Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conformal group -> Poincaré group

  1. Aug 4, 2010 #1
    Hi. I want to talk about the derivation of the form of space-time transformations T between inertial references. As it's known, this is the Poincaré group, defined as the group that leaves invariant the (+,-,-,-) metric. This derives from two things: c is constant and space-time is homogeneous and isothropic. Now, I understand that the fact that c is constant defines the conformal group, that leaves the metric invariant up to a generic multiplicative number f(x,T) (that depends on the coordinates and, of course, on the transformation considered). But I don't understand why the homogeneity and isotropy implies f(x,T) = 1. Normally, by homogeneity I mean that if T(x) is in the group, then so does T(x) + a (a is a generic vector), and by isotropy I mean that also T(R(x)) (R is a generic rotation) is in the group. But this doesn't imply f = 1, as translations and rotations are already in the conformal group.

    Somebody can explain?
    What is the exact mathematical meaning of homogeneity and isotropy?
  2. jcsd
  3. Aug 4, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The whole business of "deriving" such results is a bit of a mess, because people always want to take non-mathematical statements like Einstein's postulates as the starting point. It obviously isn't possible to derive something mathematical from something non-mathematical, so the first thing you have to do is to choose a set of (mathematical) axioms. If you don't know exactly what axioms you need when you start, you will of course have to be prepared to modify your list as you go along, but it's essential that you have a list, because every theorem is a statement of the form "if A then B", so if you don't keep track of what your A is, you have nothing.

    You really just have to specify what you're looking for. For example, you can look for a connected topological group of smooth bijections on [tex]\mathbb R^4[/tex] that take straight lines to straight lines...and (insert something about isotropy and whatever else you need here).

    Isotropy can enter the picture in many ways, I think. This article assumes that the group you're looking for has the rotation group as a subgroup. That can be interpreted as a statement about isotropy. That's probably the most elegant way to do it.
  4. Aug 4, 2010 #3
    Thanks for the article, I will read it more carefully later, but at a first glance, it seems to me that he just assumes f = 1. If I assume it as well, then everything is nice, I get wat it seems to be the invariance group of (that kind of) physics, with rotations and translations, and I don't have to masturbate my mind too much.

    I intended something different. Is there a reason to select the f = 1 subgroup of the conformal group? You can say: yes, the reason is that you get the correct Poincaré group and you discard, for example, dilations, which don't mean anything in the framework of special relativity. That's good for me, but what has Homogeneity and Isotropy to do with all this?
  5. Aug 4, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your point of view on the relationship between physics and mathematics is idiosyncratic, and, IMO, completely wrong. Einstein derived mathematical results from physical axioms in his 1905 paper on SR.
  6. Aug 4, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You've said so before, but your arguments are empty IMO. Last time, you said that "derivation" means something different to physicists than it does to mathematicians. I agree, but I would say that this is because 99.99% of all physicists have never bothered to really think these things through. And just because Einstein did something in a certain way back in 1905 doesn't mean that we can't do better in 2010.
  7. Aug 4, 2010 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sorry about that. I thought he would cover that part in more detail. I haven't actually studied that part of the article very closely. I don't have time to really think about this now, but my first thought is that it can't have anything to do with homogeneity and isotropy. I think what you need to use is (a mathematical version of an aspect of) the principle of relativity, specifically the two observers will both predict that the other observer's clock is slow by the same factor.
  8. Aug 4, 2010 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In answer to the OP, I have a fairly straightforward proof that in 1+1 dimensions, the Lorentz transformation must preserve area. The proof is given in appendix 1 of this book http://www.lightandmatter.com/area1book6.html . The axioms that are required for the proof are axioms 1, 2, and 3 from the list given in section 1.2. (In this treatment, constancy of c is a theorem rather than an axiom.) If area is preserved, then that also rules out arbitrary dilations.

