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Nuclear Astrophysics Lecture 6 Thurs. Nov. 29, 2011 Prof. Shawn Bishop, Office 2013, Ex. 12437 1.

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Presentation on theme: "Nuclear Astrophysics Lecture 6 Thurs. Nov. 29, 2011 Prof. Shawn Bishop, Office 2013, Ex. 12437 1."— Presentation transcript:

1 Nuclear Astrophysics Lecture 6 Thurs. Nov. 29, 2011 Prof. Shawn Bishop, Office 2013, Ex

2 THE ROAD TO NUCLEAR REACTION RATES Thermonuclear Reaction Rate in Stars 2

3 3 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:

4 4 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.

5 5 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 :

6 6 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:

7 7 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

8 8 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.

9 9 The rate integral now becomes:

10 10 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

11 11 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 the following: Number of photons per unit volume between and :

12 12 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.

13 Reaction Rate Summary 13 Reaction rate for charged particles: Reaction rate for photodisintegration (photon in entrance channel):

14 THE PATH TO CROSS SECTIONS 14

15 15 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:

16 16 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?

17 17 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.

18 18 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:

19 19 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:

20 Recall from page 17: Let’s use this to get the scattered current density from. 20 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

21 21 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?

22 22 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:

23 23 Rate of particles scattered into Incident flux Cross section


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