Presentation is loading. Please wait.

Presentation is loading. Please wait.

Nuclear Astrophysics Lecture 4 Thurs. Nov. 11, 2011 Prof. Shawn Bishop, Office 2013, Extension 12437 1

Similar presentations

Presentation on theme: "Nuclear Astrophysics Lecture 4 Thurs. Nov. 11, 2011 Prof. Shawn Bishop, Office 2013, Extension 12437 1"— Presentation transcript:

1 Nuclear Astrophysics Lecture 4 Thurs. Nov. 11, 2011 Prof. Shawn Bishop, Office 2013, Extension 12437 1

2 Energy generation from nuclear reactions is a function of temperature, density and the set of composition parameters. So, and is the energy rate per unit mass of stellar material. 2 Let be the energy rate flowing outward through a spherical surface of radius r. This is the 3 rd stellar structure equation for a static star. We have two more to derive. It can also be expressed as a mass derivative quite simply: Energy generated in dm + energy entering dm = energy flowing out of dm. Or,


4 4 We have just derived the relation between the luminosity gradient in a star. It depends on the local energy generation rate and the local density of the material This, in turn, is connected to the nuclear reactions occurring in that material. Before we can get to nuclear reaction physics, however, we require one more stellar structure equation. It is clear that the energy generation rate in the star, being density dependent (and temperature dependent because nuclear reaction rates are highly sensitive to temperature), must produce a temperature gradient in the star. We know this, of course: the center of the star is hottest and the surface is where the energy escapes to space. Heat flows from hot to cold, so the star has a temperature gradient. This temperature gradient is responsible for the transport of heat to the surface. The carriers of the thermal energy, for a star consisting of a mixture of ideal gas and radiation (like Main Sequence stars), are photons. Our final stellar structure equation must somehow connect the luminosity of the star with the temperature gradient. Our system of stellar structure equations will, then, be closed.

5 Reviewing your Lecture 2 notes, you will find on page 18 (after some algebra), that the energy density of a photon gas is: This suggests defining a spectral energy density (energy per unit volume per unit frequency interval) as: We know also, from our work in Lecture 2, that the total photon gas pressure is: (*) This result further suggests a definition for the “spectral pressure”, which can be thought of as the fractional contribution to the total pressure by photons of angular frequency : 5

6 Using the last expression and Equation (*) we can obviously write: From which, it trivially follows: (**) Remember this last result. With it, we can now derive the stellar temperature gradient in terms of the photon luminosity. 6

7 7 Difference in radiation force across the slab: Furthermore, a photon flux, when passing through a thin slab of material, will suffer an attenuation in flux. The change in flux of this beam, for normal incidence (as is the case with our geometry) is given by: Now, is the energy per frequency interval carried by all photons with frequency crossing through per unit time. Dividing it by c will yield the momentum flux (total momentum) carried by all photons with angular frequency crossing through per unit time. Thus:

8 Equate the bottom expression to what we have at the top of previous slide, and take the limit as : Now we must integrate over all photon frequencies. LHS is just, (by definition) (***) 8

9 9 We are almost done: Remember the identity (**) on page 6. Multiply (***) by it. We have: Finally, the luminosity of our wedge, once integrated over a spherical shell, is just:. We have, therefore, that: Or, more compact: Averaged over distribution of. is the average opacity s

10 10 Recall from lecture 2 the condition for hydrostatic equilibrium: Also recall for a polytrope star with particle and radiation pressure the definitions of lecture 3 Use the above two expressions to get dP tot /dr and substitute into L(r) from previous page For massive stars, Thomson scattering dominates the opacity (Compton scattering on free electrons). The cross section for this process is a constant. For complete ionization it leads to a opacity of 0.4 cm 2 /g

11 11 Eddington’s Quartic Equation: we derived the mass of a radiation + particle gas star is almost always close to 1 except for the most massive of stars Substitute this into the previous luminosity expression: Assuming is constant over entire star, then Mass luminosity relation of Lecture 1

12 Sources of Opacity 12 Bound-Bound Absorption: Absorption of a photon by an atom, causing an upward transition to electron orbital of higher energy. It is a true-absorption process; its inverse is normal emission via downward transitions. Bound-Free Absorption: Absorption of a photon by an atom causing a bound electron to make a transition to the continuum. True-absorption process; inverse is radiative recombination. Free-Free Absorption: Absorption of a photon by a continuum electron as is passes an ion and makes a transition to another continuum state at higher energy. A true- absorption process; inverse is bremsstrahlung. Scattering from free electrons: Scattering of photons by individual free electrons in the gas, and known as Compton Scattering; in non-relativistic limit, called Thomson scattering. Not true-absorption as the photon energy remains unchanged.

