Physics 161 Homework 7 - Solutions Wednesday November 16, 2011

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1 Physics 161 Homework 7 - s Wednesday November 16, 2011 Make sure your name is on every page, and please box your final answer Because we will be giving partial credit, be sure to attempt all the problems, even if you don t finish them The homework is due at the beginning of class on Wednesday, November 23rd Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions during the discussion session or during office hours 1 (a) Assuming that the surface temperature of the Sun is T = 5800 K, use Stefan s law to determine the rest mass lost per second to radiation by the Sun Take the Sun s radius to be = meters (b) What fraction of the Sun s rest mass is lost each year from electromagnetic radiation? Take the Sun s rest mass to be M = kg (a) The intensity of radiation (in watts per square meter) from the Sun is I = P/A = σt 4, where σ is the Stefan-Boltzman constant and T is the temperature Thus the power is P = AI = 4π 2 σt 4, where is the radius of the Sun Now, the power is P = Ė, and since the energy is coming from the rest mass, E = mc2, then Ė = ṁc2 Thus, the rate of mass loss is So, plugging in the constants gives dm dt = 4π2 σ c 2 T 4 dm dt = 4π2 σ c 2 T 4 = 4π (7 108 ) 2 ( ) ( ) 2 (5800) 4 = kg/sec (b) So, in one year the Sun burns through This is a fraction m = dm dt t = ( ) ( ) = kg m M = = , meaning that the Sun loses roughly % of its mass per year

2 2 Emission lines of hydrogen, Hβ (n = 4 2 and λ rest = 4861Å) are observed in the spectrum of a spiral galaxy at redshift z = 09 The galaxy disk is inclined by 45 to the line of sight (if the inclination was 0 then we d see the galaxy from a top view, while if it was 90, we d see it edge-on) The Hβ wavelength of lines from one side of the galaxy are shifted to the blue by 5Å relative to the emission line from the center of the galaxy, and to the red by 5Å on the other side (a) What is the galaxy s rotation speed? (b) Find the age of the Universe when the light was emitted from this galaxy (a) There are two Doppler shifts in this problem, one from the redshift z, and the other from the rotational motion of the galaxy The redshift from the galaxy shifts the wavelength λ obs = 1 + z = 19 λ emit Thus, the observed wavelength to the center of the galaxy is λ obs = (1 + z) λ emit = 19λ emit = 9236 Å Next, the rotation of the galaxy adds an additional redshift, depending on the rotational velocity This velocity also depends on the angle of orientation If the angle was zero, giving a top view of the galaxy, then we wouldn t measure any rotational velocity, since it would be rotating perpendicular to the line of sight If the angle was 90, then we d see the galaxy edge-on, and would measure the full rotational velocity Thus, we see that we care about the sine component of the velocity, v sin θ Now, the Doppler effect in the light emitted from the rotational velocity is (taking the specific case of the edge that is moving away from us), λ obs λ obs = c + v sin θ c v sin θ, where λ obs is the wavelength seen at the edge of the galaxy, and λ obs is the wavelength at the center of the galaxy ((1 + z) λ emit ) Since the rotation velocity is much smaller that light we can expand λ obs λ obs ( 1 + v ) 2 c sin θ v = 1 + sin θ c So, λ obs λ obs λ obs = λ = v sin θ, λ obs c

3 where λ is the difference in wavelength between the edge and center of the galaxy (5 Å) Now, λ obs = (1 + z) λ emit, and so, all together the velocity is v = c λ sin θ (1 + z) λ emit Plugging in the numbers gives v = 230 km/s (which shows that we were justified in assuming v small in our derivation) (b) The redshift is defined in terms of the scale factor as a (t) a 0 = z The galaxy emitted light in the matter-dominated era, and so a t 2/3 Thus, ( ) 2/3 a (t) t = = 1 a 0 t z, where t 0 is the current age of the Universe Solving for the emission time, t we find t = t 0 (1 + z) 3/2 Taking t 0 = 137 billion years, then t = t = = 523 billion years 3/2 3/2 (1 + z) (1 + 09)

