dark matter mass density as a function of \(r\), \(R_g\), \(v_0\), \(n\), and \(m_s\) for \(r<R_g\) and \(r\ge R_g\). 1.5 pt.

C. Interstellar Gas and Dark Matter

Now consider a young galaxy whose mass is dominated by interstellar gas and dark matters (neglect the mass of the stars). The interstellar gas is assumed to consist of identical particles of mass \(m_p\). The number density \(n(r)\) and temperature \(T(r)\) of the gas depend on the distance from the center of the galaxy \(r\). Although many physical processes happen in the gas, we can assume the gas is in a hydrostatic equilibrium due to its pressure and the galaxy gravitational attraction.

C.1 Find the pressure gradient of the gas \(dP/dr\), in terms of \(m'(r)\), \(r\) and \(n(r)\). Here, \(m'(r)\) is the total mass of gas and dark matters within a sphere of radius \(r\) from the galaxy center. 0.5 pt.
C.2 Assuming the interstellar gas as an ideal gas, find \(m'(r)\) in terms of \(n(r)\), \(T(r)\) and their derivatives with respect to \(r\). 0.5 pt.

Next for simplicity assume that the gas is in isothermal distribution at temperature \(T_0\) and the interstellar gas number density is given by

\[ n(r)=\frac{\alpha}{r(\beta+r)^2}, \]

where \(\alpha\) and \(\beta\) are some constants.

C.3 Find the dark matter mass density as a function of \(r\) inside the galaxy. 1.0 pt.

Official English text of the problem.

Answers

Let \[ \bar f=\frac1N\sum_{i=1}^N f_i. \] Here \(G\) is the gravitational constant and \(k_B\) is Boltzmann’s constant. The Doppler results retain first order in speed divided by \(c\).

  • A.1 \[ \boxed{U=-\frac{3GM^2}{5R}}. \]

  • A.2 \[ \boxed{V_{cr}\simeq c\left(1-\frac{\bar f}{f_0}\right)}. \]

  • A.3 \[ \boxed{v_{rms}\simeq \frac{c}{f_0}\sqrt{\frac3N\sum_{i=1}^N(f_i-\bar f)^2}}, \qquad \boxed{\langle K_{\mathrm{gal}}\rangle=\frac12m v_{rms}^2}. \]

  • A.4 \[ \boxed{\gamma=\frac12}. \]

  • A.5 \[ \boxed{M_{\mathrm{DM}}=\frac{5Rv_{rms}^2}{3G}-Nm_g}. \]

  • B.1 \[ \boxed{ v(r)= \begin{cases} \displaystyle r\sqrt{\frac{4\pi Gnm_s}{3}},&0\le r<R_g,\\[6pt] \displaystyle \sqrt{\frac{4\pi Gnm_sR_g^3}{3r}},&r\ge R_g. \end{cases}} \] The graph rises linearly from the origin to its maximum at \(R_g\), then decreases as \(r^{-1/2}\), approaching zero.

  • B.2 \[ \boxed{m_R=\frac{v_0^2R_g}{G}}. \]

  • B.3 \[ \boxed{ \rho_{\mathrm{DM}}(r)= \begin{cases} \displaystyle \frac{3v_0^2}{4\pi GR_g^2}-nm_s,&r<R_g,\\[7pt] \displaystyle \frac{v_0^2}{4\pi Gr^2},&r\ge R_g. \end{cases}} \]

  • C.1 \[ \boxed{\frac{dP}{dr}=-\frac{Gm'(r)m_p n(r)}{r^2}}. \]

  • C.2 \[ \boxed{ m'(r)=-\frac{k_Br^2}{Gm_p} \left[\frac{dT}{dr}+\frac{T(r)}{n(r)}\frac{dn}{dr}\right]}. \]

  • C.3 \[ \boxed{ \rho_{\mathrm{DM}}(r)= \frac{k_BT_0(\beta^2+6\beta r+3r^2)} {4\pi Gm_p r^2(\beta+r)^2} -\frac{\alpha m_p}{r(\beta+r)^2} }\qquad(r>0). \]

Solution

A.1

Choose zero gravitational potential energy when all the mass is dispersed at infinity. The uniform mass density and the mass inside radius \(r\) are

\[ \rho=\frac{3M}{4\pi R^3}, \qquad M(r)=\frac{4\pi}{3}\rho r^3 =M\frac{r^3}{R^3}. \]

