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I'm trying to understand how the translation of coordinate system works in physics, (for example in the Galilean transformations).

When I talk about vectors, I usually mean quantities that can be added or multiplied by a scalar in a manner that the axioms of vector space are verified. What makes me feel confused are the illustrations drawn in physics textbooks, like this one: translation of coordinate system pic

For example, if I have to model a situation where there are two different observers, each one using a different coordinate system, I think of each observer as an (ortonormal) basis (of $\mathbb{R}^3$), and then I express the position vector with respect the more convenient one for the description of the motion according to the change of basis matrix. In the picture above, the vectors of the "new" coordinate system $O'x'y'z'$ are given, by definition, by the transformation: $$(1)\quad \tau : r \mapsto r - r_0$$ But where I can find, in this example, the concept of change of basis? I'm confused when I find things like this in physics: $$(2)\quad \overbrace{r'}^{\text{"new" coordinate system}} = \overbrace{r-r_0}^{\text{"original" coordinate system}}$$: here we are saying that $\hat\imath' x' + \hat\jmath' y' + \hat k' z' = \hat\imath (x-x_0)+\hat\jmath (y-y_0)+\hat k (z-z_0)$, but in this case who are $\hat\imath'$,$\hat\jmath'$ and $\hat k'$? Are they determined by our transformation law $\tau$?

I hope someone could clarify me the general concept of position vectors and change of frame of reference (or coordinate system), or link me a good resource where this stuff is treated from the mathematician's viewpoint.

Edit: The thing that I don't get is how to build the transformation matrix: if (I'm talking about translations for abbreviating notation, but it can be surely generalized to include rotations) we know the law $T(\mathbf{v}) = \mathbf{v}' = \mathbf{R}\mathbf{v}+\mathbf{t}$, then we can, calling $\mathbf{e}^i$ the $i$-th vector of the original basis, write $T(\mathbf{e}^i) = \mathbf{e}^i + \mathbf{t} = \sum_{i=1}^n{\alpha_{i}' {\mathbf{e}^{i}}'}$, where $\{{\mathbf{e}^i}'\}_{i=1,\dots,n}$ are the vectors of our "translated" basis.

If $\mathbf{t} = {^t({\alpha_j}_0)}_{\{\mathbf{e}^i\}}$ if I take $T(\mathbf{e}^i)$ I get $\mathbf{e}^i + \mathbf{t} = {^t({\alpha_j}_0 + \delta_{i,j})}_{\{\mathbf{e}^i\}} = {^t({\alpha_1}_0,\dots,{\alpha_i}_0+1, \dots, {\alpha_n}_0)}_{\{\mathbf{e}^i\}}$, for $i=1,\dots,n$. But how do I interpret this result in term of space transformation? It is still related to the original frame of reference, and doesn't tell me nothing about the $\{{\mathbf{e}^i}'\}$.

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  • $\begingroup$ From your diagram it looks like your transformation is more than just a translation. It looks like it also involves a rotation. But you have written the transformation as just a translation. Are you just asking about translations, or any general transformation that could involve rotations or even "stretching/compression" of the axes? $\endgroup$ – Aaron Stevens Aug 10 '18 at 12:48
  • $\begingroup$ I'm asking about a general transformation, that involves also rotation. Here $\tau$ is a translation, but i could have written a more general (linear) transformation involving also a rotation matrix. In $(2)$ for example, the "new" basis vectors could be an ortonormal triple, roto-translated with respect to the original one. $\endgroup$ – user178093 Aug 10 '18 at 13:00
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I hope someone could clarify me the general concept of position vectors and change of frame of reference (or coordinate system), or link me a good resource where this stuff is treated from the mathematician's viewpoint.

First off, a better name for a position vector is a displacement vector. Displacement vectors are not free vectors. They aren't quite vectors, period, in the sense of the mathematical concept of a vector space. They instead are members of an affine space, with the transformation between two affine spaces given by an affine transformation $\boldsymbol x' = \boldsymbol {\mathrm M} \boldsymbol x + \boldsymbol b$, where $\boldsymbol {\mathrm M}$ is an invertible matrix (or a proper orthogonal matrix if you want to keep things simple) and $\boldsymbol b$ is the displacement vector from one origin to another.

