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<math>e^{i + \pi}+1=0<math> | <math>e^{i + \pi}+1=0<math> | ||
Versionen från 13 maj 2013 kl. 16.04
I forumet så blir: [tex]\mathbf{B}(\mathbf{r},t)=-\frac{\mu_0q}{4\pi}\left[\frac{c\,\hat{\mathbf{n}}\times \vec{\beta}}{\gamma^2R^2(1-\vec{\beta}\mathbf{\cdot}\hat{\mathbf{n}})^3}+ \frac{\hat{\mathbf{n}}\times[\,\dot{\vec{\beta}}+\hat{\mathbf{n}}\times(\vec{\beta} \times\dot{\vec{\beta}})]}{R\,(1-\vec{\beta}\mathbf{\cdot}\hat{\mathbf{n}})^3} \right]_{\mathrm{retarded}} \qquad (1)[/tex]
Till:
![{\displaystyle \mathbf {B} (\mathbf {r} ,t)=-{\frac {\mu _{0}q}{4\pi }}\left[{\frac {c\,{\hat {\mathbf {n} }}\times {\vec {\beta }}}{\gamma ^{2}R^{2}(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}+{\frac {{\hat {\mathbf {n} }}\times [\,{\dot {\vec {\beta }}}+{\hat {\mathbf {n} }}\times ({\vec {\beta }}\times {\dot {\vec {\beta }}})]}{R\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right]_{\mathrm {retarded} }\qquad (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/babf643cc68fb75c6908d4082c15cc6cb47ce27d)
[tex]\sqrt{2}[/tex] ger
[c]e^{i + \pi} + 1 = 0[/c] ger <tex>\sqrt{4}</tex>
