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}
d
V
{\displaystyle dV}
(
R
≠
0
)
{\displaystyle (R\neq 0)}
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}
d
v
{\displaystyle dv}
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}}
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}
ψ
(
t
−
R
c
)
{\displaystyle \psi \left(t-{\frac {R}{c}}\right)}
Zaryadga cheksiz yaqin turgan nuqtalar uchun
R
c
{\displaystyle {\frac {R}{c}}}
φ
(
t
)
=
ψ
(
t
)
R
{\displaystyle \varphi (t)={\frac {\psi (t)}{R}}}
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}
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}
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}
t
′
=
t
−
R
c
{\displaystyle t^{\prime }=t-{\frac {R}{c}}}
zaryad zichligi bilan aniqlanadi. Zaryad turgan joyda vaqtning
t
′
=
t
−
R
c
{\displaystyle t^{\prime }=t-{\frac {R}{c}}}
R
{\displaystyle R}
R
c
{\displaystyle {\frac {\textbf {R}}{c}}}
R
c
{\displaystyle {\frac {R}{c}}}
Zaryadlar sistemasining potensiali
tahrir
Zaryadlar sistemasining potensialini aniqlash uchun (13) formulani sistemaning barcha zaryadlari joylashgan hajm boʻyicha integrallash lozim:
φ
(
t
)
=
∫
ρ
(
t
−
R
c
)
R
d
V
;
(
14
)
{\displaystyle \varphi (t)=\int {\frac {\rho \left(t-{\frac {R}{c}}\right)}{R}}\,dV;\ \ \ \ \ \ (14)}
Xuddi shuningdek,
A
(
t
)
=
1
c
∫
j
(
t
−
R
c
)
R
d
V
;
(
15
)
{\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
r
′
{\displaystyle {\textbf {r}}^{\prime }}
r
{\displaystyle {\textbf {r}}}
R
=
r
−
r
′
;
(
16
)
{\displaystyle {\textbf {R}}={\textbf {r}}-{\textbf {r}}^{\prime };\ \ \ \ \ \ (16)}
boʻladi. Yangi koordinatalar boshiga nisbatan kechikuvchi potensiallar quyidagicha yoziladi:
φ
(
r
,
t
)
=
∫
ρ
(
r
′
,
t
−
|
r
−
r
′
|
c
)
|
r
−
r
′
|
d
V
;
(
17
)
{\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)}
A
(
r
,
t
)
=
1
c
∫
j
′
(
r
′
,
t
−
|
r
−
r
′
|
c
)
|
r
−
r
′
|
d
V
;
(
18
)
{\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 
