Lorentz and Poincare Groups

Main Question or Discussion Point

I'm new here and I have checked the FAQs. I'm not sure if this question has been posted before.

This may actually be a silly question.

Why do we study Lorentz and Poincare Groups?

I have studied a bit of the theory but was wondering what exactly are we talking about when we study the theories, representations and generators of Lorentz and Poincare Groups.

Also, is there a way to represent Lorentz transformations, boosts and translations diagrammatically (explicitly - in the sense that one can look at the diagram and understand what's going on)?

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Dale
Mentor
Hi Shakthi, welcome to PF!

Why do we study Lorentz and Poincare Groups?
The main reason that we study them is because it appears that the Poincare group is the symmetry group of all the known fundamental laws of physics. The laws of physics are unchanged under certain transformations, those transformations are the transformations corresponding to the Poincare group.

Also, is there a way to represent Lorentz transformations, boosts and translations diagrammatically (explicitly - in the sense that one can look at the diagram and understand what's going on)?
Yes, they are called spacetime diagrams. Just label the vertical axis "t" and the horizontal axis "x" and use units where c=1. Probably the simplest way to start is simply to plot the lines of constant t' and x' as obtained from the standard Lorentz transform. Such a diagram will help you understand geometrically the concepts of time dilation, length contraction, and relativity of simultaneity.

Thank you very much.

Should I use the same method for representing translations too?

Dale
Mentor
You certainly could do that, but the diagram would not be any different from the "boost-only" diagram except for a re-labeling of the lines.

Thanks again.

Why do we use five-by-five matrices to represent the generators of Poincare Group while the generators of Lorentz Group are represented by 4-by-4 matrices?

Fredrik
Staff Emeritus
Gold Member
A Poincaré transformation is a map

$$x\mapsto\Lambda x+a$$

from ℝ4 into ℝ4, such that Λ is linear and satisfies

$$\Lambda^T\eta\Lambda=\eta$$

where

$$\eta=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$

This set is closed under composition of functions. In fact, it's easy to verify that

$$T(\bar\Lambda,\bar a)\circ T(\Lambda,a)=T(\bar\Lambda\Lambda,\bar\Lambda a+\bar a)$$

which implies that this set is a group with identity element $T(I,0)$ and inverse elements given by $T(\Lambda,a)^{-1}=T(\Lambda^{-1},-\Lambda^{-1}a)$. This is one of several ways to define the Poincaré group. As you can see, a typical member is identified by a 4×4 matrix $\Lambda$ and a 4×1 matrix $a$.

This group is isomorphic to the group of 5×5 matrices of the form

$$\begin{pmatrix}\Lambda & a \\ 0 & 1 \end{pmatrix}$$

where $\Lambda$ satisfies the same condition as before. I hope it's obvious enough what I mean by this notation. For example, the "0" should be interpreted as 0 0 0. So we can represent Poincaré transformations as 5×5 matrices, but it's more natural to represent them as pairs $(\Lambda,a)$. The isomorphism between these two groups is useful because it implies that theorems that hold for matrix Lie groups can be applied to the Poincaré group.

I suppose I should also mention that the Lorentz group is the subgroup with $a=0$, but you probably knew that already.

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Fredrik
Staff Emeritus
Gold Member
Why do we study Lorentz and Poincare Groups?
Physicists are interested in theories of motion. To describe motion mathematically, we need to choose a mathematical structure to represent space and time. The obvious choice is the set ℝ4. But we also need to be able to translate one observer's description of what's going on into another one's. It's natural to try to solve this problem for the "inertial" observers first. They are the ones that never accelerate, and never rotate. Each observer should describe the motion of any of the other inertial observer as a straight line in ℝ4, right? It seems at least reasonable to try the assumption that they will be able to do that.

This seemingly harmless assumption together with the very natural requirement that the set of functions that represent a coordinate change from one inertial observer's point of view to another's, is a group, is (along with some minor technical assumptions) enough to narrow down our options to two choices: Either the set of coordinate change functions is the Galilei group, or it's the Poincaré group. Non-relativistic classical mechanics and special relativistic classical mechanics are both defined by a choice of which one of these two groups to use.

If that isn't enough to explain why they're important, you should look into symmetries, in particular Noether's theorem in. Invariance of the laws of physics under translations in time imply that there's a conserved quantity, which we call "energy". Invariance under translations in space imply conservation of "momentum", and so on.

These groups play an even larger role in the corresponding quantum theories, because each particle species is associated with an irreducible representation. The generators of the appropriate one-parameter subgroups (e.g. translations in time) are identified with the measuring devices that measure some quantity (e.g. energy). (These identifications should be considered part of the axioms of quantum mechanics, even though they're never described that way in QM books).