Assemble the sphere by bringing successive thin shells from infinity. A shell of mass

\[ dM=4\pi r^2\rho\,dr=\frac{3Mr^2}{R^3}\,dr \]

has interaction energy with the previously assembled interior mass

\[ dU=-\frac{GM(r)}r\,dM. \]

This construction counts every pair interaction once. Therefore,

\[ U=-\int_0^R \frac{G}{r}\left(\frac{Mr^3}{R^3}\right) \left(\frac{3Mr^2}{R^3}\right)dr =-\frac{3GM^2}{R^6}\frac{R^5}{5}. \]

Thus,

\[ \boxed{U=-\frac{3GM^2}{5R}}. \]

A.2

Let \(u_i\) be the line-of-sight velocity of galaxy \(i\) relative to Earth, positive for recession. For \(u_i\ll c\), the Doppler relation to first order is

\[ f_i\simeq f_0\left(1-\frac{u_i}{c}\right), \]

so

\[ u_i\simeq c\frac{f_0-f_i}{f_0}. \]

The internal velocities average to zero in the cluster rest frame. Hence the cluster recession speed is the mean of these measured line-of-sight velocities:

\[ V_{cr}=\frac1N\sum_{i=1}^N u_i \simeq \frac{c}{Nf_0}\sum_{i=1}^N(f_0-f_i). \]

Consequently,

\[ \boxed{ V_{cr}\simeq c\left(1-\frac{1}{Nf_0}\sum_{i=1}^Nf_i\right) =c\left(1-\frac{\bar f}{f_0}\right)}. \]

A.3

The line-of-sight component of galaxy \(i\)'s velocity relative to the cluster centre is

\[ w_i=u_i-V_{cr} \simeq-\frac{c}{f_0}(f_i-\bar f). \]

For an isotropic velocity distribution, the mean squared velocity is shared equally among three mutually perpendicular directions:

\[ \langle v_x^2\rangle =\langle v_y^2\rangle =\langle v_z^2\rangle. \]

Taking the \(z\)-direction along the line of sight gives

\[ v_{rms}^2 =\langle v_x^2+v_y^2+v_z^2\rangle =3\langle w^2\rangle =\frac3N\sum_{i=1}^Nw_i^2. \]

Thus,

\[ \boxed{ v_{rms}\simeq\frac{c}{f_0} \sqrt{\frac3N\sum_{i=1}^N(f_i-\bar f)^2}}. \]

Equivalently,

\[ v_{rms}^2\simeq \frac{3c^2}{f_0^2} \left[ \frac1N\sum_{i=1}^N f_i^2 -\left(\frac1N\sum_{i=1}^Nf_i\right)^2 \right]. \]

Using the representative total galaxy mass \(m\), the mean kinetic energy in the cluster rest frame is

\[ \boxed{\langle K_{\mathrm{gal}}\rangle =\frac12m\langle v^2\rangle =\frac12m v_{rms}^2}. \]

A.4

Work in the centre-of-mass frame. Differentiating the given quantity gives

\[ \frac{d\Gamma}{dt} =\sum_i\left( \frac{d\vec p_i}{dt}\cdot\vec r_i +\vec p_i\cdot\frac{d\vec r_i}{dt} \right) =\sum_i\vec F_i\cdot\vec r_i+2K. \]

Let \(\vec F_{ij}\) be the force on particle \(i\) due to particle \(j\). Newton’s third law allows the force term to be grouped into unordered pairs:

\[ \sum_i\vec r_i\cdot\vec F_i =\sum_{i<j} \left(\vec r_i\cdot\vec F_{ij} +\vec r_j\cdot\vec F_{ji}\right) =\sum_{i<j}(\vec r_i-\vec r_j)\cdot\vec F_{ij}. \]

For Newtonian gravity,

\[ \vec F_{ij} =-Gm_i m_j\frac{\vec r_i-\vec r_j}{r_{ij}^3}, \qquad r_{ij}=|\vec r_i-\vec r_j|. \]

