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courses:ast301:2 [2024/03/15 02:18] – [1. Maxwell-Boltzmann statistics (MBS)] asadcourses:ast301:2 [2024/03/15 02:26] (current) – [3.1 Degeneracy in 1D] asad
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 ==== - Momentum space ==== ==== - Momentum space ====
-Momentum $p_x=mv_x$ and the probability $P(p_x)dp_x=P(v_x)dv_x$. So+Momentum $p_x=mv_x$ and the probability $\mathsf{P}(p_x)dp_x=P(v_x)dv_x$. So
  
-$$ P(p_x)\ dp_x = \left(\frac{1}{2\pi mkT}\right)^{1/2} e^{-p_x^2/(2mkT)} dp_x $$+$$ \mathsf{P}(p_x)\ dp_x = \left(\frac{1}{2\pi mkT}\right)^{1/2} e^{-p_x^2/(2mkT)} dp_x $$
  
 which is plotted in the figure (a) below. We can use figure (b) to transform this into three dimensions. which is plotted in the figure (a) below. We can use figure (b) to transform this into three dimensions.
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 {{:courses:ast301:mbd.webp?nolink&750|}} {{:courses:ast301:mbd.webp?nolink&750|}}
  
-In three dimensions $P(p)\ d^3p = P(p_x)dp_x\ P(p_y)dp_y\ P(p_z)dp_z$ or+In three dimensions $\mathsf{P}(p)\ d^3p = \mathsf{P}(p_x)dp_x\ \mathsf{P}(p_y)dp_y\ \mathsf{P}(p_z)dp_z$ or
  
-$$ P(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p_x^2+p_y^2+p_z^2}{2mkT}\right) \ dp_x dp_y dp_z $$+$$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p_x^2+p_y^2+p_z^2}{2mkT}\right) \ dp_x dp_y dp_z $$
  
 which becomes which becomes
  
-$$ P(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) \ d^3p $$+$$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) \ d^3p $$
  
-which is the most used form of **MBD**. The quantity is dimensionless: $P(p)$ has units (Ns)$^{-3}$ which is canceled by the unit (Ns)$^3$ of $d^3p$. That means the probability represents the number of particles expected per momentum cube.+which is the most used form of **MBD**. The quantity is dimensionless: $\mathsf{P}(p)$ has units (Ns)$^{-3}$ which is canceled by the unit (Ns)$^3$ of $d^3p$. That means the probability represents the number of particles expected per momentum cube.
  
 MBD can also be written in terms of energy because $E=mv^2/2=p^2/(2m)$ which means MBD can also be written in terms of energy because $E=mv^2/2=p^2/(2m)$ which means
  
-$$ P(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} e^{-E/(kT)} \ d^3p $$+$$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/2} e^{-E/(kT)} \ d^3p $$
  
-which means $P(p) \propto e^{-E/(kT)}$ and the exponential becomes $e^{-1}$ when $E=kT$ . This form is also known as **Maxwell-Boltzmann statistics** or MBS.+which means $\mathsf{P}(p) \propto e^{-E/(kT)}$ and the exponential becomes $e^{-1}$ when $E=kT$ . This form is also known as **Maxwell-Boltzmann statistics** or MBS.
  
 Now as shown in figure (c), for a spherical shell in isotropic momentum space $d^3p=4\pi p^2\ dp$ and, hence, the probability of finding a particle in a thin shell of thickness $dp$ at a radius $p$ Now as shown in figure (c), for a spherical shell in isotropic momentum space $d^3p=4\pi p^2\ dp$ and, hence, the probability of finding a particle in a thin shell of thickness $dp$ at a radius $p$
  
-$$ P(p)\ dp = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) 4\pi p^2\ dp $$+$$ \mathsf{P}(p)\ dp = \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) 4\pi p^2\ dp $$
  
-which is more convenient because here the function is in 1D even though it is capable of describing a 3D gas. As in any probability, $\int P(p)dp=1$ if the limit is from zero to infinity. This function is plotted in figure (d) above at two different temperatures $T_1$ and $T_2$.+which is more convenient because here the function is in 1D even though it is capable of describing a 3D gas. As in any probability, $\int \mathsf{P}(p)dp=1$ if the limit is from zero to infinity. This function is plotted in figure (d) above at two different temperatures $T_1$ and $T_2$.
  
