# How can the Gallilean transformations form a group?

In class my professor said the Galilean transformations form a group of order 10. $$x'=x-vt\\ y'=y\\ z'=z\\ t'=t\\$$ But how do these form a group? I don't see 10 things to interpret as elements. I also don't see what the group operation would be. What are the elements of the Galilean group and what is its operation?

• Sets of transformations (if inversible) always form a group - they are the most canonical example of a group. Every group may be naturally interpreted as a set of transformations. In your 1-dimensional example, the transformations are labeled by 1 parameter $v$. The whole Galileo-group is 10-dimensional - you incorrectly reproduced what's the role of 10, it surely has more than 10 (infinitely many) elements. It's 10 dimensions parameterized as 3 components of $v$, 3 components of rotation $\Omega$, 3 components of translation in $\Delta x$ and 1 $\Delta t$. – Luboš Motl Dec 4 '14 at 6:18
• The whole transformation identified by these 10 numbers - how much you shift the velocity, how much you rotate around which axis, and how much you move in space and time - is an element of the group. The composition of the transformations is the group operation. You just transform the spacetime twice and you get another transformation that may be identified as one of those I have already described, and parameterized by those 10 numbers. There is also the identity element - transformation that keeps $t,x,y,z$ fixed - and inverse transformations. – Luboš Motl Dec 4 '14 at 6:20
• @LubošMotl that should probably be an answer – David Z Dec 4 '14 at 7:10
• @LubošMotl Minor comment re. "sets of transformations (if inversible) always form a group," just for OP for avoid confusion (this is not directed towards you -- I know you know this even better than I): strictly speaking the set of transformations must include the identity transformation (which you refer to in comment 2), and must close under composition of transformations (if composition is chosen as the binary group operation), so it's not quite always the case that an arbitrarily chosen set of transformations forms a group. – joshphysics Dec 4 '14 at 7:25
• This question shows no research effort, the Galilean group is easily found on Wikipedia and elsewhere. – ACuriousMind Dec 4 '14 at 13:40

Your prof is using slightly wrong words: the group is a Lie group of dimension 10, not order 10. A group's order is the number of its elements, which is here uncountably infinite. A group of order 10 is a finite group (and indeed there are only two possible groups with 10 elements).

I'm not sure how much continuous group theory (Lie group) theory you have done, but what "Lie group of dimension 10" means roughly is that any transformation "near enough" to the identity can be specified by exactly 10 independent parameters.

The easiest way to concretely form the group is as a matrix group of $5\times 5$ matrices, with, of course, matrix multiplication as the group operation. The group elements act on $5\times 1$ column vectors called homogeneous co-ordinates; they are of the form:

$$\left(\begin{array}{c}x\\y\\z\\t\\1\end{array}\right)\tag{1}$$

where $x,\,y,\,z,\,t$ are the spatial co-ordinates and time of a point in one co-ordinate system, to be transformed to their values in another inertial frame by one of our group members.

The first kind of group elements are of the following form:

$$T(X,\,Y,\,Z,\,\tau) = \left(\begin{array}{ccccc}1&0&0&0&X\\0&1&0&0&Y\\0&0&1&0&Z\\0&0&0&1&\tau\\0&0&0&0&1\end{array}\right)\tag{2}$$

and these are the translations (in space as well as time). Try multiplying a column vector of the form (1) on the left by one of the matrices in (2). You will find that it represents a translation in space and time as the values in (1) transform by $x\mapsto x+X,\,y\mapsto y+Y,\,z\mapsto z+Z,\,t\mapsto t+\tau$.

The second kind of group elements is the "boosts" (transformations between relatively moving frames):

$$B(v_x,\,v_y,\,v_z) = \left(\begin{array}{ccccc}1&0&0&v_x&0\\0&1&0&v_y&0\\0&0&1&v_z&0\\0&0&0&1&0\\0&0&0&0&1\end{array}\right)\tag{3}$$

(I put the name "boosts" in quotes because their nearest analogies are called this in the Lorentz and Poincaré groups and I suspect they are called something else here but I don't know what exactly; I'm rather ashamed to say that given that I claim Galileo as my most admired scientist of all time - maybe someone can help me out here).

The third kind of group members is the rotations:

$$R(\theta\,\gamma_x,\,\theta\,\gamma_y,\,\theta\,\gamma_z) = \left(\begin{array}{c|c}\exp\left(\theta\left(\begin{array}{ccc}0&-\gamma_z&\gamma_y\\\gamma_z&0&-\gamma_x\\-\gamma_y&\gamma_x&0\end{array}\right)\right)&\mathbf{0}_{3\times2}\\\hline\mathbf{0}_{2\times3}&\mathrm{id}_2\end{array}\right)\tag{4}$$

where $\mathrm{id}_2$ is the $2\times 2$ identity matrix, $\mathbf{0}_{3\times2}$ three rows of two columns of noughts and $\mathbf{0}_{2\times3}$ two rows of three columns of noughts. (4) represents a rotation of angle $\theta$ about an axis with direction cosines $\gamma_x,\,\gamma_y,\,\gamma_z$ (note that only three of these parameters, namely the products $\theta\,\gamma_x,\,\theta\,\gamma_y,\,\theta\,\gamma_z$.

Any Galilean transformation can be represented as a product of the form $T(X,\,Y,\,Z,\,\tau)\,B(v_x,\,v_y,\,v_z)\,R(\theta\,\gamma_x,\,\theta\,\gamma_y,\,\theta\,\gamma_z)$ and there are precisely ten independent real parameters needed to specify it - three for rotations, three for "boosts" and four for translations.

• I've heard them called "Galilean boosts" before. – Nathaniel Dec 4 '14 at 11:06