- #1
Petar Mali
- 290
- 0
If we have case [tex]\rho=0[/tex], [tex]\vec{j}=\vec{0}[/tex]
then we have this equations
[tex]\Delta\vec{E}-\frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2}[/tex]
[tex]\Delta\vec{B}-\frac{1}{c^2}\frac{\partial^2 \vec{B}}{\partial t^2}[/tex]
Particular solutions of this equation
[tex]\vec{E}=\vec{E}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
[tex]\vec{B}=\vec{B}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
If we have coordinate frames [tex]S,S'[/tex]
System [tex]S'[/tex] has relative velocity [tex]\vec{u}[/tex] in [tex]x[/tex] direction compared with [tex]S[/tex]
System [tex]S'[/tex]
[tex]\vec{E}'=\vec{E}_0' e^{i(w't'-\vec{k}'\cdot\vec{r}')}[/tex]
[tex]\vec{B}'=\vec{B}'_0 e^{i(w't'-\vec{k}'\cdot\vec{r}')}[/tex]
System [tex]S[/tex]
[tex]\vec{E}=\vec{E}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
[tex]\vec{B}=\vec{B}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
Why phase
[tex]wt-\vec{k}\cdot\vec{r}[/tex]
must be invariant?
then we have this equations
[tex]\Delta\vec{E}-\frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2}[/tex]
[tex]\Delta\vec{B}-\frac{1}{c^2}\frac{\partial^2 \vec{B}}{\partial t^2}[/tex]
Particular solutions of this equation
[tex]\vec{E}=\vec{E}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
[tex]\vec{B}=\vec{B}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
If we have coordinate frames [tex]S,S'[/tex]
System [tex]S'[/tex] has relative velocity [tex]\vec{u}[/tex] in [tex]x[/tex] direction compared with [tex]S[/tex]
System [tex]S'[/tex]
[tex]\vec{E}'=\vec{E}_0' e^{i(w't'-\vec{k}'\cdot\vec{r}')}[/tex]
[tex]\vec{B}'=\vec{B}'_0 e^{i(w't'-\vec{k}'\cdot\vec{r}')}[/tex]
System [tex]S[/tex]
[tex]\vec{E}=\vec{E}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
[tex]\vec{B}=\vec{B}_0 e^{i(wt-\vec{k}\cdot\vec{r})}[/tex]
Why phase
[tex]wt-\vec{k}\cdot\vec{r}[/tex]
must be invariant?