What are change of frame and change of coordinates?

Question

What's the difference between a change of frame and a change of coordinates? I feel like both are transformations on the coordinates but change of frame changes also the vectors.

Answer 1

There's a difference because your coordinates do not always specify your frame. It's true that coordinates always give you one particular choice of a frame, but you can choose to use a different set of basis vectors to represent the vector space at any particular point.

The first example of this you'll probably run into is the difference between the coordinate basis and the orthonormal basis in spherical coordinates. They're based on the exact same coordinate system, but they're totally different frames. If $\mathbf{p}(r, \theta, \phi)$ is the position vector to the point with spherical coordinates $r, \theta, \phi$, then the coordinate basis vectors are defined as \begin{align} \mathbf{r} &= \frac{\partial \mathbf{p}} {\partial r} \\ \boldsymbol{\theta} &= \frac{\partial \mathbf{p}} {\partial \theta} \\ \boldsymbol{\phi} &= \frac{\partial \mathbf{p}} {\partial \phi}. \end{align} Now, remember that we can express the point in terms of the usual orthonormal Cartesian basis as \begin{equation} \mathbf{p} = r\sin\theta\cos\phi\, \hat{\mathbf{x}} + r\sin\theta\sin\phi\, \hat{\mathbf{y}} + r\cos\theta\, \hat{\mathbf{z}}, \end{equation} which we can differentiate to find \begin{align} \mathbf{r} &= \sin\theta\cos\phi\, \hat{\mathbf{x}} + \sin\theta\sin\phi\, \hat{\mathbf{y}} + \cos\theta\, \hat{\mathbf{z}} \\ \boldsymbol{\theta} &= r\cos\theta\cos\phi\, \hat{\mathbf{x}} + r\cos\theta\sin\phi\, \hat{\mathbf{y}} - r\sin\theta\, \hat{\mathbf{z}} \\ \boldsymbol{\phi} &= -r\sin\theta\sin\phi\, \hat{\mathbf{x}} + r\sin\theta\cos\phi\, \hat{\mathbf{y}}. \end{align} So already, we see that there are two different bases for the vector space at this point: one that uses the usual $(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})$ basis, and another that uses $(\mathbf{r}, \boldsymbol{\theta}, \boldsymbol{\phi})$. Any vector can be expanded in either basis, even though they correspond to the same point.

In fact, we can even find another basis based on spherical coordinates. Using the expressions above, it's a simple exercise to see that we have \begin{align} \mathbf{r} \cdot \mathbf{r} &= 1 \\ \boldsymbol{\theta} \cdot \boldsymbol{\theta} &= r^2 \\ \boldsymbol{\phi} \cdot \boldsymbol{\phi} &= r^2 \sin^2 \theta. \end{align} So this frame is not orthonormal. If you want an orthonormal basis, you need \begin{align} \mathbf{e}_r &= \mathbf{r} \\ \mathbf{e}_\theta &= \frac{1}{r}\boldsymbol{\theta} \\ \mathbf{e}_\phi &= \frac{1}{r\sin\theta} \boldsymbol{\phi}. \end{align} You'll see and use both of these frames. They are clearly different frames, even though they're based on exactly the same coordinates.

More generally, you're free to define your frame at any point in any way you want (as long as you keep the same number of vectors, and keep them linearly independent). So a change of coordinates implies a change in coordinate frame, but it is not equivalent to a change in frame generally.

Answer 2

It depends what you mean with 'change'. If 'change' is 'rate of change' i.e. the operator $\frac{d}{dt}$ acting on a vector $\vec v$, then the following is the case:

Any vector that is the element of a vector space can be split into its basis vectors $\vec e_i$ and its scalar components $v^i$, so that $\vec v = \sum_{i=1}^{n=3} v^i \vec e_i \equiv v^i \vec e_i$, where the last definition just uses the summation convention for indices.

If you now take $\frac{d}{dt}\vec v$ then according to the product rule you get $$\frac{d}{dt}\vec v = \left(\frac{d}{dt}v^i\right) \vec e_i + v^i\left(\frac{d}{dt}\vec e_i\right).$$ Now $\left(\frac{d}{dt}v^i\right)$ is what I would call a change of coordinates, and $\left(\frac{d}{dt}\vec e_i\right)$ a change of frame. The change of coordinates is just the change of the scalar numbers denoting some position in the given coordinate system $\vec e_i$. Should the coordinate system be accelerated, or, you just wish to change the 'grid' with which the $v^i$ are computed, then you need to change the basis vectors accordingly.

Mathematically there is no difference between a change of coordinates at a given time, or an accelerated coordinate system w.r.t time, they give you only different transformation Jacobians.

On another note, the change in basis vectors can impact our understanding of physics fundamentally. Simply changing from cartesian to spherical coordinates in Newtons equation of motion (at a given time) already introduces a term that behaves like a centrifugal force, although we only changed our 'viewpoint'.