 We see that the function goes to zero at $p=0$ because of the $4\pi p^2$ term. At high momenta, $E > kT$, and the function again goes to zero because the exponential overcomes the $4\pi p^2$ term. So, unlike before, the maximum probability is not found at $p=0$. Also the $T^{-3/2}$ term ensures that the higher temperature curve has a lower amplitude. We see that the function goes to zero at $p=0$ because of the $4\pi p^2$ term. At high momenta, $E > kT$, and the function again goes to zero because the exponential overcomes the $4\pi p^2$ term. So, unlike before, the maximum probability is not found at $p=0$. Also the $T^{-3/2}$ term ensures that the higher temperature curve has a lower amplitude.
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 If $N$ particles are distributed uniformly in volume $V$, then there position is described by $(x,y,z)$ in physical space (figure a), and their momentum is described by $(p_x,p_y,p_z)$ in momentum space (figure b). The number density of particles $n=N/V$ and hence, the phase space distribution function If $N$ particles are distributed uniformly in volume $V$, then there position is described by $(x,y,z)$ in physical space (figure a), and their momentum is described by $(p_x,p_y,p_z)$ in momentum space (figure b). The number density of particles $n=N/V$ and hence, the phase space distribution function
  
-$$ f_{MB} = nP(p) = \frac{N}{V} \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) $$+$$ f_{MB} = n\mathsf{P}(p) = \frac{N}{V} \left(\frac{1}{2\pi mkT}\right)^{3/2} \exp\left(-\frac{p^2}{2mkT}\right) $$
  
 where $f_{MB}$ has the units m$^{-3}$ (Ns)$^{-3}$, which results in (Js)$^{-3}$. So the phase space function gives probabilities of finding particles per joule-second cube. This $f$ is sometimes called //phase-space density// or just //the distribution function//. where $f_{MB}$ has the units m$^{-3}$ (Ns)$^{-3}$, which results in (Js)$^{-3}$. So the phase space function gives probabilities of finding particles per joule-second cube. This $f$ is sometimes called //phase-space density// or just //the distribution function//.
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 ===== - Ideal gas ===== ===== - Ideal gas =====
-Now we can derive the **equation of state (EOS)** of an ideal gas where the particles follow MB statistics. The EOS is the pressure $\mathcal{P}(\rho,T)$ as a function of density $\rho$ and temperature $T$.+Now we can derive the **equation of state (EOS)** of an ideal gas where the particles follow MB statistics. The EOS is the pressure $P(\rho,T)$ as a function of density $\rho$ and temperature $T$.
  
 {{:courses:ast301:idealgas.webp?nolink&700|}} {{:courses:ast301:idealgas.webp?nolink&700|}}
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 So $+2m\mathbf{v}$ amount of momentum is transferred to the wall due to the collision. Force is the rate of change of momentum and pressure is the force per unit area of the wall. So the total pressure transferred to the wall due a total of $N$ particles So $+2m\mathbf{v}$ amount of momentum is transferred to the wall due to the collision. Force is the rate of change of momentum and pressure is the force per unit area of the wall. So the total pressure transferred to the wall due a total of $N$ particles
  
-$$ \mathcal{P} = N\frac{\Delta p}{\Delta t} \frac{1}{\Delta A} $$+$$ P = N\frac{\Delta p}{\Delta t} \frac{1}{\Delta A} $$
  
 where $N=V(n/6)$ and $n$ is the number of particles per unit volume $V=v\Delta t \Delta A$. The factor $6$ arises because there are six walls in the box containing the cylinder and only one-sixth of the particles hit a particular wall that we are considering. So where $N=V(n/6)$ and $n$ is the number of particles per unit volume $V=v\Delta t \Delta A$. The factor $6$ arises because there are six walls in the box containing the cylinder and only one-sixth of the particles hit a particular wall that we are considering. So
  
-$$ \mathcal{P} = \frac{nv}{6} 2mv = \frac{nmv^2}{3} = \frac{2}{3} n \frac{1}{2} mv^2 = \frac{2}{3} n E $$+$$ P = \frac{nv}{6} 2mv = \frac{nmv^2}{3} = \frac{2}{3} n \frac{1}{2} mv^2 = \frac{2}{3} n E $$
  
 where $E$ is the kinetic energy. So total pressure is proportional to the //kinetic energy density// $nE$. This is true if all the particles have the same mass and speed. However, if the particles have a distribution speeds, the general form of the pressure would be where $E$ is the kinetic energy. So total pressure is proportional to the //kinetic energy density// $nE$. This is true if all the particles have the same mass and speed. However, if the particles have a distribution speeds, the general form of the pressure would be
  
