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IOAA 2019, Теория — Задача 13. Distance to the Coma galaxy cluster

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Theory problem sheet · 40 т.

Условие

The Coma galaxy cluster (Abell 1656) has an angular diameter on the sky of about 100 arcminutes, and contains more than 1000 individual galaxies, most of which are dwarf and giant ellipticals orbiting the common center of mass of the cluster in approximately circular orbits. The table below lists the measured radial velocities of a few individual cluster member galaxies.

No.	$v_{\mathrm{r}}$ (km/s)	No.	$v_{\mathrm{r}}$ (km/s)	No.	$v_{\mathrm{r}}$ (km/s)	No.	$v_{\mathrm{r}}$ (km/s)
1	6001	6	7116	11	7156	16	7111
2	7666	7	7004	12	7522	17	8292
3	6624	8	4476	13	7948	18	5358
4	5952	9	6954	14	4951	19	4957
5	5596	10	8953	15	7797	20	7183

a) Derive the distance of the cluster from the mean radial velocity of the galaxies listed in the table. [8 т.]

b) Estimate the physical diameter of the cluster (in Mpc). [4 т.]

c) The virial theorem states that if the galaxy cluster is in dynamic equilibrium, then the mean kinetic energy, $\langle K\rangle$ , and the mean gravitational potential energy, $\langle U\rangle$ , are related by

-2\langle K\rangle = \langle U\rangle

assuming the Coma cluster is spherical.

For simplicity, assume that each galaxy has approximately the same mass, $m$ .

Use the virial theorem to prove that, in this case, the cluster mass $M$ (also called as the virial mass) can be expressed as

M = \frac{5R}{G}\,\sigma_{\mathrm{r}}^2

where $\sigma_{\mathrm{r}}^2$ is the velocity dispersion of the cluster.

The formula for standard deviation:

\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}

[10 т.]

d) Using the data in the table, estimate the virial mass of the Coma cluster in solar masses. [12 т.]

e) The total luminosity of the Coma cluster (in solar luminosity, $L_\odot$ ) is $L \approx 5\times10^{12}L_\odot$ . Calculate the mass-luminosity ratio of the cluster in solar mass per solar luminosity units. [2 т.]

f) Which of the following statements is true (more than one answer is possible)? Mark the letter of your answer with × on the answer sheet.

(A) The mass-luminosity ratio of the Coma cluster is much higher than that of a typical spiral galaxy, like the Milky Way.
(B) The mass-luminosity ratio of the Coma cluster is similar to that of a typical spiral galaxy.
(C) The mass-luminosity ratio of the Coma cluster is much less than that of a typical spiral galaxy.
(D) The Coma cluster contains much more dark matter than a typical spiral galaxy.
(E) The Coma cluster contains much less dark matter than a typical spiral galaxy. [4 т.]

Решение

a)

The mean radial velocity of the cluster, from the data given in the table, is

\langle v_{\mathrm r}\rangle \approx 6731\ \mathrm{km\,s^{-1}}

(5 p)

Using the Hubble law in the following form

v_{\mathrm r} = H_0 D,

(2 p)

where $v_{\mathrm r}$ is the mean radial velocity in km/s, $D$ is the distance in Mpc, and $H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$ is the Hubble constant, we can get

D = \frac{v_{\mathrm r}}{H_0}

With numerical values:

D = \boxed{96\ \mathrm{Mpc}}

(1 p)

b)

The angular diameter of the cluster is $\theta = 100' = 6000''$ . Applying the definition of the parsec, $1''$ angular distance from 1 pc distance corresponds to 1 au (Astronomical Unit, $1\ \mathrm{au} \approx 1.5 \times 10^{8}\ \mathrm{km}$ ), i.e.

d\,[\mathrm{AU}] = \theta\,[''] \times D\,[\mathrm{pc}]

(3 p)

Therefore, $\theta = 6000''$ from $D = 96\ \mathrm{Mpc} = 96 \times 10^{6}\ \mathrm{pc}$ gives (with $1\ \mathrm{Mpc} = 3.086 \times 10^{22}\ \mathrm{m}$ ):

d = 5.77 \times 10^{11}\ \mathrm{au} = 8.63 \times 10^{22}\ \mathrm{m} = \boxed{2.80\ \mathrm{Mpc}}

(1 p)

Alternative solution: $100' = 1.67^{\circ} = 0.0291\ \mathrm{rad}$ . For such small angles $d \approx \theta D$ . If $\theta$ is expressed in radians, then $d$ and $D$ have the same units. Thus, if $D = 96\ \mathrm{Mpc}$ , then

d = 0.0291 \times 96\ \mathrm{Mpc} = \boxed{2.80\ \mathrm{Mpc}}

c)

In the virial theorem

(13.1)\quad \langle K \rangle = \frac{1}{N}\sum_{i=1}^{N} \frac{1}{2} m_i (v_i - v_{\mathrm m})^2

is the mean kinetic energy of $N$ galaxies. The $i^{\mathrm{th}}$ galaxy has mass $m_i$ and velocity $v_i$ , and $v_{\mathrm m}$ is the mean velocity of the cluster.

