Doppler Shift

At each point the emitter is at the center of a circular wavefront extending out from its present location. Spectroscopy, the Doppler Shift and Masses of Binary Stars

http://apod.nasa.gov/apod/astropix.html

The Doppler Shift

c since ν = λ

⎛ v⎞ λ = λ + λ obs emit emit ⎜ c⎟ ⎝ ⎠ The Doppler Shift Astronomical Examples of Doppler Shift c since ν = λ ⎛ v⎞ λ = λ + λ obs emit emit ⎜ c⎟ ⎝ ⎠ • A star or (nearby) galaxy moves towards you or ⎛ v ⎞ = 1 if moving away from you with speed v away from you (cant measure transverse motion) λobs λemit ⎜ + ⎟ ⎝ c ⎠ ⎛ v ⎞ λ = λ 1 − if moving toward you with speed v obs emit ⎝⎜ c ⎠⎟ • Motion of stars in a binary system

Δλ =λobs − λemit v = λ (1 ± − 1) • Thermal motion in a hot gas emit c Δλ v = ± if v is away from you Δλ >0 red shift • Rotation of a star λemit c if v is toward you Δλ < 0 blue shift This formula can only be used when v << c Otherwise, without proof, 1/2 ⎛ 1+ v/c⎞ λ = λ obs emit ⎜ 1 v/c⎟ ⎝ − ⎠

Doppler Shift:

E.g. A H atom in a star is moving away from you at 3.0 x 107 cm s-1 = 0.001 times c.

At what wavelength will you see H ?

λobs = 6562.8 (1+ 0.001) = 6569.4 A

Note that the Doppler shift only measures that part of the velocity that is directed towards or Note – different from a cosmological red shift! away from you. A binary star pair

Again, only see a shift due to motion along our line of sight

Thermal Line Broadening Thermal Line Broadening

In a gas with some temperature T atoms will be moving The full range of wavelengths, hence the width around in random directions. Their average speed will of the spectral line will be depend upon the temperature. Recall that the definition

of temperature, T, is Δλ vaverage 2 3kT 1 3 = 2 = m 〈v2 〉 = k T λ c c m 2 atom 2 atom The mass of an atom is the mass of a neutron or proton −16 −1 (they are about the same) times the total number of both towards away where k =1.38 × 10 erg K Here 〈〉 means "average". Some atoms in the nucleus, this is an integer "A". will be moving faster than the average, 1/2 A = 1 for hydrogen ⎛ 3 1.38×10−16 T ⎞ 0 Δλ ( )( )( ) ⎛ 1 ⎞ 4 for helium others hardly at all. Some will be moving = 2⎜ ⎟ 10 12 for carbon speed λ 1.66×10−24 A ⎝⎜ 2.99×10 ⎠⎟ towards you, others away, still others ⎝ ( )( ) ⎠ 16 for oxygen etc. 3kT across your line of sight. Δλ −6 T v = = 1.05× 10 where T is in K; nb depends on A average λ A matom Thermal Line Broadening ROTATION (as viewed in the plane of the equator) T(in K) Full width = Δλ = 1.05 × 10−6 λ A Eg. H at 5800 K (roughly the photospheric α temperature of the sun)

A = 1 T= 5800 λ = 6563 A

1/2 ⎛ 5800⎞ Δλ = 1.05 x 10−6 (6563) = 0.53 A ⎝⎜ 1 ⎠⎟

This is (another) way of measuring a star's temperature Wien's law (wavelength where most emission comes out) Spectral class (O,B, F, G, K, M and subsets thereof) 1/4 ⎛ L ⎞ L= 4 π R2 σ T4 ⇒ T = ⎝⎜ 4πR2σ ⎠⎟ Thermal line broadening

Note: Potential complications:

1) Star may have both thermal and rotational broadening

2) May see the star at some other angle than in its equatorial plane.

