In Schutz's General Relativity Chapter 5, after he defines vectors in the modern view as tangents to a particular curve, he states the relationship between bases of different coordinates as:

$$\vec{e}_{\alpha'}=\Lambda^\beta_{~~\alpha'}\vec{e}_{\beta}=\frac{\partial x^\beta}{\partial x^{\alpha'}}\vec{e}_{\beta} \tag{1}$$

I recognise this equation from the earlier chapters in SR, i.e. the basis vectors must transform oppositely to the components. The problem is that I don't know if this is still valid given the new definition of vectors, since he doesn't explain why the above equation is true and only explains how the vector components transform.

In doing some of his exercises, I also found that the tangent vectors to the coordinate curves are the same as the basis vectors of that coordinate system! For example:

In polar coordinates, the coordinate curve parameterised by $r$, keeping $\theta$ constant, gives a tangent vector:

$$\vec{V}=\vec{e}_x\frac{\partial x}{\partial r}+\vec{e}_y\frac{\partial y}{\partial r}=\vec{e}_x\cos\theta + \vec{e}_y\sin\theta \tag{2}$$

which is exactly the same as $\vec{e}_r$ given by $(1)$.

  • How exactly does Schutz make the deduction for equation $(1)$?
  • Why does $(2)$ give the basis vectors for the coordinate system? Is it possible to construct basis vectors which aren't the same as the corresponding tangent vectors?
  • Could $(2)$ be taken as the definition for basis vectors?

4 Answers 4


A vector is a tangent to some sufficiently smooth curve at some point.

So let us imagine an $n$-dimensional space (manifold) $M$ which can be parametrized by coordinates $\{x^i\}_{i=1\dots n}$ in some neighbourhood $U\subset M$. We can formalize this by saying that there is a map $\phi:U\to \phi\left(U\right)\subset\mathbb{R}^n$ such that $\phi^i\left(m\in U\right)=x^i\:\forall i=1\dots n$.

Next we also define another set of coordinates that parametrize the same neighbourhood $U\subset M$: $\{\bar{x}^i\}_{i=1\dots N}$, and a corresponding map $\bar{\phi}:U\to \bar{\phi}\left(U\right)\subset\mathbb{R}^n$

To define vector at point $P\in U\subset M$ we need to define a curve $\lambda$, that is a map from real-values to manifold. $\lambda:\mathbb{R}\to M$.

Based on this we can construct $\phi\circ\lambda:\mathbb{R}\to\mathbb{R}^n$, such that (common abuse of notation) $x^i=x^i\left(\lambda\right)$, and $\bar{\phi}\circ\lambda:\mathbb{R}\to\mathbb{R}^n$, such that $\bar{x}^i=\bar{x}^i\left(\lambda\right)$.

After that, define the vector to be $V=\frac{d}{d\lambda}$. We can then use chain rule to show that:

$$ V=\frac{d}{d\lambda}=\frac{dx^i}{d\lambda}\:\partial_{x^i}=\frac{d\bar{x}^i}{d\lambda}\:\partial_{\bar{x}^i} $$

To have $V$ be the basis vector in the first coordinate system we can choose $\lambda=x^j$ for some specific $j$, such that $dx^i/d\lambda=\delta^i_j$. Then $V\equiv e_j$ and:

$$ e_j=\partial_{x^j}=\frac{d\bar{x}^i}{dx^j}\partial_{\bar{x}^i} $$

The full derivative here ($\frac{d\bar{x}^i}{dx^j}$) actually means differentiate along a curve where all $x^{i\neq j}=const$, so it is essentially partial derivative, hence:

$$ e_j=\partial_{x^j}=\frac{\partial\bar{x}^i}{\partial x^j}\partial_{\bar{x}^i} $$

Regarding your last question. An easy way to construct basis in 3d is to use the gradient operator. Define basis for spherical coordinates as $\mathbf{e}_r=\boldsymbol{\nabla}r,\,\mathbf{e}_\theta=\boldsymbol{\nabla}\theta,\mathbf{e}_\phi=\boldsymbol{\nabla}\phi$