(13.2)\quad \langle U \rangle = -\frac{3}{5}\,\frac{1}{N}\sum_{i=1}^{N} \frac{GM}{R}\, m_i

is the mean gravitational potential energy of $N$ galaxies (having individual masses $m_i$ as above) filling in a sphere of $R$ radius, while $G$ is the gravitational constant and $M = \sum_{i=1}^{N} m_i$ is the total mass of the cluster.

If the orbital velocity vectors are distributed randomly, then

(13.3)\quad \frac{1}{N}\sum_{i=1}^{N} (v_i - v_{\mathrm m})^2 = \frac{1}{N}\sum_{i=1}^{N} 3(v_{i,\mathrm r} - v_{\mathrm m,r})^2 = \frac{N-1}{N}\sum_{i=1}^{N} \frac{3(v_{i,\mathrm r} - v_{\mathrm m,r})^2}{N-1} \to 3\sigma_{\mathrm r}^2,\ \text{if } N \to \infty,

where $v_{i,\mathrm r}$ is the radial velocity for the $i^{\mathrm{th}}$ galaxy, and $\sigma_{\mathrm r}$ is called the velocity dispersion of the cluster. Combining Eq. (13.1) and Eq. (13.3) one can get:

\langle K \rangle = \frac{3}{2} m \sigma_{\mathrm r}^2

(2 p)

Inserting $\langle K \rangle$ into the equation of virial theorem gives:

-2 \times \frac{3}{2} m \sigma_{\mathrm r}^2 = -3m\sigma_{\mathrm r}^2 = \langle U \rangle

(2 p)

Using Eq. (13.2) to express $\langle U \rangle$ one can get

\langle U \rangle = -\frac{3}{5}\, m\, \frac{GM}{R}\, \frac{1}{N}\sum_{i=1}^{N} 1 = -\frac{3}{5}\, m\, \frac{GM}{R},

(2 p)

because $\sum_{i=1}^{N} 1 = N$ .

Inserting this expression into the one above, we get:

-3m\sigma_{\mathrm r}^2 = -\frac{3}{5}\, m\, \frac{GM}{R}

(2 p)

Dividing by $-3m$ both sides, and expressing $M$ we finally arrive at

M = \boxed{\frac{5R\sigma_{\mathrm r}^2}{G}},

(2 p)

which is the expression of the virial mass.

d)

The radial velocity dispersion of the data given in the table is the same as the standard deviation of the data around the mean. The mean velocity, $\langle v_{\mathrm r}\rangle = 6731\ \mathrm{km\,s^{-1}}$ , was already derived in part a). Thus (depending on how we define the standard deviation):

\sigma_{\mathrm r,N} = \sqrt{\frac{1}{N}\sum_{i=1}^{N} (v_{\mathrm r} - \langle v_{\mathrm r}\rangle)^2} = 1184\ \mathrm{km\,s^{-1}}

\sigma_{\mathrm r,N-1} = \sqrt{\frac{1}{N-1}\sum_{i=1}^{N} (v_{\mathrm r} - \langle v_{\mathrm r}\rangle)^2} = 1214\ \mathrm{km\,s^{-1}}

(10 p)

Inserting this value into the expression of the virial mass and using $R \approx 1.40\ \mathrm{Mpc}$ as the cluster radius (from b)), the result is:

M_N = \frac{5 \times (1.40 \times 3.086 \times 10^{22}\ \mathrm{m}) \times (1.184 \times 10^{6}\ \mathrm{m\,s^{-1}})^2}{6.67 \times 10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}}

M_N = \boxed{4.53 \times 10^{45}\ \mathrm{kg}} \approx \boxed{2.28 \times 10^{15}\ \mathrm{M}_{\odot}}

M_{N-1} = \frac{5 \times (1.40 \times 3.086 \times 10^{22}\ \mathrm{m}) \times (1.214 \times 10^{6}\ \mathrm{m\,s^{-1}})^2}{6.67 \times 10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}}

M_{N-1} = \boxed{4.77 \times 10^{45}\ \mathrm{kg}} \approx \boxed{2.40 \times 10^{15}\ \mathrm{M}_{\odot}}

(2 p)

e)

Using the virial mass from d) and the given total luminosity of the cluster one can get:

\frac{M_N}{L} = \frac{2.28 \times 10^{15}\ \mathrm{M}_{\odot}}{5 \times 10^{12}\ \mathrm{L}_{\odot}} = 455\ \mathrm{M}_{\odot}/\mathrm{L}_{\odot} \approx 460\ \mathrm{M}_{\odot}/\mathrm{L}_{\odot}

\frac{M_{N-1}}{L} = \frac{2.40 \times 10^{15}\ \mathrm{M}_{\odot}}{5 \times 10^{12}\ \mathrm{L}_{\odot}} = 479\ \mathrm{M}_{\odot}/\mathrm{L}_{\odot} \approx 480\ \mathrm{M}_{\odot}/\mathrm{L}_{\odot}

(1 p)

This mass-luminosity ratio is much higher than that of the individual galaxies which is typically in between 1 and 100. (1 p)

f)

(A) and (D) (4 p)

Оригинал в Архива: TH-problem-final.pdf · официални решения: TH-solution-final.pdf · допълнителни данни: TH-anssheet.pdf