-$$ \mathcal{P} = \frac{2}{3} n E_{av}. $$+$$ P = \frac{2}{3} n E_{av}. $$
  
 Pressure and kinetic energy density even have the same units. Note that pressure is a **tensor** because it requires taking into account two vectors: the flow of particles and the normal force resulting from the bounce of the particles on a wall. But for an isotropic gas in equilibrium, the tensor becomes a scalar. Pressure and kinetic energy density even have the same units. Note that pressure is a **tensor** because it requires taking into account two vectors: the flow of particles and the normal force resulting from the bounce of the particles on a wall. But for an isotropic gas in equilibrium, the tensor becomes a scalar.
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 For stars, the Equation of State (**EOS**) for the gas inside a star is found by substituting $E_{av}$ in the equation of pressure: For stars, the Equation of State (**EOS**) for the gas inside a star is found by substituting $E_{av}$ in the equation of pressure:
  
-$$ \mathcal{P} = nkT = \frac{\rho}{m_{av}} kT $$+$$ P = nkT = \frac{\rho}{m_{av}} kT $$
  
-where the average mass $m_{av}=m_p/2$ for a hydrogen plasma. This is the preferred form of the **ideal gas law**. The more familiar form $\mathcal{P}V=\mathsf{n}RT$ results from $n=\mathsf{n}N_0/V$ where $N_0$ is the Avogadro number.+where the average mass $m_{av}=m_p/2$ for a hydrogen plasma. This is the preferred form of the **ideal gas law**. The more familiar form $PV=\mathsf{n}RT$ results from $n=\mathsf{n}N_0/V$ where $N_0$ is the Avogadro number.
  
  
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 To explain degeneracy, consider a gas of 37 electrons occupying the states shown in the 1D phase-space diagram of Fig. (a) following MBS. They are not degenerate. Most of them are occupying the lowest states as dictated by 1D MBD. The momentum at which kientic energy is equal to $E_{av}=kT/2$ is indicated using a dashed line. To explain degeneracy, consider a gas of 37 electrons occupying the states shown in the 1D phase-space diagram of Fig. (a) following MBS. They are not degenerate. Most of them are occupying the lowest states as dictated by 1D MBD. The momentum at which kientic energy is equal to $E_{av}=kT/2$ is indicated using a dashed line.
  
-If this gas loses energy and cools down, the electrons will lose energy and settle down at the bottom of the diagram as in Fig. (c). All the low states are completely filled. Many of the electrons thus have a higher energy than they would normally have at this low temperature. This unusually higher energy creates the degeneracy pressure. The momentum at the upper end of the FDD is called the Fermi momentum $p_F$. The FD distribution $P(p_x)$ looks almost rectangular, there are almost no electrons with $p>p_F$.+If this gas loses energy and cools down, the electrons will lose energy and settle down at the bottom of the diagram as in Fig. (C). All the low states are completely filled. Many of the electrons thus have a higher energy than they would normally have at this low temperature. This unusually higher energy creates the degeneracy pressure. The momentum at the upper end of the FDD is called the Fermi momentum $p_F$. The FD distribution $\mathsf{P}(p_x)$ looks almost rectangular, there are almost no electrons with $p>p_F$.
  
 We can make a gas degenerate by squeezing it (decreasing $\Delta X$) instead of cooling, i. e. by moving from Fig. (a) to (b). Moving from (b) to (d) we see that the electrons remain almost in the same states. That means in a completely degenerate gas, the pressure becomes independent of temperature. We can make a gas degenerate by squeezing it (decreasing $\Delta X$) instead of cooling, i. e. by moving from Fig. (a) to (b). Moving from (b) to (d) we see that the electrons remain almost in the same states. That means in a completely degenerate gas, the pressure becomes independent of temperature.
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 In a degenerate gas, the protons and the nuclei are responsible for the inward gravitational pull and the electrons are almost totally responsible for the outward pressure that balances gravity. This is because pressure In a degenerate gas, the protons and the nuclei are responsible for the inward gravitational pull and the electrons are almost totally responsible for the outward pressure that balances gravity. This is because pressure
  
-$$ \mathcal{P} \approx nE_{av} $$+$$ P \approx nE_{av} $$
  
 and the kinetic energy of degenerate electrons are much higher than their normal average. In non-degenerate matter, the average kinetic energy of electrons and protons are equal dictated by the equipartition of energy. and the kinetic energy of degenerate electrons are much higher than their normal average. In non-degenerate matter, the average kinetic energy of electrons and protons are equal dictated by the equipartition of energy.
courses/ast301/2.1710490722.txt.gz · Last modified: 2024/03/15 02:18 by asad
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