Lets say I want to express $\mathbf{\hat{x}}=\boldsymbol{\nabla}x$ in terms of spherical basis. Then $x=x\left(r,\theta\,\phi\right)$ so

$$ \mathbf{\hat{x}}=\boldsymbol{\nabla}x=\frac{\partial x}{\partial r} \boldsymbol{\nabla}r + \dots = \frac{\partial x}{\partial r}\mathbf{e}_r+\frac{\partial x}{\partial \theta}\mathbf{e}_\theta+\frac{\partial x}{\partial \phi}\mathbf{e}_\phi $$

Or in the opposite direction:

$$ \mathbf{e}_r=\boldsymbol{\nabla}r\left(x,y,z\right)=\frac{\partial r}{\partial x}\mathbf{\hat{x}}+\dots $$

If you have a curve $f^i=f^i\left(s\right)$ then its tangent is given by:

$$ \mathbf{V}=\frac{df^i}{ds}\mathbf{e}_i $$

Let $f^i=\{h\left(r\right),\Theta=const,\Phi=const\}$ in spherical coordinate system. So that the tangent in spherical basis is:

$$ V=h'\partial_r \quad\implies\quad \mathbf{V}=h'\mathbf{e}_r $$

In Cartesian coordinates

$\mathbf{V}=h'\mathbf{e}_r=h'\cdot\left(\frac{\partial x}{\partial r}\mathbf{\hat{x}}+\dots\right)$

Using results from before. I think this is what you had

  • $\begingroup$ Realized the last part of question was wrong, changed it. I think it agrees with TC original expression now $\endgroup$
    – Cryo
    Commented Jan 24, 2021 at 13:14

The simple answer is that this follows from the chain rule. First identify $$\vec e_{(\mu)}\cong \frac{\partial}{\partial x^\mu},\quad\quad{ \vec e}\,'_{(\mu)}\cong \frac{\partial}{\partial x'^\mu}$$ where the prime indicates the transformed frame and brackets indicate that you should see $(\mu)$ as a label, not as an index. The reason we can identify derivatives with vectors is that derivatives obey all the axioms needed for a vector space. Intuitively you can identify a partial derivative with the vector you would get by acting with the derivative on a general position vector. Here are two examples from Cartesian and polar coordinates: \begin{align} \partial_x\cong \vec e_{(x)}&=\partial_x(x,y)^T=(1,0)^T\\ \partial_r\cong\vec e_{(r)}&=\partial_r(r\cos\theta,r\sin\theta)^T=(\cos\theta,\sin\theta)^T \end{align}

Using this the transformation law is just the chain rule: $$\frac{\partial}{\partial x'^\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial}{\partial x^\nu}$$ or equivalently $${ \vec e}\,'_{(\mu)}=\frac{\partial x^\nu}{\partial x'^\mu}{\vec e}_{(\nu)}={\Lambda^\nu}_\mu{\vec e}_{(\nu)}.$$ I like to think as these basis vectors as the fundamental quantities and how vector components transform follows from the fact that vectors should stay invariant under change of basis (coordinate transformations in this case). \begin{align} V'&=V\\ { \vec e}\,'_{(\mu)}V'^\mu &={ \vec e}_{(\nu)}V^\nu\\ {\vec e}_{(\nu)}\frac{\partial x^\nu}{\partial x'^\mu}V'^\mu&={ \vec e}_{(\nu)}V^\nu \end{align} So to have invariance we must have $\frac{\partial x^\nu}{\partial x'^\mu}V'^\mu=V^\nu$ which can be shown to be equivalent to $${V'}^\mu=\frac{\partial {x'}^\mu}{\partial x^\nu}V^\nu.$$ So the inverse transformation corresponds to the inverse Jacobian associated with the coordinate transformation.

So as a summary we can identify partial derivatives with vectors and the nice thing about this is that it automatically equips you with a transformation/inverse transformation: the Jacobian of the coordinate transformation.

Edit: you can generalize this easily to a general change of basis. If you have two bases $\{\vec e_{(\mu)}\}$ and $\{\vec e'_{(\mu)}\}$ which span the entire space you can always write a vector as a linear combination of these basis vectors. So you can expand $\vec e'_{(\mu)}$ as a linear combination of $\{\vec e_{(\mu)}\}$: $$\vec e'_{(\mu)}={\Lambda^\nu}_\mu \vec e_{(\nu)}$$ This expansion _defines_ the transformation matrix ${\Lambda^\nu}_\mu$. Using a derivation similar to where I set $V'=V$ you can deduce how vector components should transform: $$V'^{\mu}={(\Lambda^{-1})^\mu}_\nu V^\nu$$ where ${\Lambda^\mu}_\lambda {(\Lambda^{-1})^\lambda}_\nu=\delta^\mu_\nu$. This is quite a general derivation. If you now pick as basis vectors the coordinate tangent vectors you get all the nice properties I mentioned above. So now I'll try to answer some of your questions more explicitly.
  • How does Schutz make the deduction for equation (1)?