Example: Hα in a star with equatorial rotational speed 100 km/s = 107 cm/s

⎛ v⎞ Full width = Δλ = 2 λ ⎝⎜ c⎠⎟ ⎛ 107 ⎞ =(2)(6563)⎜ 10 ⎟ = 4.4 A ⎝ 3 × 10 ⎠ Average rotational velocities (main sequence stars) 3 sources of spectral line broadening

Stellar v equator Stellar winds and magnetic Class (km/s) torques are thought to be 1) Pressure or Stark broadening (Pressure)

involved in slowing the rotation O5 190 of stars of class G, K, and M. 2) Thermal broadening (Temperature) B0 200

B5 210 Stars hotter than F5 have A0 190 3) Rotational broadening (w, rotation rate) stable surfaces and perhaps A5 160 low magnetic fields. F0 95

F5 25 The sun, a G2 star rotates at One can distinguish rotational broadening from G0 12" 2 km/s at its equator thermal broadening because rotation affects all ions equally, but in a hot gas the heavier ions move slower. -1/2 . Red giant stars rotate very slowly. Single white dwarfs in Thermal width depends on A hours to days. Neutron stars may rotate in milliseconds

SUMMARY

planets? Hyperfine Splitting The 21 cm Line

8) Magnetic fields

From Zeeman splitting

Less tightly bound 9) Expansion speeds in stellar winds and explosions

Supernovae, novae, planetary nebulae

10) From 21 cm - rotation rates of galaxies. Distribution of neutral hydrogen in galaxies. Suns motion in More tightly bound the Milky Way.

Merits: 21 cm (radio) • Hydrogen is the most abundant element in the universe and a lot of it is in neutral atoms - H I  = 21 cm λ =cν  = 1.4 x 109 Hz h = (6.63 x 10-27)(1.4 x 109) = 9.5 x 10-18 erg • It is not so difficult to build big radio telescopes = 5.6 x 10-6 eV • The earths atmosphere is transparent at 21 cm Must have neutral H I

Emission collisionally excited

Lifetime of atom in excited state about 107 yr

Galaxy is transparent to 21 cm

Aerecibo - 305 m radio telescope - Puerto Rico Binary and Multiple Stars (about half of all stars are found in multiple systems)

Getting Masses in Binary Systems

Beta-Cygnus (also known as Alberio) Alpha Ursa Minoris (Polaris) Separation 34.6. Magnitudes 3.0 and 5.3. Separation 18.3. Magnitudes Yellow and blue. 380 ly away. Bound? 2.0 and 9.0. Now known to be a triple. P > 75000 y. The brighter yellow component Separation ~2000 AU for distant pair. is also a (close) binary. P ~ 100 yr. When the star system was born it apparently had too much angular momentum to end up as a single star.

Polaris

1.2 Msun Polaris Ab Type F6 - V 4.5 Msun Polaris A Cepheid

Period 30 yr

Polaris B is F3 - V > 105 years 585 years 1165 years

Castor A and B complete an orbit every 400 years. In their elliptical orbits their separation varies from 1.8 to 6.5. The mean separation is 8 billion miles. Each star is actually a double with period only a few Epsilon Lyra – a double double. days (not resolvable with a telescope). There is actually a C component The stars on the left are separated by 2.3 about 140 AU; that orbits A+B with a period of of about 10,000 years (distance 11,000 AU). those on the right by 2.6 . The two pairs are separated Castor C is also a binary. 6 stars in total by about 208 (13,000 AU separation, 0.16 ly between the two pairs, all about 162 ly distant). Each pair would be about as bright as the quarter moon viewed from the other.

An optical double Stars not really bound. Center of Mass

m r =m r 1 1 2 2

5000 years for Mizar A and B to complete an orbit

6 months

20.5 d

For constant total separation, 20 AU, vary the masses Circular Orbit – Unequal masses Two stars of similar mass but eccentric orbits

Two stars of unequal mass and an eccentric orbit Aside E.g. A binary consisting of

a F0v and M0v star The actual separation between the stars is obviously not constant in the general case shown of non-circular orbits.

The separation at closest approach is the sum of the

semi-major axes of the two elliptical orbits, a = a1+a2, times (1-e) where e is the eccentricity.

At the most distant point the separation is a times (1+e).

