ε
i
j
k
=
e
→
i
(
e
→
j
×
e
→
k
)
=
∑
a
=
1
3
e
i
,
a
(
e
→
j
×
e
→
k
)
a
=
∑
a
=
1
3
∑
b
,
c
=
1
3
ε
a
b
c
e
i
,
a
e
j
,
b
e
k
,
c
=
{\displaystyle \varepsilon _{ijk}={\vec {e}}_{i}({\vec {e}}_{j}\times {\vec {e}}_{k})=\sum _{a=1}^{3}e_{i,a}({\vec {e}}_{j}\times {\vec {e}}_{k})_{a}=\sum _{a=1}^{3}\sum _{b,c=1}^{3}\varepsilon _{abc}e_{i,a}e_{j,b}e_{k,c}=}
=
e
i
,
1
(
e
j
,
2
e
k
,
3
−
e
j
,
3
e
k
,
2
)
−
e
i
,
2
(
e
j
,
3
e
k
,
1
−
e
j
,
1
e
k
,
3
)
+
e
i
,
3
(
e
j
,
1
e
k
,
2
−
e
j
,
2
e
k
,
1
)
=
{\displaystyle \!\,=e_{i,1}(e_{j,2}e_{k,3}-e_{j,3}e_{k,2})-e_{i,2}(e_{j,3}e_{k,1}-e_{j,1}e_{k,3})+e_{i,3}(e_{j,1}e_{k,2}-e_{j,2}e_{k,1})=}
=
e
i
,
1
|
e
j
,
2
e
j
,
3
e
k
,
2
e
k
,
3
|
+
e
i
,
2
|
e
j
,
1
e
j
,
3
e
k
,
1
e
k
,
3
|
+
e
i
,
3
|
e
j
,
1
e
j
,
2
e
k
,
1
e
k
,
2
|
=
|
e
i
,
1
e
i
,
2
e
i
,
3
e
j
,
1
e
j
,
2
e
j
,
3
e
k
,
1
e
k
,
2
e
k
,
3
|
{\displaystyle =e_{i,1}{\begin{vmatrix}e_{j,2}&e_{j,3}\\e_{k,2}&e_{k,3}\end{vmatrix}}+e_{i,2}{\begin{vmatrix}e_{j,1}&e_{j,3}\\e_{k,1}&e_{k,3}\end{vmatrix}}+e_{i,3}{\begin{vmatrix}e_{j,1}&e_{j,2}\\e_{k,1}&e_{k,2}\end{vmatrix}}={\begin{vmatrix}e_{i,1}&e_{i,2}&e_{i,3}\\e_{j,1}&e_{j,2}&e_{j,3}\\e_{k,1}&e_{k,2}&e_{k,3}\end{vmatrix}}}
1. Möglichkeit
ε
i
j
k
ε
i
l
m
=
|
e
i
,
1
e
i
,
2
e
i
,
3
e
j
,
1
e
j
,
2
e
j
,
3
e
k
,
1
e
k
,
2
e
k
,
3
|
⋅
|
e
i
,
1
e
i
,
2
e
i
,
3
e
l
,
1
e
l
,
2
e
l
,
3
e
m
,
1
e
m
,
2
e
m
,
3
|
=
{\displaystyle \varepsilon _{ijk}\varepsilon _{ilm}={\begin{vmatrix}e_{i,1}&e_{i,2}&e_{i,3}\\e_{j,1}&e_{j,2}&e_{j,3}\\e_{k,1}&e_{k,2}&e_{k,3}\end{vmatrix}}\cdot {\begin{vmatrix}e_{i,1}&e_{i,2}&e_{i,3}\\e_{l,1}&e_{l,2}&e_{l,3}\\e_{m,1}&e_{m,2}&e_{m,3}\end{vmatrix}}=}
=
|
e
i
,
1
e
i
,
1
+
e
i
,
2
e
j
,
1
+
e
i
,
3
e
k
,
1
e
i
,
1
e
i
,
2
+
e
i
,
2
e
j
,
2
+
e
i
,
3
e
k
,
2
e
i
,
1
e
i
,
3
+
e
i
,
2
e
j
,
3
+
e
i
,
3
e
k
,
3
e
l
,
1
e
i
,
1
+
e
l
,
2
e
j
,
1
+
e
l
,
3
e
k
,
1
e
l
,
1
e
i
,
2
+
e
l
,
2
e
j
,
2
+
e
l
,
3
e
k
,
2
e
l
,
1
e
i
,
3
+
e
l
,
2
e
j
,
3
+
e
l
,
3
e
k
,
3
e
m
,
1
e
i
,
1
+
e
m
,
2
e
j
,
1
+
e
m
,
3
e
k
,
1
e
m
,
1
e
i
,
2
+
e
m
,
2
e
j
,
2
+
e
m
,
3
e
k
,
2
e
m
,
1
e
i
,
3
+
e
m
,
2
e
j
,
3
+
e
m
,
3
e
k
,
3
|
