帶電粒子的移動軌道。 假設，從源頭位置
r
′
{\displaystyle \mathbf {r} '\,\!}
r
{\displaystyle \mathbf {r} \,\!}
t
{\displaystyle t\,\!}
r
{\displaystyle \mathbf {r} \,\!}
t
r
{\displaystyle t_{r}\,\!}
電磁波 傳播於真空 的速度是有限的，觀測者檢驗到電磁波的檢驗時間
t
{\displaystyle t\,\!}
t
r
{\displaystyle t_{r}\,\!}
推遲時間
t
r
{\displaystyle t_{r}\,\!}
t
{\displaystyle t\,\!}
電磁波 傳播的時間：
t
r
=
d
e
f
t
−
|
r
−
r
′
|
c
{\displaystyle t_{r}\ {\stackrel {def}{=}}\ t-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\,\!}
其中，
c
{\displaystyle c\,\!}
光速 。
推遲時間的概念意味著電磁波的傳播不是瞬時的。電磁波從發射位置傳播到終點位置，需要一段傳播期間，稱為時間延遲 。與日常生活的速度來比，電磁波傳播的速度相當快。因此，對於小尺寸系統，這時間延遲，通常很難察覺。例如，從開啟電燈泡到這電燈泡的光波抵達到觀測者的雙眼，所經過的時間延遲，只有幾兆分之一秒。但是，對於大尺寸系統，像太陽照射陽光到地球，時間延遲大約為8分鐘，可以經過實驗偵測察覺。
假設，一個移動中的帶電粒子，所帶電荷為
q
{\displaystyle q\,\!}
t
{\displaystyle t\,\!}
w
(
t
)
{\displaystyle \mathbf {w} (t)\,\!}
R
{\displaystyle {\boldsymbol {\mathfrak {R}}}\,\!}
r
′
=
w
(
t
)
{\displaystyle \mathbf {r} '=\mathbf {w} (t)\,\!}
r
{\displaystyle \mathbf {r} \,\!}
R
=
r
−
r
′
=
r
−
w
(
t
)
{\displaystyle {\boldsymbol {\mathfrak {R}}}=\mathbf {r} -\mathbf {r} '=\mathbf {r} -\mathbf {w} (t)\,\!}
則黎納-維謝純量勢
Φ
(
r
,
t
)
{\displaystyle \Phi (\mathbf {r} ,\,t)\,\!}
A
(
r
,
t
)
{\displaystyle \mathbf {A} (\mathbf {r} ,\,t)\,\!}
Φ
(
r
,
t
)
=
1
4
π
ϵ
0
q
c
R
c
−
R
⋅
v
{\displaystyle \Phi (\mathbf {r} ,\,t)={\frac {1}{4\pi \epsilon _{0}}}\ {\frac {qc}{{\mathfrak {R}}c-{\boldsymbol {\mathfrak {R}}}\cdot \mathbf {v} }}\,\!}
A
(
r
,
t
)
=
v
c
2
Φ
(
r
,
t
)
{\displaystyle \mathbf {A} (\mathbf {r} ,\,t)={\frac {\mathbf {v} }{c^{2}}}\Phi (\mathbf {r} ,\,t)\,\!}
其中，
ϵ
0
{\displaystyle \epsilon _{0}\,\!}
真空電容率 ，
v
{\displaystyle \mathbf {v} \,\!}
v
(
t
)
=
d
w
d
t
{\displaystyle \mathbf {v} (t)={\frac {d\mathbf {w} }{dt}}\,\!}
雖然黎納-維謝純量勢
Φ
(
r
,
t
)
{\displaystyle \Phi (\mathbf {r} ,\,t)\,\!}
A
(
r
,
t
)
{\displaystyle \mathbf {A} (\mathbf {r} ,\,t)\,\!}
t
{\displaystyle t\,\!}
r
′
{\displaystyle \mathbf {r} '\,\!}
v
{\displaystyle \mathbf {v} \,\!}
t
r
{\displaystyle t_{r}\,\!}
r
′
=
w
(
t
r
)
{\displaystyle \mathbf {r} '=\mathbf {w} (t_{r})\,\!}
