Kechikuvchi potensiallar
Elektromagnit toʻlqinlar quyidagi differensial tenglamalarga boʻysunadi: div A + 1 c ∂ φ ∂ t = 0 ;             ( 1 ) {\displaystyle {\textrm {div}}{\textbf {A}}+{\frac {1}{c}}{\dfrac {\partial \varphi }{\partial t}}=0;\ \ \ \ \ \ (1)} Δ A − 1 c 2 ∂ 2 A ∂ t 2 = − 4 π 1 c ;             ( 2 ) {\displaystyle \Delta {\textbf {A}}-{\frac {1}{c^{2}}}{\dfrac {\partial ^{2}{\textbf {A}}}{\partial t^{2}}}=-4\pi {\frac {1}{c}};\ \ \ \ \ \ (2)} Δ φ − 1 c 2 ∂ 2 φ ∂ t 2 = − 4 π ρ ;               ( 3 ) {\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 δ e = ρ d V {\displaystyle \delta e=\rho dV} boʻladi. Koordinatalar boshini shu zaryadda joylashgan deb hisoblaymiz. U vaqtda, d V {\displaystyle dV} hajmdan tashqaridagi ( R ≠ 0 ) {\displaystyle (R\neq 0)} nuqtalarda zaryad yoʻq, demak (2 — 3) ga muvofiq div A + 1 c ∂ φ ∂ t = 0 ;             ( 4 ) {\displaystyle {\textrm {div}}{\textbf {A}}+{\frac {1}{c}}{\dfrac {\partial \varphi }{\partial t}}=0;\ \ \ \ \ \ (4)} Δ A − 1 c 2 ∂ 2 A ∂ t 2 = 0 ;             ( 5 ) {\displaystyle \Delta {\textbf {A}}-{\frac {1}{c^{2}}}{\dfrac {\partial ^{2}{\textbf {A}}}{\partial t^{2}}}=0;\ \ \ \ \ \ (5)} Δ φ − 1 c ∂ 2 φ ∂ t 2 = 0 ;             ( 6 ) {\displaystyle \Delta \varphi -{\frac {1}{c}}{\dfrac {\partial ^{2}\varphi }{\partial t^{2}}}=0;\ \ \ \ \ \ (6)} δ e {\displaystyle \delta e} zaryadning d v {\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 1 R 2 ∂ ∂ R ( R 2 ∂ φ ∂ R ) − 1 c 2 ∂ 2 φ ∂ t 2 = 0 ;             ( 7 ) {\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: φ ( t ) = ψ ( R , t ) R ;             ( 8 ) {\displaystyle \varphi (t)={\frac {\psi (R,t)}{R}};\ \ \ \ \ \ (8)} Koʻrinib turibdiki, ∂ φ ∂ R = 1 R ∂ ψ ∂ R − ψ R 2 , {\displaystyle {\frac {\partial \varphi }{\partial R}}={\frac {1}{R}}{\frac {\partial \psi }{\partial R}}-{\frac {\psi }{R^{2}}},} R 2 ∂ φ ∂ R = R ∂ ψ ∂ R − ψ , {\displaystyle R^{2}{\frac {\partial \varphi }{\partial R}}=R{\frac {\partial \psi }{\partial R}}-\psi ,} ∂ ∂ R ( R 2 ∂ φ ∂ R ) = R ∂ 2 ψ ∂ R 2 + ∂ ψ R − ∂ ψ R = R ∂ 2 ψ ∂ R 2 , {\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}}},} 1 R 2 ∂ ∂ R ( R 2 ∂ φ ∂ R ) = 1 R ∂ 2 ψ ∂ 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}}}} 1 c 2 ∂ 2 φ ∂ t 2 = 1 c 2 ∂ 2 ψ ∂ t 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 ∂ 2 ψ ∂ R 2 − 1 c ∂ 2 ψ ∂ t 2 = 0 ;             ( 9 ) {\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, ψ = ψ 1 ( t + R c ) + ψ 2 ( t − R c ) .               ( 10 ) {\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 ψ 2 {\displaystyle \psi _{2}} bilan cheklanamiz: ψ = ψ ( t − R c ) .             ( 11 ) {\displaystyle \psi =\psi \left(t-{\frac {R}{c}}\right).\ \ \ \ \ \ (11)} U holda, (8) ga muvofiq φ ( t ) = ψ ( t − R c ) R .             ( 12 ) {\displaystyle \varphi (t)={\frac {\psi \left(t-{\frac {R}{c}}\right)}{R}}.\ \ \ \ \ \ (12)} boʻladi. Koʻrinib turibdiki, vaqt oʻtishi bilan ( t > 0 {\displaystyle t>0} ) toʻlqin radiusning orta borish tomoniga qarab tarqaladi. Nomaʼlum ψ ( t − R c ) {\displaystyle \psi \left(t-{\frac {R}{c}}\right)} funksiyani aniqlash lozim. Zaryadga cheksiz yaqin turgan nuqtalar uchun R c {\displaystyle {\frac {R}{c}}} cheksiz kichik. Demak, φ ( t ) = ψ ( t ) R {\displaystyle \varphi (t)={\frac {\psi (t)}{R}}} . Maʼlumki, zaryadning Kulon potensiali: φ ( t ) = δ e ( t ) R = ρ ( t ) d V R {\displaystyle \varphi (t)={\frac {\delta e(t)}{R}}={\frac {\rho (t)dV}{R}}} Demak, ψ ( t ) = ρ ( t ) d V {\displaystyle \psi (t)=\rho (t)dV} , bu yerdan ψ ( t − R c ) = ρ ( t − R c ) d V {\displaystyle \psi \left(t-{\frac {R}{c}}\right)=\rho \left(t-{\frac {R}{c}}\right)dV} , u vaqtda (12) ga muvofiq φ ( t ) = ρ ( t − R c ) d V R .             ( 13 ) {\displaystyle \varphi (t)={\frac {\rho \left(t-{\frac {R}{c}}\right)dV}{R}}.\ \ \ \ \ \ (13)} Koʻramizki, (13) ga asosan, kuzatish nuqtasida vaqtning t {\displaystyle t} momentidagi potensial vaqtning oldingi t ′ = t − R c {\displaystyle t^{\prime }=t-{\frac {R}{c}}} momentidagi zaryad zichligi bilan aniqlanadi. Zaryad turgan joyda vaqtning t ′ = t − R c {\displaystyle t^{\prime }=t-{\frac {R}{c}}} momentida paydo boʻlgan potensial R {\displaystyle R} masofani R c {\displaystyle {\frac {\textbf {R}}{c}}} vaqtda oʻtib, kuzatish nuqtasiga R c {\displaystyle {\frac {R}{c}}} vaqt kechikish bilan yetib keladi. Shuning uchun (13) bilan ifodalangan potensial kechikuvchi potensial deb yuritiladi.