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Fermi Gas Distinguishable <---> Indistinguishable Classical <----> degenerate depend on density. If the wavelength.

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Presentation on theme: "Fermi Gas Distinguishable <---> Indistinguishable Classical <----> degenerate depend on density. If the wavelength."— Presentation transcript:

1 Fermi Gas Distinguishable <---> Indistinguishable Classical <----> degenerate depend on density. If the wavelength similar to the separation than degenerate Fermi gas larger temperatures have smaller wavelength  need tighter packing for degeneracy to occur degenerate electron examples conductors and semiconductors pressure at Earth’s core (at least some of it) aids in initiating transition from Main Sequence stars to Red Giants (allows T to grow electron pressure is independent of T) white dwarves and Iron core of massive stars Neutron and proton examples nuclei with Fermi momentum = 250 MeV/c neutron stars

2 Degenerate vs non-degenerate

3 Conduction electrons Most electrons in a metal are attached to individual atoms. But 1-2 are “free” to move through the lattice. Can treat them as a “gas” (in a 3D box) more like a finite well but energy levels (and density of states) similar (not bound states but “vibrational” states of electrons in box) depth of well V = W (energy needed for electron to be removed from metal’s surface - photoelectric effect) Fermi Energy at T = 0 all states up to EF are filled W V EF Filled levels

4 Conduction electrons Can then calculate the Fermi energy for T=0 (and it doesn’t usually change much for higher T) Ex. Silver 1 free electron/atom n T=0 E

5 Conduction electrons Can determine the average energy at T = 0
for silver  eV can compare to classical statistics Pauli exclusion forces electrons to much higher energy levels at “low” temperatures. (why e’s not involved in specific heat which is a lattice vibration/phonons)

6 Conduction electrons

7 Conduction electrons Similarly, from T-dependent
the terms after the 1 are the degeneracy terms….large if degenerate. For silver atoms at T=300 K not until the degeneracy term is small will the electron act classically. Happens at high T The Fermi energy varies slowly with T and at T=300 K is almost the same as at T=0 You obtain the Fermi energy by normalization. Quark-gluon plasma (covered later) is an example of a high T Fermi gas

8 Fermi Gases in Stars Equilibrium: balance between gravitational pressure and “gas” (either normal or degenerate) pressure total gravitational Energy: density varies in normal stars (in Sun: average is 1 g/cm3 but at r=0 is 100 g/cm3). More of a constant in white dwarves or neutron stars will have either “normal” gas pressure of P=nkT (P=n<E>) or pressure due to degenerate particles. Normal depends on T, degenerate (mostly) doesn’t n = particle density in this case

9 Degenerate Fermi Gas Pressure
Start with p = n<E> non-relativistic relativistic P depends ONLY on density

10 Degenerate Fermi Gas Pressure
non-relativistic relativistic P depends ONLY on density Pressure decreases if, for a given density, particles become relativistic if shrink star’s radius by  density increases by  gravitational E increases by 2 if non-relativistic. <E> increases by (N/V)2/3 = 4 if relativistic <E> increases by (N/V)1/3 = 2  non-relativistic stable but relativistic is not. can aid in collapse of white dwarf

11 Older Sun-like Stars SKIP
Density of core increases as HHe. He inert (no fusion yet). Core contracts electrons become degenerate. 4 e per He nuclei. Electrons have longer wavelength than He electrons move to higher energy due to Pauli exclusion/degeneracy. No longer in thermal equilibrium with p, He nuclei pressure becomes dominated by electrons. No longer depends on T allows T of p,He to increase rapidly without “normal” increase in pressure and change in star’s equilibrium. Onset of 3HeC fusion and Red Giant phase (helium flash when T = 100,000,000 K)

12 White Dwarves Leftover cores of Red Giants made (usually) from C + O nuclei and degenerate electrons cores of very massive stars are Fe nuclei plus degenerate electrons and have similar properties gravitational pressure balanced by electrons’ pressure which increases as radius decreases  radius depends on Mass of star Determine approximate Fermi Energy. Assume electron density = 0.5(p+n) density electrons are in this range and often not completely relativistic or non-relativistic  need to use the correct E2 = p2 + m2 relationship

13 White Dwarves + Collapse
If the electron energy is > about 1.4 MeV can have: any electrons > ET “disappear”. The electron energy distribution depends on T (average E) the “lost” electrons cause the pressure from the degenerate electrons to decrease the energy of the neutrinos is also lost as they escape  “cools” the star as the mass increases, radius decreases, and number of electrons above threshold increases #e’s EF ET

14 White Dwarves+Supernovas SKIP
another process - photodisentegration - also absorbs energy “cooling” star. Similar energy loss as e+p combination At some point the not very stable equilibrium between gravity and (mostly) electron pressure doesn’t hold White Dwarf collapses and some fraction (20-50% ??) of the protons convert to neutrons during the collapse gives Supernovas

15 White Dwarves+Supernovas

16 Neutron Stars-approx. numbers SKIP
Supernovas can produce neutron stars radius ~ 10 km mass about that of Sun. always < 3 mass Sun relative n:p:e ~ 99:1:1 gravity supported by degenerate neutrons plug into non-relativistic formula for Fermi Energy  140 MeV (as mass =940 MeV, non-rel OK) look at wavelength can determine radius vs mass (like WD) can collapse into black hole

17 Neutron Stars SKIP 3 separate Fermi gases: n:p:e p+n are in the same potential well due to strong nuclear force assume independent and that p/n = 0.01 (depends on star’s mass) so need to use relativistic for electrons but not independent as p <---> n plus reactions with virtual particles free neutrons decay. But in a neutron star they can only do so if there is an available unfilled electron state. So suppresses decay

18 Neutron Stars SKIP Will end up with an equilibrium between n-p-e which can best be seen by matching up the Fermi energy of the neutrons with the e-p system neutrons with E > EF can then decay to p-e-nu (which raises electron density and its Fermi energy thus the balance) need to include rest mass energies. Also density of electrons is equal to that of protons can then solve for p/n ratio (we’ll skip algebra) gives for typical neutron star:


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