轉換方式
向量：反變轉換
標記法說明：向量
v
{\displaystyle \mathbf {v} \,\!}
向量空間
V
{\displaystyle V\,\!}
e
1
,
e
2
,
.
.
.
,
e
n
{\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},...,\mathbf {e} _{n}\,\!}
x
1
,
x
2
,
.
.
.
,
x
n
{\displaystyle x^{1},x^{2},...,x^{n}\,\!}
v
{\displaystyle \mathbf {v} \,\!}
v
1
,
v
2
,
.
.
.
,
v
n
{\displaystyle v^{1},v^{2},...,v^{n}\,\!}
v
=
∑
i
v
i
e
i
{\displaystyle \textstyle \mathbf {v} =\sum _{i}v^{i}\mathbf {e} _{i))
（註：
v
2
{\displaystyle v^{2}\,\!}
2
{\displaystyle 2}
平方 ，而是代表第二個坐標，在較基礎的數學上，常寫作
v
2
{\displaystyle v_{2}\,\!}
愛因斯坦求和約定 。）
向量空間
V
{\displaystyle V}
e
¯
1
,
.
.
.
,
e
¯
n
{\displaystyle {\bar {\mathbf {e} ))_{1},...,{\bar {\mathbf {e} ))_{n}\,\!}
x
¯
1
,
.
.
.
,
x
¯
n
{\displaystyle {\bar {x))^{1},...,{\bar {x))^{n}\,\!}
v
{\displaystyle \mathbf {v} \,\!}
v
¯
1
,
v
¯
2
,
.
.
.
,
v
¯
n
{\displaystyle {\bar {v))^{1},{\bar {v))^{2},...,{\bar {v))^{n}\,\!}
v
=
∑
i
v
¯
i
e
¯
i
{\displaystyle \textstyle \mathbf {v} =\sum _{i}{\bar {v))^{i}{\bar {\mathbf {e} ))_{i))
對於1...n之間任意整數
μ
{\displaystyle \mu \,\!}
v
¯
μ
{\displaystyle {\bar {v))^{\mu }\,\!}
v
1
,
v
2
,
.
.
.
,
v
n
{\displaystyle v^{1},v^{2},...,v^{n}\,\!}
v
¯
μ
=
∂
x
¯
μ
∂
x
1
v
1
+
∂
x
¯
μ
∂
x
2
v
2
+
.
.
.
+
∂
x
¯
μ
∂
x
n
v
n
{\displaystyle {\bar {v))^{\mu }={\frac {\partial {\bar {x))^{\mu )){\partial x^{1))}v^{1}+{\frac {\partial {\bar {x))^{\mu )){\partial x^{2))}v^{2}+...+{\frac {\partial {\bar {x))^{\mu )){\partial x^{n))}v^{n}\,\!}
使用愛因斯坦求和約定 可寫成：
v
¯
μ
=
∂
x
¯
μ
∂
x
i
v
i
{\displaystyle {\bar {v))^{\mu }={\frac {\partial {\bar {x))^{\mu )){\partial x^{i))}v^{i}\,\!}
余向量：共變轉換
假設對偶空間
V
∗
{\displaystyle V^{*))
d
x
1
,
d
x
2
,
.
.
.
,
d
x
n
{\displaystyle {\mathbf {dx} ^{1},\mathbf {dx} ^{2},...,\mathbf {dx} ^{n))\,\!}
d
x
¯
1
,
d
x
¯
2
,
.
.
.
,
d
x
¯
n
{\displaystyle \mathbf {d{\bar {x))} ^{1},\mathbf {d{\bar {x))} ^{2},...,\mathbf {d{\bar {x))} ^{n}\,\!}
[1] :289-297
假設
ω
∈
V
∗
,
ω
=
∑
i
η
i
d
x
i
=
∑
j
η
¯
j
d
x
¯
j
{\displaystyle \textstyle {\boldsymbol {\omega ))\in V^{*},{\boldsymbol {\omega ))=\sum _{i}\mathbf {\eta } _{i}\mathbf {dx} ^{i}=\sum _{j}{\bar {\mathbf {\eta } ))_{j}d{\bar {\mathbf {x} ))^{j))
1
{\displaystyle 1}
n
{\displaystyle n}
μ
{\displaystyle \mu \,\!}
η
¯
μ
{\displaystyle {\bar {\mathbf {\eta } ))_{\mu }\,\!}
η
1
,
η
2
,
.
