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Conversion of Cartesian Coordinate System Unit Vectors to Spherical Coordinate System Unit Vectors 📂Mathematical Physics

Conversion of Cartesian Coordinate System Unit Vectors to Spherical Coordinate System Unit Vectors

Formulas

The expression for converting the unit vectors from the Cartesian coordinate system to the spherical coordinate system is as follows. x^=cos⁡ϕsin⁡θr^+cos⁡ϕcos⁡θθ^−sin⁡ϕϕ^y^=sin⁡ϕsin⁡θr^+sin⁡ϕcos⁡θθ^+cos⁡ϕϕ^z^=cos⁡θr^−sin⁡θθ^ \begin{align*} \hat{ \mathbf{x} }&= \cos \phi \sin \theta \hat{ \mathbf{r} } + \cos \phi \cos \theta \hat{ \boldsymbol{\theta} } - \sin\phi\hat{ \boldsymbol{\phi} } \\ \hat{ \mathbf{y} } &= \sin\phi\sin\theta \hat{ \mathbf{r} } + \sin\phi\cos\theta\hat{ \boldsymbol{\theta} } + \cos\phi\hat{ \boldsymbol{\phi} } \\ \hat{ \mathbf{z} } &= \cos\theta\hat{ \mathbf{r} } - \sin\theta\hat{ \boldsymbol{\theta} } \end{align*}

The expression for converting the unit vectors from the spherical coordinate system to the Cartesian coordinate system is as follows. (Relation between spherical and Cartesian coordinate systems)

r^=cos⁡ϕsin⁡θx^+sin⁡ϕsin⁡θy^+cos⁡θz^θ^=cos⁡ϕcos⁡θx^+sin⁡ϕcos⁡θy^−sin⁡θz^ϕ^=−sin⁡ϕx^+cos⁡ϕy^ \begin{align*} \hat{ \mathbf{r} } &= \cos\phi \sin\theta \hat{ \mathbf{x} } + \sin\phi \sin\theta\hat{ \mathbf{y} } + \cos\theta\hat{ \mathbf{z} } \\ \hat{ \boldsymbol{\theta} } &= \cos\phi \cos\theta \hat{ \mathbf{x} } + \sin\phi \cos\theta\hat{ \mathbf{y} } - \sin\theta\hat{ \mathbf{z} } \\ \hat{ \boldsymbol{\phi} } &= -\sin\phi \hat{ \mathbf{x} } + \cos\phi \hat{ \mathbf{y} } \end{align*}

Derivation

Unit vector x^\hat{\mathbf{x}}

To find x^\hat{ \mathbf{x} }, you need to multiply certain terms by r^, θ^, ϕ^\hat{ \mathbf{r} } ,\ \hat{ \boldsymbol{\theta} } ,\ \hat{ \boldsymbol{\phi} } and add them so that the y^, z^\hat{ \mathbf{y} } ,\ \hat{ \mathbf{z} } term disappears. It can be seen that multiplying r^\hat{ \mathbf{r} } by sin⁡θ\sin \theta and adding that to θ^\hat{ \boldsymbol{\theta} } multiplied by cos⁡θ\cos \theta makes the z^\hat{ \mathbf{z} } term vanish.

sin⁡θr^+cos⁡θθ^=(cos⁡ϕsin⁡2θ+cos⁡ϕcos⁡2θ)x^+(sin⁡ϕsin⁡2θ+sin⁡ϕcos⁡2θ)y^=cos⁡ϕx^+sin⁡ϕy^ \begin{align*} & \sin \theta \hat{ \mathbf{r} } + \cos \theta \hat{ \boldsymbol{\theta} } \\ &= (\cos \phi \sin^2 \theta + \cos \phi \cos^2\theta)\hat{ \mathbf{x} } + (\sin \phi \sin^2 \theta + \sin \phi \cos^2 \theta)\hat{ \mathbf{y} } \\ &= \cos \phi \hat{ \mathbf{x} } +\sin \phi \hat{ \mathbf{y} } \end{align*}

Further, by multiplying this result by cos⁡ϕ\cos \phi and adding it to ϕ^\hat{ \boldsymbol{\phi} } multiplied by −sin⁡ϕ-\sin \phi, you can neatly obtain x^\hat{ \mathbf{x} }.

cos⁡ϕ(sin⁡θr^+cos⁡θθ^)−sin⁡ϕϕ^=(cos⁡2ϕx^+cos⁡ϕsin⁡ϕy^)+(sin⁡2ϕx^−sin⁡ϕcos⁡ϕy^)=x^ \begin{align*} & \cos \phi (\sin \theta \hat{ \mathbf{r} } + \cos \theta \hat{ \boldsymbol{\theta} })-\sin \phi \hat{ \boldsymbol{\phi} } \\ &= (\cos^2\phi \hat{ \mathbf{x} } + \cos \phi \sin \phi \hat{ \mathbf{y} } ) + (\sin^2\phi \hat{ \mathbf{x} } -\sin \phi \cos \phi \hat{ \mathbf{y} } ) \\ &= \hat{ \mathbf{x} } \end{align*}

Therefore, it is summarized as follows.