Also, is there a way to represent Lorentz transformations, boosts and translations diagrammatically (explicitly - in the sense that one can look at the diagram and understand what's going on)?
As DaleSpam said, spacetime diagrams! A 1+1-dimensional (proper) Lorentz transformation (which is always a boost) can be expressed in the form

$$\gamma\begin{pmatrix}-v & 1\\ 1 & -v\end{pmatrix}$$

where

$$\gamma=\frac{1}{\sqrt{1-v^2}}$$

As you can verify by having this matrix act on $$\begin{pmatrix}1\\ 0\end{pmatrix}$$ and $$\begin{pmatrix}0\\ 1\end{pmatrix}$$, a boost tilts the "t" axis in one direction, and the "x" axis by the same angle in the opposite direction. This is (the main part of) what ensures that a line with slope 1 represents motion at the invariant speed.

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dextercioby
Homework Helper
Thanks again.

Why do we use five-by-five matrices to represent the generators of Poincare Group while the generators of Lorentz Group are represented by 4-by-4 matrices?
Group isomorphisms:

1. The set of Lorentz transformations together with composition form a group which is isomorphic to the group formed by the set of 4x4 matrices denoted generically $\Lambda$ subject to the condition

$$\Lambda^{T}\eta\Lambda = \eta$$,

where

$$\eta = \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$$

together with the matrix multiplication.

2. The set of Poincare transformations together with composition form a group which is isomorphic to the group formed by the set of 5x5 matrices denoted generically $\mbox{P}$ together with the matrix multiplication.

Since

$$\mathbb{R}^4 \simeq \left\{\begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \\ 1 \end{pmatrix} \left|\right x^{\mu}\in \mathbb{R} \right\}$$

as abelian groups (together with addition),

$\mbox{P}$ has the form

$$\mbox{P} = \begin{pmatrix} \left(\Lambda^{\mu}_{~\nu}\right)_{4\times 4} & \left(a^{\mu}\right)_{1\times 4} \\ 0_{4\times 1} & 1 \end{pmatrix}$$

with $\Lambda$ a Lorentz transformation and $a^{\mu} \in \mathbb{R}^{4}$.

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Fredrik
Staff Emeritus
Gold Member
For example, the "0" should be interpreted as 0 0 0.
Oops, I meant 0 0 0 0. I forgot that I was dealing with a 5×5 matrix for a while.

Thanks a lot friends

The information really helped me.

Thanks again.

One more doubt:

I read that the Lorentz group requires a 6-dimensional space. Does that mean it needs 6 real parameters to be specified?

Do those six parameters come from the 3 rotations and 3 boosts? How do the boosts contribute to these real parameters if they do not form a group themselves?

By extension, is the 6-D space sufficient for the Poincare Group too?

Fredrik
Staff Emeritus
Gold Member
I read that the Lorentz group requires a 6-dimensional space. Does that mean it needs 6 real parameters to be specified?
That's right. (Note however that parameters don't take arbitrary values. The fact that the maximum speed of a Lorentz transformation of 1 puts restrictions on the boost parameters, and you never need to rotate anything by an angle greater than 2π).

Do those six parameters come from the 3 rotations and 3 boosts?
Yes.

How do the boosts contribute to these real parameters if they do not form a group themselves?
Any Lorentz transformation can be expressed as a product of a boost matrix and a rotation matrix.

By extension, is the 6-D space sufficient for the Poincare Group too?
No. This should be apparent from the form T(Λ,a) discussed above. If you need to specify 6 real numbers to specify Λ, you're going to need 10 for T(Λ,a), since a is an arbitrary 4×1 matrix.

Sorry I forgot about the 4-vector of the Poincare Group.

I am still confused as to why we need a 6 dimensional space while we need just 4 co-ordinates to specify the Lorentz Transformation. Wouldn't a 4-D Space be enough?

Fredrik
Staff Emeritus
Gold Member
6 for the Lorentz transformations (a 3-component vector representing a velocity difference, and three Euler angles) and 4 for the translations, not the other way round.

6 for the Lorentz transformations (a 3-component vector representing a velocity difference, and three Euler angles) and 4 for the translations, not the other way round.
Now, I get it. It sounds similar to what we do in Small Oscillations in Classical Mechanics. The 6 dimensions here refer to the degrees of freedom and not the co-ordinates, right?