.
.
,
η
n
{\displaystyle \mathbf {\eta } _{1},\mathbf {\eta } _{2},...,\mathbf {\eta } _{n}\,\!}
η
¯
μ
=
∂
x
1
∂
x
¯
μ
η
1
+
∂
x
2
∂
x
¯
μ
η
2
+
.
.
.
+
∂
x
n
∂
x
¯
μ
η
n
{\displaystyle {\bar {\mathbf {\eta } ))_{\mu }={\frac {\partial x^{1)){\partial {\bar {x))^{\mu ))}\mathbf {\eta } _{1}+{\frac {\partial x^{2)){\partial {\bar {x))^{\mu ))}\mathbf {\eta } _{2}+...+{\frac {\partial x^{n)){\partial {\bar {x))^{\mu ))}\mathbf {\eta } _{n}\,\!}
或使用愛因斯坦求和約定寫成：
η
¯
μ
=
∂
x
i
∂
x
¯
μ
η
i
{\displaystyle {\bar {\mathbf {\eta } ))_{\mu }={\frac {\partial x^{i)){\partial {\bar {x))^{\mu ))}\mathbf {\eta } _{i}\,\!}
向量的共變分量和反變分量
在歐幾里得空間
V
{\displaystyle V\,\!}
內積 運算從向量求得餘向量；對於所有餘向量
w
{\displaystyle \mathbf {w} \,\!}
v
{\displaystyle \mathbf {v} \,\!}
線性泛函
α
(
w
)
{\displaystyle \alpha (\mathbf {w} )\,\!}
w
{\displaystyle \mathbf {w} \,\!}
α
(
w
)
=
v
⋅
w
{\displaystyle \alpha (\mathbf {w} )=\mathbf {v} \cdot \mathbf {w} \,\!}
逆過來，通過上述方程式，線性泛函
α
(
w
)
{\displaystyle \alpha (\mathbf {w} )\,\!}
v
{\displaystyle \mathbf {v} \,\!}
給予
V
{\displaystyle V\,\!}
f
=
(
X
1
,
X
2
,
…
,
X
n
)
{\displaystyle {\mathfrak {f))=(X_{1},X_{2},\dots ,X_{n})\,\!}
f
♯
=
(
Y
1
,
Y
2
,
…
,
Y
n
)
{\displaystyle {\mathfrak {f))^{\sharp }=(Y^{1},Y^{2},\dots ,Y^{n})\,\!}
Y
i
⋅
X
j
=
δ
j
i
{\displaystyle Y^{i}\cdot X_{j}=\delta _{j}^{i}\,\!}
其中，
δ
j
i
{\displaystyle \delta _{j}^{i}\,\!}
克羅內克函數 。
以這兩種基底，任意向量
v
{\displaystyle \mathbf {v} \,\!}
v
=
∑
i
v
i
[
f
]
X
i
=
f
v
[
f
]
=
∑
i
v
i
[
f
]
Y
i
=
f
♯
v
[
f
♯
]
{\displaystyle {\begin{aligned}v&=\sum _{i}v^{i}[{\mathfrak {f))]X_{i}={\mathfrak {f))\,\mathbf {v} [{\mathfrak {f))]\\&=\sum _{i}v_{i}[{\mathfrak {f))]Y^{i}={\mathfrak {f))^{\sharp }\,\mathbf {v} [{\mathfrak {f))^{\sharp }]\end{aligned))\,\!}
； 其中，
v
i
[
f
]
{\displaystyle v^{i}[{\mathfrak {f))]\,\!}
v
{\displaystyle \mathbf {v} \,\!}
f
{\displaystyle {\mathfrak {f))\,\!}
v
i
[
f
]
{\displaystyle v_{i}[{\mathfrak {f))]\,\!}
v
{\displaystyle \mathbf {v} \,\!}
f
{\displaystyle {\mathfrak {f))\,\!}
歐幾里得空間
將向量
a
{\displaystyle \mathbf {a} \,\!}
投影 於坐標軸
e
i