Therefore,

\[ \sum_i\vec r_i\cdot\vec F_i =-\sum_{i<j}\frac{Gm_i m_j}{r_{ij}} =U, \]

where \(U\) is the total pairwise gravitational potential energy. It follows that

\[ \frac{d\Gamma}{dt}=2K+U. \]

Since \(\Gamma\) remains bounded,

\[ \left\langle\frac{d\Gamma}{dt}\right\rangle_t =\lim_{\tau\to\infty} \frac{\Gamma(t+\tau)-\Gamma(t)}{\tau}=0. \]

Taking the time average consequently yields

\[ 2\langle K\rangle_t+\langle U\rangle_t=0, \qquad \langle K\rangle_t=-\frac12\langle U\rangle_t. \]

Comparison with the stated theorem gives

\[ \boxed{\gamma=\frac12}. \]

A.5

Treat the cluster as being in stationary virial equilibrium, so its observed bulk properties represent the time averages. Because the dark matter and galaxies have the same rms speed, the total kinetic energy is

\[ K=\frac12 Mv_{rms}^2. \]

Combining the virial theorem with A.1,

\[ 2K=-U \quad\Longrightarrow\quad Mv_{rms}^2=\frac{3GM^2}{5R}. \]

For \(M>0\),

\[ M=\frac{5Rv_{rms}^2}{3G}. \]

The total visible mass is \(Nm_g\). Subtracting it from the total mass gives

\[ \boxed{M_{\mathrm{DM}} =\frac{5Rv_{rms}^2}{3G}-Nm_g}. \]

B.1

The stellar mass density is

\[ \rho_\star=nm_s. \]

Neglect the total mass of the sparse stars beyond \(R_g\), while retaining them as test particles whose orbits can trace the outer gravitational field. The enclosed stellar mass is then

\[ M_\star(r)= \begin{cases} \displaystyle \frac{4\pi}{3}nm_s r^3,&r<R_g,\\[5pt] \displaystyle \frac{4\pi}{3}nm_s R_g^3,&r\ge R_g. \end{cases} \]

By the shell theorem, only enclosed mass contributes to the gravitational acceleration at \(r\). Equating this acceleration to the centripetal acceleration gives

\[ \frac{v(r)^2}{r}=\frac{GM_\star(r)}{r^2}, \qquad v(r)^2=\frac{GM_\star(r)}r. \]

Thus,

\[ \boxed{ v(r)= \begin{cases} \displaystyle r\sqrt{\frac{4\pi Gnm_s}{3}},&0\le r<R_g,\\[7pt] \displaystyle \sqrt{\frac{4\pi Gnm_sR_g^3}{3r}},&r\ge R_g. \end{cases}} \]

For the sketch, define

\[ v_g=R_g\sqrt{\frac{4\pi Gnm_s}{3}}. \]

The curve is a straight line from \((0,0)\) to \((R_g,v_g)\). It is continuous and has its maximum at that point. Beyond \(R_g\), it follows

\[ v(r)=v_g\sqrt{\frac{R_g}{r}}, \]

decreasing toward the horizontal asymptote \(v=0\).

B.2

At \(r=R_g\), the observed circular speed is \(v_0\). The circular-orbit condition gives

\[ \frac{v_0^2}{R_g}=\frac{Gm_R}{R_g^2}. \]

Therefore, the total enclosed mass, including dark matter, is

\[ \boxed{m_R=\frac{v_0^2R_g}{G}}. \]

B.3

For any spherically symmetric mass distribution traced by circular orbits,

\[ M_{\mathrm{enc}}(r)=\frac{rv(r)^2}{G}. \]

The stated approximation to the observed rotation curve is

\[ v(r)= \begin{cases} v_0r/R_g,&r<R_g,\\ v_0,&r\ge R_g. \end{cases} \]

Hence,

\[ M_{\mathrm{enc}}(r)= \begin{cases} \displaystyle \frac{v_0^2r^3}{GR_g^2},&r<R_g,\\[6pt] \displaystyle \frac{v_0^2r}{G},&r\ge R_g. \end{cases} \]

A spherical shell has mass \(dM_{\mathrm{enc}}=4\pi r^2\rho_{\mathrm{tot}}\,dr\), so

\[ \rho_{\mathrm{tot}}(r) =\frac{1}{4\pi r^2}\frac{dM_{\mathrm{enc}}}{dr} = \begin{cases} \displaystyle \frac{3v_0^2}{4\pi GR_g^2},&r<R_g,\\[7pt] \displaystyle \frac{v_0^2}{4\pi Gr^2},&r\ge R_g. \end{cases} \]