{\displaystyle ={\begin{vmatrix}e_{i,1}e_{i,1}+e_{i,2}e_{j,1}+e_{i,3}e_{k,1}&e_{i,1}e_{i,2}+e_{i,2}e_{j,2}+e_{i,3}e_{k,2}&e_{i,1}e_{i,3}+e_{i,2}e_{j,3}+e_{i,3}e_{k,3}\\e_{l,1}e_{i,1}+e_{l,2}e_{j,1}+e_{l,3}e_{k,1}&e_{l,1}e_{i,2}+e_{l,2}e_{j,2}+e_{l,3}e_{k,2}&e_{l,1}e_{i,3}+e_{l,2}e_{j,3}+e_{l,3}e_{k,3}\\e_{m,1}e_{i,1}+e_{m,2}e_{j,1}+e_{m,3}e_{k,1}&e_{m,1}e_{i,2}+e_{m,2}e_{j,2}+e_{m,3}e_{k,2}&e_{m,1}e_{i,3}+e_{m,2}e_{j,3}+e_{m,3}e_{k,3}\end{vmatrix}}}
Für ein rechtsdrehendes, orthonormales Koordinatensystem mit den Einheitsvektoren , folgt:
Mit:
e
→
i
=
(
1
0
0
)
,
e
→
j
=
(
0
1
0
)
,
e
→
k
=
(
0
0
1
)
{\displaystyle {\vec {e}}_{i}={\begin{pmatrix}1\\0\\0\end{pmatrix}},{\vec {e}}_{j}={\begin{pmatrix}0\\1\\0\end{pmatrix}},{\vec {e}}_{k}={\begin{pmatrix}0\\0\\1\end{pmatrix}}}
=
|
1
⋅
1
+
0
⋅
0
+
0
⋅
0
1
⋅
0
+
0
⋅
1
+
0
⋅
0
1
⋅
0
+
0
⋅
0
+
0
⋅
1
e
l
,
1
⋅
1
+
e
l
,
2
⋅
0
+
e
l
,
3
⋅
0
e
l
,
1
⋅
0
+
e
l
,
2
⋅
1
+
e
l
,
3
⋅
0
e
l
,
1
⋅
0
+
e
l
,
2
⋅
0
+
e
l
,
3
⋅
1
e
m
,
1
⋅
1
+
e
m
,
2
⋅
0
+
e
m
,
3
⋅
0
e
m
,
1
⋅
0
+
e
m
,
2
⋅
1
+
e
m
,
3
⋅
0
e
m
,
1
⋅
0
+
e
m
,
2
⋅
0
+
e
m
,
3
⋅
1
|
{\displaystyle ={\begin{vmatrix}1\cdot 1+0\cdot 0+0\cdot 0&1\cdot 0+0\cdot 1+0\cdot 0&1\cdot 0+0\cdot 0+0\cdot 1\\e_{l,1}\cdot 1+e_{l,2}\cdot 0+e_{l,3}\cdot 0&e_{l,1}\cdot 0+e_{l,2}\cdot 1+e_{l,3}\cdot 0&e_{l,1}\cdot 0+e_{l,2}\cdot 0+e_{l,3}\cdot 1\\e_{m,1}\cdot 1+e_{m,2}\cdot 0+e_{m,3}\cdot 0&e_{m,1}\cdot 0+e_{m,2}\cdot 1+e_{m,3}\cdot 0&e_{m,1}\cdot 0+e_{m,2}\cdot 0+e_{m,3}\cdot 1\end{vmatrix}}}
=
|
e
i
,
1
e
i
,
1
e
i
,
2
e
j
,
2
e
i
,
3
e
k
,
3
e
l
,
1
e
i
,
1
e
l
,
2
e
j
,
2
e
l
,
3
e
k
,
3
e
m
,
1
e
i
,
1
e
m
,
2
e
j
,
2
e
m
,
3
e
k
,
3
|
=
|
1
0
0
e
l
,
1
e
i
,
2
e
l
,
2
e
j
,
2
e
l
,
3
e
k
,
3
e
m
,
1
e
i
,
1
e
m
,
2
e
j
,
2
e
m
,
3
e
k
,
3
|
{\displaystyle ={\begin{vmatrix}e_{i,1}e_{i,1}&e_{i,2}e_{j,2}&e_{i,3}e_{k,3}\\e_{l,1}e_{i,1}&e_{l,2}e_{j,2}&e_{l,3}e_{k,3}\\e_{m,1}e_{i,1}&e_{m,2}e_{j,2}&e_{m,3}e_{k,3}\end{vmatrix}}={\begin{vmatrix}1&0&0\\e_{l,1}e_{i,2}&e_{l,2}e_{j,2}&e_{l,3}e_{k,3}\\e_{m,1}e_{i,1}&e_{m,2}e_{j,2}&e_{m,3}e_{k,3}\end{vmatrix}}}
Wir haben
ε
i
j
k
=
e
→
i
(
e
→
j
×
e
→
k
)
{\displaystyle \varepsilon _{ijk}={\vec {e}}_{i}({\vec {e}}_{j}\times {\vec {e}}_{k})}
ε