13 13

14 14 Energy generation rate per unit mass of material average opacity coefficient in the material The 4 Equations of Stellar Structure

15 Ancillary Equations: Equations of State 15 Internal energy of photon gas

16 First, let’s go back to the First Law of Thermodynamics and something already familiar: Generalized Adiabatic Coefficients Take the internal energy to be functions of T and V: Then, by definition: 16

17 For an ideal gas: So, we have:and: Heat Capacity at constant volume: When dP = 0 Summarizing: Heat Capacity at constant pressure: Ideal Gas adiabatic exponent: 17

18 18 Let’s go back to first law, now, for ideal gas: using For an adiabatic change in the gas, dQ = 0 From EOS we have:. Use this above to also get two more:

19 19 When integrated, these 3 equations lead to the familiar adiabatic formulae for an ideal gas:

20 Mixture of Ideal and Photon Gases 20 On page 17 we had the general result: The total pressure of the gas: The total internal energy of the gas: Start doing the partial derivatives:

21 21 But, remember that: So, we have: And finally:

22 Back on page 19, for the case of an Ideal Gas only, we found 3 differential equations that related the adiabatic exponent to the temperature, volume and pressure of the gas. Here, we have a mixed gas of particles and photons, but let us use the structure of the equations on page 19 as a guide for building analogous equations for the case of a photon + ideal gas. Here they are: 22 For adiabatic changes to the gas, we require dQ = 0. With this condition in the previous equation, we have: This equation ( * ) is central to what follows, so “get to know” it well!

23 23 The adiabatic exponents need to be determined. Let’s proceed. Go back to equation ( * ) on page 23. Compare it directly to equation ( 3 ) of page 23. Thus, we have the following result: In the gas, some fraction of the total pressure is from the Ideal part. Call it. Photon part carries a fraction of the total pressure. Substitute these in above to get: Now to get.

24 Now to equation ( 1 ) of page 23: First, we have the total gas pressure: 24 Substitute this into equation ( 1 ):

25 25 Using the relationships: and, we have: Now, we know from equation ( 3 ) the adiabatic exponent (solved it on page 24). And, above, we can replace d  /  using equation ( 3 ). Homework: Substitute into ( * * * ) for d  /  and use the result for on page 24 to determine the formula for. Result:

26 26 We now have, and we still need. There are different ways to get it. One way is this: Take the difference between equations ( 1 ) and ( 2 ) and add that result to equation ( 3 ). Should get: You’ve got and. The remaining algebra is for you to do. Show:

27 Summarizing Adiabatic Exponents 27

28 Some Theorems of Stellar Equilibrium 28 Let’s take two of the 4 stellar structure equations and run with them; see where they take us. Hydrostatic equilibrium Radiative Transport. (Equation ( D ) on page 2, rearranged) Dividing the 2 nd by the 1 st, leaves us: Remember:, which is the opacity averaged over all photon frequencies.

29 29 Next, let’s define the following quantity: This is just the average energy rate per unit mass interior to the point “r” divided by the total energy generation rate per unit mass of the entire star. By its definition, We can now write: Nothing has changed here; it’s all still exact. All we have done is buried the dependence on radial coordinate, r, into the term. But, now we can derive a simple theorem, that limits the value of the stellar opacity at all points in the star, from this result. Let’s see.

30 30 We have: and, also: Also, Therefore, we have: Also, because Finally:Or, a limit on opacity:

31 31 We have: So, after reversing the integral limits, and dividing by unity [P(r)/P(r)], we have: Pressure-averaged quantity. Call it: Let’s return back to:. And let’s integrate it. So, in terms of the total mass of the star, and its total luminosity, we can write:. Now use: And now we can write:

32 32 The previous result is Strömgren’s Theorem. In words, we have: “The ratio of the radiation pressure to the total pressure at a point inside a star in radiative equilibrium is proportional to the average value of for the regions exterior to the point r, with the average being taken with respect to dP.” Explaining this with a diagram (picture worth 1000 words): The ratio of radiation pressure to total pressure on surface of pink sphere (radius r) is proportional to the pressure average of in the region of the blue shell of radius R-r.

33 The Concept of Convection 33 We can use the Strömgren Theorem equation to write as: But, we have:, so we can write: Radiation pressure: Sub this into ( * * * *): Compare with Adiabatic Exponent equation 2, on pages 10 and 15!

34 34 We just derived: And 2 nd Adiabatic Equation (rearranged): Consider a mass element dm. Suppose it undergoes an increase in temperature relative to its surroundings. The temperature increase will cause dm to expand, and its density to become less than the surroundings. It will, therefore, make a displacement to a region of lower density. Assume: 1.Pressure exerted by dm on surroundings is equal to pressure surroundings exert on dm. 2.Expansion (or contraction) of dm occurs adiabatically. 3.Friction can be neglected.

35 35 By the adiabatic assumption: And is what we (or what you will) derived on page 14. By the first assumption, the dP/P terms will be the same for the surroundings and for dm. This means: the temperature change of dm is different from the change in temperature of the surroundings. If, then dm moves outward until it has the same temp. and density as the surroundings. The converse means dm will sink until its temperature and density are the same as surroundings.

36 36 Thus, if the mass bubble, dm, has internal conditions such that then the star will be convective; the energy is carried away by convection, which is the movement of hot material to regions of lower temperature. When the bubble and surroundings have the same temperature, at that point, the star carries the energy by by radiative transport. When convective transport is dominant, then we have: Now, we have: and:

Download ppt "Nuclear Astrophysics Lecture 4 Thurs. Nov. 11, 2011 Prof. Shawn Bishop, Office 2013, Extension 12437 1"

Similar presentations

Ads by Google