Galaxy Dynamics part 2Physics of Galaxies 2012part 10

1

Orbits of stars in spherical systems

In a time-independent gravitational potential, energy is conserved:

because

In a spherical potential, the angular momentum is also conserved:

so

but the force always points towards the center, so does not change

��/�t = 0

E � 12mv2 + m�(x) = constant

L ⇥ x�mv

��L

dLdt

= x⇥ m dvdt

= �mx⇥⇤�

2

Therefore the motion of a star is restricted to the orbital plane

and only two coordinates are needed to describe the location of the star

typically, polar coordinates in the plane are used to describe the motion

(r, �)

3

In an axisymmetric system, like a disk, we use (of course) a cylindrical coordinate system where z=0 corresponds to the symmetry plane (e.g., the mid-plane of the disk)

In an axisymmetric system, the mass distribution and therefore the potential is independent of the angular coordinate:

no force in the ϕ-direction: stars conserve angular momentum about the z-axis

neglect non-axisymmetric features like the bar and the spiral arms!

Orbits in an axisymmetric galaxy

(R,�, z)

⇥�/⇥� = 0

4

The equations of motion for a star in the disk are

In each direction, using , we have

Equation (3) implies conservation of angular momentum around the z-axis:

d2rdt2

= ���

r = RR + zz

Lz � R2 d�

dt= constant

d2R

dt2�R

�d�

dt

�2

= �⇥�⇥R

(1)

d2z

dt2= �⇥�

⇥z(2)

d(R2d�/dt)dt

= �⇥�⇥�

= 0 (3)remember, no force in the φ-direction!

5

In the R-direction, we can rewrite Equation (1) as

where the effective potential is

If we multiply Equation (4) by and integrate with respect to t, we find

a sort of energy conservation law!

(4)

�e� � �(R, z) +L2

z

2R2

�e�(R, z;Lz)

d2R

dt2= R

�d�

dt

�2

� ⇥�⇥R

= �⇥�e�

⇥R

dR/dt

12

�dR

dt

�2

+ �e�(R, z;Lz) = constant

6

Therefore the effective potential acts as a potential energy for the star’s motion in R and z

The effective potential is constant if

and

The second equation is satisfied for motion in the mid-plane, so that z=0; combined with , this implies a circular orbit in the disk mid-plane

The radius of this circular orbit is Rg, where

�e�(R, z;Lz)

��e�

�z=

���z

= 0

dR/dt = 0

⇥�dR

����Rg

=L2

z

R3g

= Rg

�d�

dt

�2

@�e↵

@R= 0 and thus

@�

@R� L2

z

R3= 0

7

Since , the term in the effective potential acts as an angular momentum barrier preventing a star coming closer to R=0 than some perigalactic radius where

The circular orbit is that with the least energy for a given angular momentum Lz

Effective potential for a star with Lz=0.595 in a Plummer potential

R2 � 0 L2z

R = 0

8

EpicyclesLet’s now derive approximate solutions for the equations of motion for stars on (nearly) circular orbits in the disk (symmetry) plane

Define a distance around Rg, , and expand the effective potential around (Rg,0):

x = R�Rg

�e↵(R, z) ⇠ �e↵(Rg, 0) +@�e↵

@R

����Rg,0

x+@�e↵

@z

����Rg,0

z + · · ·

9

Let’s define two quantities:

Then the equations of motion become

These are the equations of motion of two decoupled harmonic oscillators with frequencies κ and ν

κ is the epicyclic frequency:

ν is the vertical frequency:

�2 =⇤2�e�

⇤R2

����Rg,0

and ⇥2 =⇤2�e�

⇤z2

����Rg,0

�2(Rg) =⇥2�⇥R2

����Rg,0

+ 3L2

z

R4g

�2 =⇥2�⇥z2

����Rg,0

d2z

dt2= �⇥�e�

⇥zso

d2z

dt2= �z

⇥2�e�

⇥z2

����Rg,0

=� d2z

dt2= ��2z

d2R

dt2= �⇥�e�

⇥Rso

d2x

dt2= �x

⇥2�e�

⇥R2

����Rg,0

=� d2x

dt2= ��2x

10

The solution to the equations of motion is then

The motion of a star in the disk can be described as the oscillation about a guiding center at Rg moving on a circular orbit

x = X0 cos(⇥t + �) for ⇥2 > 0z = Z0 cos(⇤t + �) for ⇤2 > 0

2⇥X/�

�(Rg)

