Nuclear Astrophysics Lecture 9 Thurs. Dec. 22, 2011 Prof. Shawn Bishop, Office 2013, Ex

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

Return to the Standard (Stellar) Model 3 The stellar gas is a mixture of photons and Ideal particles. Thus, Total Pressure: In thermodynamic equilibrium, these two gases have the same temperature. And let and Then, we have: Polytrope of type 3

Polytrope Solutions 4 n = 0 n = 2 n = 1 n = 5 n = 3 n = 4

Mass Luminosity Relation 5 Take the following equations from the 4 Structure Equations: And use our friend: Sub this into first equation above Small for all but the most massive of stars.

Eddington’s Quartic Equation 6 Parametric plot of versus Product of the root of  and its 1 st derivate at the root. In 3 rd Lecture, pages 13 & 20, it was shown that the stellar mass can be written as, For all, but the most massive stars, is a small quantity. So, rearrange the above equation to isolate and sub result into the previous equation for to get L(R * )

7 After the algebra (you should check, to make sure I’m right ), we arrive at the Mass Luminosity Relation for Main Sequence stars! Numerically: We now have our first theoretical prediction of the relationship between two observable properties of stars. The Luminosity of Main Sequence Stars (H-burning, hydrostatic, up to ~15 solar masses) should be proportional to the 3 rd power of the stellar mass. (first order result, we can do better). To do better, we have to deal with that annoying opacity,.

8 Kramer’s Opacity: Varies (crudely) as and almost as Empirical relation given as: This factor of 2 is required to get agreement for curves with Thomson’s Opacity: Constant, and has a value of The exponent in Kramer’s Opacity is also: Total Opacity of Solar Composition Material We need to simplify the Kramer’s opacity so that it is “averaged” over all of the stars we are considering in the Mass- Luminosity relationship.

9 Recall, from page 3, we found (and this is also in Lecture 2/3 pages 36, 37) that: Get his from squaring and rearranging the stellar mass formula on page 6. You have all the ingredients here to now relate temperature to density. The final result, after doing the algebra: The first step: get a formula that expresses temperature in terms of density. This will give us a Kramer’s formula that is now only a function of density.

10 Use this last result in the Kramer’s Opacity formula by replacing : Now, no two stars are alike, so we have to start doing some reasonable averages of T and. First thing is to average over the stellar temperature. And remember, the temperature of the ideal gas and photon gas are the same. However, the photon gas pressure depends only on the temperature (not on density), so there seems like a good place to start. Try a volume average of the radiation temperature

11 From the polytrope formalism, the solution to the Lane-Emden equation gives the run of density as a function of radial coordinate, r. For n=3 polytrope, we had (page 29 of Lec. 2/3). On previous slide, we had. The above integrals can be done numerically (Mathematica), using numerical. Result is: We now have an “average” temperature (weighted over volume) in terms of central temperature. Next, we need the density that corresponds with this T av Two slides ago (slide 9) we had the following result: Need to eliminate this We will need T c

12 And from the Polytrope formalism in, you(!) should have found the following result: Where, From the table on page 4. Finally, for the Sun (a Main Sequence Star), What have we got now: and And we need to complete: We still need the central temperature, and then we are DONE!

13 The central temperature: From the last slide, we had: And: For a Solar-type star: and Collecting all the numbers, we finally have: Assuming fully ionized

14 Finally, Kramer’s Opacity becomes simplified to: Total Opacity: where And, for fully ionized material: And X = 0.71 for Solar. The total Opacity is now: And we had for Luminosity:

15 Calling we finally have the function for Luminosity: Assuming fully ionized With, as before, given by its Solar value: The function above is parametric in. We work the function by choosing a value for m, and then solving Eddington’s Quartic equation for, to evaluate the RHS. What does it look like when plotted against REAL Main Sequence data?? Or, using

Mass-Luminosity: Main Sequence 16 Data are from: G. Torres et al., Astron. Astrophys. Rev. (2009)

THE ROAD TO NUCLEAR REACTION RATES Thermonuclear Reaction Rate in Stars 17

18 Some basic kinematics: We have two particles with masses and with velocities and The velocity of their common centre of mass is: The velocity of particle 1 relative to the CoM velocity is just: And v is just the relative velocity between 1 and 2. Similarly, particle 2 has a velocity relative to CoM velocity:

19 Before the collision, the total incident kinetic energy is: Using the previous two vector equations, we can substitute in for v 1 and v 2 in terms of v and V. (An exercise for you) The first term is the kinetic energy of the center of mass itself; while the second term is the kinetic energy of the reduced mass as it moves in the center of mass frame.

