Различия
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| Предыдущая версия справа и слева Предыдущая версия Следующая версия | Предыдущая версия | ||
| optics:res4.4 [2020/05/14 11:06] – root | optics:res4.4 [2020/05/14 11:17] (текущий) – root | ||
|---|---|---|---|
| Строка 44: | Строка 44: | ||
| \text{rot}{}_{\alpha}\vec{A}=\frac{1}{r}\frac{\partial\left(rA_{\theta}\right)}{\partial r}-\frac{1}{r}\frac{\partial A_{r}}{\partial\theta}, | \text{rot}{}_{\alpha}\vec{A}=\frac{1}{r}\frac{\partial\left(rA_{\theta}\right)}{\partial r}-\frac{1}{r}\frac{\partial A_{r}}{\partial\theta}, | ||
| то ненулевая компонента | то ненулевая компонента | ||
| - | $$B_{\alpha}=\text{rot}{}_{\alpha}\vec{A}=\frac{1}{r}\frac{i\omega}{c}p_{0}\frac{\partial\left(e^{i\omega\left(t-\frac{r}{c}\right)}\right)\sin\theta}{\partial r}-\frac{i\omega}{cr^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\frac{\partial\cos\theta}{\partial\theta}=$$ | + | $$B_{\alpha}=\text{rot}{}_{\alpha}\vec{A}=\frac{1}{r}\frac{i\omega}{c}p_{0}\frac{\partial\left(e^{i\omega\left(t-\frac{r}{c}\right)}\right)\sin\theta}{\partial r}+\frac{i\omega}{cr^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\frac{\partial\cos\theta}{\partial\theta}=$$ |
| - | $$\frac{1}{r}\frac{\omega^{2}}{c^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\sin\theta+\frac{i\omega}{cr^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\sin\theta= | + | $$\frac{1}{r}\frac{\omega^{2}}{c^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\sin\theta-\frac{i\omega}{cr^{2}}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\sin\theta= |
| $$ | $$ | ||
| - | $$\left(\frac{\omega}{c}+\frac{i}{r}\right)\frac{\omega}{cr}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\sin\theta.$$ | + | $$\left(\frac{\omega}{c}-\frac{i}{r}\right)\frac{\omega}{cr}p_{0}e^{i\omega\left(t-\frac{r}{c}\right)}\sin\theta.$$ |
| Электрическое поле в силу $$\frac{1}{c}\frac{\partial\vec{E}}{\partial t}=\text{rot }\vec{H}$$ | Электрическое поле в силу $$\frac{1}{c}\frac{\partial\vec{E}}{\partial t}=\text{rot }\vec{H}$$ | ||
| будем искать в виде $$\vec{E}=\vec{E}_{0}e^{i\omega\left(t-\frac{r}{c}\right)}, | будем искать в виде $$\vec{E}=\vec{E}_{0}e^{i\omega\left(t-\frac{r}{c}\right)}, | ||
| 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}}, |
| - | $$E_{\theta}=\left(\frac{1}{r^{2}}+\frac{\omega^{2}}{c^{2}}+\frac{i\omega}{rc}\right)\frac{p_{0}\sin\theta e^{i\omega\left(t-\frac{r}{c}\right)}}{r}.$$ | + | $$E_{\theta}=\left(\frac{1}{r^{2}}+\frac{\omega^{2}}{c^{2}}-\frac{i\omega}{rc}\right)\frac{p_{0}\sin\theta e^{i\omega\left(t-\frac{r}{c}\right)}}{r}.$$ |