Kechikuvchi potensiallar: Versiyalar orasidagi farq

Elektromagnit toʻlqinlar quyidagi differensial tenglamalarga boʻysunadi:

${\displaystyle \textrm{div}\textbf{A}+\frac{1}{c}\dfrac{\partial\varphi}{\partial t}=0;\ \ \ \ \ \ (1)}$

${\displaystyle \Delta\textbf{A}-\frac{1}{c^{2}}\dfrac{\partial^{2}\textbf{A}}{\partial t^{2}}=-4\pi\frac{1}{c};\ \ \ \ \ \ (2)}$

${\displaystyle \Delta\varphi-\frac{1}{c^{2}}\dfrac{\partial^{2}\varphi}{\partial t^{2}}=-4\pi\rho;\ \ \ \ \ \ \ (3)}$

Koʻrinib turibdiki, xususiy hosilalardan iborat chiziqli differensial tenglamalar sistemasi bilan amal qilishga toʻgʻri keladi.

Cheksiz kichik hajmdagi zaryad ${\displaystyle \delta e=\rho dV}$ boʻladi. Koordinatalar boshini shu zaryadda joylashgan deb hisoblaymiz. U vaqtda, ${\displaystyle dV}$ hajmdan tashqaridagi ${\displaystyle (R\ne 0)}$ nuqtalarda zaryad yoʻq, demak (2 — 3) ga muvofiq

${\displaystyle \textrm{div}\textbf{A}+\frac{1}{c}\dfrac{\partial\varphi}{\partial t}=0;\ \ \ \ \ \ (4)}$

${\displaystyle \Delta \textbf{A}-\frac{1}{c^{2}}\dfrac{\partial^{2}\textbf{A}}{\partial t^{2}}=0;\ \ \ \ \ \ (5)}$

${\displaystyle \Delta\varphi-\frac{1}{c}\dfrac{\partial^{2}\varphi}{\partial t^{2}}=0;\ \ \ \ \ \ (6)}$

${\displaystyle \delta e}$ zaryadning ${\displaystyle dv}$ hajmdan tashqaridagi nuqtalarda hosil qilgan maydoni koordinatalar boshigacha boʻlgan masofa va vaqtgagina bogʻliq, yaʼni maydon sferik simmetriyaga ega. Endi Laplas operatorini sferik koordinatalarda yozib koʻrsak, (6) ga muvofiq

${\displaystyle \frac{1}{R^{2}}\dfrac{\partial}{\partial R}\left(R^{2}\frac{\partial\varphi}{\partial R}\right)-\frac{1}{c^{2}}\dfrac{\partial^{2}\varphi}{\partial t^{2}}=0;\ \ \ \ \ \ (7)}$

Quyidagi koʻrinishda yangi funksiya kiritamiz:

${\displaystyle \varphi(t)=\frac{\psi(R,t)}{R};\ \ \ \ \ \ (8)}$

Koʻrinib turibdiki,

${\displaystyle \frac{\partial\varphi}{\partial R}=\frac{1}{R}\frac{\partial\psi}{\partial R}-\frac{\psi}{R^{2}},}$

${\displaystyle R^{2}\frac{\partial\varphi}{\partial R}=R\frac{\partial\psi}{\partial R}-\psi,}$

${\displaystyle \dfrac{\partial}{\partial R}\left(R^{2}\frac{\partial\varphi}{\partial R}\right)=R\frac{\partial^{2}\psi}{\partial R^{2}}+\frac{\partial\psi}{R}-\frac{\partial\psi}{R}=R\frac{\partial^{2}\psi}{\partial R^{2}},}$

${\displaystyle \frac{1}{R^{2}}\dfrac{\partial}{\partial R}\left(R^{2}\frac{\partial\varphi}{\partial R}\right)=\frac{1}{R}\frac{\partial^{2}\psi}{\partial R^{2}}}$

${\displaystyle \frac{1}{c^{2}}\dfrac{\partial^{2}\varphi}{\partial t^{2}}=\frac{1}{c^{2}}\frac{\partial^{2}\psi}{\partial t^{2}}}$

Demak, (7) ga muvofiq

${\displaystyle \dfrac{\partial^{2}\psi}{\partial R^{2}}-\frac{1}{c}\dfrac{\partial^{2}\psi}{\partial t^{2}}=0;\ \ \ \ \ \ (9)}$

boʻladi. Bu yassi elektromagnit to'lqin tenglamasidir. Shunday qilib,

${\displaystyle \psi=\psi_{1}\left(t+\frac{R}{c}\right)+\psi_{2}\left(t-\frac{R}{c}\right).\ \ \ \ \ \ \ (10)}$