v
=
v
(
t
r
)
{\displaystyle \mathbf {v} =\mathbf {v} (t_{r})\,\!}
從推遲勢 ，可以推導出黎納-維謝勢。推遲純量勢
Φ
(
r
,
t
)
{\displaystyle \Phi (\mathbf {r} ,\,t)\,\!}
推遲向量勢
A
(
r
,
t
)
{\displaystyle \mathbf {A} (\mathbf {r} ,\,t)\,\!}
推遲勢 ）
Φ
(
r
,
t
)
=
d
e
f
1
4
π
ϵ
0
∫
V
′
ρ
(
r
′
,
t
r
)
R
d
3
r
′
{\displaystyle \Phi (\mathbf {r} ,\,t)\ {\stackrel {def}{=}}\ {\frac {1}{4\pi \epsilon _{0}}}\int _{{\mathcal {V}}'}{\frac {\rho (\mathbf {r} ',\,t_{r})}{\mathfrak {R}}}\,d^{3}\mathbf {r} '\,\!}
A
(
r
,
t
)
=
d
e
f
μ
0
4
π
∫
V
′
J
(
r
′
,
t
r
)
R
d
3
r
′
{\displaystyle \mathbf {A} (\mathbf {r} ,\,t)\ {\stackrel {def}{=}}\ {\frac {\mu _{0}}{4\pi }}\int _{{\mathcal {V}}'}{\frac {\mathbf {J} (\mathbf {r} ',\,t_{r})}{\mathfrak {R}}}\,d^{3}\mathbf {r} '\,\!}
其中，
ρ
(
r
′
,
t
r
)
{\displaystyle \rho (\mathbf {r} ',\,t_{r})\,\!}
J
(
r
′
,
t
r
)
{\displaystyle \mathbf {J} (\mathbf {r} ',\,t_{r})\,\!}
V
′
{\displaystyle {\mathcal {V}}'\,\!}
d
3
r
′
{\displaystyle d^{3}\mathbf {r} '\,\!}
R
{\displaystyle {\mathfrak {R}}\,\!}
t
r
{\displaystyle t_{r}\,\!}
帶電粒子運動軌道的電荷密度 可以用狄拉克δ函數 表達為
ρ
(
r
,
t
)
=
q
δ
(
r
−
w
(
t
)
)
{\displaystyle \rho (\mathbf {r} ,\,t)=q\delta (\mathbf {r} -\mathbf {w} (t))\,\!}
其中，
δ
(
r
−
w
(
t
)
)
{\displaystyle \delta (\mathbf {r} -\mathbf {w} (t))\,\!}
代入推遲純量勢
Φ
(
r
,
t
)
{\displaystyle \Phi (\mathbf {r} ,\,t)\,\!}
Φ
(
r
,
t
)
=
q
4
π
ϵ
0
∫
V
′
δ
(
r
′
−
w
(
t
r
)
)
R
d
3
r
′
{\displaystyle \Phi (\mathbf {r} ,\,t)={\frac {q}{4\pi \epsilon _{0}}}\int _{{\mathcal {V}}'}{\frac {\delta (\mathbf {r} '-\mathbf {w} (t_{r}))}{\mathfrak {R}}}\,d^{3}\mathbf {r} '\,\!}
由於狄拉克δ函數
δ
(
r
′
−
w
(
t
r
)
)
{\displaystyle \delta (\mathbf {r} '-\mathbf {w} (t_{r}))\,\!}
r
′
{\displaystyle \mathbf {r} '\,\!}
r
′
=
w
(
t
r
)
{\displaystyle \mathbf {r} '=\mathbf {w} (t_{r})\,\!}
r
′
=
w
(
t
r
)
{\displaystyle \mathbf {r} '=\mathbf {w} (t_{r})\,\!}
Φ
(
r
,
t
)
=
q
4
π
ϵ
0
R
∫
V
′
δ
(
r
′
−
w
(
t
r
)
)
d
3
r
′
{\displaystyle \Phi (\mathbf {r} ,\,t)={\frac {q}{4\pi \epsilon _{0}{\mathfrak {R}}}}\int _{{\mathcal {V}}'}\delta (\mathbf {r} '-\mathbf {w} (t_{r}))\,d^{3}\mathbf {r} '\,\!}
