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Context

In [1], cowlicks asks, ``How can the Gallilean transformations form a group?''. In [1] Selene Routely explains that the Galilean transformations form a group of dimension 10. Routely explains that any transformation "near enough" to the identity can be specified by exactly 10 independent parameters. In this answer, these 10 independent parameters are represented by $\theta\,\gamma_x$, $\theta\,\gamma_y$, $\theta\,\gamma_z$, $v_x$, $v_y$, $v_z$, $X$, $Y$, $Z$, and $\tau$. In [1], Routely explains that the Galilean group can be represented with a set of $5\times 5$ matrices as the group elements, with matrix multiplication as the group operation. Further, she explains that group's elements act on $5\times 1$ column vectors called the homogeneous co-ordinates. Finally, in [1] Routely writes down three three kinds of group elements: Galilean translations, Galilean boosts, and Galilean rotations.

In [2], I ask what type of transformations are we talking about when we talk about Lorentz transformations. The motivation for my question in [2] was to find out what the group is consistent with the postulates of special relativity and that is analogue to the Galilean transformations. In [2] robphy explains that there are two names that are relevant. These names the homogeneous Lorentz group, which is a group of dimension 6, and the inhomogeneous Lorentz group, which is also referred to as the Poincaré group, and which is a group of dimension 10.

Questions

Just as Routely gives the concrete representations of the Galilean translations, Galilean boosts, and Galilean rotations in [1], I want to know their analogues, respectively, in the realm of the Poincaré group.

(1) Translations: What is a concrete representation of those elements of the Poincaré group that describes the transformation between two frames of reference that are neither in relative motion nor rotated with respect to each other?

(2) Boosts: What is a concrete representation of those elements of the Poincaré group that describes the transformation between two frames of reference that are in relative motion but are not rotated with respect to each other?

(3) Rotations: What is a concrete representation of those elements of the Poincaré group that describes the transformation between two frames of reference that are rotated with respect to each other, but are not in relative motion?

My Answer

I have searched Wikipedia. I have found three wikis ([3], [4], and [5]) that are relevant, but none offer a specific representation. In 5, the authors explain that the Poincaré group "is a semidirect product of the translations and the Lorentz group, $$\mathbf{R}^{1,3} \rtimes \operatorname{O}(1, 3) .$$ This semi-direct product gives the structural properties of the group. That said, it does not offer the explicit representations that I am seeking.

With the help of lpz and ZeroTheHero, I gather that the Poincare group's elements might act on $5\times 1$ column vectors of the form:

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

In (1) $x,\,y,\,z $ are the three spatial co-ordinates, $t$ is the one and only time co-ordinate, and $c$ is the speed of light.

In the case of translations, where there are two frames of reference that are neither in relative motion nor rotated with respect to each other, it seems to me that the situation is exactly like the Galilean case. Thus, adapting from [1], the translations, $T$, are transformations between two frames that are not in relative motion and not rotated with respect to each other. The translations are represented by 5×5 matrices of the following form:

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

In the case of rotations, I gather that the situation is similar to pure rotations about an arbitrary axis [6]. Thus, for two frames rotated by a fixed angle $ \theta$ about an axis defined by the unit vector [7]: $\hat{\mathbf{u}} = \left(u_1, u_2, u_3\right), $ the rotations, $R$, are represented by $5\times 5$ matrices of the following form:

\begin{align} R( \theta\,u_1,\,\theta\,u_2,\,\theta\,u_3) &= \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0& 1 & -\exp(\theta\,u_3) & \exp(\theta\,u_2) \\ 0&0& \exp(\theta\,u_3) & 1 & -\exp(\theta\,u_1) \\ 0&0& -\exp(\theta\,u_2) & \exp(\theta\,u_1) & 1 \end{array} \right) \tag{3}. \end{align}

In the case of boosts, I gather that the situation is similar to Pure boosts in an arbitrary direction [6]. Thus, for two frames moving at constant relative three-velocity $\mathbf{v} = (v_1,v_2,v_3)$ the boosts, $\Lambda$, are represented by $5\times 5$ matrices of the following form:

\begin{align} \Lambda(v_1,v_2,v_3) &= \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0 & \gamma & -\gamma\frac{v_1}{c} & -\gamma\frac{v_2}{c} & -\gamma\frac{v_3}{c} \\ 0& -\gamma\,\frac{v_1}{c} & (\gamma - 1)\frac{v_1^2}{v^2} + 1 & (\gamma - 1)\frac{v_1\, v_2}{v^2} & (\gamma - 1)\frac{v_1\,v_3}{v^2} \\ 0 & -\gamma\,\frac{v_2}{c} & (\gamma - 1)\frac{v_1\, v_2}{v^2} & (\gamma - 1)\frac{v_2^2}{v^2} + 1 & (\gamma - 1)\frac{v_2\, v_3}{v^2} \\ 0 & -\gamma\,\frac{v_3}{c} & (\gamma - 1)\frac{v_1\, v_3}{v^2} & (\gamma - 1)\frac{v_2\, v_3}{v^2} & (\gamma - 1)\frac{v_3^2}{v^2} + 1 \end{array} \right) \tag{3}. \end{align}

Discussion

Just as the Galilean transformation form a group of dimension 10, so too does the the Poincaré transformation form a group of dimension 10. The same 10 parameters are required to represent the elements of the group. These are: $\tau$, $X$, $Y$, $Z$, $v_1$, $v_2$, $v_3$, $\theta\,u_1$, $\theta\,u_2$, $\theta\,u_3$.