Subtracting the stellar density \(nm_s\) inside the visible edge, and neglecting the stellar density outside it, gives

\[ \boxed{ \rho_{\mathrm{DM}}(r)= \begin{cases} \displaystyle \frac{3v_0^2}{4\pi GR_g^2}-nm_s,&r<R_g,\\[7pt] \displaystyle \frac{v_0^2}{4\pi Gr^2},&r\ge R_g. \end{cases}} \]

The enclosed mass is continuous at \(R_g\), although the density can jump there in this idealized model.

C.1

Here \(m'(r)\) denotes the enclosed mass specified in the problem; the prime is part of its notation, not a derivative.

Consider a small gas element of radial thickness \(dr\) and cross-sectional area \(dA\). Its mass is

\[ dm=m_p n(r)\,dA\,dr. \]

The net outward pressure force is

\[ dF_P=-\frac{dP}{dr}\,dA\,dr. \]

The inward gravitational force has magnitude

\[ dF_G=\frac{Gm'(r)}{r^2}\,m_p n(r)\,dA\,dr. \]

Hydrostatic equilibrium requires these forces to balance. Therefore,

\[ \boxed{\frac{dP}{dr} =-\frac{Gm'(r)m_p n(r)}{r^2}}. \]

The negative sign expresses the outward decrease of pressure needed to support the gas against inward gravity.

C.2

For an ideal gas,

\[ P(r)=n(r)k_BT(r). \]

Differentiating,

\[ \frac{dP}{dr} =k_B\left[T(r)\frac{dn}{dr} +n(r)\frac{dT}{dr}\right]. \]

Substituting into the hydrostatic equilibrium equation and solving for the enclosed mass gives

\[ m'(r) =-\frac{k_Br^2}{Gm_p n(r)} \left[T(r)\frac{dn}{dr} +n(r)\frac{dT}{dr}\right]. \]

Thus,

\[ \boxed{ m'(r)=-\frac{k_Br^2}{Gm_p} \left[\frac{dT}{dr}+\frac{T(r)}{n(r)}\frac{dn}{dr}\right]}. \]

An equivalent form is

\[ m'(r) =-\frac{k_BT(r)r}{Gm_p} \left[ \frac{d\ln n}{d\ln r} +\frac{d\ln T}{d\ln r} \right]. \]

C.3

For the isothermal gas, \(T(r)=T_0\), so C.2 becomes

\[ m'(r) =-\frac{k_BT_0r^2}{Gm_p}\frac{d\ln n}{dr}. \]

The prescribed number density satisfies

\[ \ln n=\ln\alpha-\ln r-2\ln(\beta+r), \]

and therefore

\[ \frac{d\ln n}{dr} =-\frac1r-\frac{2}{\beta+r}. \]

The total enclosed mass is consequently

\[ m'(r) =\frac{k_BT_0}{Gm_p} \left(r+\frac{2r^2}{\beta+r}\right) =\frac{k_BT_0}{Gm_p} \frac{r(\beta+3r)}{\beta+r}. \]

Differentiate this enclosed mass:

\[ \frac{d\,m'(r)}{dr} =\frac{k_BT_0}{Gm_p} \frac{\beta^2+6\beta r+3r^2}{(\beta+r)^2}. \]

Thus the total mass density is

\[ \rho_{\mathrm{tot}}(r) =\frac{1}{4\pi r^2}\frac{d\,m'(r)}{dr} =\frac{k_BT_0(\beta^2+6\beta r+3r^2)} {4\pi Gm_p r^2(\beta+r)^2}. \]

The gas mass density is

\[ \rho_{\mathrm{gas}}(r)=m_p n(r) =\frac{\alpha m_p}{r(\beta+r)^2}. \]

Subtracting it from the total density yields

\[ \boxed{ \rho_{\mathrm{DM}}(r) = \frac{k_BT_0(\beta^2+6\beta r+3r^2)} {4\pi Gm_p r^2(\beta+r)^2} -\frac{\alpha m_p}{r(\beta+r)^2} }\qquad(r>0). \]

The solution was checked against the official solution.