x^=cos⁡ϕsin⁡θr^+cos⁡ϕcos⁡θθ^−sin⁡ϕϕ^ \hat{ \mathbf{x} } = \cos \phi \sin \theta \hat{ \mathbf{r} } + \cos \phi \cos \theta \hat{ \boldsymbol{\theta} } - \sin\phi\hat{ \boldsymbol{\phi} }

Unit vector y^\hat{\mathbf{y}}

The method is the same as in the case of x^\hat{ \mathbf{x} }, so I will just write the equations without explanation. Removing the z^\hat{ \mathbf{z} } term is the same as in the case of x^\hat{ \mathbf{x} }.

sin⁡θr^+cos⁡θθ^=cos⁡ϕx^+sin⁡ϕy^  ⟹  sin⁡ϕ(sin⁡θr^+cos⁡θθ^)=sin⁡ϕcos⁡ϕx^+sin⁡2ϕy^ \begin{align*} &&\sin \theta \hat{ \mathbf{r} } + \cos \theta \hat{ \boldsymbol{\theta} } &= \cos \phi \hat{ \mathbf{x} } + \sin \phi \hat{ \mathbf{y} } \\ \implies&& \sin\phi (\sin\theta\hat{ \mathbf{r} }+\cos\theta \hat{ \boldsymbol{\theta} } ) &= \sin \phi \cos \phi \hat{ \mathbf{x} } + \sin^2\phi \hat{ \mathbf{y} } \end{align*}

And

cos⁡ϕϕ^=−sin⁡ϕcos⁡ϕx^+cos⁡2ϕy^ \cos \phi \hat{ \boldsymbol{\phi} }=-\sin\phi\cos\phi\hat{ \mathbf{x} }+\cos^2\phi\hat{ \mathbf{y} }

Therefore, it is as follows.

sin⁡ϕ(sin⁡θr^+cos⁡θθ^)+cos⁡ϕϕ^=(cos⁡2ϕ+sin⁡2ϕ)y^=y^ \begin{align*} && \sin\phi (\sin\theta\hat{ \mathbf{r} }+\cos\theta \hat{ \boldsymbol{\theta} } )+\cos \phi \hat{ \boldsymbol{\phi} } &= (\cos^2\phi + \sin^2\phi )\hat{ \mathbf{y} } \\ && &= \hat{ \mathbf{y} } \end{align*}

∴y^=sin⁡ϕsin⁡θr^+sin⁡ϕcos⁡θθ^+cos⁡ϕϕ^ \therefore \hat{ \mathbf{y} } = \sin\phi\sin\theta \hat{ \mathbf{r} } + \sin\phi\cos\theta \hat{ \boldsymbol{\theta} } + \cos\phi \hat{ \boldsymbol{\phi} }

Unit vector z^\hat{\mathbf{z}}

Since the z^\hat{ \mathbf{z} } term must remain, multiplying r^\hat{ \mathbf{r} } by cos⁡θ\cos\theta and adding that to θ^\hat{ \boldsymbol{\theta} } multiplied by −sin⁡θ-\sin\theta results in the following.

cos⁡θr^−sin⁡θθ^=cos⁡θ(cos⁡ϕsin⁡θx^+sin⁡ϕsin⁡θy^+cos⁡θz^)−sin⁡θ(cos⁡ϕcos⁡θx^+sin⁡ϕcos⁡θy^−sin⁡θz^)=(sin⁡θcos⁡θcos⁡ϕx^+sin⁡θcos⁡θsin⁡ϕy^+cos⁡2θz^)+(−sin⁡θcos⁡θcos⁡ϕx^−sin⁡θcos⁡θsin⁡ϕy^+sin⁡2θz^)=(cos⁡2θ+sin⁡2θ)z^=z^ \begin{align*} \cos\theta\hat{ \mathbf{r} } -\sin\theta \hat{ \boldsymbol{\theta} } &=\cos\theta (\cos \phi \sin \theta \hat{\mathbf{x}}+\sin \phi \sin \theta \hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}}) \\ &\quad -\sin \theta (\cos \phi \cos \theta \hat{\mathbf{x}} + \sin \phi \cos \theta \hat{\mathbf{y}} -\sin \theta \hat{\mathbf{z}}) \\ &= (\sin\theta\cos\theta\cos\phi \hat{ \mathbf{x} } + \sin\theta\cos\theta\sin\phi \hat{ \mathbf{y} }+ \cos^2\theta \hat{ \mathbf{z} }) \\ &\quad +(-\sin\theta\cos\theta\cos\phi \hat{ \mathbf{x} }-\sin\theta\cos\theta\sin\phi \hat{ \mathbf{y} } + \sin^2{\theta} \hat{ \mathbf{z} }) \\ &= (\cos^2\theta + \sin ^2 \theta ) \hat{ \mathbf{z} } \\ &= \hat{ \mathbf{z} } \end{align*}

∴z^=cos⁡θr^−sin⁡θθ^ \therefore \hat{ \mathbf{z} } = \cos\theta\hat{ \mathbf{r} } - \sin\theta \hat{ \boldsymbol{\theta} }