Understanding the Galilean transformation

[etotheipi]

I got a bit confused, and hoped someone could clarify a few things. As far as I am aware, a change of basis is an identity transformation $I_V$ on the vector space (pg. 113) and we can write the relationship between the components of some vector $v$ in the different bases $\beta$ and $\beta'$ in matrix form, like$$[\mathbf{v}]^{\beta'} = [I_V]_{\beta}^{\beta'} [\mathbf{v}]^{\beta}$$where the matrix representation of the identity transformation $[I_V]_{\beta}^{\beta'}$ is known as the change of basis matrix.

For a Galilean transformation $G$, between two given coordinate systems, with matrix representation $G(R, \mathbf{v}, \mathbf{a}, b)$ where $R$ is the rotation transformation, $\mathbf{v}$ is the relative velocity, $\mathbf{a}$ is a translation, $b$ is a time boost, we can write the matrix form of the transformation like $$(\mathbf{x}', t', 1)^T = \begin{pmatrix}
R & \mathbf{v} & \mathbf{a}\\
0 & 1 & b \\
0 & 0 & 1
\end{pmatrix} (\mathbf{x}, t, 1)^T
$$I had a few questions about this. Firstly, the elements of the vector space on which the Galilean transformations act look like vectors with coordinate matrices in the form $(\mathbf{x}, t, 1)^T$; what exactly are these vectors (e.g. is the vector space on which the Galilean transformation acts the space of 'events', or something?). Are $(\mathbf{x}', t', 1)^T$ and $(\mathbf{x}, t, 1)^T$ two different coordinate forms of the same vector in this underlying space, in which case the Galilean transformation is an identity linear transformation on this underlying space of events (i.e. similar in concept to the transformation $I_V$ in the example above, whose matrix representation is $[I_V]_{\beta}^{\beta'}$)?

If that's sort of along the right lines, does the same apply for the Lorentz transformations $\Lambda$ (i.e. would Lorentz transformations be identity linear transformations on the space of events?). That would again make sense, because a Lorentz transformation amounts to a re-labelling of the coordinates between inertial frames, but we're still referring to the same event at the end of the day...

Sorry if I made a mistake... thanks!

 

[vanhees71]

I think this linked manuscript is a bit confusing in its terminology but the math looks right. Also not that it is of utmost importance to keep not only the vertical but also the horizontal placement of the indices of the matrix correct. Otherwise it get's even more confusing. So in standard notation you have
$$v=v^k \vec{b}_k=v^{\prime j} \vec{b}_j'=v^k {I^j}_{k} \vec{b}_j' \; \Rightarrow \; v^{\prime j}={I^j}_k v^k.$$
The notation for the general Galilei transformation in terms of a five-dimensional column vector is just a trick to bring it into matrix-vector notation. Note that the last entry of these vectors is always $1$. Thus these columns don't form a vector space.

Minkowski spacetime forms a affine space and fixing an origin in the usual way let's you describe all points uniquely by vectors. For these four-vectors the Lorentz transformation is a transformation from one basis to another in the sense of linear algebra (with the additional property that it leaves the Minkowski product invariant).

 

[etotheipi]

The notation for the general Galilei transformation in terms of a five-dimensional column vector is just a trick to bring it into matrix-vector notation. Note that the last entry of these vectors is always $1$. Thus these columns don't form a vector space.

Minkowski spacetime forms a affine space and fixing an origin in the usual way let's you describe all points uniquely by vectors. For these four-vectors the Lorentz transformation is a transformation from one basis to another in the sense of linear algebra (with the additional property that it leaves the Minkowski product invariant).

Ah, that makes sense, thanks! So if I understand correctly, the Galilean transformations are not linear transformations and the vectors $(\mathbf{x}, t, 1)^T$ do not form a vector space. However the Lorentz transformations are identity linear transformations in the sense of linear algebra, and the $(x^0, x^1, x^2, x^3)^T$ do form a vector space.

 

[vanhees71]

Well, the matrices representing the Galilei transformation are of course matrices acting in the usual way on $\mathbb{R}^5$. They form a linear matrix representation of the Galilei group.

I don't know, what "identity linear transformations" are. That I find a very confusing terminology. In linear algebra you have a vector space $V$ with vectors $\vec{v}$. They do not depend in any way on a basis. You can, however, choose an arbitrary basis $\vec{b}_k$ and uniquely map $\vec{v}$ to $\mathbb{R}^n$ (if you have an $n$-dimensional real vector space), because the vector components $v^k$ wrt. this basis are by definition uniquely defined by
$$\vec{v}=v^k \vec{b}_k.$$
You can now express this same vector in terms of a different basis $\vec{b}_j'$, and you can express these new basis vectors in terms of the old ones and vice versa,
$$\vec{b}_j'={T^k}_j \vec{b}_k, \quad \vec{b}_k={U^j}_k \vec{b}_j'.$$
It's clear that the $n \times n$-matrices $\hat{T}=({T^k}_j)$ and $\hat{U}=({U^j}_k)$ are inverse to each other, i.e., $\hat{T} \hat{U}=\hat{1}=\hat{U} \hat{T}$. They are the matrices of "basis change", and they provide the components of any vector wrt. one basis in terms of the other. E.g.,
$$\vec{v}=v^k \vec{b}_k = v^k {U^j}_k \vec{b}_j'=v^{\prime j} \vec{b}_j' \; \Rightarrow \; v^{\prime j}={U^j}_k v^k.$$
In the same way you get
$$v^k={T^k}_j v^{\prime j}.$$
One says the basis vectors transform in the covariant, the vector components in the contravariant way.

Vectors (and tensors) do not depend on any basis. They are invariant objects. Physicists unfortunately call not only the $\vec{v}$ (invariant objects) but also the components $v^k$ wrt. a basis a vector. More carefully you call the vector components wrt. the basis, and it's clear that changing the basis implies a corresponding change of the vector components of the given vector.