i
l
m
=
e
→
i
(
e
→
l
×
e
→
m
)
{\displaystyle \varepsilon _{ilm}={\vec {e}}_{i}({\vec {e}}_{l}\times {\vec {e}}_{m})}
e
→
i
,
e
→
l
,
e
→
m
{\displaystyle {\vec {e}}_{i},{\vec {e}}_{l},{\vec {e}}_{m}}
e
→
i
⊥
e
→
l
=>
e
l
,
1
e
i
,
2
=
0
{\displaystyle {\vec {e}}_{i}\perp {\vec {e}}_{l}=>e_{l,1}e_{i,2}=0}
e
→
i
⊥
e
→
m
=>
e
m
,
1
e
i
,
1
=
0
{\displaystyle {\vec {e}}_{i}\perp {\vec {e}}_{m}=>e_{m,1}e_{i,1}=0}
|
1
0
0
0
e
l
,
2
e
j
,
2
e
l
,
3
e
k
,
3
0
e
m
,
2
e
j
,
2
e
m
,
3
e
k
,
3
|
=
|
1
0
0
0
δ
j
l
δ
k
l
0
δ
j
m
δ
k
m
|
=
1
⋅
|
δ
j
l
δ
k
l
δ
j
m
δ
k
m
|
=
δ
j
l
δ
k
m
−
δ
j
m
δ
k
l
{\displaystyle {\begin{vmatrix}1&0&0\\0&e_{l,2}e_{j,2}&e_{l,3}e_{k,3}\\0&e_{m,2}e_{j,2}&e_{m,3}e_{k,3}\end{vmatrix}}={\begin{vmatrix}1&0&0\\0&\delta _{jl}&\delta _{kl}\\0&\delta _{jm}&\delta _{km}\end{vmatrix}}=1\cdot {\begin{vmatrix}\delta _{jl}&\delta _{kl}\\\delta _{jm}&\delta _{km}\end{vmatrix}}=\delta _{jl}\delta _{km}-\delta _{jm}\delta _{kl}}
2. Möglichkeit
Für j=k oder l=m verschwindet das Kreuzprodukt. Für j≠k und l≠m gilt:
Das doppelte Kreuzprodukt verschwindet, wenn nicht j=l oder k=m
Aufgabenstellung:
Zeige:
c
→
(
a
→
×
b
→
)
=
a
→
(
b
→
×
c
→
)
=
b
→
(
c
→
×
a
→
)
{\displaystyle {\vec {c}}({\vec {a}}\times {\vec {b}})={\vec {a}}({\vec {b}}\times {\vec {c}})={\vec {b}}({\vec {c}}\times {\vec {a}})}
Lösung:
c
→
(
a
→
×
b
→
)
=
ε
i
j
k
c
i
a
j
b
k
=
ε
{\displaystyle {\vec {c}}({\vec {a}}\times {\vec {b}})=\varepsilon _{ijk}c_{i}a_{j}b_{k}=\varepsilon }
da
ε
i
j
k
=
ε
j
k
i
=
ε
k
i
j
{\displaystyle \varepsilon _{ijk}=\varepsilon _{jki}=\varepsilon _{kij}}
ε
i
j
k
c
i
a
j
b
k
=
ε
j
k
i
c
i
a
j
b
k
{\displaystyle \varepsilon _{ijk}c_{i}a_{j}b_{k}=\varepsilon _{jki}c_{i}a_{j}b_{k}}
ε
i
j
k
c
i
a
j
b
k
=
ε
k
i
j
c
i
a
j
b
k
{\displaystyle \varepsilon _{ijk}c_{i}a_{j}b_{k}=\varepsilon _{kij}c_{i}a_{j}b_{k}}
Durch Vertauschen der Indizes folgt:
ε
j
k
i
c
i
a
j
b
k
=
ε
i
j
k
c
k
a
i
b
j
=
c
k
a
i
b
j
(
e
→
j
×
e
→
k
)
e
→
i
=
a
→
(
b
→
×
c
→
)
{\displaystyle \varepsilon _{jki}c_{i}a_{j}b_{k}=\varepsilon _{ijk}c_{k}a_{i}b_{j}=c_{k}a_{i}b_{j}({\vec {e}}_{j}\times {\vec {e}}_{k}){\vec {e}}_{i}={\vec {a}}({\vec {b}}\times {\vec {c}})}
bzw.
