# How are tangent space basis vectors affected under a change of coordinates?

I'm studying 'Core Principles of Special and General Relativity' by Luscombe - the chapter on tensors.

Quoting:

Consider an arbitrary three-dimensional coordinate system where point $$P$$ is at the intersection of three coordinate curves labeled by $$(u,v,w)$$. For a nearby point $$Q$$ define the vector $$\Delta \mathbf{s}\equiv\vec{PQ}$$; $$\Delta \mathbf{s}$$ is also the vector $$\Delta \mathbf{s}=(\mathbf{r}+\Delta\mathbf{r})-\mathbf{r}$$, where $$\mathbf{r}+\Delta\mathbf{r}$$ and $$\mathbf{r}$$ are the position vectors for $$Q$$ and $$P$$ relative to the origin. To first order in small quantities,$$d\mathbf{s}=\frac{\partial\mathbf{r}}{\partial u}du+\frac{\partial\mathbf{r}}{\partial v}dv+\frac{\partial\mathbf{r}}{\partial w}dw$$ where the derivatives (with respect to the coordinates) are evaluated at $$P$$. The derivatives $$\mathbf{e}_u\equiv \frac{\partial \mathbf{r}}{\partial u}, \mathbf{e}_v\equiv \frac{\partial \mathbf{r}}{\partial v}, \mathbf{e}_w\equiv \frac{\partial \mathbf{r}}{\partial w}$$ form a local basis - an arbitrary $$d\mathbf{s}$$ in the neighborhood of $$P$$ can be expressed as a linear combination of them - and they're tangent to the coordinate curves.

The book goes on to talk about a switch to the spherical coordinate system, in which $$\mathbf{r}$$ is specified as: $$\mathbf{r}=r\sin\theta\cos\phi\ \mathbf{\hat x}+r\sin\theta\sin\phi\ \mathbf{\hat y}+r\cos\theta\ \mathbf{\hat z}$$

And from this we get the expressions: $$$$\begin{split}\mathbf{e}_r&=\frac{\partial \mathbf{r}}{\partial r}&=\sin\theta\cos\phi\ \mathbf{\hat x} + \sin\theta\sin\phi\ \mathbf{\hat y} + \cos\theta\ \mathbf{\hat z} \\\mathbf{e}_{\theta}&=\frac{\partial \mathbf{r}}{\partial \theta}&=r\cos\theta\cos\phi\ \mathbf{\hat x} + r\cos\theta\sin\phi\ \mathbf{\hat y} - r\sin\theta\ \mathbf{\hat z} \\\mathbf{e}_{\phi}&=\frac{\partial \mathbf{r}}{\partial \phi}&=-r\sin\theta\sin\phi\ \mathbf{\hat x} + r\sin\theta\cos\phi\ \mathbf{\hat y} \end{split}$$$$ Fair enough so far. I'm simultaneously reading 'A Visual Introduction to Differential Forms and Calculus on Manifolds' by Fortney. In that, a vector $$v_p$$ in the tangent space $$T_p(\mathbb{R}^3)$$ is defined as an operator acting on a real function $$f$$ defined on the manifold (i.e. $$f:\mathbb{R}^3\to\mathbb{R}$$). The operator gives the directional derivative of $$f$$ in the direction $$v_p$$ at point $$p$$:

$$v_p[f]=\sum_{i=1}^3v_i\frac{\partial f}{\partial x^i}\ \bigg|_p$$

From this, we can identify the basis vectors of $$T_p(\mathbb{R}^3)$$ as: $$\frac{\partial}{\partial x^1}\ \bigg|_p,\frac{\partial}{\partial x^2}\ \bigg|_p,\frac{\partial}{\partial x^3}\ \bigg|_p$$

which, as you can see, are quite different from the basis vectors $$\mathbf{e}_u,\mathbf{e}_v,\mathbf{e}_w$$ that I defined at the start of the question.

So now let's say I switch to spherical coordinate system and want to specify the tangent space basis vectors in the spherical coordinate representation. And I want to reconcile those basis vectors' spherical representation with the $$\mathbf{e}_r,\mathbf{e}_{\theta},\mathbf{e}_{\phi}$$ formulas from the Luscombe book that I listed at the beginning.

Now you've seen that the Luscombe book formulas for $$\mathbf{e}_r,\mathbf{e}_{\theta},\mathbf{e}_{\phi}$$, that I listed at the start, contain $$\mathbf{r}$$. My interpretation of $$\mathbf{r}$$ is that it's a function from $$\mathbb{R}^3\to\mathbb{R}$$, and it's a triple consisting of the three coordinate functions required to specify coordinates of any point $$p$$ in the manifold $$\mathbb{R}^3$$. We can specify $$\mathbf{r}=(x,y,z)$$, where $$$$\begin{split}x&=x(r,\theta,\phi)&=r\sin\theta\cos\phi \\y&=y(r,\theta,\phi)&=r\sin\theta\sin\phi \\z&=z(r,\theta,\phi)&=r\cos\theta \end{split}$$$$ are the individual coordinate functions. Since we've switched to spherical coordinates, we're now using the parameters $$r,\theta,\phi$$ to specify any point $$p$$ in the manifold $$\mathbb{R}^3$$.

Using the operator definition of tangent vectors, if I take $$f=\mathbf{r}=(x,y,z)$$ and $$v_p=(\mathbf{e}_r)_p\implies v_r=1,v_{\theta}=0,v_{\phi}=0$$, then $$$$\begin{split}(\mathbf{e}_r)_p[\mathbf{r}]&=v_r\frac{\partial \mathbf{r}}{\partial r}\ \bigg|_p+ v_{\theta}\frac{\partial \mathbf{r}}{\partial \theta}\ \bigg|_p+ v_{\phi}\frac{\partial \mathbf{r}}{\partial \phi}\ \bigg|_p=\frac{\partial \mathbf{r}}{\partial r}\ \bigg|_p \\&=\frac{\partial}{\partial r}(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)\ \bigg|_p \\&=(\sin\theta\cos\phi\,\sin\theta\sin\phi,\cos\theta)\ |_p \end{split}$$$$ which matches the definition mentioned in the Luscombe book. Similarly we can calculate for the $$\theta$$ and $$\phi$$ basis vectors.

Two questions:

1. Does the above procedure of reconciling the two definitions for basis vectors seem correct?

2. If correct, to reconcile the definitions, I had to make a specific assumption about $$f$$ and act the math book version of the basis vector (as an operator) on $$f=\mathbf{r}$$ in order to actually get the physics book version of the same basis vector.

Does that mean the physics book versions (formulas) of basis vectors are restrictive and will be incorrect in some scenarios? Or can I just take the physics book versions as the standard definition without worrying too much?

If you've read this far, thanks so much for the time and I'd appreciate any help!

• I'm not familiar with some terminology. What's the "parameterization function" in the context of my question? Do you mean $x(r, \theta, \phi)$ and so on? Commented Jun 17, 2020 at 1:21
• It's your $\mathbf r$. Commented Jun 17, 2020 at 1:22