moved to markup; some problem with last equation?

1 files changed, 19 insertions, 20 deletions

diff --git a/physics/quantum/fermigas.page b/physics/quantum/fermigas.page
index 0114b43..de66ee1 100644
--- a/physics/quantum/fermigas.page
+++ b/physics/quantum/fermigas.page
@@ -1,35 +1,34 @@
-===============
-Fermi Gas
-===============
+# Fermi Gas
 
 Derivation of the Fermi Energy
----------------------------------
+-------------------------------
+
 Consider a crystal lattice with an electron gas as a 3 dimensional infinite
-square well with dimensions :m:`$l_{x}, l_{y}, l_z$`. The wavefunctions of 
+square well with dimensions $l_{x}, l_{y}, l_z$. The wavefunctions of 
 individual fermions (pretending they are non-interacting) can be seperated
-as :m:`$\psi(x,y)=\psi_{x}(x)\psi_{y}(y)\psi_{z}(z)$`. The solutions will be
+as $\psi(x,y)=\psi_{x}(x)\psi_{y}(y)\psi_{z}(z)$. The solutions will be
 the usual ones to the Schrodinger equation:
 
-:m:`$$\frac{-\hbar^2}{2m}\frac{d^2 \psi_x}{dx}=E_x \psi_x$$`
+$$\frac{-\hbar^2}{2m}\frac{d^2 \psi_x}{dx}=E_x \psi_x$$
 
-with the usual wave numbers :m:`$k_x=\frac{\sqrt{2mE_x}}{\hbar}$`, and quantum
-numbers satisfying the boundry conditions :m:`$k_x l_x = n_x \pi$`. The full
+with the usual wave numbers $k_x=\frac{\sqrt{2mE_x}}{\hbar}$, and quantum
+numbers satisfying the boundry conditions $k_x l_x = n_x \pi$. The full
 wavefunction for each particle will be:
 
-:m:`$$\psi_{n_{x}n_{y}n_{z}}(x,y,z)=\sqrt{\frac{4}{l_{x}l_{y}}}\sin\left(\frac{n_{x}\pi}{l_{x}}x\right)\sin\left(\frac{n_{y}\pi}{l_{y}}y\right)\sin\left(\frac{n_{z}\pi}{l_{z}}z\right)$$`
+$$\psi_{n_{x}n_{y}n_{z}}(x,y,z)=\sqrt{\frac{4}{l_{x}l_{y}}}\sin\left(\frac{n_{x}\pi}{l_{x}}x\right)\sin\left(\frac{n_{y}\pi}{l_{y}}y\right)\sin\left(\frac{n_{z}\pi}{l_{z}}z\right)$$
 
-and the associated energies (with :m:`$E = E_x + E_y + E_z$`):
+and the associated energies (with $E = E_x + E_y + E_z$):
 
-:m:`$$E_{n_{x}n_{y}n_z}=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{n_{x}^{2}}{l_{x}^{2}}+\frac{n_{y}^{2}}{l_{y}^{2}}+\frac{n_{z}^{2}}{l_{z}^{2}}\right)=\frac{\hbar^2|\vec{k}|^2}{2m}$$`
+$$E_{n_{x}n_{y}n_z}=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{n_{x}^{2}}{l_{x}^{2}}+\frac{n_{y}^{2}}{l_{y}^{2}}+\frac{n_{z}^{2}}{l_{z}^{2}}\right)=\frac{\hbar^2|\vec{k}|^2}{2m}$$
 
-where :m:`$|\vec{k}|^2$` is the magnitude of the particle's k-vector in k-space.
+where $|\vec{k}|^2$ is the magnitude of the particle's k-vector in k-space.
 This k-space can be imagined as a grid of blocks, each representing a possible
 particle state (with a double degeneracy for spin). Positions on this grid have
-coordinates :m:`$(k_{x},k_{y},k_z)$` corresponding to the positive integer 
+coordinates $(k_{x},k_{y},k_z)$ corresponding to the positive integer 
 quantum numbers. These blocks will be filled
 from the lowest energy upwards: for large numbers of occupying particles, 
 the filling pattern can be approximated as an expanding spherical shell with 
-radius :m:`$|\vec{k_F}|^2$`. 
+radius $|\vec{k_F}|^2$. 
 
 Note that we're "over counting" the number of occupied states because the
 "sides" of the quarter sphere in k-space (where one of the associated quantum
@@ -41,11 +40,11 @@ side-surface), u.s.w.
 
 The surface of this shell is called the Fermi surface and represents the most
 excited states in the gas. The radius can be derived by calculating the total
-volume enclosed: each block has volume :m:`$\frac{\pi^3}{l_x l_y
-l_z}=\frac{\pi^3}{V}$` and there are N/2 blocks occupied by N fermions, so:
+volume enclosed: each block has volume $\frac{\pi^3}{l_x l_y
+l_z}=\frac{\pi^3}{V}$ and there are N/2 blocks occupied by N fermions, so:
 
-:m:`$$\frac{1}{8}(\frac{4\pi}{3} |k_{F}|^{3})&=&\frac{Nq}{2}(\frac{\pi^{3}}{V})\\|k_{F}|&=&\sqrt{\frac{3Nq\pi^2}{V}}^3=\sqrt{3\pi^2\rho}^3$$`
+$$\frac{1}{8}(\frac{4\pi}{3} |k_{F}|^{3})&=&\frac{Nq}{2}(\frac{\pi^{3}}{V})\\|k_{F}|&=&\sqrt{\frac{3Nq\pi^2}{V}}^3=\sqrt{3\pi^2\rho}^3$$
 
-:m:`$\rho$` is the "free fermion density". The corresponding energy is:
+$\rho$ is the "free fermion density". The corresponding energy is:
 
-:m:`$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$`
+$$E_{F}=\frac{\hbar^{2}}{2m}|k_{F}|^{2}=\frac{\hbar^{2}}{2m}(3\rho \pi)^{2/3}$$