ε
k
i
j
c
i
a
j
b
k
=
ε
i
j
k
c
j
a
k
b
i
=
c
j
a
k
b
i
(
e
→
j
×
e
→
k
)
e
→
i
=
b
→
(
c
→
×
a
→
)
{\displaystyle \varepsilon _{kij}c_{i}a_{j}b_{k}=\varepsilon _{ijk}c_{j}a_{k}b_{i}=c_{j}a_{k}b_{i}({\vec {e}}_{j}\times {\vec {e}}_{k}){\vec {e}}_{i}={\vec {b}}({\vec {c}}\times {\vec {a}})}
[
a
→
×
(
b
→
×
c
→
)
]
k
=
∑
i
,
j
ε
i
j
k
a
i
(
b
→
×
c
→
)
j
=
∑
i
,
j
∑
l
,
m
ε
i
j
k
ε
j
l
m
a
i
b
l
c
m
{\displaystyle [{\vec {a}}\times ({\vec {b}}\times {\vec {c}})]_{k}=\sum _{i,j}\varepsilon _{ijk}a_{i}({\vec {b}}\times {\vec {c}})_{j}=\sum _{i,j}\sum _{l,m}\varepsilon _{ijk}\varepsilon _{jlm}a_{i}b_{l}c_{m}}
Da wir unsere Freunde so nennen dürfen wie wir wollen, vertauschen wir die Indizes i und j damit der nächste Schritt "offensichtlich" ist:
∑
j
,
i
∑
l
,
m
ε
j
i
k
ε
i
l
m
a
j
b
l
c
m
{\displaystyle \sum _{j,i}\sum _{l,m}\varepsilon _{jik}\varepsilon _{ilm}a_{j}b_{l}c_{m}}
∑
i
ε
i
j
k
ε
i
l
m
=
δ
j
l
δ
k
m
−
δ
j
m
δ
k
l
{\displaystyle \sum _{i}\varepsilon _{ijk}\varepsilon _{ilm}=\delta _{jl}\delta _{km}-\delta _{jm}\delta _{kl}}
ε
j
i
k
=
−
ε
i
j
k
{\displaystyle \varepsilon _{jik}=-\varepsilon _{ijk}}
∑
j
,
i
∑
l
,
m
ε
j
i
k
ε
i
l
m
a
j
b
l
c
m
=
−
∑
j
,
i
∑
l
,
m
ε
i
j
k
ε
i
l
m
a
j
b
l
c
m
=
−
∑
j
∑
l
,
m
a
j
b
l
c
m
(
δ
j
l
δ
k
m
−
δ
j
m
δ
k
l
)
{\displaystyle \sum _{j,i}\sum _{l,m}\varepsilon _{jik}\varepsilon _{ilm}a_{j}b_{l}c_{m}=-\sum _{j,i}\sum _{l,m}\varepsilon _{ijk}\varepsilon _{ilm}a_{j}b_{l}c_{m}=-\sum _{j}\sum _{l,m}a_{j}b_{l}c_{m}(\delta _{jl}\delta _{km}-\delta _{jm}\delta _{kl})}
Und ziehen das Minus vor den Summen zu den Deltas in die Klammer und lösen die Kronecker-Deltas zu Einheitsvektoren auf:
∑
j
∑
l
,
m
a
j
b
l
c
m
(
δ
j
m
δ
k
l
−
δ
j
l
δ
k
m
)
=
∑
j
∑
l
,
m
a
j
b
l
c
m
(
(
e
→
j
,
e
→
m
)
(
e
→
k
,
e
→
l
)
−
(
e
→
j
,
e
→
l
)
(
e
→
k
,
e
→
m
)
)
{\displaystyle \sum _{j}\sum _{l,m}a_{j}b_{l}c_{m}(\delta _{jm}\delta _{kl}-\delta _{jl}\delta _{km})=\sum _{j}\sum _{l,m}a_{j}b_{l}c_{m}(({\vec {e}}_{j},{\vec {e}}_{m})({\vec {e}}_{k},{\vec {e}}_{l})-({\vec {e}}_{j},{\vec {e}}_{l})({\vec {e}}_{k},{\vec {e}}_{m}))}
Dann lösen wir die Summe auf:
(
(
a
→
,
c
→
)
(
e
→
k
,
b
→
)
−
(
a
→
,
b
→
)
(
e
→
k
,
c
→
)
)
=
[
b
→
(
a
→
,
c
→
)
−
c
→
(
a
→
,
b
→
)
]
k
{\displaystyle (({\vec {a}},{\vec {c}})({\vec {e}}_{k},{\vec {b}})-({\vec {a}},{\vec {b}})({\vec {e}}_{k},{\vec {c}}))=[{\vec {b}}({\vec {a}},{\vec {c}})-{\vec {c}}({\vec {a}},{\vec {b}})]_{k}}
