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方程式TRYING

$\int _{o}^{a}\,\,{\frac {sin^{2}(n\pi x/a)}{2mE-2m{\frac {x}{a}}}}\,\,dx$

$\int _{r=0}^{r=r}\,\,{\sqrt {2mE}}\,\,{\sqrt {1-{\frac {V_{o}}{E}}ln({\frac {r}{a}})}}dr$

$H\,=\,{\frac {1}{2m}}\,\,[{\vec {p}}-{\frac {e}{c}}{\vec {A}}]^{2}\,\,+W({\vec {r}})\,\,+\,\,e\phi$

$\int {\frac {dx}{(1-e^{-2\pi x})^{2}}}$

量子力學好文章

[1]

Newtonian Fluids

${\mathcal {\,}}insertsomethings\,$

${\mathcal {\,}}{\frac {du_{x}}{dy}}={\frac {d({\frac {dx}{dt}})}{dy}}={\frac {d\gamma }{dt}}\,$

${\mathcal {\,}}{\frac {dx}{dy}}\,$位平面B上的剪應變. ${\mathcal {\,}}{\frac {du_{x}}{dy}}={\dot {\gamma }}\,$

剪應力 ${\mathcal {\,}}\tau _{yx}={\frac {F}{A}}\,$

F為依照牛頓第三運動定律推測之內側作用力，A為被作用的平面之面積

在牛頓流體中,剪應力 ${\mathcal {\,}}\tau _{yx}=-\mu {\frac {du_{x}}{dy}}=-\mu {\frac {d\gamma }{dt}}=-\mu {\dot {\gamma }}\,$

Non-Newtonian Fluids

• 賓漢流體模式

${\mathcal {\,}}\tau _{yx}=-{\frac {\mu _{o}}{g_{c}}}{\frac {du_{x}}{dy}}\,$ ${\mathcal {\,}}\tau _{o}\,$ 若

若

${\mathcal {\,}}\tau _{yx}={\frac {du_{x}}{dy}}=0\,$

• Ostwald-de Waele Model: (指數模式)

${\mathcal {\,}}\tau _{yx}=-{\frac {K}{g_{c}}}\,\,\,{\mid }{\frac {du_{x}}{dy}}{\mid }^{n-1}\,\,\,{\frac {du_{x}}{dy}}\,$

n,K皆為參數

當n=1, 指數模式與牛頓流體流動行為相同, 且K=\mu 故由n之大小與1之偏差可看出該流體與牛頓流體流動行為之偏差程度. 當n<1時, 流體位擬塑性流體

Diffusions in physcal view

• Fick's 1st law:

${\mathcal {\,}}{\frac {\partial {C}}{\partial {t}}}=\nabla \cdot n_{A}\,+{\frac {d\rho _{A}}{dt}}\,+\,r_{A}\,$

${\mathcal {\,}}N_{A}=\,\,-CD_{AB}\,\,\nabla y_{A}\,+y_{A}(N_{A}+N_{B})\,$

${\mathcal {\,}}N_{AB}\,\,\int _{Z_{1}}^{Z_{2}}\,dZ\,=\,C\cdot \,D_{AB}\,\int _{y_{A1}}^{y_{A2}}\,{\frac {d(1-y_{A})}{1-y_{A}}}\,$

Japan Papers

${\frac {\lambda _{s}}{2sin\theta _{s}}}={\frac {\lambda _{s}}{{\sqrt {2}}sin\theta _{s}}}\times {\frac {1}{\sqrt {2}}}$

$Fig3b\,\,\,:\,\,\,{\frac {\lambda _{s}}{2sin\theta _{s}}}$

$Fig3a\,\,\,:\,\,\,{\frac {\lambda _{s}}{{\sqrt {2}}sin\theta _{s}}}$

Japanese papers

${\frac {\lambda _{s}}{2sin\theta _{s}}}={\frac {\lambda _{air}}{2sin\theta _{air}}}$

how to get curl gradient = 0

由stoke定理

$\int _{s}\int (\nabla \times \nabla \Psi )\cdot {\vec {n}}dA=\oint _{c}\nabla \Psi \cdot {\vec {dr}}=\Psi {\mid }_{A}^{A}=0$

其中S為任意曲面 所以$\nabla \times \nabla \Psi =0$

對於保守場的等價描述:

前提:若E在積分區域內為一階連續向量數

1.

4.

$\oint _{c}{\vec {E}}\cdot {\vec {dr}}=\int _{ABC}{\vec {E}}\cdot {\vec {dr}}+\int _{-ADC}{\vec {E}}\cdot {\vec {dr}}=0$

所以$\int _{ABC}{\vec {E}}\cdot {\vec {dr}}=\int _{ADC}{\vec {E}}\cdot {\vec {dr}}$表示與路徑無

光電相關

http://en.wikipedia.org/wiki/Polarization [2]

http://en.wikipedia.org/wiki/Polarization_of_classical_electromagnetic_waves [3]

http://en.wikipedia.org/wiki/Electromagnetic_wave [4]

$H(t)\,\,\phi (t)\,\,={\frac {i\,\,\hbar d\phi (t)}{dt}}$

$dR(x,y,z)={\frac {\partial {R}}{\partial {x}}}dR+{\frac {\partial {R}}{\partial {y}}}dR+{\frac {\partial {R}}{\partial {z}}}dR$

${\frac {e\,\,g}{\hbar \,\,c}}\,\,=\,\,n$

${\frac {d\psi (\,\,R(t)\,\,)}{dt'}}\,\,=\,\,{\frac {\partial {\psi (R)}}{\partial {R}}}\,\,{\frac {dR(t)}{dt}}$ d Ψ/ dt' = (dΨ/dR) *(dR/dt)

${\vec {B}}\,\,={\frac {q}{r^{2}}}\,\,{\vec {r}}$

$v(t=0)\,\,=\,\,{\sqrt {\frac {E}{m}}}>>c$

$m_{e}={\frac {\hbar ^{2}}{\,\,\,\,\,\,\,\,\,{\frac {d^{2}\epsilon }{dk^{2}}}|_{k=k_{0}}\,\,\,\,\,\,\,\,\,}}$

love

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