Because he picks tangent vectors as basis vectors the transformation law automatically follows from the chain rule.

  • How does (2) give basis vectors for the coordinate system?

You can construct coordinate tangent vectors by taking the derivative of the position vector with respect to a coordinate of choice. You essentially performed $$\vec e_r=\partial _r \vec r(x,y)=\partial_r \vec r(x(r,\theta),y(r,\theta))=\vec e_x\cos\theta+\vec e_y\sin\theta$$ where $\vec r(x,y)=\vec e_x x+\vec e_y y$. This is just an implementation of equation (1).

  • Could (2) be taken as the definition for basis vectors?

I think this would be a good definition. There's probably a more suave mathematical definition but when doing calculations this is what you should do.

  • $\begingroup$ I have seen this usage of partial derivatives as basis vectors before. In fact, Schutz does mention it in the text. However, is there no way to understand the transformation law without imposing this conceptual change? I would like to understand it in the traditional sense as well. $\endgroup$ Commented Jan 30, 2021 at 1:54
  • $\begingroup$ @HexiangChang I'm not sure what you mean by traditional sense. But taking tangent vectors as basis vectors (differential geometry) is just a special case of a general change of basis (linear algebra). I'll add a little bit to my answer to expand on this. $\endgroup$ Commented Jan 30, 2021 at 14:14


In $\boldsymbol{\S \rm A }$ we give some notes from the theory of 2d-surfaces in $\mathbb{R}^3$ useful for the interpretation of the corresponding theory in higher dimensions. In $\boldsymbol{\S \rm B}$ we make the connection with the question and in $\boldsymbol{\S \rm C}$ we give a $n$-dimensional generalization.

$\boldsymbol{\S \rm A. }$ FROM THE THEORY OF 2D-SURFACES

A real 3-vector function of two real parameters $(u,v)$ \begin{equation} \mathbf{x}\left(u,v\right)\boldsymbol{=}f_1\left(u,v\right)\mathbf{e}_1\boldsymbol{+}f_2\left(u,v\right)\mathbf{e}_2\boldsymbol{+}f_3\left(u,v\right)\mathbf{e}_3 \tag{01}\label{01} \end{equation} where $\left(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\right)$ is a basis in $\mathbb{R}^3$, represents a two-dimensional surface. Its partial derivatives are denoted by \begin{align} \mathbf{x}_{u} & \boldsymbol{=}\dfrac{\partial \mathbf{x}}{\partial u}\,,\quad \mathbf{x}_{uu}\boldsymbol{=}\dfrac{\partial}{\partial u}\left(\dfrac{\partial \mathbf{x}}{\partial u}\right)\boldsymbol{=}\dfrac{\partial^2 \mathbf{x}}{\partial u^2}\,,\quad \mathbf{x}_{uv}\boldsymbol{=}\dfrac{\partial}{\partial v}\left(\dfrac{\partial \mathbf{x}}{\partial u}\right)\boldsymbol{=}\dfrac{\partial^2 \mathbf{x}}{\partial v\partial u}\quad \texttt{etc} \nonumber\\ \mathbf{x}_{v} & \boldsymbol{=}\dfrac{\partial \mathbf{x}}{\partial v}\,,\quad \mathbf{x}_{vv}\boldsymbol{=}\dfrac{\partial}{\partial v}\left(\dfrac{\partial \mathbf{x}}{\partial v}\right)\boldsymbol{=}\dfrac{\partial^2 \mathbf{x}}{\partial v^2}\,,\quad \mathbf{x}_{vu}\boldsymbol{=}\dfrac{\partial}{\partial u}\left(\dfrac{\partial \mathbf{x}}{\partial v}\right)\boldsymbol{=}\dfrac{\partial^2 \mathbf{x}}{\partial u\partial v}\quad \texttt{etc} \tag{02}\label{02} \end{align} At a point $\mathrm P$ on the surface with curvilinear coordinates $\left(u,v\right)$, see Figure-01, the vectors \begin{equation} \mathbf{x}_{u} \boldsymbol{\equiv}\dfrac{\partial \mathbf{x}}{\partial u}\qquad \mathbf{x}_{v} \boldsymbol{\equiv}\dfrac{\partial \mathbf{x}}{\partial v} \tag{03}\label{03} \end{equation} are a basis with respect to the curvilinear coordinates $\left(u,v\right)$. The vector $\mathbf{x}_{u}$ is tangent to the $u\boldsymbol{-}$parametric curve $v\boldsymbol{=}\texttt{constant}$ while the vector $\mathbf{x}_{v}$ is tangent to the $v\boldsymbol{-}$parametric curve $u\boldsymbol{=}\texttt{constant}$. The vectors $\mathbf{x}'_{u},\mathbf{x}'_{v}$ tangent to the parametric curves on a second point $\mathrm P'$ are a basis on this point as shown in the same Figure-01.

