Hauptseite

MathTEST

Google: <m>\sin^2(x)+\cos(x)^2+e^{i \pi}=0</m>

Mediawiki-Math: $\sin(x)^{2}+\cos(x)^{2}+e^{i\pi }=0$

${\begin{aligned}&{\frac {d}{dt}}{\frac {\partial L({\bar {r}},{\bar {v}},t)}{\partial {{v}_{k}}}}=m{{\ddot {x}}_{k}}+q\left({\frac {\partial }{\partial t}}{{A}_{k}}({\bar {r}},t)+{\frac {\partial {{A}_{k}}({\bar {r}},t)}{\partial {{x}_{l}}}}{\frac {\partial {{x}_{l}}}{\partial t}}\right)=m{{\ddot {x}}_{k}}+q\left({\frac {\partial }{\partial t}}+{\bar {v}}\cdot \nabla \right){{A}_{k}}({\bar {r}},t)\\&{\frac {\partial L({\bar {r}},{\bar {v}},t)}{\partial {{x}_{k}}}}=q\left[{\frac {\partial }{\partial {{x}_{k}}}}\left({\bar {v}}{\bar {A}}\right)-{\frac {\partial }{\partial {{x}_{k}}}}\Phi \right]\\&\Rightarrow 0={\frac {d}{dt}}{\frac {\partial L({\bar {r}},{\bar {v}},t)}{\partial {{v}_{k}}}}-{\frac {\partial L({\bar {r}},{\bar {v}},t)}{\partial {{x}_{k}}}}=m{{\ddot {x}}_{k}}+q\left({\frac {\partial }{\partial t}}+{\bar {v}}\cdot \nabla \right){{A}_{k}}({\bar {r}},t)-q\left[{\frac {\partial }{\partial {{x}_{k}}}}\left({\bar {v}}{\bar {A}}\right)-{\frac {\partial }{\partial {{x}_{k}}}}\Phi \right]\\&=m{{\ddot {x}}_{k}}+q{\frac {\partial }{\partial t}}{{A}_{k}}({\bar {r}},t)+q\left[\left({\bar {v}}\cdot \nabla \right){{A}_{k}}({\bar {r}},t)-{\frac {\partial }{\partial {{x}_{k}}}}\left({\bar {v}}{\bar {A}}\right)\right]+q{\frac {\partial }{\partial {{x}_{k}}}}\Phi \\&\left[\left({\bar {v}}\cdot \nabla \right){{A}_{k}}({\bar {r}},t)-{\frac {\partial }{\partial {{x}_{k}}}}\left({\bar {v}}{\bar {A}}\right)\right]=-{{\left[{\bar {v}}\times \left(\nabla \times {\bar {A}}\right)\right]}_{k}}\\&\Rightarrow 0=m{\ddot {\bar {r}}}+q{\frac {\partial }{\partial t}}A({\bar {r}},t)-q\left[{\bar {v}}\times \left(\nabla \times {\bar {A}}\right)\right]+q\nabla \Phi =m{\ddot {\bar {r}}}+q\left[{\frac {\partial }{\partial t}}A({\bar {r}},t)+\nabla \Phi -\left[{\bar {v}}\times \left(\nabla \times {\bar {A}}\right)\right]\right]\\\end{aligned}}$

<m> \begin{align} & \frac{d}{dt}\frac{\partial L(\bar{r},\bar{v},t)}{\partial {{v}_{k}}}=m{{{\ddot{x}}}_{k}}+q\left( \frac{\partial }{\partial t}{{A}_{k}}(\bar{r},t)+\frac{\partial {{A}_{k}}(\bar{r},t)}{\partial {{x}_{l}}}\frac{\partial {{x}_{l}}}{\partial t} \right)=m{{{\ddot{x}}}_{k}}+q\left( \frac{\partial }{\partial t}+\bar{v}\cdot \nabla \right){{A}_{k}}(\bar{r},t) \\ & \frac{\partial L(\bar{r},\bar{v},t)}{\partial {{x}_{k}}}=q\left[ \frac{\partial }{\partial {{x}_{k}}}\left( \bar{v}\bar{A} \right)-\frac{\partial }{\partial {{x}_{k}}}\Phi \right] \\ & \Rightarrow 0=\frac{d}{dt}\frac{\partial L(\bar{r},\bar{v},t)}{\partial {{v}_{k}}}-\frac{\partial L(\bar{r},\bar{v},t)}{\partial {{x}_{k}}}=m{{{\ddot{x}}}_{k}}+q\left( \frac{\partial }{\partial t}+\bar{v}\cdot \nabla \right){{A}_{k}}(\bar{r},t)-q\left[ \frac{\partial }{\partial {{x}_{k}}}\left( \bar{v}\bar{A} \right)-\frac{\partial }{\partial {{x}_{k}}}\Phi \right] \\ & =m{{{\ddot{x}}}_{k}}+q\frac{\partial }{\partial t}{{A}_{k}}(\bar{r},t)+q\left[ \left( \bar{v}\cdot \nabla \right){{A}_{k}}(\bar{r},t)-\frac{\partial }{\partial {{x}_{k}}}\left( \bar{v}\bar{A} \right) \right]+q\frac{\partial }{\partial {{x}_{k}}}\Phi \\ & \left[ \left( \bar{v}\cdot \nabla \right){{A}_{k}}(\bar{r},t)-\frac{\partial }{\partial {{x}_{k}}}\left( \bar{v}\bar{A} \right) \right]=-{{\left[ \bar{v}\times \left( \nabla \times \bar{A} \right) \right]}_{k}} \\ & \Rightarrow 0=m\ddot{\bar{r}}+q\frac{\partial }{\partial t}A(\bar{r},t)-q\left[ \bar{v}\times \left( \nabla \times \bar{A} \right) \right]+q\nabla \Phi =m\ddot{\bar{r}}+q\left[ \frac{\partial }{\partial t}A(\bar{r},t)+\nabla \Phi -\left[ \bar{v}\times \left( \nabla \times \bar{A} \right) \right] \right] \\ \end{align}</m>

${\underline {b}}$

${\begin{aligned}a&+b\\\\c&+d\end{aligned}}$

${\text{Magnetfeld}}\quad$

${\underline {\nabla }}$ $c+c*c^{2}+c+2c+8+\cos ^{2}x+y$

${\begin{aligned}{\text{Magnetfeld}}\quad {\underline {B}}&={\underline {\nabla }}\times {\underline {A}}\\{\text{elektrisches Feld}}\quad {\underline {E}}&=-{\underline {\nabla }}\phi -{\frac {1}{c}}{{\partial }_{t}}{\underline {A}}\\\end{aligned}}$