2nd Week: Properties of Astrophysical Plasmas B: Difference between revisions
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===Bose-Einstein=== | ===Bose-Einstein=== | ||
Consider a gas of bosons in which the particles satisfy equations | Consider a gas of bosons in which the particles satisfy the same two equations for N and <math>E_{Total}</math>. Similarly as in the Fermi-Dirac case, we can write the thermodynamic potential for a particular energy <math>\epsilon_k</math> as | ||
<math>\Omega_k=-kT \log (\sum_{n_k=0}^\infty e^{n_k(\mu-\epsilon_k)/kT})</math> | <math>\Omega_k=-kT \log (\sum_{n_k=0}^\infty e^{n_k(\mu-\epsilon_k)/kT})</math> | ||
In this case there is no restriction on the number of particles occupying the same state <math>\epsilon_k</math>. This is because the particles have integer spin, and therefore they are not limited by Pauli exclusion principle. Notice that the sum in the previous equation is convergent only if <math>(\mu-\epsilon_k)<0 </math>. This condition must be satisfied for all the values of <math>\epsilon_k</math>. If we choose <math>\epsilon_k>=0</math> (we can do this because this is a non-relativistic theory, so that the energy is defined up to an arbitrary additive constant) then we must have <math>\mu<0</math>. Therefore, in our particular energy scale, the chemical potential in a bosonic gas is always negative. Recall that for <math>0<x<1</math> | In this case there is no restriction on the number of particles occupying the same state <math>\epsilon_k</math>. This is because the particles have integer spin, and therefore they are not limited by Pauli exclusion principle. Notice that the sum in the previous equation is convergent only if <math>(\mu-\epsilon_k)<0 </math>. This condition must be satisfied for all the values of <math>\epsilon_k</math>. If we choose <math>\epsilon_k>=0</math> (we can do this because this is a non-relativistic theory, so that the energy is defined up to an arbitrary additive constant) then we must have <math>\mu<0</math>. Therefore, in our particular energy scale, the chemical potential in a bosonic gas is always negative. Recall that for <math>0<=x<1</math> | ||
<math>\sum_{n=0}^\infty x^n={1\over {1-x}}</math> | <math>\sum_{n=0}^\infty x^n={1\over {1-x}}</math> | ||
Using this in the thermodynamic potential, with <math>x=e^{(\mu-\epsilon_k)/kT}</math> we get | |||
<math>\Omega_k=-kT \log ({1 \over {1-e^{(\mu-\epsilon_k)/kT}}} )=kT \log (1-e^{(\mu-\epsilon_k)/kT} )</math> | |||
And as in the Fermi-Dirac case | |||
<math>f(\epsilon_k)=-{\partial \Omega_k \over \partial \mu}=...</math> | |||
<math>f(\epsilon_k)={1\over e^{(\epsilon_k-\mu)/kT}-1}</math> | |||
Revision as of 14:52, 29 January 2009
Before an in-depth analysis of nuclear astrophysics can begin, one must review the basics of nuclear physics. This begins with thermodynamics.
Basics of Thermodynamics
Here are the definitions of some of the basic quantities.
The particle density: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n = \int_{0}^{\infty}{\omega(p)f(p)dp}}
The energy density: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = \int_{0}^{\infty}{E\omega(p)f(p)dp}}
The pressure: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P = {1\over3} \int_{0}^{\infty}{pv\omega(p)f(p)dp}}
where v is the velocity, p is the momentum, E is the energy, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(p)} is the probability function, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega(p)} is the state density.
Wave State Derivation
A brief derivation of a wave state density Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega(E)} .