The members of a basis $\{\mathbf{x}_{u},\mathbf{x}_{v}\}$ are not in general unit vectors. Also they are not in general perpendicular to each other. In Figure-02 we see two points $\mathrm P\left(u,v\right)$ and $\mathrm P\left(u\boldsymbol{+}\mathrm du,v\boldsymbol{+}\mathrm dv\right)$ on the surface infinitesimally close to each other.

The infinitesimal vector \begin{equation} \mathrm d\mathbf{x}\boldsymbol{=} \mathbf{x}_u \mathrm du\boldsymbol{+} \mathbf{x}_v \mathrm dv \tag{04}\label{04} \end{equation} has the property \begin{equation} \mathbf{x}\left(u\boldsymbol{+}\mathrm du,v\boldsymbol{+}\mathrm dv\right)\boldsymbol{=}\mathbf{x}\left(u,v\right)\boldsymbol{+}\mathrm d\mathbf{x}\boldsymbol{+}\mathbf{o}\left(\left(\mathrm du^2\boldsymbol{+}\mathrm dv^2\right)^{1/2}\right) \tag{05}\label{05} \end{equation} Thus the vector $\mathrm d\mathbf{x}$ is a first order approximation to the vector $\mathbf{x}\left(u\boldsymbol{+}\mathrm du,v\boldsymbol{+}\mathrm dv\right)\boldsymbol{-} \mathbf{x}\left(u, v\right)$ from the point $\mathrm P\left(u,v\right)$ to the neighboring point $\mathrm P\left(u\boldsymbol{+}\mathrm du,v\boldsymbol{+}\mathrm dv\right)$ on the surface.

