看板 Physics 關於我們 聯絡資訊
3 1/3 Kf = 3π^2 N/V => Kf = ( 3π^2 N/V ) 2 2 2 hbar Kf (h/2π) 1/3 2 Ef(0) = k Tf = ──── = ──── [ ( 3π^2 N/V ) ] 2m 2m 2 2/3 = h [ 3π^2 N/(8V) ] /2m 2 2/3 2 2/3 -2/3 => Tf = h [ 3π^2 N/(8V) ] /2mk = h [ 3π^2 N/8 ] /2mk V 2 2/3 -5/3 dTf/dV = - 2/3 h [ 3π^2 N/8 ] /2mk V 2 2/3 -1 = - 2/3 h [ 3π^2 N/(8V) ] /2mk V -2 = ─ Tf 3V -1 -2 -2 -2 2 -1 d(Tf )/dV = - Tf dTf/dV = - Tf ( ─ Tf ) = ─ Tf 3V 3V -3 -4 -4 -2 2 -3 d(Tf )/dV = - 3 Tf dTf/dV = - 3 Tf ( ─ Tf ) = - Tf 3V V 3 π^2 T 2 π^4 T 4 F(T) = N(0) k Tf [ - - ── ( ─ ) + ── ( ─ ) + … ] 5 4 Tf 80 Tf 3 Tf π^2 T π^4 T 3 = N(0) k T [ - ( ─ ) - ── ( ─ ) + ── ( ─ ) + … ] 5 T 4 Tf 80 Tf The Fermion Gas Pressure P(T): 因為 P = - ( ∂ F/∂ V ) T,N 3 1 -2 π^2 2 -1 = - N(0) k T [ - - ( ─ Tf ) - ── T ( ─ Tf ) 5 T 3V 4 3V π^4 3 2 -3 + ── T ( - Tf ) + … ] 80 V N(0) k T 2 Tf π^2 T π^4 T 3 = ──── [ - ( ─ ) + ── ( ─ ) - ── ( ─ ) + … ] V 5 T 6 Tf 40 Tf N(0) k Tf 2 π^2 T 2 π^4 T 4 = ───── [ - + ── ( ─ ) - ── ( ─ ) + … ] V 5 6 Tf 40 Tf 2 N(0) k Tf 5 π^2 T 2 π^4 T 4 = - ───── [ 1 + ─── ( ─ ) - ── ( ─ ) + … ] 5 V 12 Tf 16 Tf 2 N(0) k Tf 5 π^2 T 2 π^4 T 4 P(T) = - ───── [ 1 + ─── ( ─ ) - ── ( ─ ) + … ] 5 V 12 Tf 16 Tf 2 N(0)Ef(0) 5 π^2 T 2 π^4 T 4 = - ───── [ 1 + ─── ( ─ ) - ── ( ─ ) + … ] 5 V 12 Tf 16 Tf 3 5 π^2 T 2 π^4 T 4 U(T) = - N(0) Ef(0) [ 1 + ─── ( ─ ) - ── ( ─ ) + … ] 5 12 Tf 16 Tf P(T) 2 2U ── = ─ => P = ─ U(T) 3V 3V -- Van Der Waals' equation: ( P + a/υ^2 )( υ - b ) = R T -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 118.160.213.211
feynman511:雖然我從沒好好看過你打的 可是真的打得很認真XD 03/25 02:03
Frobenius:其實我本來以為很簡單,沒想到會用到這麼多近似的技巧XD 03/25 02:20
ANUBISANKH:沒想到今天碰到同樣的習題 XDD 剛好在這板上看過! 04/02 12:06
alfadick:從數學板跑過來。外行給推! 05/28 09:47
alfadick:可以用LaTeX打!打好再外連pdf網址 05/28 09:47