Hauptseite

MathTEST

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

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

$\begin{array}{rl}& \frac{d}{dt}\frac{\partial L(\overline{r},\overline{v},t)}{\partial v_k}=m{\ddot{x}}_k+q(\frac{\partial }{\partial t}{A}_k(\overline{r},t)+\frac{\partial A_k(\overline{r},t)}{\partial x_l}\frac{\partial x_l}{\partial t})=m{\ddot{x}}_k+q(\frac{\partial }{\partial t}+\overline{v}\cdot \mathrm{\nabla })A_k(\overline{r},t)\\ & \frac{\partial L(\overline{r},\overline{v},t)}{\partial x_k}=q[\frac{\partial }{\partial x_k}\left(\overline{v}\overline{A}\right)-\frac{\partial }{\partial x_k}\mathrm{\Phi }]\\ & \Rightarrow 0=\frac{d}{dt}\frac{\partial L(\overline{r},\overline{v},t)}{\partial v_k}-\frac{\partial L(\overline{r},\overline{v},t)}{\partial x_k}=m{\ddot{x}}_k+q(\frac{\partial }{\partial t}+\overline{v}\cdot \mathrm{\nabla })A_k(\overline{r},t)-q[\frac{\partial }{\partial x_k}\left(\overline{v}\overline{A}\right)-\frac{\partial }{\partial x_k}\mathrm{\Phi }]\\ & =m{\ddot{x}}_k+q\frac{\partial }{\partial t}{A}_k(\overline{r},t)+q[(\overline{v}\cdot \mathrm{\nabla })A_k(\overline{r},t)-\frac{\partial }{\partial x_k}\left(\overline{v}\overline{A}\right)]+q\frac{\partial }{\partial x_k}\mathrm{\Phi }\\ & [(\overline{v}\cdot \mathrm{\nabla })A_k(\overline{r},t)-\frac{\partial }{\partial x_k}\left(\overline{v}\overline{A}\right)]=-{[\overline{v}\times (\mathrm{\nabla }\times \overline{A})]}_k\\ & \Rightarrow 0=m\ddot{\overline{r}}+q\frac{\partial }{\partial t}A(\overline{r},t)-q[\overline{v}\times (\mathrm{\nabla }\times \overline{A})]+q\mathrm{\nabla }\mathrm{\Phi }=m\ddot{\overline{r}}+q[\frac{\partial }{\partial t}A(\overline{r},t)+\mathrm{\nabla }\mathrm{\Phi }-[\overline{v}\times (\mathrm{\nabla }\times \overline{A})]]\end{array}$

<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>

$\underset{\_}{b}$

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

$\text{Magnetfeld}$

$\underset{\_}{\mathrm{\nabla }}$ $c+c*c^2+c+2c+8+{\cos }^{2}x+y$

$\begin{array}{rl}\text{Magnetfeld}\underset{\_}{B}& =\underset{\_}{\mathrm{\nabla }}\times \underset{\_}{A}\\ \text{elektrisches Feld}\underset{\_}{E}& =-\underset{\_}{\mathrm{\nabla }}\varphi -\frac{1}{c}{\partial }_t\underset{\_}{A}\\ \text{what about spaces}\end{array}$