For the infinitesimal length $\mathrm ds^2$ we have \begin{align} \mathrm ds^2 & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{\cdot}\mathrm d\mathbf{x}\boldsymbol{=}\left(\mathbf{x}_{u}\mathrm du\boldsymbol{+}\mathbf{x}_{v}\mathrm dv\right)\boldsymbol{\cdot}\left(\mathbf{x}_{u}\mathrm du\boldsymbol{+}\mathbf{x}_{v}\mathrm dv\right) \nonumber\\ & \boldsymbol{=}\left(\mathbf{x}_{u}\boldsymbol{\cdot}\mathbf{x}_{u}\right)\mathrm du^2\boldsymbol{+}2\left(\mathbf{x}_{u}\boldsymbol{\cdot}\mathbf{x}_{v}\right)\mathrm du \mathrm dv\boldsymbol{+}\left(\mathbf{x}_{v}\boldsymbol{\cdot}\mathbf{x}_{v}\right)\mathrm dv^2 \nonumber\\ &\boldsymbol{=}g_{uu}\mathrm du^2\boldsymbol{+}2g_{uv}\mathrm du \mathrm dv\boldsymbol{+}g_{vv}\mathrm dv^2 \tag{06}\label{06} \end{align} where \begin{align} g_{uu} & \boldsymbol{=} \left(\mathbf{x}_{u}\boldsymbol{\cdot}\mathbf{x}_{u}\right)\boldsymbol{=}\Vert\mathbf{x}_{u} \Vert^2 \tag{07a}\label{07a}\\ g_{uv} & \boldsymbol{=}\left(\mathbf{x}_{u}\boldsymbol{\cdot}\mathbf{x}_{v}\right)\boldsymbol{=}\left(\mathbf{x}_{v}\boldsymbol{\cdot}\mathbf{x}_{u}\right)\boldsymbol{=}g_{vu} \tag{07b}\label{07b}\\ g_{vv} & \boldsymbol{=}\left(\mathbf{x}_{v}\boldsymbol{\cdot}\mathbf{x}_{v}\right)\boldsymbol{=}\Vert\mathbf{x}_{v} \Vert^2 \tag{07c}\label{07c} \end{align} the elements of the metric tensor \begin{equation} g_{ij} \boldsymbol{=} \begin{bmatrix} g_{uu} & g_{uv}\vphantom{\dfrac{a}{b}} \\ g_{vu} & g_{vv}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{08}\label{08} \end{equation} In Figure-03 we have a vector $\boldsymbol{\xi}$ in the plane tangent to the point $\mathrm P\left(u,v\right)$ expressed in terms of the basis $\{\mathbf{x}_{u},\mathbf{x}_{v}\}$
\begin{equation} \boldsymbol{\xi} \boldsymbol{=} \xi_u\mathbf{x}_{u}\boldsymbol{+}\xi_v\mathbf{x}_{v} \tag{09}\label{09} \end{equation} So $\boldsymbol{\xi}\boldsymbol{=}\left(\xi_u,\xi_v\right)$ is the representation of this vector with respect to the curvilinear system of coordinates $\left(u,v\right)$ with magnitude squared \begin{equation} \Vert\boldsymbol{\xi}\Vert^2 \boldsymbol{=} g_{uu}\vert\xi_u\vert^2\boldsymbol{+}2g_{uv}\xi_u \xi_v\boldsymbol{+}g_{vv}\vert\xi_v\vert^2 \tag{10}\label{10} \end{equation}

Suppose now that $\mathbf{x}\boldsymbol{=}\mathbf{x}^{\boldsymbol{*}}\left(\theta,\phi\right)$ is an other parametric representation of the surface, that is an other system of curvilinear coordinates \begin{equation} \theta \boldsymbol{=}\theta \left(u,v\right)\,,\qquad \phi \boldsymbol{=}\phi \left(u,v\right) \tag{11}\label{11} \end{equation} At a point $\mathrm P$ on the surface having curvilinear coordinates $\left(u,v\right)$ and $\left(\theta,\phi\right)$ in the two systems respectively we have for the tangent vectors of latter system \begin{align} \mathbf{x}^{\boldsymbol{*}}_{\theta} & \boldsymbol{=} \dfrac{\partial \mathbf{x}^{\boldsymbol{*}}}{\partial \theta}\boldsymbol{=}\dfrac{\partial \mathbf{x}}{\partial u}\dfrac{\partial u}{\partial \theta} \boldsymbol{+}\dfrac{\partial \mathbf{x}}{\partial v}\dfrac{\partial v}{\partial \theta}\boldsymbol{=}u_\theta\mathbf{x}_u \boldsymbol{+} v_\theta\mathbf{x}_v \tag{12a}\label{12a}\\ \mathbf{x}^{\boldsymbol{*}}_{\phi} & \boldsymbol{=} \dfrac{\partial \mathbf{x}^{\boldsymbol{*}}}{\partial \phi}\boldsymbol{=}\dfrac{\partial \mathbf{x}}{\partial u}\dfrac{\partial u}{\partial \phi} \boldsymbol{+}\dfrac{\partial \mathbf{x}}{\partial v}\dfrac{\partial v}{\partial \phi}\boldsymbol{=}u_\phi\mathbf{x}_u \boldsymbol{+} v_\phi\mathbf{x}_v \tag{12b}\label{12b} \end{align}
as shown in Figure-04. So \begin{equation} \begin{bmatrix} \mathbf{x}^{\boldsymbol{*}}_{\theta} \vphantom{\dfrac{\tfrac{a}{b}}{\dfrac{a}{b}}}\\ \mathbf{x}^{\boldsymbol{*}}_{\phi} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \:\:\dfrac{\partial u}{\partial \theta} \:& \:\dfrac{\partial v}{\partial \theta}\:\:\vphantom{\dfrac{a}{\dfrac{a}{b}}} \\ \:\:\dfrac{\partial u}{\partial \phi} \:& \:\dfrac{\partial v}{\partial \phi}\:\:\vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \mathbf{x}_{u} \vphantom{\dfrac{\tfrac{a}{b}}{\dfrac{a}{b}}} \\ \mathbf{x}_{v} \vphantom{\dfrac{a}{\tfrac{a}{b}}} \end{bmatrix} \tag{13}\label{13} \end{equation} Note that $\{\mathbf{x}^{\boldsymbol{*}}_{\theta},\mathbf{x}^{\boldsymbol{*}}_{\phi}\}$ is a basis at point $\mathrm P$ with respect to the new system of curvilinear coordinates $\left(\theta,\phi\right)$.