So if one measures the separation at closest approach and

farthest separation, adds and divides by 2, one has a1 + a2 http://www.astronomy.ohio-state.edu/~pogge/Ast162/Movies/ - visbin For circular orbits e = 0 and the separation is constant, r1+r2 . farthest separation (a1 +a2 ) (1+ e) (a +a ) (1− e) 1 + 2 closest separation

2 (a +a ) / 2 = (a + a ) 1 2 1 2

Period = 50.1 years semi-major axis of A is 6.4 AU and B is 13.4 AU (In Kepler’s equation use the sum of the semimajor axes)

ASSUME CIRCULAR ORBITS Some things to note: x r 1 r2 m • A binary star system has only one period. The time for CM 2 star A to go round B is the same as for B to go round A m1 both stars feel the same gravitational 2 • A line connecting the centers of A and B always m1v1 Gm1m2 2πr1 2πr2 attraction and thus = = = Period passes through the center of mass of the system 2 both have the same r1 (r1 + r2 ) v1 v2

centrifugal force m v 2 r v • The orbits of the two stars are similar ellipses with = 2 2 ∴ v = 1 2 r 1 r the center of mass at a focal point for both ellipses 2 2 2 2 2 m1r1 v2 m2v2 2 = • For the case of circular orbits, the distance from r1r2 r2 More massive star is r1 m2 the center of mass to the star times the mass of each m r = m r = closer to the center 1 1 2 2 r m star is a constant. (next page) of mass and moves 2 1 slower. Circular Orbit – Unequal masses

r v 1 = 1 r2 v2

So

r m since 1 = 2 r m 2 1 m v 2 = 1 m1 v2

E.g.. Motion of the sun because of Jupiter; Note: “wobble” of Roughly the same as two stars in circular orbits the star is bigger if the planet is d = radius of sun's bigger or closer m1r1 = m2r2  orbit around center to the star (hence Md = M J dJ of mass has a shorter period).

dJ = Jupiter's orbital radius M J d = dJ = 5.20 AU M = 7.80 x 1013 cm -4 13 = (9.95 x 10 )(7.80 x 10 ) 30 MJ = 1.90 x 10 gm 10 = 7.45 x 10 cm -4 = 9.55 x 10 M 12.5 m/s  11.86 years

Can ignore the influence of P = 11.86 years the other planets. Doppler shift

As of today –2052 extra solar planets in 1300 stellar systems and the number is growing rapidly.

About 40 mph Many were detected by their Doppler shifts. Many more by the “transits” they produce as they cross the stellar disk.

http://exoplanet.eu/catalog.php

KEPLERS THIRD LAW FOR BINARIES Circular Orbit – Unequal masses

2 GM1M2 M1v1 2 = r +r r1 ( 1 2 ) r M1 x 1 M2 2 r2 GM1M2 M2v2 2 = + (r1+r2 ) r2

G(M +M ) v2 v2 4π2r2 4π2r2 1 2 = 1 + 2 = 1 + 2 (r +r )2 r r P2r P2r 2πr 1 2 1 2 1 2 v = 4π2 P = (r1+r2 ) P2 2 2 3 4π P = K (r1+r2 ) K = G(M1+M2 ) In the general case (r + r ) (a a ) i.e., just like before but 1 2 → 1 + 2

where a1 and a2 are the semimajor axes M → M1+M2 R→ r1+r2 of the two stars in ellipical orbits around each other 2 3 4π 3 ⎛ r r ⎞ ( M + M ) = (r + r ) M + M ( 1 + 2 ) 1 2 GP2 1 2 1 2 = ⎜ AU ⎟ M ⎜ P2 ⎟  ⎝ yr ⎠ 4π 2 M = ( AU )3 for the earth  2 M1 r2 M1 v2 G(1yr) = or = M r M v 2 1 2 1

Divide the two equations If you know r1, r2 in AU, or v1, v2, and P in years you can solve for the two masses.