One can begin with the stationary Schroedinger equation.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H \psi_{1,2,...n} = \left(T + V\right) \psi_{1,2,...n} = E \psi_{1,2,...n} }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = \sum_{i=1}^{n}t_i = \sum_{i=1}^{n} -{\hbar^2\over2m}\nabla_{i}^2}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V = \sum_{i=1}^{n}\sum_{j>i}^{n} v_{ij}}
However, for noninteracting particles: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\left(i,j\right) = v\left(i,j\right) = 0}
Also since these are stationary states, H describes a particle in a box of size d. So then breaking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} into Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_x\psi_y\psi_z} one gets the solution:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_x = \sqrt{2\over d} sin\left({\sqrt{2mE}\over \hbar}x\right)} and that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\sqrt{2mE}\over \hbar} = n_x\pi}
This has a corresponding energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_x = {\pi^2\hbar^2 \over 2md^2} n_{x}^2}
One will get similar solutions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_y} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_z} .
This means that the total energy of the particle can be found by: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i = {\pi^2\hbar^2 \over 2md^2} \left(n_{x}^2 + n_{y}^2 + n_{z}^2\right) = {\pi^2\hbar^2 \over 2md^2} R_{\rho}^2}
One does not want all of the eigenstates but only those within a certain range. These can be described as:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi(E) = {1 \over 8} {4\pi \over 3} R_{\rho}^3 = {4\pi \over 3} {V(2m)^{3/2} \over \hbar^3} E^{3/2}}
Then taking the derivative with respect to E and dividing by the volume will give the wave state density.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega(E) = {1 \over V}{\partial \phi \over \partial E} = {2\pi(2m)^{3/2} \over \hbar^3} E^{1/2} }
Occupation probabilities
The 1st law of Thermodynamics in a system (or subsystem) with variable number of particles is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dE=TdS-PdV+\mu dN}
Other thermodynamical identities can be found in the lecture notes on Blackboard.
Maxwell-Boltzmann
The probability distribution can be found by:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\epsilon_k) = \left( e^{ {\epsilon_k - \mu}/{kT} } \right)^{-1}}
Fermi-Dirac
Suppose that our system has discrete energies and that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_k} is the number of particles occupying the energy level Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} . This two quantities must satisfy
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N=\sum_{k}n_k }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{Total}=\sum_{k}n_k\epsilon}
Since we are dealing with fermions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_k} can be 0 or 1. The thermodynamic (Landau) potential for a particular energy sate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} can be written as
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_k=-kT \log (\sum_{n_k=0}^1 e^{n_k(\mu-\epsilon_k)/kT})}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_k=-kT \log (1 + e^{(\mu-\epsilon_k)/kT})}
Recall that, the mean particle number in a certain energy state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} is minus the derivative of the thermodynamic potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_k} with respect to the chemical potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu} , at V and T constant. Therefore
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\epsilon_k)=-{\partial \Omega_k \over \partial \mu}= ...}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\epsilon_k)={1 \over {e^{(\epsilon_k-\mu)/kT}+1}}}
Bose-Einstein
Consider a gas of bosons in which the particles satisfy the same two equations for N and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{Total}} . Similarly as in the Fermi-Dirac case, we can write the thermodynamic potential for a particular energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} as
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_k=-kT \log (\sum_{n_k=0}^\infty e^{n_k(\mu-\epsilon_k)/kT})}
In this case there is no restriction on the number of particles occupying the same state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} . This is because the particles have integer spin, and therefore they are not limited by Pauli exclusion principle. Notice that the sum in the previous equation is convergent only if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mu-\epsilon_k)<0 } . This condition must be satisfied for all the values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k} . If we choose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_k>=0} (we can do this because this is a non-relativistic theory, so that the energy is defined up to an arbitrary additive constant) then we must have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu<0} . Therefore, in our particular energy scale, the chemical potential in a bosonic gas is always negative. Recall that for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<=x<1}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty x^n={1\over {1-x}}}
Using this in the thermodynamic potential, with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=e^{(\mu-\epsilon_k)/kT}} we get
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega_k=-kT \log ({1 \over {1-e^{(\mu-\epsilon_k)/kT}}} )=kT \log (1-e^{(\mu-\epsilon_k)/kT} )}
And as in the Fermi-Dirac case
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\epsilon_k)=-{\partial \Omega_k \over \partial \mu}=...}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\epsilon_k)={1\over e^{(\epsilon_k-\mu)/kT}-1}}