User talk:GyBlop

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]

to write in

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

http://wiki.riteme.site/wiki/Polarization_of_classical_electromagnetic_waves [3]

http://wiki.riteme.site/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

趕搭公車時有看到你^^ 我認得你的鞋下週同一時間我會在校門等你在茫茫人海中能遇到你是一種緣分我一定要等你