All vectors $\mathbf{x}_{u},\mathbf{x}_{v},\mathbf{x}^{\boldsymbol{*}}_{\theta},\mathbf{x}^{\boldsymbol{*}}_{\phi}$ are coplanar : they belong to the plane tangent to the point $\mathrm P$.

$\boldsymbol{\S \rm B. }$ CONNECTION WITH THE QUESTION

\begin{equation} \vec{e}_{\alpha'}\boldsymbol{=}\Lambda^\beta_{~\alpha'}\vec{e}_{\beta}\boldsymbol{=}\frac{\partial x^\beta}{\partial x^{\alpha'}}\vec{e}_{\beta} \tag{1 of the question}\label{1 of the question} \end{equation} To interpret above equation \eqref{1 of the question} for the two-dimensional case let make the following correspondences
\begin{align} & u \boldsymbol{=} x^1\qquad v\boldsymbol{=} x^2\qquad \theta\boldsymbol{=} x^{1'}\qquad \phi\boldsymbol{=} x^{2'} \tag{14a}\label{14a}\\ &\mathbf{x}_{u}\boldsymbol{=}\vec{e}_{1}\qquad\mathbf{x}_{v}\boldsymbol{=}\vec{e}_{2} \qquad\mathbf{x}^{\boldsymbol{*}}_{\theta}\boldsymbol{=}\vec{e}_{1'}\qquad\mathbf{x}^{\boldsymbol{*}}_{\phi}\boldsymbol{=}\vec{e}_{2'} \tag{14b}\label{14b}\\ &\dfrac{\partial u}{\partial \theta}\boldsymbol{=}\frac{\partial x^1}{\partial x^{1'}}\qquad\dfrac{\partial v}{\partial \theta}\boldsymbol{=}\frac{\partial x^2}{\partial x^{1'}} \qquad\dfrac{\partial u}{\partial \phi}\boldsymbol{=}\frac{\partial x^1}{\partial x^{2'}}\qquad\dfrac{\partial v}{\partial \phi}\boldsymbol{=}\frac{\partial x^2}{\partial x^{2'}} \tag{14c}\label{14c} \end{align}
Then equation \eqref{13} corresponds to the following \begin{equation} \vec{e}_{\alpha'}\boldsymbol{=}\frac{\partial x^\beta}{\partial x^{\alpha'}}\vec{e}_{\beta} \qquad \alpha'\boldsymbol{=}1',2' \qquad \beta\boldsymbol{=}1,2 \tag{15}\label{15} \end{equation}


Consider that $\mathbf{x}\boldsymbol{=}\mathbf{x}\left(x^1,x^2,\cdots,x^\beta,\cdots,x^n\right)\in \mathbb R^{n+1}$ is a real $(n+1)$-vector where $x^\beta\in \mathbb R$ are $n$ real parameters. Then $\mathbf{x}$ is a parametric representation of a $n$-dimensional $^{\prime\prime}$surface$^{\prime\prime}$(manifold) in $\mathbb R^{n+1}$.

At a point $\mathrm P$ we define the following basis of $n$-vectors \begin{equation} \vec{e}_{\beta}\boldsymbol{=}\dfrac{\partial \mathbf{x}}{\partial x^{\beta}}\qquad \beta\boldsymbol{=}1,2,3,\cdots,n \tag{16}\label{16} \end{equation} The vector $\vec{e}_{\beta}$ is tangent to the $x^{\beta}$-parametric curve with $x^{\rho}\boldsymbol{=}\texttt{constant},\rho\boldsymbol{\ne}\beta $. The linear $n$-dimensional space generated by this basis $\{\vec{e}_{1},\vec{e}_{2},\cdots,\vec{e}_{n}\}$ is the tangent space to the point $\mathrm P$.

