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| courses:ast301:2 [2024/03/12 11:30] – [3.4 Relativistic EOS] asad | courses:ast301:2 [2024/03/15 02:26] (current) – [3.1 Degeneracy in 1D] asad | ||
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| MB statistics describes the distribution of speed or momentum of the particles in a gas constrained by three conditions: (i) all speeds in all three dimensions are equally probable, (ii) the total energy is limited and (iii) the total number of particles is limited. Number (i) ensures that the particles have a wide range of speeds and (ii) and (iii) ensure that there is enough variation in the speeds. | MB statistics describes the distribution of speed or momentum of the particles in a gas constrained by three conditions: (i) all speeds in all three dimensions are equally probable, (ii) the total energy is limited and (iii) the total number of particles is limited. Number (i) ensures that the particles have a wide range of speeds and (ii) and (iii) ensure that there is enough variation in the speeds. | ||
| - | Consider a gas made of a limited number of particles. $P(v_x)\ dv_x$ is the probability of finding a particle with speed $v_x$ in the interval $dv_x$, which can be shown to be | + | Consider a gas made of a limited number of particles. $\mathsf{P}(v_x)\ dv_x$ is the probability of finding a particle with speed $v_x$ in the interval $dv_x$, which can be shown to be |
| - | $$ P(v_x)\ dv_x = \left(\frac{m}{2\pi kT}\right)^{1/ | + | $$ \mathsf{P}(v_x)\ dv_x = \left(\frac{m}{2\pi kT}\right)^{1/ |
| where $m$ is the mass of the particle, $T$ the temperature of the gas and $k$ Boltzmann constant. | where $m$ is the mass of the particle, $T$ the temperature of the gas and $k$ Boltzmann constant. | ||
| ==== - 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 |
| - | $$ P(p_x)\ dp_x = \left(\frac{1}{2\pi mkT}\right)^{1/ | + | $$ \mathsf{P}(p_x)\ dp_x = \left(\frac{1}{2\pi mkT}\right)^{1/ |
| 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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| - | 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\ |
| - | $$ P(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/ | + | $$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/ |
| which becomes | which becomes | ||
| - | $$ P(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/ | + | $$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/ |
| - | which is the most used form of **MBD**. The quantity is dimensionless: | + | which is the most used form of **MBD**. The quantity is dimensionless: |
| MBD can also be written in terms of energy because $E=mv^2/ | MBD can also be written in terms of energy because $E=mv^2/ | ||
| - | $$ P(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/ | + | $$ \mathsf{P}(p)\ d^3p = \left(\frac{1}{2\pi mkT}\right)^{3/ |
| - | which means $P(p) \propto e^{-E/ | + | which means $\mathsf{P}(p) \propto e^{-E/ |
| 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 $p$ | ||
| - | $$ P(p)\ dp = \left(\frac{1}{2\pi mkT}\right)^{3/ | + | $$ \mathsf{P}(p)\ dp = \left(\frac{1}{2\pi mkT}\right)^{3/ |
| - | 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, | + | 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, |
| 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, | 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, | ||
| - | $$ f_{MB} = nP(p) = \frac{N}{V} \left(\frac{1}{2\pi mkT}\right)^{3/ | + | $$ f_{MB} = n\mathsf{P}(p) = \frac{N}{V} \left(\frac{1}{2\pi mkT}\right)^{3/ |
| where $f_{MB}$ has the units m$^{-3}$ (Ns)$^{-3}$, | where $f_{MB}$ has the units m$^{-3}$ (Ns)$^{-3}$, | ||
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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 $T$. |
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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, | 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, | ||
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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/ | + | where the average mass $m_{av}=m_p/ |
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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/ | 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/ | ||
| - | 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, | + | 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, |
| 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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| which is plotted in Fig. (b) above. Note that at zero temperature the factor $(E-E_F)/ | which is plotted in Fig. (b) above. Note that at zero temperature the factor $(E-E_F)/ | ||
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| + | < | ||
| In Fig. (C), the Fermi function is plotted for two more temperatures $T_1$ and $T_2$. As temperature increases, more particles become available at higher energies. For $T>0$, the function becomes $1/2$ at $E=E_F$ because the exponential becomes unity ($e^0=1$). | In Fig. (C), the Fermi function is plotted for two more temperatures $T_1$ and $T_2$. As temperature increases, more particles become available at higher energies. For $T>0$, the function becomes $1/2$ at $E=E_F$ because the exponential becomes unity ($e^0=1$). | ||
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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. | ||
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| meaning the pressure is proportional to $\rho^{4/ | meaning the pressure is proportional to $\rho^{4/ | ||
| - | When the electrons inside a white dwarf reach relativistic speeds, the pressure varies //softly// with density and cannot be high enough to sustain the dead star against gravity leading to a gravitational collapse. This can occur when the mass of a white dwarf approaches the Chandrasekhar limit of $\sim 1.4$ M$_\odot$. | + | When the electrons inside a white dwarf reach relativistic speeds, the pressure varies //softly// with density and cannot be high enough to sustain the dead star against gravity leading to a gravitational collapse. This can occur when the mass of a white dwarf approaches the [[uv:Chandrasekhar]] limit of $\sim 1.4$ M$_\odot$. |
| ===== - Summary of the EOS ===== | ===== - Summary of the EOS ===== | ||
| We have discussed three different EOS for non-degenerate gas following MBS and the relativistic and nonrelativistic degenerate gas following FDS. Photons follow a different statistics, the **Bose-Einstein Statistics** (BES) and we will see in a later chapter that in that case the EOS is $P=aT^4/3$, meaning the pressure depends only on temperature. | We have discussed three different EOS for non-degenerate gas following MBS and the relativistic and nonrelativistic degenerate gas following FDS. Photons follow a different statistics, the **Bose-Einstein Statistics** (BES) and we will see in a later chapter that in that case the EOS is $P=aT^4/3$, meaning the pressure depends only on temperature. | ||
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| The four EOS are depicted above in a logarithmic temperature-density ($T-\rho$) plot. At low density and high temperature, | The four EOS are depicted above in a logarithmic temperature-density ($T-\rho$) plot. At low density and high temperature, | ||
| + | |||
| + | Neutron stars are mainly supported by repulsive nuclear forces, but neutron degeneracy pressure also plays a role. When all support mechanisms fail, gravity overtakes and turns a star into a black hole. | ||
courses/ast301/2.1710264658.txt.gz · Last modified: 2024/03/12 11:30 by asad