{\displaystyle \mathbf {e} ^{i}\,\!}
a
i
{\displaystyle a^{i}\,\!}
a
{\displaystyle \mathbf {a} \,\!}
坐標曲面 的法線
e
i
{\displaystyle \mathbf {e} _{i}\,\!}
a
i
{\displaystyle a_{i}\,\!}
在歐幾里得空間 Ｒ3 裏，使用內積 運算，能夠從向量求得餘向量。給予一組可能不是標準正交基 的基底，其基底向量為
e
1
{\displaystyle \mathbf {e} _{1}\,\!}
e
2
{\displaystyle \mathbf {e} _{2}\,\!}
e
3
{\displaystyle \mathbf {e} _{3}\,\!}
e
1
=
e
2
×
e
3
τ
;
e
2
=
e
3
×
e
1
τ
;
e
3
=
e
1
×
e
2
τ
{\displaystyle \mathbf {e} ^{1}={\frac {\mathbf {e} _{2}\times \mathbf {e} _{3)){\tau ));\qquad \mathbf {e} ^{2}={\frac {\mathbf {e} _{3}\times \mathbf {e} _{1)){\tau ));\qquad \mathbf {e} ^{3}={\frac {\mathbf {e} _{1}\times \mathbf {e} _{2)){\tau ))\,\!}
其中，
τ
=
e
1
⋅
(
e
2
×
e
3
)
{\displaystyle \tau =\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})\,\!}
e
1
{\displaystyle \mathbf {e} _{1}\,\!}
e
2
{\displaystyle \mathbf {e} _{2}\,\!}
e
3
{\displaystyle \mathbf {e} _{3}\,\!}
平行六面體 的體積。
反過來計算，
e
1
=
e
2
×
e
3
τ
′
;
e
2
=
e
3
×
e
1
τ
′
;
e
3
=
e
1
×
e
2
τ
′
{\displaystyle \mathbf {e} _{1}={\frac {\mathbf {e} ^{2}\times \mathbf {e} ^{3)){\tau '));\qquad \mathbf {e} _{2}={\frac {\mathbf {e} ^{3}\times \mathbf {e} ^{1)){\tau '));\qquad \mathbf {e} _{3}={\frac {\mathbf {e} ^{1}\times \mathbf {e} ^{2)){\tau '))\,\!}
其中，
τ
′
=
e
1
⋅
(
e
2
×
e
3
)
=
1
/
τ
{\displaystyle \tau '=\mathbf {e} ^{1}\cdot (\mathbf {e} ^{2}\times \mathbf {e} ^{3})=1/\tau \,\!}
e
1
{\displaystyle \mathbf {e} ^{1}\,\!}
e
2
{\displaystyle \mathbf {e} ^{2}\,\!}
e
3
{\displaystyle \mathbf {e} ^{3}\,\!}
雖然
e
i
{\displaystyle \mathbf {e} _{i}\,\!}
e
j
{\displaystyle \mathbf {e} ^{j}\,\!}
e
i
⋅
e
j
=
δ
i
j
{\displaystyle \mathbf {e} _{i}\cdot \mathbf {e} ^{j}=\delta _{i}^{j}\,\!}
這樣，任意向量
a
{\displaystyle \mathbf {a} \,\!}
a
1
=
a
⋅
e
1
;
a
2
=
a
⋅
e
2
;
a
3
=
a
⋅
e
3
{\displaystyle a^{1}=\mathbf {a} \cdot \mathbf {e} ^{1};\qquad a^{2}=\mathbf {a} \cdot \mathbf {e} ^{2};\qquad a^{3}=\mathbf {a} \cdot \mathbf {e} ^{3}\,\!}
類似地，共變坐標為
a
1
=
a
⋅
e
1
;
a
2
=
a
⋅
e
2
;
a
3
=
a
⋅
e
3
{\displaystyle a_{1}=\mathbf {a} \cdot \mathbf {e} _{1};\qquad a_{2}=\mathbf {a} \cdot \mathbf {e} _{2};\qquad a_{3}=\mathbf {a} \cdot \mathbf {e} _{3}\,\!}