Getting Stellar Masses – Visual binaries E.g., Sirius A and B; distance 2.67 pc) Measure:

• Period

• Separation (if circular orbit; sum of semi-major axes if elliptical; average of max and min separations)

• Ratio of speeds or separations from center of mass if circular (or ratio of semi-major axes if elliptical) Period = 50.1 years semi-major axis of A is 6.4 AU and B is 13.4 AU (In Kepler’s equation use the sum of the semimajor axes) Calculation In the case of circular orbits seen face on M + M the separation is a constant. As shown on the next A B P 2(yr) = A3(AU) M page the separation in AU is just the distance in pc  A= A + A = 6.4 +13.4 =19.8 AU P =51 yr times the angular separation in arc seconds. This A B is a consequence of the way the pc is defined. M + M 19.83 A B = = 2.99 M 512  One can also measure, in the same way the separation So the sum of the masses is 2.98 solar masses. between star 1 and the center of mass, r1, and star 2 and the center of mass, r2.

The ratio of the masses is AA / AB = 6.4 / 13.4=0.478 =MB / MA M 0.478 M 2.99 M 2.02 M The total mass is then given by the usual equation A + A = A =  M M 3 M = 0.478 M = 0.97M 1 + 2 2 3 B A  P (yr) = R (AU) = r + r M ( 1 2 ) and since we see the orbits face on these are the actual masses.  M r and 1 = 2 Sirius B only has a radius 0.0084 times that of the sun. What is it? M r 2 1

Spectroscopic Binaries s % s (in pc) =r (in pc) θ (in radians) r s (in AU) = r (in AU) θ (in radians) ⎛ number AU⎞ r (in AU) = r (in pc) ⎝⎜ 1 pc ⎠⎟ ⎛ 1 radian ⎞ θ in radians = θ (in arc sec) ⎝⎜ number arc sec⎠⎟ ⎛ number AU⎞ ⎛ 1 radian ⎞ s in AU = r (in pc) θ (in arc sec) ⎜ ⎟ ⎝⎜ 1 pc ⎠⎟ ⎝ number arc sec⎠ e.g. The separtion of Sirius A and B varies from 3.5 to 11.5 arc sec. The average is therefore 7.5 arc sec. Sirius is 2.67 pc away so

a1 + a2 = (2.67)(7.5)= 20 AU (away)

Star 2

Star 1 (toward)

Complication:

The viewing angle

Getting Stellar Masses For Sprectroscopic Binaries For spectroscopic binaries measure: 2πr Pv r P= ⇒ r = • Period v 2π • Velocity of each star CM v • Inclination will be unknown so mass measured will be a lower bound (TBD)

CALCULATION 2π r P = Assume circular orbits v

First get r1 and r2 from v1 and v2 v P r = i i 2π Example: −1 −1 v1 = 75 km s v2 = 25 km s P= 17.5 days Note - the bigger the speeds measured for a given P the bigger the masses Complication – The Inclination Angle Eclipsing Binary

Let i be the angle of the observer relative to Only if i = 90 degrees the rotation axis, i.e., i = 0 if we re along the axis. do we measure the full velocity.

Measure v Sin i which is a lower bound to v. 2 To 2 4π 3 P = (r1 + r2 ) Observer G(M1 + M 2 ) P v r = i i 2π but measure v =vSin i, so we end up measuring i r = r Sin i and calculate 3 4π 2 v + v M M ⎛ 1 2 ⎞ P3 1 + 2 = 2 ⎜ ⎟ measured GP ⎝ 2π ⎠ v when the actual mass is

2 3 3 4π v + v 3 i 0.59 M M ⎛ 2 2 ⎞ P actual Sin = 1 + 2 = 2 ⎜ ⎟ GP ⎝ 2π ⎠ hence the measurement gives a low bound on the actual mass But we tend to discover   3 For an eclipsing binary you know you are viewing (M1 + M 2 ) = (M1 + M 2 )Sin i more edge on binaries Since Sin i < 1, the measurement is a lower bound. so 2/3 is perhaps better the system in the plane of the orbit. I.e., Sin i = 1

Limits of stellar mass:

Observed stars end up having masses between

0.08 M and about 150 M.

The upper number is uncertain (130? 200?). The lower number will be derived later in class (minimum mass to ignite H burning before becoming degenerate). to 4