Da gilt, dass
(AxB) C = A (BxC)
ist
(AxB),(CxD)= AB(CxD) = A(B x (CxD))
(B x (CxD)) = (BD)C-(BC)D
A(B x (CxD)) = A ((BD)C-(BC)D)=(AC)(BD)-(AD)(BC)
q.e.d.
Hinweis:
Ich glaube ich habe möglicherweise im ersten Lösungsweg einen Denkfehler:
Wenn ich die Indizes i und j vertausche, müsste ich sie ggf. auch in der
Ausgangsgleichung, mit der ich sie vergleiche, tauschen.
Zeige:
d
i
v
r
o
t
A
→
=
0
{\displaystyle \operatorname {div\ } \operatorname {rot\ } {\vec {A}}=0}
1. Lösung:
d
i
v
r
o
t
A
→
=
∇
(
∇
×
A
→
)
=
ε
i
j
k
∇
i
∇
j
A
k
{\displaystyle \operatorname {div\ } \operatorname {rot\ } {\vec {A}}=\nabla (\nabla \times {\vec {A}})=\varepsilon _{ijk}\nabla _{i}\nabla _{j}A_{k}}
Nach dem Satz von Schwarz kann man bei zweifach stetig differenzierbaren Funktionen die Differentialoperatoren vertauschen:
ε
i
j
k
∇
i
∇
j
A
k
=
ε
i
j
k
∇
j
∇
i
A
k
{\displaystyle \varepsilon _{ijk}\nabla _{i}\nabla _{j}A_{k}=\varepsilon _{ijk}\nabla _{j}\nabla _{i}A_{k}}
Mit einer ungeraden Permutation des Levi-Civita-Symbols
ε
i
j
k
=
−
ε
j
i
k
{\displaystyle \varepsilon _{ijk}=-\varepsilon _{jik}}
ε
i
j
k
∇
j
∇
i
A
k
=
−
ε
j
i
k
∇
j
∇
i
A
k
{\displaystyle \varepsilon _{ijk}\nabla _{j}\nabla _{i}A_{k}=-\varepsilon _{jik}\nabla _{j}\nabla _{i}A_{k}}
Da ich meine Indize so nennen kann wie ich will, vertausche ich alle i und j:
−
ε
i
j
k
∇
i
∇
j
A
k
{\displaystyle -\varepsilon _{ijk}\nabla _{i}\nabla _{j}A_{k}}
Jetzt haben wir:
ε
i
j
k
∇
i
∇
j
A
k
=
−
ε
i
j
k
∇
i
∇
j
A
k
{\displaystyle \varepsilon _{ijk}\nabla _{i}\nabla _{j}A_{k}=-\varepsilon _{ijk}\nabla _{i}\nabla _{j}A_{k}}
Daraus folgt:
d
i
v
r
o
t
A
→
=
0
{\displaystyle \operatorname {div\ } \operatorname {rot\ } {\vec {A}}=0}
Alternative Lösung:
Mit der in Übungsaufgabe 1.2 gezeigten Identität:
d
i
v
r
o
t
A
→
=
∇
1
(
∇
2
×
A
→
)
=
A
→
(
∇
1
×
∇
2
)
{\displaystyle \operatorname {div\ } \operatorname {rot\ } {\vec {A}}=\nabla _{1}(\nabla _{2}\times {\vec {A}})={\vec {A}}(\nabla _{1}\times \nabla _{2})}
Nach dem Satz von Schwarz folgt
(
∇
1
×
∇
2
)
=
0
→
{\displaystyle (\nabla _{1}\times \nabla _{2})={\vec {0}}}
d
i
v
r
o
t
A
→
=
0
{\displaystyle \operatorname {div\ } \operatorname {rot\ } {\vec {A}}=0}
Zeige:
r
o
t
g
r
a
d
B
=
0