20 Nuclear reaction rate: The reaction rate is proportional to the number density of particle species 1, the flux of particle species 2 that collide with 1, and the reaction cross section. This v is the relative velocity between the two colliding particles. Flux of N 2 as seen by N 1 : Reaction cross section: Important: this reaction rate formula only holds when the flux of particles has a mono-energetic velocity distribution of just Flux of N 1 as seen by N 2 :

21 Inside a star, the particles clearly do not move with a mono-energetic velocity distribution. Instead, they have their own velocity distributions. We must generalize the previous rate formula for the stellar environment. From Lecture 2,3 the particles 1 and 2 will have velocity distributions given by Maxwell- Boltzmann distributions. We have the 6-D integral: The fraction of particles 1 with velocities between is therefore, And similarly for particle species 2. Let’s take a closer look at:

22 From equations on page 4, we can write the argument in [...] in terms of the center of mass velocity and relative velocity. So in terms of the CoM parameters, The reaction rate now becomes (6-D integral): And we note: and

23 We now need to change the differential variables into the new CoM variables. From page 3, in component form, we have: Jacobian: And this is the same for the case of y and z components.

24 The rate integral now becomes:

25 Note: the product N 1 N 2 is the number of unique particle pairs (per unit volume). If it should happen that 1 and 2 are the same species, then we must make a small correction to the rate formula to avoid double-counting of particle pairs. Kronecker delta

26 We can extend the previous result to the case when one of the particles in the entrance channel is a photon. So reaction is: The rate: As before, we generalize this by integrating over the number density distributions: A Maxwell-Boltzmann for species 1, and for photons we recall from Lecture 2,3 page 18 the following: Number of photons per unit volume between and :

27 The Einstein postulate of Special Relativity: speed of light is the same in all reference frames. Therefore, the relative velocity = N 1 because N 1 is M-B And is the photo-disintegration cross section.

Reaction Rate Summary 28 Reaction rate for charged particles: Reaction rate for photodisintegration (photon in entrance channel):

THE PATH TO CROSS SECTIONS 29

30 3-Dimensional SWE, after separation of variables, will produce a radial equation of the following type: Make the substitution: and show that the above equation becomes For, and outside the interaction zone, the above equation asymptotically becomes:

31 Incident beam along z-axis has a solution:, where For a beam incident from the left, the solution outside the potential zone at large r is just: Outgoing wave function? Some of the incident beam transmits through the interaction zone, the rest scatters or undergoes a reaction. How to quantify this?

32 Outgoing wave = Incident wave + Scattering wave Now we need to relate this function and the incident wave function to the reality of what we measure in a nuclear physics experiment. Consider:. This is a (local) probability density for the existence of a particle at the spatial coordinates. Its time derivative tells us how the probability of there being a particle at these coordinates evolves in time. We must find a way to relate this quantity to the reality of an experiment.

33 Flux in Transmitted Beam Scattered Beam Differential cross section is defined as: Rate of particles scattered into Incident flux The total cross section is obviously:

34 The general SWE is: Its complex-conjugate is: Exercise for the student: Multiply 1. on the left by and 2. on the left by. Then take the difference of the resulting two equations. The result should be: Another exercise: Use some vector calculus to show the above result can be written: Where the “current” j is:

Recall from page 17: Let’s use this to get the scattered current density from. 35 Are we any closer to reality?? Yes. Look: The incident wave function was just Its current density j is This is the incident flux. All that remains now is to get the result for the scattered current density

36 In spherical coordinates: In the limit that the only terms that matter to us is the first one. Student exercise: Apply this result to the scattered wavefuction to determine the scattered current as: This is a vector quantity, direct radially outward from the interaction zone. Where does it go experimentally?

37 Flux in Transmitted Beam Scattered Beam It goes into our detector, which occupies a solid angle and has an area of The rate of particles entering the detector is therefore:

38 Rate of particles scattered into Incident flux Cross section