Fredrik
Staff Emeritus
Gold Member
Right. For example, the three boost parameters represent the velocity of one inertial observer relative to another. This post has some of the details:

Any 4×4 matrix $$\Lambda$$ can be written as

$$\Lambda=\gamma\begin{pmatrix}1 & \alpha^T \\ -v & \beta\end{pmatrix}$$

where $$v$$ and $$\alpha$$ are 3×1 matrices, $$\beta$$ is a 3×3 matrix and $$\gamma$$ is a number. The reason why we have taken the 00 component outside the matrix is that the v defined this way, i.e. the v defined by

$$v=-\frac{1}{\Lambda_{00}}\begin{pmatrix}\Lambda_{10}\\ \Lambda_{20}\\ \Lambda_{30}\end{pmatrix}$$

has a natural interpretation as a velocity. To understand why, note that all Lorentz transformations take straight lines through the origin to straight lines through the origin. In particular, the 0 axis is mapped to some straight line through the origin. We can find out which one by letting $$\Lambda$$ act on (1,0,0,0).

$$\Lambda\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix} =\begin{pmatrix}\Lambda_{00}\\ \Lambda_{10}\\ \Lambda_{20}\\ \Lambda_{30}\end{pmatrix} = \Lambda_{00}\begin{pmatrix}1\\ \Lambda_{10}/\Lambda_{00}\\ \Lambda_{20}/\Lambda_{00}\\ \Lambda_{30}/\Lambda_{00}\end{pmatrix} =\Lambda_{00}\begin{pmatrix}1\\ -v_1\\ -v_2\\ -v_3\end{pmatrix}$$

We see that the 0 axis is mapped to the line

$$\tau\mapsto\tau\begin{pmatrix}1\\ -v_1\\ -v_2\\ -v_3\end{pmatrix}$$

When $$\Lambda$$ represents a coordinate change from an inertial frame S to another inertial frame S', this line can be thought of as the world line of S in S', or if you prefer, as the world line of the physical observer who's using the coordinates S. We choose to call the velocity represented by this line -v, to make v the velocity of S' in S. We will refer to v as the velocity of $$\Lambda$$.

This brings up another question. what is meant by 'orthocronus'? I read that the first element of $$\Lambda$$ is of consequence when dividing the Lorentz Group into four different classes. And the restricted Lorentz Group, namely the Proper Orthrocronus Group is the one mapped to SL(2,C) to bring in Spinor notations.

Fredrik
Staff Emeritus
Gold Member
This brings up another question. what is meant by 'orthocronus'?
A Lorentz transformation $\Lambda$ is said to be orthochronous if $\Lambda_{00}\geq 1$. See this thread for more.

Also note that if you take the determinant of both sides of $\Lambda^T\eta\Lambda=\eta$, you see that $\det\Lambda=\pm 1$. The set of all Lorentz transformations with positive determinant is a subgroup called the "proper" Lorentz group.

Back again...

What exactly are Casimirs? Why do we need them and why are we studying them?

Dale
Mentor
You may want to start a new thread.

dextercioby
Homework Helper
It's ok in this thread as well. One can give examples for the restricted Poincare group. The wikipedia article seems pretty clearly written and I have not spotted any mistakes.

We use them because they help us write down a consistent theory of symmetries in quantum mechanics.

http://en.wikipedia.org/wiki/Casimir_operator#cite_note-0

Actually, I wanted to ask about Casimirs of the Poincare Group, the Lorentz Group and those of SUSY Algebra.

I am kind of lost in this topic. How do we find the Casimirs for a particular group (for the Poincare Group in particular)? Is there any restriction for the number of Casimirs for a Group? Are the Casimirs of a Group unique?

My doubts can be ridiculous but there are so many things I don't understand when it comes to High Energy Physics...

I'm back to square one here.

I was reading about Supersymmetry and basic Group Theory. I need some clarifications.

A proper and orthochronous Poincare transformation excludes space inversion and time inversion, right?

Can this group (proper and orthrochronous Poincare group) have generators that are 4-by-4 matrices?

In a book I read on Supersymmetry, the translations were mentioned to be four vectors. That's where I got confused. Is it possible to represent the Poincare generators as 4-by-4 matrices in this case?

Fredrik
Staff Emeritus
Gold Member
A proper and orthochronous Poincare transformation excludes space inversion and time inversion, right?
Right.

Can this group (proper and orthrochronous Poincare group) have generators that are 4-by-4 matrices?
You either consider the proper and orthochronous Lorentz group and make all your matrices 4×4, or the proper and orthochronous Poincaré group and make all your matrices 5×5.

In a book I read on Supersymmetry, the translations were mentioned to be four vectors. That's where I got confused. Is it possible to represent the Poincare generators as 4-by-4 matrices in this case?
No. Translations aren't linear maps. That's why we had to go from 4×4 to 5×5 to include them.