optics:res4.4

Различия

Показаны различия между двумя версиями страницы.

Ссылка на это сравнение

Предыдущая версия справа и слева Предыдущая версия
Следующая версия
Предыдущая версия
optics:res4.4 [2020/05/14 11:09] – rootoptics:res4.4 [2020/05/14 11:17] (текущий) – root
Строка 53: Строка 53:
 $$\frac{i\omega}{c}E_{r}=\text{rot}{}_{r}\vec{B}=\frac{1}{r\sin\theta}\left(\frac{\partial\left(\sin\theta B_{\alpha}\right)}{\partial\theta}-\frac{\partial B_{\theta}}{\partial\alpha}\right)=$$ $$\frac{i\omega}{c}E_{r}=\text{rot}{}_{r}\vec{B}=\frac{1}{r\sin\theta}\left(\frac{\partial\left(\sin\theta B_{\alpha}\right)}{\partial\theta}-\frac{\partial B_{\theta}}{\partial\alpha}\right)=$$
 $$ $$
-\frac{1}{r\sin\theta}\left(\frac{\omega}{c}+\frac{i}{r}\right)\frac{\omega}{cr}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\frac{\partial\left(\sin^{2}\theta\right)}{\partial\theta}=$$+\frac{1}{r\sin\theta}\left(\frac{\omega}{c}-\frac{i}{r}\right)\frac{\omega}{cr}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\frac{\partial\left(\sin^{2}\theta\right)}{\partial\theta}=$$
 $$ $$
-2\left(\frac{\omega}{c}+\frac{i}{r}\right)\frac{\omega}{cr^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\cos\theta.$$+2\left(\frac{\omega}{c}-\frac{i}{r}\right)\frac{\omega}{cr^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\cos\theta.$$
 $$\frac{i\omega}{c}E_{\theta}=\text{rot}{}_{\theta}\vec{B}=\frac{1}{r\sin\theta}\frac{\partial B_{r}}{\partial\alpha}-\frac{1}{r}\frac{\partial\left(rB_{\alpha}\right)}{\partial r}=$$ $$\frac{i\omega}{c}E_{\theta}=\text{rot}{}_{\theta}\vec{B}=\frac{1}{r\sin\theta}\frac{\partial B_{r}}{\partial\alpha}-\frac{1}{r}\frac{\partial\left(rB_{\alpha}\right)}{\partial r}=$$
 $$ $$
--\frac{\frac{\omega}{c}p_{0}\sin\theta}{r}\frac{\partial\left(\left(\frac{\omega}{c}+\frac{i}{r}\right)e^{i\omega\left(t-\frac{r}{c}\right)}\right)}{\partial r}=$$ +-\frac{\frac{\omega}{c}p_{0}\sin\theta}{r}\frac{\partial\left(\left(\frac{\omega}{c}-\frac{i}{r}\right)e^{i\omega\left(t-\frac{r}{c}\right)}\right)}{\partial r}=$$ 
-$$\frac{i\frac{\omega}{c}p_{0}\sin\theta}{r^{3}}e^{i\omega\left(t-\frac{r}{c}\right)}+\frac{i\frac{\omega^{2}}{c^{2}}p_{0}\sin\theta}{r}\left(\frac{\omega}{c}+\frac{i}{r}\right)e^{i\omega\left(t-\frac{r}{c}\right)}=$$+$$\frac{i\frac{\omega}{c}p_{0}\sin\theta}{r^{3}}e^{i\omega\left(t-\frac{r}{c}\right)}+\frac{i\frac{\omega^{2}}{c^{2}}p_{0}\sin\theta}{r}\left(\frac{\omega}{c}-\frac{i}{r}\right)e^{i\omega\left(t-\frac{r}{c}\right)}=$$
 $$ $$
-\left(\frac{1}{r^{2}}+\frac{\omega^{2}}{c^{2}}+\frac{i\omega}{rc}\right)i\frac{\omega}{rc}p_{0}\sin\theta e^{i\omega\left(t-\frac{r}{c}\right)}.$$+\left(\frac{1}{r^{2}}+\frac{\omega^{2}}{c^{2}}-\frac{i\omega}{rc}\right)i\frac{\omega}{rc}p_{0}\sin\theta e^{i\omega\left(t-\frac{r}{c}\right)}.$$
 Тогда Тогда
-$$E_{r}=2\left(\frac{1}{r}-\frac{i\omega}{c}\right)\frac{p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\cos\theta}{r^{2}},$$ +$$E_{r}=-2\left(\frac{1}{r}+\frac{i\omega}{c}\right)\frac{p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\cos\theta}{r^{2}},$$ 
-$$E_{\theta}=\left(\frac{1}{r^{2}}+\frac{\omega^{2}}{c^{2}}+\frac{i\omega}{rc}\right)\frac{p_{0}\sin\theta e^{i\omega\left(t-\frac{r}{c}\right)}}{r}.$$+$$E_{\theta}=\left(\frac{1}{r^{2}}+\frac{\omega^{2}}{c^{2}}-\frac{i\omega}{rc}\right)\frac{p_{0}\sin\theta e^{i\omega\left(t-\frac{r}{c}\right)}}{r}.$$