Lecture Eight: Galactic Dynamics: motion under gravity

Lecture Eight:

Galactic Dynamics: motion under gravity

Sparke & Gallagher, chapter 313th May 2014

http://www.astro.rug.nl/~etolstoy/pog14

Material for this lecture was also taken from book: Binney & Tremaine Galactic Dynamics (2nd edition), chapter 2

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Stellar DynamicsPlanets orbit stars, stars and gas travel around the galaxies, and galaxies move within groups and clusters all under the force of gravity.

Stars are so much denser than any interstellar gas they move through that neither gas pressure nor forces from the embedded magnetic fields can impede their travels. It is all and only gravity that determines their orbits.

If we know how the mass is distributed in a galaxy then we can calculate the gravitational forces, and from this we can predict how the positions and velocities of stars will change with time.

This means that knowing the positions and velocities of stars allows the derivation of the mass distribution of the galaxy, irrespective of whether the mass is visible or not.

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We can consider stars as point masses since their sizes are much smaller than their separations.

Newton's law of gravity tells us that a point mass M attracts a second mass m separated from it by distance r, causing a the velocity v of m to change according to:

d

dt(mv) = �GmM

r3r

Gravitational constant

G = 6.67384⇥ 10�11m3kg�1s�2

The mass, m , cancels out of this equation, so the acceleration (dv/dt) is independent of the star's mass.

Newton's Law of Gravity & methods for computing gravitational forces

For N stars with masses mα (α=1,2,...N), at positions xα, we can add the forces on star mass m from all the other stars:

d

dt(m�v�) = �

�

�,� �=�

Gm�m�

|x� � x� |3 (x� � x�)�X

↵

Gm m↵

|x� x↵|3(x� x↵)

This is the principle of equivalence, between gravitational and inertial mass.

F = GmM

r2F = ma

This is the reason that light and heavy objects fall equally fast on Earth.

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The principle of equivalence means that we can write the force from a cluster of stars on a star of mass m at position x as the gradient of the gravitational potential Φ(x).

Φ(x) ➛ 0 at large distances

To compute the gravitational potential of a large collection of stars, we can simply add all the point-mass potentials of the stars together.


the gradient of a scalar is a vector, rf =@f

@x

x+@f

@y

y +@f

@z

z

d

dt(mv) = �m��(x) with �(x) = �

�

�

Gm�

|x� x�| for x �= x�d

dt(mv) = �m��(x) with �(x) = �

�

�

Gm�

|x� x�| for x �= x�where

This is not practical for the large number (≈ 1011) of stars in a typical galaxy, and so the potential is modeled by smoothing the mass density on a scale that is small compared to the size of the galaxy, but large compared to the mean distance between stars.

d

dt(mv) = �

X

↵

Gmm↵

|x� x↵|3(x� x↵)

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if we define the gravitational potential:


To calculate the force F(x) on a particle of mass ms at position x that is generated by the gravitational attraction of a distribution of mass ρ(x′) we use Newton’s inverse-square law of gravitation.

⇢ =M

V

this is the gravitational field, the force per unit mass

The potential Φ is useful because it is a scalar field that is easier to visualize than the vector gravitational field g(x) but contains the same information. In many situations the easiest way to obtain g is first to calculate Φ and then to take its gradient.

F = GmM

r2

summing the small contributions to the overall force from each small element of volume d3x' located at x'.

so we can rewrite g(x):

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If we take the divergence of


Therefore, any contribution to the integral must come from the point x′ = x, and we may restrict the volume of integration to a small sphere of radius h centered on this point. Since, for sufficiently small h, the density will be effectively constant through this volume, we can take ρ(x′) out of the integral.

Uses the divergence theorem to convert a volume integral into a surface integral.

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On the sphere |x′ − x| = h we have

d2S′ = (x′ − x)h d2Ω,

where d2Ω is a small element of solid angle.

Poisson's equation is a differential equation that can be solved for Φ(x) given ρ(x).

If we integrate both sides of the Poisson Eqn over an arbitrary volume containing a mass M, and then apply the divergence theorem

Gauss' theorem: the integral of the normal component of ∇Φ over any closed surface equals 4πG times the mass contained within that surface.


To chose an approximation for the density ρ(x), we can select a mathematically convenient form for the potential Φ(x) and then calculate the corresponding density. We must take care that ρ(x) ≥ 0 everywhere in a potential.

we obtain Poisson’s equation, relating the potential Φ to the density ρ

If we substitute g(x) from this equation

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Since g is determined by the gradient of a potential, the gravitational field is conservative, that is, the work done against gravitational forces in moving two stars from infinity to a given configuration is independent of the path along which they are moved, and is defined to be the potential energy of the configuration.

Similarly, the work done against gravitational forces in assembling an arbitrary continuous distribution of mass ρ(x) is independent of the details of how the mass distribution was assembled, and is defined to be equal to the potential energy of the mass distribution.