To'lqin tenglamaning xususiy yechimigina bizni qiziqtiradi. Shu sababli ${\displaystyle \psi_{2}}$ bilan cheklanamiz:

${\displaystyle \psi=\psi\left(t-\frac{R}{c}\right).\ \ \ \ \ \ (11)}$

U holda, (8) ga muvofiq

${\displaystyle \varphi (t)=\frac{\psi\left(t-\frac{R}{c}\right)}{R}.\ \ \ \ \ \ (12)}$

boʻladi. Koʻrinib turibdiki, vaqt oʻtishi bilan (${\displaystyle t>0}$) toʻlqin radiusning orta borish tomoniga qarab tarqaladi. Nomaʼlum ${\displaystyle \psi\left(t-\frac{R}{c}\right)}$ funksiyani aniqlash lozim.

Zaryadga cheksiz yaqin turgan nuqtalar uchun ${\displaystyle \frac{R}{c}}$ cheksiz kichik. Demak, ${\displaystyle \varphi(t)=\frac{\psi(t)}{R}}$. Maʼlumki, zaryadning Kulon potensiali:

${\displaystyle \varphi(t)=\frac{\delta e(t)}{R}=\frac{\rho(t)dV}{R}}$

Demak, ${\displaystyle \psi(t)=\rho(t)dV}$, bu yerdan ${\displaystyle \psi\left(t-\frac{R}{c}\right)=\rho\left(t-\frac{R}{c}\right)dV}$, u vaqtda (12) ga muvofiq

${\displaystyle \varphi(t)=\frac{\rho\left(t-\frac{R}{c}\right)dV}{R}.\ \ \ \ \ \ (13)}$

Koʻramizki, (13) ga asosan, kuzatish nuqtasida vaqtning ${\displaystyle t}$ momentidagi potensial vaqtning oldingi ${\displaystyle t^{\prime}=t-\frac{R}{c}}$ momentidagi zaryad zichligi bilan aniqlanadi. Zaryad turgan joyda vaqtning ${\displaystyle t^{\prime}=t-\frac{R}{c}}$ momentida paydo boʻlgan potensial ${\displaystyle R}$ masofani ${\displaystyle \frac{\textbf{R}}{c}}$ vaqtda oʻtib, kuzatish nuqtasiga ${\displaystyle \frac{R}{c}}$ vaqt kechikish bilan yetib keladi. Shuning uchun (13) bilan ifodalangan potensial kechikuvchi potensial deb yuritiladi.

Zaryadlar sistemasining potensialini aniqlash uchun (13) formulani sistemaning barcha zaryadlari joylashgan hajm boʻyicha integrallash lozim:

${\displaystyle \varphi(t)=\int \frac{\rho\left(t-\frac{R}{c}\right)}{R}\,dV;\ \ \ \ \ \ (14)}$

Xuddi shuningdek,

${\displaystyle \textbf{A}(t)=\frac{1}{c}\int \frac{\textbf{j}\left(t-\frac{R}{c}\right)}{R}\,dV;\ \ \ \ \ \ (15)}$

Ushbu tenglamalarda zaryad turgan nuqtani koordinatalar boshi sifatida olingan. Agar koordinatalar boshini sistemaning ichidagi boshqa biror nuqtaga koʻchirilsa, yangi koordinatalar boshiga nisbatan zaryad turgan nuqtaning radius-vektori ${\displaystyle \textbf{r}^{\prime}}$ va kuzatish nuqtasining radius-vektori ${\displaystyle \textbf{r}}$ bilan belgilanadi. Demak,

${\displaystyle \textbf{R}=\textbf{r}-\textbf{r}^{\prime};\ \ \ \ \ \ (16)}$

boʻladi. Yangi koordinatalar boshiga nisbatan kechikuvchi potensiallar quyidagicha yoziladi:

${\displaystyle \varphi(\textbf{r},t)=\int \frac{\rho\left(\textbf{r}^{\prime},\ t-\frac{|\textbf{r}-\textbf{r}^{\prime}|}{c}\right)}{|\textbf{r}-\textbf{r}^{\prime}|}\,dV;\ \ \ \ \ \ (17)}$

${\displaystyle \textbf{A}(\textbf{r},t)=\frac{1}{c}\int \frac{\textbf{j}^{\prime}\left(\textbf{r}^{\prime},\ t-\frac{|\textbf{r}-\textbf{r}^{\prime}|}{c}\right)}{|\textbf{r}-\textbf{r}^{\prime}|}\,dV;\ \ \ \ \ \ (18)}$

• R.X.Mallin, Klassik elektrodinamika, Toshkent, 1974

• Zaryadlar sistemasining uzoq masofalardagi kechikuvchi potensiallari
• Elektr dipolning elektromagnit maydoni va nurlanishi
• Nurlanish reaksiyasi

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