Bibliography

[1] How can the Gallilean transformations form a group?

[2] What we talk about when we talk about Lorentz transformations?

[3] https://en.wikipedia.org/wiki/Lorentz_group

[4] https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group#Infinite-dimensional_representations

[5] https://en.wikipedia.org/wiki/Poincar%C3%A9_group#Poincar%C3%A9_algebra

[6] https://en.wikipedia.org/w/index.php?title=Four-vector&action=edit&section=5

[7] https://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula

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    $\begingroup$ It's a very interesting topic, but when asking questions here you should stick to a "one post one question" format. You should split this post into several (with links between them if necessary) to avoid having your question closed for lacking focus. Thank you. $\endgroup$
    – Miyase
    Commented Jul 19, 2022 at 7:54
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    $\begingroup$ A sentence like “ this product is not the explicit representations that I am seeking” makes no sense. The semi-direct product is a structural property: space time translations are Abelian, and transform into other translations under $SO(3,1)$. This will be baked into any representation (by matrices) of the group. $\endgroup$ Commented Jul 19, 2022 at 11:48
  • $\begingroup$ In fact the form of your (1) where you added an extra 5th column even if you have 4 spacetime coordinates) is one way of accommodating the translation part. $\endgroup$ Commented Jul 19, 2022 at 11:50

1 Answer 1

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In general, there are two classic ways to represent the Poincaré group. The first one comes from its definition. They are the "rigid" motions for a 4D Minkowski spacetime. Identifying space-time with $\mathbb R^4$ equipped with the metric: $$\eta = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{pmatrix} \tag{1} $$ you can view the Poincaré transformations as the affine transformation $f: x\in\mathbb R^4 \to\Lambda x+u$ with $u$ a constant fourvector and $\Lambda$ the analogue of a rotation matrix satisfying: $$ \Lambda \eta \Lambda^T = \eta $$ ie a linear transformation preserving the metric, more commonly known as a Lorentz transformation.

Translations are just the case when $u=0$ and Lorentz transformations about the origin when $\Lambda=I_4$. Within the Lorentz transformations, you typically distinguish rotations and boosts. This is done by fixing a time coordinate (note that the distinction is observer dependent). In our case, let's fix time to be the first coordinate $x^0$, with corresponding basis vector $e_0$.

Rotations are the subgroup of the Lorentz group that fix $e_0$ ie don't mess with time. You can show that you can write: $$ \Lambda = \begin{pmatrix} 1 & 0 \\ 0 & R \end{pmatrix} \tag{2} $$ with $R$ a rotation matrix ie a $3\times 3$ satisfying $RR^T = I_3$.

For boosts, you can't rigorously talk about a representation as they do not form a subgroup of the Lorentz group (this is the origin of the famous Thomas precession). This subset can be parametrised by a $3$ vector $v$ with corresponding Lorentz factor $\gamma = (1-v^Tv)^{-1/2}$ by: $$ \Lambda = \begin{pmatrix} \gamma & \gamma v^T \\ \gamma v & I_3+\frac{-1+\gamma}{v^T v}v v^T \end{pmatrix} \tag{3} $$

Note that there is another convenient way of representing the Lorentz group as a matrix subgroup. It is the mathematical trick of homogenising The idea is to add an extra dimension. You can view them as $5\times 5$ the matrices of the form: $$ \begin{pmatrix} 1 & 0 \\ u & \Lambda \end{pmatrix} \tag{4} $$ To prove the equivalence with the previous one, you can easily show it has the same action on the affine subspace $\begin{pmatrix}1 \\ x\end{pmatrix}$ with $x\in\mathbb R^4$. In a similar way, you can view the Galilean transform more explicitly as a degenerate Lorentz group by viewing the group of boosts and rotations as a matrix subgroup. In fact, this representation shows that it is in some sense a degenerate version of a bigger group, which leads to AdS and dS spaces.

Hope this helps

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  • $\begingroup$ Check out the edit for clarity, sorry, made some mistakes. For translations, the lower $4\times4$ block must be an identity matrix $\endgroup$
    – LPZ
    Commented Jul 19, 2022 at 9:01
  • $\begingroup$ I don't know if you are aware, by representation is a mathematically loaded term. By default, it's a group representation hence my remark. They are indeed a subset of the Poincaré group, and you can easily parametrised them as I have shown. $\endgroup$
    – LPZ
    Commented Jul 19, 2022 at 9:03
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    $\begingroup$ It should be clarified that boosts about different axes do not form a group, whereas boosts about the same axis do. $\endgroup$ Commented Jul 19, 2022 at 11:43
  • $\begingroup$ @lpz I hope you will allow me to add equation numbers to make reference to parts of your answer easier. $\endgroup$ Commented Jul 19, 2022 at 12:12
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    $\begingroup$ @MichaelLevy Eq.(3) is as compact as you can get for the Lorentz part. You can unpack this into a full $4\times 4$ matrix but the entries are not terribly intuitive. Using this (4) gives “the simplest” full irrep of Poincaré. See also physics.stackexchange.com/q/342504/36194 $\endgroup$ Commented Jul 19, 2022 at 12:15

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