由於推遲時間
t
r
{\displaystyle t_{r}\,\!}
t
{\displaystyle t\,\!}
r
{\displaystyle \mathbf {r} \,\!}
r
′
{\displaystyle \mathbf {r} '\,\!}
換元積分法 [ 4]
η
=
r
′
−
w
(
t
r
)
{\displaystyle {\boldsymbol {\eta }}=\mathbf {r} '-\mathbf {w} (t_{r})\,\!}
雅可比行列式
J
{\displaystyle {\mathfrak {J}}\,\!}
J
=
∂
η
∂
r
′
=
|
∂
η
x
∂
x
′
∂
η
x
∂
y
′
∂
η
x
∂
z
′
∂
η
y
∂
x
′
∂
η
y
∂
y
′
∂
η
y
∂
z
′
∂
η
z
∂
x
′
∂
η
z
∂
y
′
∂
η
z
∂
z
′
|
{\displaystyle {\mathfrak {J}}={\cfrac {\partial {\boldsymbol {\eta }}}{\partial \mathbf {r} '}}={\begin{vmatrix}{\cfrac {\partial \eta _{x}}{\partial x'}}&{\cfrac {\partial \eta _{x}}{\partial y'}}&{\cfrac {\partial \eta _{x}}{\partial z'}}\\{\cfrac {\partial \eta _{y}}{\partial x'}}&{\cfrac {\partial \eta _{y}}{\partial y'}}&{\cfrac {\partial \eta _{y}}{\partial z'}}\\{\cfrac {\partial \eta _{z}}{\partial x'}}&{\cfrac {\partial \eta _{z}}{\partial y'}}&{\cfrac {\partial \eta _{z}}{\partial z'}}\\\end{vmatrix}}\,\!}
。 行列式內分量很容易計算，例如：
∂
η
x
∂
x
′
=
1
−
∂
w
x
∂
x
′
=
1
−
∂
w
x
∂
t
r
∂
t
r
∂
x
′
=
1
−
v
x
∂
t
r
∂
x
′
{\displaystyle {\cfrac {\partial \eta _{x}}{\partial x'}}=1-{\cfrac {\partial w_{x}}{\partial x'}}=1-{\cfrac {\partial w_{x}}{\partial t_{r}}}\ {\cfrac {\partial t_{r}}{\partial x'}}=1-v_{x}{\cfrac {\partial t_{r}}{\partial x'}}\,\!}
∂
η
y
∂
x
′
=
∂
w
y
∂
x
′
=
∂
w
y
∂
t
r
∂
t
r
∂
x
′
=
v
y
∂
t
r
∂
x
′
{\displaystyle {\cfrac {\partial \eta _{y}}{\partial x'}}={\cfrac {\partial w_{y}}{\partial x'}}={\cfrac {\partial w_{y}}{\partial t_{r}}}\ {\cfrac {\partial t_{r}}{\partial x'}}=v_{y}{\cfrac {\partial t_{r}}{\partial x'}}\,\!}
按照上述方法，經過一番計算，可以得到
J
=
1
−
v
⋅
∇
′
t
r
=
1
−
R
^
⋅
v
/
c
{\displaystyle {\mathfrak {J}}=1-\mathbf {v} \cdot \nabla 't_{r}=1-{\hat {\boldsymbol {\mathfrak {R}}}}\cdot \mathbf {v} /c\,\!}
所以，推遲純量勢
Φ
(
r
,
t
)
{\displaystyle \Phi (\mathbf {r} ,\,t)\,\!}
Φ
(
r
,
t
)
=
q
4
π
ϵ
0
R
∫
V
′
δ
(
η
)
∂
r
′
∂
η
d
3
η
=
q
4
π
ϵ
0
R
∫
V
′
δ
(
η
)
J
d
3
η
=
q
4
π
ϵ
0
R
∫
V
′
δ
(
η
)
1
−
R
^
⋅
v
/
c
d
3
η