11

In order to conserve angular momentum Lz, the azimuthal speed must also vary:

Integrating, we find

The first two terms give the guiding center motion (ϕ0 is an arbitrary constant); the third term is harmonic motion with the same frequency as the x oscillation but 90º out of phase and with a different amplitude

d�

dt=

Lz

R2=

⇥(Rg)R2

(Rg + x)2⇥ ⇥(Rg)

�1� 2x

Rg+ · · ·

�

�(t) = �0(t) + !gt�1

Rg

2!g

X0 sin(t+ )

12

This motion is known as the epicyclic motion

It is retrograde because it is in the opposite sense to the guiding center’s motion: speeds the star up closer to the center and slows it down farther out

2⇥X/�

�(Rg)

13

Note that the approximation to second order in z in the effective potential is only valid if the density is constant in the z-direction (since ). But the disk density decreases exponentially away from the mid-plane — so the approximation is valid at most one scale height away from the plane (z<300 pc). Since a good fraction of the disk stars move to greater heights, the motion in the z-direction is not well-described as a harmonic oscillation...

(�e� � z2)�2� � �

14

There is a relation between the epicyclic frequency κ and the angular frequency ω:

First, the centrifugal force=gravitation pull, so

and

so

In general,

for a sphere of uniform density,for the Kepler problem (a point mass),

R�2 =⇥�⇥R

�2 =L2

z

R4

�2 =�

Rd⇥2

dR+ 4⇥2

�

Rg

� � � � 2�

⇥(R) = cnst and � = 2⇥

⇥ � r�3/2and � = ⇥

15

So the epicyclic frequency is also related to the Oort constants...

Recall that and

so at the Sun,

Using the measured value of B,

So the Sun makes 1.3 radial oscillations in the time it takes to make one complete revolution around the Galactic Center...

A = �12R

d�

dR

����R0

B = ��

� +12R

d�

dR

�����R0

�20 = �4B(A�B) = �4B⇥0

�0/⇥0 � 1.3± 0.2

16

All of these equations can actually be derived from the “collisionless Boltzmann equation” (or the Liouville equation), which describes the paths of stars in phase space

Let’s assume we have a large number of stars moving in a smooth potential

Then phase space is well-populated and a distribution function applies: is the number of particles per unit volume in phase space

The Collisionless Boltzmann Equation (CBE)

f(x,v, t)d3x d3v

(x,v)

17

Since the potential is smooth, neighboring particles in phase space (at the same x,v) move together, and we can use the fluid approximation:

Finally, the flow is smooth: stars do not jump discontinuously from one region of phase space to another — no collisions

Then f has to satisfy a continuity equation:

w = (x,v)w = (v,���)

phase space divergence

phase space “current”

(5)�f

�t+�6 · (fw) = 0

18

Expand equation (5):

Consider the term . This is the divergence of the flow in phase space. Let’s now show that this is incompressible:

Flow in phase space is incompressible as long as the forces do not depend on particle velocity

�f

�t+ f�6 · (w) + w ·�6f = 0

�6 · (w)

�6 · (w) =6�

i=1

�wi

�wi=

3�

i=1

��vi

�xi+

�vi

�vi

�= 0

vi and xi are independent assumed a potential, so�vi/�vi = 0

19

The result is the CBE for incompressible flow in phase space:

or

This means that the convective derivative, the change in the local density of phase-space particles seen by an observer moving with the phase-space fluid at a velocity is zero:

�f

�t+ w ·�6f = 0

�f

�t+ v ·⇥f �⇥� · �f

�v= 0

(6)

w Df

Dt=

�f

�t+

6�

i=1

wi�f

�wi= 0

Note that this is also called the “Lagrangian derivative”

20

This Liouville’s theorem: phase-space densities along particle trajectories are constant

Two key assumptions:

No small scale structure in the potential inside a volume element and no discontinuous jumps out of the volume element, so stars in the element are conserved: collisionless system

No frictional drag or encounters where energy and/or momentum can be exchanged with other stars: no entropy increase

21

We are going to “take moments” of the CBE by multiplying f by powers of v

First, let’s define the space density (of the stellar population component we’re interested in) as