{\displaystyle \operatorname {rot\ } \operatorname {grad\ } B=0}
Lösung: (in Arbeit - mit Sicherheit noch falsch ;-) )
r
o
t
g
r
a
d
B
=
∇
×
(
∇
,
B
)
=
ε
i
j
k
∇
j
(
∇
,
B
)
k
e
→
i
=
ε
i
j
k
∇
j
∇
k
B
e
→
i
{\displaystyle \operatorname {rot\ } \operatorname {grad\ } B=\nabla \times (\nabla ,B)=\varepsilon _{ijk}\nabla _{j}(\nabla ,B)_{k}{\vec {e}}_{i}=\varepsilon _{ijk}\nabla _{j}\nabla _{k}B{\vec {e}}_{i}}
=
∇
j
∇
k
B
e
→
i
(
e
→
j
×
e
→
k
)
e
→
i
=
B
(
∇
×
∇
)
δ
i
i
=
B
⋅
0
→
=
0
{\displaystyle =\nabla _{j}\nabla _{k}B{\vec {e}}_{i}({\vec {e}}_{j}\times {\vec {e}}_{k}){\vec {e}}_{i}=B(\nabla \times \nabla )\delta _{ii}=B\cdot {\vec {0}}=0}
Zeige:
d
i
v
g
r
a
d
B
=
0
{\displaystyle \operatorname {div\ } \operatorname {grad\ } B=0}
Lösung:
![{\displaystyle [{\vec {a}}\times ({\vec {b}}\times {\vec {c}})]_{k}=\sum _{i,j}\varepsilon _{ijk}a_{i}({\vec {b}}\times {\vec {c}})_{j}=\sum _{i,j}\sum _{l,m}\varepsilon _{ijk}\varepsilon _{jlm}a_{i}b_{l}c_{m}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6efff3a4b4b80a782733de9678cd169fb24f278)
![{\displaystyle (({\vec {a}},{\vec {c}})({\vec {e}}_{k},{\vec {b}})-({\vec {a}},{\vec {b}})({\vec {e}}_{k},{\vec {c}}))=[{\vec {b}}({\vec {a}},{\vec {c}})-{\vec {c}}({\vec {a}},{\vec {b}})]_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0c62c0e8659325da560740c91961abf3e40b228)
![{\displaystyle {\dot {x}}(t)=v_{0}+{\frac {-e{\vec {E}}_{0}}{2m}}{\biggl [}{\frac {1}{-\alpha +i\omega }}e^{-\alpha t'+i\omega t'}{\biggl ]}_{0}^{t}+{\frac {-e{\vec {E}}_{0}}{2m}}{\biggl [}{\frac {1}{-\alpha -i\omega }}e^{-\alpha t'-i\omega t'}{\biggl ]}_{0}^{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c69bf948f956e60209cd46dbd363b22d1f0b12e1)
![{\displaystyle x(t)=x_{0}+{\biggl [}v_{0}t'{\biggl ]}_{0}^{t}+{\biggl [}{\frac {-e{\vec {E}}_{0}e^{-\alpha t'+i\omega t'}}{2m(-\alpha +i\omega )^{2}}}{\biggl ]}_{0}^{t}-{\biggl [}{\frac {-e{\vec {E}}_{0}t'}{2m(-\alpha +i\omega )}}{\biggl ]}_{0}^{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b57823f9ccbecf580364882f8e3c49a867591543)
![{\displaystyle +{\biggl [}{\frac {-e{\vec {E}}_{0}e^{-\alpha t-i\omega t}}{2m(-\alpha -i\omega )^{2}}}{\biggl ]}_{0}^{t}-{\biggl [}{\frac {-e{\vec {E}}_{0}t'}{2m(-\alpha -i\omega )}}{\biggl ]}_{0}^{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ac566c87d2998491618ae57d48f628dc9c976a2)