The divergence of the gravitational field gives the density distribution (or source) of the mass which determines the gravitational field.

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Newton proved two results that enable us to easily calculate the gravitational potential of any spherically symmetric distribution of matter:

Newton's first theorem: a body that is inside a spherical shell of matter experiences no net gravitational force from that shell.

Newton's second theorem: the gravitational force on a body that lies outside a spherical shell of matter is the same as it would be if all the shell's matter were concentrated into a point at its centre.


Proof of First Theorem: A star at S experiences a gravitational pull from the matter at A within a narrow cone of solid angle ΔΩ, and a force in the opposite direction from mass within cone towards B. From symmetry the line AB makes the same angle with OA and OB. Thus the ratio of mass enclosed is (SA/SB)2; by the inverse square law, the forces are exactly equal, and thus cancel each other out. Thus there is no force on star at S, and the potential Φ(x) must be constant within the shell.

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Proof of Second Theorem:

Compare the potential Φ at a point p located a distance r from the centre of a spherical inner shell of mass M and radius a (r > a), with the potential Φ′ at a point p′ located a distance a from the center of an outer shell of mass M and radius r. The contribution δΦ to the potential at p from the portion of the inner shell with solid angle δΩ located at q′ is


Because |p-q'| = |p'-q|, by symmetry, it follows that δΦ = δΦ' and then by summation over all points q and q' that Φ = Φ'.

The contribution δΦ′ of the matter in the outer shell near q to the potential at p′ is

We already know that Φ′ = −GM/r, therefore Φ = −GM/r, which is exactly the potential that would be generated by concentrating the entire mass of the inner shell at its center.

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From Newton’s first and second theorems, it follows that the gravitational attraction of a spherical density distribution ρ(r′) on a unit mass at radius r is entirely determined by the mass interior to r:

where

The total gravitational potential may be considered to be the sum of the potentials of spherical shells of mass dM(r) = 4πρ(r)r2dr, so we may calculate the gravitational potential at r generated by an arbitrary spherically symmetric density distribution ρ(r′) by adding the contributions to the potential produced by shells (i) with r′ < r, and (ii) with r′ > r.

In this way we obtain

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An important property of a spherical matter distribution is its circular speed vc(r), defined to be the speed of a particle of negligible mass (a test particle) in a circular orbit at radius r. We may readily evaluate vc by equating the gravitational attraction |F| to the centripetal acceleration vc2/r:

The associated angular frequency is called the circular frequency

So vc ∝ r–1/2 and since the mass increases (or at least stays flat) with radius, the circular velocity in a spherical galaxy can never fall off more steeply than r–1/2 which is Keplerian:

The circular speed and frequency measure the mass interior to r.

Another important quantity is the escape speed ve

A star at r can escape from the gravitational field Φ only if it has a speed at least as great as ve(r), as only then does its (positive) kinetic energy (1/2v2) exceed the absolute value of its (negative) potential energy Φ. The escape speed at r depends on the mass both inside and outside r.

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Putting this into practise....

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Very Large Array (VLA)

Westerbork Synthesis Radio Telescope (WSRT)

Interferometer Observations...

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HI Rotation Curve: NGC3198

Bosma 1978, PhD thesis

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Velocity field

Line profiles Doppler effect

Rotation of a galactic disc

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80 km/s

250 km/s

130 km/s

150 km/s

10 km/s

200 km/s

Channel maps showing gas at a certain velocity along the line of sight

Data Cube

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Major axis

Minor axis

If the data cubes are sliced along a spatial direction one produces

position-velocity diagrams

Minor axis

Position-Velocity Plots

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Begeman 1987

Optical disc

HI Rotation Curve

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Rotation curves of disc galaxies

Circular speed of a spherical system According to the second Newton’s theorem, the gravitational force on a test particle that lies outside a closed spherical shell of matter is the same as it would be if the matter were concentrated at its centre:

2.1) where M(r) is the mass contained inside the radius r :

2.2) . The corresponding circular speed is:

2.3) . As a consequence if the entire mass of the system is concentrated within a radius r=rK, beyond that radius M(r)=M=constant. Thus, for r > rK, the circular velocity becomes:

2.4) frequently referred to as the Keplerian fall.

Circular Speed and Spherical systems

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Spherical power-law density profile It is instructive to derive the circular speed for a spherical body with a density profile that follows the power law:

2.5) if the mass interior to r (calculated integrating eq. 2.5 between 0 and r) is:

2.6) and the corresponding circular speed is:

2.7) . Eq. 2.7 suggests that, in order to have a flat rotation velocity (as observed in the outer parts of disc galaxies), we would need , i.e. the mass density should be proportional to r -2. In this case eq. 2.5 becomes that of a so-called singular isothermal sphere.

Circular Speed of a power law density profile

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Lecture Eight: Galactic Dynamics: motion under gravity