4 3 We have seen that, for a star composed of a classical, nonrelativistic, ideal gas, E total = Eth tot + E grav = Eth tot, and therefore the star is bound epeat the derivation, but for a classical relativistic gas of particles ecall that the equation of state of a relativistic gas is P = 1 E th Show that, in this case, E 3 V grav = Eth total, and therefore E total = Eth tot + E grav = 0; ie, the star is marginally bound As a result, stars dominated by radiation pressure are unstable The pressure of a relativistic gas is P = E th /3V, which means that P = 1 3V Etot th Furthermore, we ve seen that P = 1 E grav 3 V, which tells us that E grav = Eth tot Since E tot = Eth tot + E grav = E grav + E grav = 0

5 4 Because of the destabilizing influence of radiation pressure, the most massive stars that can form are those in which the radiation pressure and the nonrelativistic kinetic pressure are approximately equal Estimate the mass of the most massive stars, as follows: (a) Assume that the gravitational binding energy of a star of mass M and radius is E grav G N M 2 / Use the virial theorem to show that where ρ is the typical density P = 1 E grav 3 V, ( ) 1/3 4π P G N M 2/3 ρ 4/3, 3 4 (b) Show that if the radiation pressure, P rad = 1aT 4, equals the kinetic pressure, then 3 the total pressure is ( ) 1/3 ( ) 4/3 3 kb ρ P = 2, a m where m = m H /2 is the mean mass (c) Equate the expressions for the pressure in parts (a) and (b), to obtain an expression for the maximal mass of a star Find its value, in solar masses, assuming a fully ionized hydrogen composition (a) The pressure P G NM 2 3V Now, 1/V = ρ/m, while = (3M/4πρ) 1/3, and so P G ( ) 1/3 ( ) 1/3 NMρ 4πρ 4π = G 3 3M 3 4 N M 2/3 ρ 4/3 (b) The total pressure is P = P grav + P rad = ρk BT m pressures are equal then at 4 If the kinetic and radiation Then, the total pressure is ρk B T m = 1 3 at 4 T = P = P g at 4 = 2 3 at 4 = 2 3 a ( 3ρkB ma ( ) 1/3 3ρkB ma ) 4/3 = 2 ( ) 1/3 ( ) 4/3 3 kb ρ a m

6 (c) Equating the two expressions gives ( ) 1/3 4π G N M 2/3 ρ 4/ ( 3 a ) 1/3 ( ) 4/3 kb ρ m This gives for the mass, after plugging in for a = 8π 5 kb 4 /15h3 c 3, ( ) hc M = kg, π 6 m 4 H G N which is M 113M So, the largest stars have a mass of roughly 100 solar masses

7 5 Suppose a star of total mass M and radius has a density profile ( ρ = ρ C 1 r ), where ρ C is the central density (assume constant) (a) Find M(r) (b) Express the total mass M in terms of and ρ C (c) Solve for the pressure profile, P (r), with the boundary condition P () = 0 (a) We can determine M(r) via the mass continuity equation, ( ) dm(r) = 4πr 2 ρ (r) = 4πρ C r 2 r3 dr Separating and integrating noting that M(r = 0) = 0, we find M(r) 0 r dm (r) = 4πρ C dr 0 ) (r 2 r3 Integrating both sides gives M (r) = 4π ( 3 ρ Cr 3 1 3r ) 4 (b) Finding the total mass is very simple We can either repeat the above derivation in part (a), integrating to r =, or just plug in r = in to our result for M(r) Then, M = M() = 4π ( 3 ρ C ) = π 4 3 ρ C 3 (c) Now, to determine the pressure, P (r), we use the hydrostatic equilibrium equation, dp (r) = G NM (r) ρ (r) dr r 2 The density ρ (r) was given, and we have just found M (r), so we find dp (r) ρ ( ) ( ) 2 C r3 1 3r 4 1 r = 4πG N dr 3r 2 ( = 4πG N ρ 2 3 C r 1 r 3r = 4πG N 3 ρ 2 C ( r 7r r3 4 2 ) ) + 3r2 4 2

8 Now, since we know that P () = 0, we integrate from r = r to r =, 0 Integrating gives or, P (r) dp (r) = 4πG N 3 P (r) = 4πG N 3 ρ 2 C P (r) = πg N ρ 2 C 2 [ ρ 2 C r dr ) (r 7r r3 4 2 ( r r ( r r 4 2 ), ) 2 7 ( r ) 3 1 ( r ) ] Notice that the pressure at the center of the star (in this model) is P (0) = 5πG N 36 ρ2 C 2