This is the zeroth moment in the velocity of the distribution function

Moments of the CBE: the Jeans Equations

� ��

f d3v

22

The first and second moments in velocity of f are

Now, the zeroth moment (in velocity) of the CBE is

note that here I’ve used the summation convention, where repeated indices are summed over

�vi� =1�

�vif d3v

�vivj� =1�

�vivjf d3v

��f

�td3v +

�vi

�f

�xid3v � ��

�xi

��f

�vid3v = 0

23

This can be rewritten as

But the last term is 0, because

And so

or

which is the continuity equation in x space (conservation of mass)

�

�t

�f d3v +

�

�xi

�vif d3v � ��

�xi

� �dv1dv2

� v3=+�

v3=���f = 0

f(vi)|+��� = 0 if limvi��

f = 0

⇥�

⇥t+

⇥

⇥xi(�vi) = 0

⇥�

⇥t+� · (�v) = 0 (7)

24

This can be rewritten as

But the last term is 0, because

And so

or

which is the continuity equation in x space (conservation of mass)

�

�t

�f d3v +

�

�xi

�vif d3v � ��

�xi

� �dv1dv2

� v3=+�

v3=���f = 0

f(vi)|+��� = 0 if limvi��

f = 0

⇥�

⇥t+

⇥

⇥xi(�vi) = 0

⇥�

⇥t+� · (�v) = 0 (7)

24

The first moment in velocity of the CBE is

Now,

so

�vj

�f

�td3v +

�vivj

�f

�xid3v � ��

�xi

�vj

�f

�vid3v = 0

�vj

⇤f

⇤vid3v =

� �dv1dv2

� v3=+�

v3=��⇤f �

� �⇤vj

⇤vi

�f d3v

= 0� �ij⇥⇥

⇥t(��vj�) +

⇥

⇥xi(��vivj�) + �

⇥�⇥xj

= 0

25

Subtracting off the zeroth moment, we have

We usually rewrite this in terms of the velocity dispersion tensor, the velocity dispersion about the mean streaming motion:

�⇥�vj�⇥t

� �vi�⇥

⇥xi(��vj�) +

⇥

⇥xi(��vivj�) + �

⇥�⇥xj

= 0

�2ij = �(vi � �vi�)(vj � �vj�)� = �vivj� � �vi��vj�

26

Substituting and rearranging, we finally have

The left-hand side of this equation is the derivative of the momentum:

The right-hand side is the sum of the gravity and a stress term, which is the anisotropic pressure

This equation is a force equation: the conservation of momentum along particle trajectories

�⇤�vj�⇤t

+ ��vi�⇤�vj�⇤xi

= ��⇤�⇤xj

�⇤(�⇥2

ij)⇤xi

(8)

�Dv/Dt

27

Since the velocity dispersion tensor is symmetric, it can be diagonalized. This diagonalized tensor is the velocity ellipsoid, which we met earlier...

So what use are the Jeans equations (moments of the CBE)?

Can relate observables like , v, and to the gravitational potential

they give us a way to “weigh” galaxies

�ij

�2ij�

��/�xi

28

But!

Jeans equations describe a massless tracer population in an external potential — need to add Poisson’s equation to get the potential from the (total) mass density ρ: Jeans equations are incomplete

There is no “equation of state” relating ν to σ (like the ideal gas law relates ρ and T in a gas). You have to assume , and thus f(v) — every different assumption leads to a different solution: Jeans equations depend on f(v) and are non-unique

The equations never close; have always to assume a higher-order tensor, i.e., some form for f(v).

�ij

29

Let’s evaluate the quantity

the amount a population lags behind the circular velocity

Write second moment of the Jeans equation in cylindrical coordinates to find:

In the Solar Neighborhood, the term in brackets is ~4

Asymmetric drift: an application of the Jeans eqn

va � vc � �v��

2vcva ⇥ ⇤v2R⌅

�⇥2

⇤v2R⌅

� 32� 2

⇤ ln �

⇤ lnR+

12⇤v2

z⌅⇤v2

R⌅±

�⇤v2

z⌅⇤v2

R⌅� 1

��

30

So the lag is

In the absence of radial streaming,

As increases, the population lags more behind the circular motion — hot populations rotate slower

Energy comes out of ordered motion and is put into disordered motion

�v2R� = �2

R

va � 4�v2R�/440 km s�1 � �v2

R�/110 km s�1

�2R

31

This is why depends on the velocity dispersion of the population used to measure it — older populations have higher velocity dispersions and rotate slower

V�

Stromberg’s asymmetric drift equation [e.g. equation (4-34) of

Binney & Tremaine (1987, hereafter BT)]. That is, V increases

systematically with S2 because the larger a stellar group’s velocity

dispersion is, the more slowly it rotates about the Galactic Centre

and the faster the Sun moves with respect to its lagging frame.