You asked for a good resource where this stuff is treated from a mathematician's point of view. The sister site, Mathemetics StackExchange has a good number of questions and answers on the topics of affine spaces and affine transformations.

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A change of basis means simply a transformation of the way you represent your vectors. In 3D space, all vectors will be represented usually as a linear combination of three 'axes', aka basis vectors: $\hat{i}$, $\hat{j}$ and $\hat{k}$ in your example. It is also important however to consider where the origin of the vectors is fixed, namely, the zero of the vector space, $O$.

Now, any change in either $\hat{i}$, $\hat{j}$, $\hat{k}$ or $O$ constitutes a change of basis. A change in $O$ is a translation; the direction of the basis vectors doesn't change, they're just moved around in space. A change in the other three is usually either a rotation or a reflection (that is, if you want to keep using an orthogonal basis, otherwise other transformations, like skewing, are possible).

The thing you're presenting seems to me like a translation of $O$ by $r_0$, with the other three remaining unchanged. Otherwise you would have a more complex expression, with the new $\hat{i}'$, $\hat{j}'$ and $\hat{k}'$ being themselves expressed as linear combinations of $\hat{i}$, $\hat{j}$, $\hat{k}$. Usually for these purposes the most convenient thing is to use a rotation matrix; the general way to write a change of basis in vector form would be:

$$ \mathbf{v}' = \mathbf{R}\mathbf{v}+\mathbf{t} $$

where $\mathbf{R}$ is the rotation matrix, $\mathbf{t}$ the translation vector, and $\mathbf{v}$ and $\mathbf{v}'$ the vector before and after the transformation.

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  • $\begingroup$ Thank you for the answer! I'll edit my question to make more clear what I'm asking, taking into account the transformation law you wrote here. $\endgroup$ – user178093 Aug 10 '18 at 14:19
  • $\begingroup$ One thing: the versors i, j and k don't translate when transformed. They don't express a specific vector, just a 'direction', and translating them makes no sense. They are merely rotated. Check out this link for how rotation matrices are constructed, and what they do represent physically: en.wikipedia.org/wiki/Rotation_matrix. $\endgroup$ – Okarin Aug 10 '18 at 14:23
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The answer by Okarin is very good (gave it a +1 myself). I just wanted to add a distinction between a change of coordinates like the one you have presented and a change in basis like one you see in quantum mechanics (in case you ever get into QM).

In the example you have posted, you are moving from one place to another in space. You might be viewing your vectors differently in this new location, but the new and old coordinate systems are still located in space.

Changing basis, however, can be more abstract. In quantum mechanics our system can be described by a state vector $| \Psi \rangle$ that we can express in different bases. For example, we can express $| \Psi \rangle$ in terms of eigenstates of the Hamiltonian (states with definite energy): $$\hat H |\psi_n \rangle=E_n |\psi _n\rangle$$ $$|\Psi \rangle=\sum_n E_n|\psi_n \rangle$$

Or we can express the same state vector in terms of states with definite position. This is usually what students first coming into QM see as the wavefunction: $$\langle x|\Psi\rangle =\psi(x)$$

i.e., each component of the state vector in position space can be determines the wavefunction.

The similarity between your change of coordinates and this more abstract change of basis is that we are seeing our vectors from "different points of view" and expressing them in a way to match that view. The difference is that you can't explain the abstract change of basis in terms of just moving/rotating/stretching/etc. the original coordinate system.

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  • $\begingroup$ I'm not an expert in abstract algebra, but I think that would be true of basically any group closed with respect to some scalar product. Vectors just happen to be a very easy one to visualise! $\endgroup$ – Okarin Aug 10 '18 at 14:11
  • $\begingroup$ @Okarin Yes I agree! And it seems like the OP really is just interested in things like Galilean transformations. I just saw basis and thought of QM. So I thought it would be nice to talk about. $\endgroup$ – Aaron Stevens Aug 10 '18 at 15:46

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