{\displaystyle \Phi (\mathbf {r} ,\,t)={\frac {q}{4\pi \epsilon _{0}{\mathfrak {R}}}}\int _{{\mathcal {V}}'}\delta ({\boldsymbol {\eta }}){\cfrac {\partial \mathbf {r} '}{\partial {\boldsymbol {\eta }}}}\,d^{3}{\boldsymbol {\eta }}={\frac {q}{4\pi \epsilon _{0}{\mathfrak {R}}}}\int _{{\mathcal {V}}'}{\cfrac {\delta ({\boldsymbol {\eta }})}{\mathfrak {J}}}\,d^{3}{\boldsymbol {\eta }}={\frac {q}{4\pi \epsilon _{0}{\mathfrak {R}}}}\int _{{\mathcal {V}}'}{\cfrac {\delta ({\boldsymbol {\eta }})}{1-{\hat {\boldsymbol {\mathfrak {R}}}}\cdot \mathbf {v} /c}}\,d^{3}{\boldsymbol {\eta }}\,\!}
這樣，可以得到黎納-維謝純量勢：
Φ
(
r
,
t
)
=
1
4
π
ϵ
0
q
c
R
c
−
R
⋅
v
{\displaystyle \Phi (\mathbf {r} ,\,t)={\frac {1}{4\pi \epsilon _{0}}}\ {\frac {qc}{{\mathfrak {R}}c-{\boldsymbol {\mathfrak {R}}}\cdot \mathbf {v} }}\,\!}
類似地，也可以推導出黎納-維謝向量勢。
從推遲勢的表達式可以看出它只依賴於推遲時刻源點的速度，而不依賴於源點的加速度，所以通過電磁勢的洛侖茲變換 也可以推導出黎納-維謝勢。考慮一個在推遲時刻瞬時速度與電荷運動速度相同的慣性系，記作
S
′
{\displaystyle S^{\prime }}
S
′
{\displaystyle S^{\prime }}
庫侖定律 給出，矢勢為零。[ 5] [ 6] :165ff
ϕ
′
=
q
4
π
ϵ
0
R
′
{\displaystyle \phi '={\frac {q}{4\pi \epsilon _{0}{\mathfrak {R}}'}}}
A
′
=
0
{\displaystyle A'=0}
標勢和矢勢從
S
′
{\displaystyle S^{\prime }}
S
{\displaystyle S}
ϕ
=
γ
(
ϕ
′
−
c
β
A
′
)
{\displaystyle \phi =\gamma (\phi '-c\beta A')}
A
=
γ
(
−
A
′
+
β
ϕ
′
/
c
)
{\displaystyle A=\gamma (-A'+\beta \phi '/c)}
其中，
γ
{\displaystyle \gamma }
洛侖茲因子 ，
β
=
v
/
c
{\displaystyle {\boldsymbol {\beta }}=\mathbf {v} /c}
代入後可以得到：
ϕ
=
γ
q
4
π
ϵ
0
R
′
{\displaystyle \phi ={\frac {\gamma q}{4\pi \epsilon _{0}{\mathfrak {R}}'}}}
A
=
γ
q
β
4
π
ϵ
0
R
′
c
{\displaystyle {\boldsymbol {A}}={\frac {\gamma q{\boldsymbol {\beta }}}{4\pi \epsilon _{0}{\mathfrak {R}}'c}}}
R
′
{\displaystyle {\mathfrak {R}}'}
R
{\displaystyle {\mathfrak {R}}}
R
′
=
c
Δ
t
′
=
c
γ
(
Δ
t
−
β
⋅
R
/
c
)
=
γ
(
R
−
β
⋅
R
)
{\displaystyle {\mathfrak {R}}'=c\Delta t'=c\gamma (\Delta t-{\boldsymbol {\beta }}\cdot {\boldsymbol {\mathfrak {R}}}/c)=\gamma ({\mathfrak {R}}-{\boldsymbol {\beta }}\cdot {\boldsymbol {\mathfrak {R}}})}
將
R
′
{\displaystyle {\mathfrak {R}}'}
對於固定不動的帶電粒子，電勢的方程式為
