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I forumet så blir:<br>
I forumet så blir:
[tex]\mathbf{B}(\mathbf{r},t)=-\frac{\mu_0q}{4\pi}\left[\frac{c\,\hat{\mathbf{n}}\times
 
'''[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}+
\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}
\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}
\times\dot{\vec{\beta}})]}{R\,(1-\vec{\beta}\mathbf{\cdot}\hat{\mathbf{n}})^3}
\right]_{\mathrm{retarded}} \qquad (1)[/tex]
\right]_{\mathrm{retarded}} \qquad (1)'''[/tex]'''


Till:<br>
Till:<br>
Rad 15: Rad 16:
[tex]\sqrt{2}[/tex] ger <math>\sqrt{2}</math>
[tex]\sqrt{2}[/tex] ger <math>\sqrt{2}</math>


"e^{i + \pi} + 1 = 0" ger  
[tex]e^{i + \pi} + 1 = 0[/tex] ger <math>e^{i + \pi}+1=0</math>
<math>e^{i + \pi}+1=0</math>


<math>\sqrt{7}</math>
[tex]\sqrt{7}[/tex] ger <math>\sqrt{7}</math><br>
<math>\sqrt{4}</math>
[tex]\sqrt{4}[/tex] ger <math>\sqrt{4}</math>


== Externa länkar ==
== Externa länkar ==

Versionen från 13 maj 2013 kl. 16.23

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:

[tex]\sqrt{2}[/tex] ger

[tex]e^{i + \pi} + 1 = 0[/tex] ger

[tex]\sqrt{7}[/tex] ger
[tex]\sqrt{4}[/tex] ger

Externa länkar