這樣，
a
{\displaystyle \mathbf {a} \,\!}
a
=
a
i
e
i
=
a
1
e
1
+
a
2
e
2
+
a
3
e
3
{\displaystyle \mathbf {a} =a_{i}\mathbf {e} ^{i}=a_{1}\mathbf {e} ^{1}+a_{2}\mathbf {e} ^{2}+a_{3}\mathbf {e} ^{3}\,\!}
或者，
a
=
a
i
e
i
=
a
1
e
1
+
a
2
e
2
+
a
3
e
3
{\displaystyle \mathbf {a} =a^{i}\mathbf {e} _{i}=a^{1}\mathbf {e} _{1}+a^{2}\mathbf {e} _{2}+a^{3}\mathbf {e} _{3}\,\!}
綜合上述關係式，
a
=
(
a
⋅
e
i
)
e
i
=
(
a
⋅
e
i
)
e
i
{\displaystyle \mathbf {a} =(\mathbf {a} \cdot \mathbf {e} _{i})\mathbf {e} ^{i}=(\mathbf {a} \cdot \mathbf {e} ^{i})\mathbf {e} _{i}\,\!}
向量
a
{\displaystyle \mathbf {a} \,\!}
a
i
=
a
⋅
e
i
=
(
a
j
e
j
)
⋅
e
i
=
(
e
j
⋅
e
i
)
a
j
=
g
j
i
a
j
{\displaystyle a_{i}=\mathbf {a} \cdot \mathbf {e} _{i}=(a^{j}\mathbf {e} _{j})\cdot \mathbf {e} _{i}=(\mathbf {e} _{j}\cdot \mathbf {e} _{i})a^{j}=g_{ji}a^{j}\,\!}
其中，
g
j
i
=
e
j
⋅
e
i
{\displaystyle g_{ji}=\mathbf {e} _{j}\cdot \mathbf {e} _{i}\,\!}
度規張量 。
向量
a
{\displaystyle \mathbf {a} \,\!}
a
i
=
a
⋅
e
i
=
(
a
j
e
j
)
⋅
e
i
=
(
e
j
⋅
e
i
)
a
j
=
g
j
i
a
j
{\displaystyle a^{i}=\mathbf {a} \cdot \mathbf {e} ^{i}=(a_{j}\mathbf {e} ^{j})\cdot \mathbf {e} ^{i}=(\mathbf {e} ^{j}\cdot \mathbf {e} ^{i})a_{j}=g^{ji}a_{j}\,\!}
其中，
g
j
i
=
e
j
⋅
e
i
{\displaystyle g^{ji}=\mathbf {e} ^{j}\cdot \mathbf {e} ^{i}\,\!}
共軛度規張量 。
共變坐標的標號是下標，反變坐標的標號是上標。假若共變基底向量組成的基底是標準正交基，或反變基底向量組成的基底是標準正交基，則共變基底與反變基底相互等價。那麼，就沒有必要分辨共變坐標和反變坐標，所有的標號都可以用下標來標記。
![{\displaystyle {\begin{aligned}v&=\sum _{i}v^{i}[{\mathfrak {f))]X_{i}={\mathfrak {f))\,\mathbf {v} [{\mathfrak {f))]\\&=\sum _{i}v_{i}[{\mathfrak {f))]Y^{i}={\mathfrak {f))^{\sharp }\,\mathbf {v} [{\mathfrak {f))^{\sharp }]\end{aligned))\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/706accb597961c64fa1ffe9c4db0bf94314c7488)
![{\displaystyle v^{i}[{\mathfrak {f))]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e403b5fdaee2e8bd8cb405934e5ff4ac90eb12d)
![{\displaystyle v_{i}[{\mathfrak {f))]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fdff97f5ab560841a1baccf4f27b6a756a1dfd7)