Φ
(
r
,
t
)
=
1
4
π
ϵ
0
q
R
{\displaystyle \Phi (\mathbf {r} ,\,t)={\frac {1}{4\pi \epsilon _{0}}}\ {\frac {q}{\mathfrak {R}}}\,\!}
這是黎納-維謝純量勢乘以雅可比行列式因子
J
{\displaystyle {\mathfrak {J}}\,\!}
多普勒效應 。[ 5]
從黎納-維謝勢，可以計算電場
E
{\displaystyle \mathbf {E} \,\!}
B
{\displaystyle \mathbf {B} \,\!}
E
=
−
∇
Φ
−
∂
A
∂
t
{\displaystyle \mathbf {E} =-\nabla \Phi -{\dfrac {\partial \mathbf {A} }{\partial t}}\,\!}
B
=
∇
×
A
{\displaystyle \mathbf {B} =\nabla \times \mathbf {A} \,\!}
求得的電場
E
{\displaystyle \mathbf {E} \,\!}
B
{\displaystyle \mathbf {B} \,\!}
[ 7]
E
(
r
,
t
)
=
q
4
π
ϵ
0
R
(
R
⋅
u
)
3
[
(
c
2
−
v
2
)
u
+
R
×
(
u
×
a
)
]
{\displaystyle \mathbf {E} (\mathbf {r} ,\,t)={\frac {q}{4\pi \epsilon _{0}}}\ {\cfrac {\mathfrak {R}}{({\boldsymbol {\mathfrak {R}}}\cdot \mathbf {u} )^{3}}}[(c^{2}-v^{2})\mathbf {u} +{\boldsymbol {\mathfrak {R}}}\times (\mathbf {u} \times \mathbf {a} )]\,\!}
B
(
r
,
t
)
=
1
c
R
^
×
E
(
r
,
t
)
{\displaystyle \mathbf {B} (\mathbf {r} ,\,t)={\frac {1}{c}}{\hat {\boldsymbol {\mathfrak {R}}}}\times \mathbf {E} (\mathbf {r} ,\,t)\,\!}
其中，向量
u
{\displaystyle \mathbf {u} \,\!}
c
R
^
−
v
{\displaystyle c{\hat {\boldsymbol {\mathfrak {R}}}}-\mathbf {v} \,\!}
加速度 是
a
=
d
v
d
t
{\displaystyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}\,\!}
檢查電場
E
{\displaystyle \mathbf {E} \,\!}
廣義庫侖場 ，又稱為速度場 ，因為這項目與加速度無關。當
v
≪
c
{\displaystyle v\ll c\,\!}
u
→
c
R
^
{\displaystyle \mathbf {u} \to c{\hat {\boldsymbol {\mathfrak {R}}}}\,\!}
庫侖方程式 ：
E
=
q
4
π
ϵ
0
R
^
R
2
{\displaystyle \mathbf {E} ={\frac {q}{4\pi \epsilon _{0}}}\ {\frac {\hat {\boldsymbol {\mathfrak {R}}}}{{\mathfrak {R}}^{2}}}\,\!}
右邊第二項稱為輻射場 ，又稱為加速度場 ，因為這項目的物理行為主要是由粒子的加速度決定。這個項目能夠描述電磁輻射 的生成程序。
![{\displaystyle \mathbf {E} (\mathbf {r} ,\,t)={\frac {q}{4\pi \epsilon _{0}}}\ {\cfrac {\mathfrak {R}}{({\boldsymbol {\mathfrak {R}}}\cdot \mathbf {u} )^{3}}}[(c^{2}-v^{2})\mathbf {u} +{\boldsymbol {\mathfrak {R}}}\times (\mathbf {u} \times \mathbf {a} )]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1127f7c3b22875b0052c7cfbb7b4cc2b4be591c5)