For very early-type stars with B ¹ V ! 0:1 mag and/or S!

15 km s¹1, the V-component of vh i decreases with increasing S,

colour, and hence age contradicting the explanation given in the last

paragraphs. However, the stars concerned are very young, and there

are several possibilities for them not to follow the general trend.

First, because of their youth these stars are unlikely to constitute a

kinematically well-mixed sample; rather, they move close to the

orbit of their parent cloud; many will belong to a handful of moving

groups. Secondly, Stromberg’s asymmetric drift relation predicts a

linear relation between V and S2 only if both the shape of the

velocity ellipsoid, i.e. the ratios of the eigenvalues of j, and the

radial density gradient are independent of S. Young stars probably

violate these assumptions, especially the latter one.

The radial and vertical components U0 and W0 of the velocity of

the Sun with respect to the LSR (the velocity of the closed orbit in

the plane that passes through the location of the Sun) can be derived

from hvi estimated for all stars together. The component in the

direction of rotation, V0, may be read off from Fig. 4 by linearly

extrapolating back to S ¼ 0. Ignoring stars blueward of

B ¹ V ¼ 0 mag we find

U0 ¼ 10:00 " 0:36 ð"0:08Þ km s¹1;

V0 ¼ 5:25 " 0:62 ð"0:03Þ km s¹1;

W0 ¼ 7:17 " 0:38 ð"0:09Þ km s¹1;

ð20Þ

where the possible effect arising from a systematic error of 0:1 mas

in the parallax is given in brackets. When we use this value of the

Local stellar kinematics 391

! 1998 RAS, MNRAS 298, 387–394

Figure 3. The components U, V and W of the solar motion with respect to stars with different colour B ¹ V. Also shown is the variation of the dispersion S with

colour.

Figure 4. The dependence of U, V and W on S2. The dotted lines correspond to the linear relation fitted (V) or the mean values (U and W) for stars bluer than

B ¹ V ¼ 0:

32

Weighing the MW diskLet’s use the Jeans eqns to see if we can measure the mass of the MW disk in the Solar Neighborhood

First, select a tracer population like K dwarfs

Now assume that the potential is constant with time

Then ν and f are also constant

High above the plane,

Then the first Jeans eqn, Eq. (7), everywhere

�vz��(z) � 0

�vz� = 0

33

Now we can use the second Jeans eqn, Eq. (8):

Next we need Poisson’s equation in cylindrical coordinates:

Then we can write

4�G⇥(R, z) =⇤2�⇤z2

+1R

⇤

⇤R

�R

⇤�⇤R

�

���R

=v2

c (R)R

d

dz[�(z)⇥2

z ] = �⇤�⇤z

�(z)

34

and so we have

Near the Sun, the rotation curve is nearly flat, so the last term can be ignored

To find ρ we need to differentiate ν twice, which is very sensitive to small errors! Let’s concentrate on the surface density instead:

4⇥G⇤(R, z) =d

dz

�� 1

�(z)d

dz[�(z)⌅2

z ]�

+1R

⇧v2c (R)⇧R

2⇥G�(< z) = 2⇥G

� +z

�z⇤(z�)dz� � � 1

�(z)d

dz[�(z)⌅2

z ]

35

Oort was the first to try this (and derive the equations), using K giants and F dwarfs

He assumed σz did not vary with distance from the plane and found

But at z>1 kpc, Σ(z) began to decrease!

σz is not constant with distance from the plane...

Using K dwarfs and allowing σz to vary, Kuijken & Gilmore (1991) found

�(< 700 pc) � 90 M� pc�2

�(< 1100 pc) = 71± 6 M� pc�2

36

The total surface density in gas and stars in this range is ~40–55 M⊙ pc–2

But not all of the surface density determined by K&G is actually in the disk (some must be in the halo), so the amount of dark matter in the disk is probably negligible

37