Under a parametric transformation \begin{equation} x^{\alpha'}\boldsymbol{=}x^{\alpha'}\left(x^1,x^2,\cdots,x^\beta,\cdots,x^n\right)\qquad \alpha'\boldsymbol{=}1',2',3',\cdots,n' \tag{17}\label{17} \end{equation} we have a new parametric representation \begin{equation} \mathbf{x}'\left(x^{1'},x^{2'},\cdots,x^{\alpha'},\cdots,x{n'}\right)\boldsymbol{=}\mathbf{x}\left(x^1,x^2,\cdots,x^\beta,\cdots,x^n\right) \tag{18}\label{18} \end{equation} and so a new basis of tangent $n$-vectors \begin{equation} \vec{e}_{\alpha'}\boldsymbol{=}\dfrac{\partial \mathbf{x}'}{\partial x^{\alpha'}}\boldsymbol{=}\dfrac{\partial \mathbf{x}}{\partial x^{\beta}}\dfrac{\partial x^\beta}{\partial x^{\alpha'}}\boldsymbol{=}\dfrac{\partial x^\beta}{\partial x^{\alpha'}}\vec{e}_{\beta}\: \left(\alpha'\boldsymbol{=}1',2',3',\cdots,n'\:\Vert\:\beta\boldsymbol{=}1,2,3,\cdots,n\right) \tag{19}\label{19} \end{equation}

  • $\begingroup$ Is your answer incomplete? $\endgroup$ Commented Jan 29, 2021 at 3:19

I am not good enought in tensor analysis to answer the frist part of the question, and can only address the part (2) and (3).

There are two ways to construct basis vectors for a curved coordinate system, which is parameterized by $\xi_1$, $\xi_2$, and $\xi_3$. Each parameter is a familly of curve surface defined by function $ f_i(x,y,z) = \xi_i$.

The first definition of the basis vector is along the normal vector of the constant $\xi_i$ surface, i.e. the gradient vector: $$ \hat{e}_i = \frac{\vec{\nabla} f_i(x,y,z)}{|\vec{\nabla}f_i|} $$ The set of bases is perpendicular to the constant surface defined by $\xi_i$. This definition is very intuitive to all physicists, but indeed harder to calculate.

Another set of bases are tangential vectors. A basis vector $\hat{e}'_i$ lies in intersection of the constant surfaces of $\xi_j$'s for $j \ne i$. This is define as $$ \hat{e}'_i \parallel \vec{r}(\xi_i + d\xi_i) - \vec{r}(\xi_i), \text{ while kept other parameters } \{ \xi_j, j\ne i\} \text{ constant. } $$ $$ \hat{e}'_i = \frac{1}{h_i} \frac{\partial \vec{r}}{\partial \xi_i}; $$ where $ h_i = | \frac{\partial \vec{r}}{\partial \xi_i} | $. This is what you wrote in the post, where $\xi_i = r$ and $h_r = 1$.

For non-orthgonal system $\hat{e}_i \cdot \hat{e}_j$ is not zero (assuming $i \ne j$), but $\hat{e}_i \cdot \hat{e}'_j = 0$. For orthogonal system, the two set of bases are indentical $\hat{e}_i = \hat{e}'_i$.

  • $\begingroup$ The normal vector definition seems to be the definition of a basis one-form, instead of a basis vector, since gradients are one-forms. $\endgroup$ Commented Jan 22, 2021 at 16:34
  • $\begingroup$ @HexiangChang In curved coordiante, all basis vectors are functions of position. The tangential vectors are also a differential form, If you say that is "a form"? I might miss your meaning. $\endgroup$
    – ytlu
    Commented Jan 22, 2021 at 16:44
  • 1
    $\begingroup$ @HexiangChang They also use a norm, so there's a metric assumed over the space, hence there is a musical iso. between vectors and forms which can be understood there. ytlu, forms are essentially covectors and so have slightly different properties than vectors. $\endgroup$ Commented Jan 22, 2021 at 17:59

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