    Your OP outlines some reasons why it seems like dilations would not contradict the axioms you've listed. Therefore it seems like there must be an ingredient in the equal-area proof that you didn't anticipate. I think what's going on here is that your argument is talking about any group of transformations at all, with any kind of parameters. But the Lorentz boosts are labeled by a certain specific continuous parameter, v, which has a definite physical significance. In particular, v is a vector (a one-dimensional vector, in the simplified 1+1 case I discuss), and that means that the dependence of the Lorentz boosts on v is constrained by the axiom of homogeneity.
  9. Aug 4, 2010 #8


    User Avatar
    Science Advisor

    I don't know the answer, but what I have heard is that for massless fields, eg. Maxwell's equations, then the full symmetry is the conformal group, but for massive fields eg. the electron field, then the full symmetry if the Poincare group.
  10. Aug 29, 2010 #9


    User Avatar
    Science Advisor

    I am sorry for not keeping my promise to you! It was beyond my control.

    Ok, I could not understand what you were trying to do. So, I will try to explain some relevant issues involved here. Because I do not know your background in physics, I might end up re-explaining few things from different angles.

    The usual starting point is to consider a differentiable manifold M and use it as a mathematical model of our space-time. We then define what we mean by event, world line and inertial motion. In this model, an observer corresponds to a specific coordinate system [itex]F: \mathbb{R}^{4}\rightarrow M[/itex] covering the whole space-time manifold. To restrict the domain of application of our mathematical model, a set of general physical principles is needed. In special relativity, these are

    (1) space and time are homogeneous (i.e., nothing special about the points of M).
    (2) space is isotropic (i.e., no preferred direction).
    (3)={(1) + (2)} the choice of coordinates on M is a matter of convenience and we speak of equivalent inertial observers.

    From the mathematical point of view, the above two principles mean that the coordinates [itex]x^{a}[/itex] and [itex]\bar{x}^{a}[/itex], assigned by two sets of inertial observers to the same event, are related by a set of linear(*) invertable(**) functions, i.e., there exists a non-singular matrix A such that

    [tex]\bar{x}^{a} = F^{a}(x) = A^{a}{}_{b}x^{b} + a^{a} \ \ (1)[/tex]

    (*) Observe that the world line of a free particle is represented by a straight line in all inertial systems. Thus the function F(x) must map straight lines onto straight lines, and we conclude that it is a linear function.
    (**) Regardless of how the equivalence of the observers (frames) is realized in practice, the equivalence relation has the structure of a group. Thus the matrix A is non-singular.
    A part from being non-singular, the transformation matrices A are still arbitrary. Indeed, nothing we have said so far requires these matrices to have certain property. Thus another physical principle is needed:
    (4) the velocity of light signal is a constant independent of the relative motion of observers.
    This means the following: suppose [itex]P(x)\equiv P(\bar{x})[/itex] and [itex]Q(x + dx) \eqiv Q(\bar{x} + d\bar{x})[/itex] are two points lying on the world line of a photon. Then the requirement that two observers find the same speed of light, means

    [tex]ds^{2} = \eta_{ab}dx^{a}dx^{b} = 0[/tex]

    if and only if

    [tex]d\bar{s}^{2} = \eta_{ab}d\bar{x}^{a}d\bar{x}^{b} = 0[/tex]

    Or, in matrix form

    [tex]ds^{2} = (dx)^{t} \eta (dx) = 0[/tex]

    if and only if

    [tex]d\bar{s}^{2} = (dx)^{t} A^{t} \eta A (dx) = 0[/tex]

    Hence the two symmetric matrices [itex]\eta[/itex] and [itex]A^{t}\eta A[/itex] generate the same light cone

    [tex]C = \{x^{a}| (x^{0})^{2} - (x^{1})^{2}- (x^{2})^{2} - (x^{3})^{2} = 0 \}[/tex]

    But then they must be proportional:

    [tex]A^{t} \eta A = \alpha \eta [/tex]

    Or, if we do not want to use the linear transformation law, just

    [tex]d\bar{s}^{2} = \alpha ds^{2}[/tex]

    So, let us organise what we said so far by saying that the theory of special relativity can be based on two allegedly equivalent sets of principles:
    SET A:
    (1)two inertial observers are related by linear transformations

    [tex]\bar{x} = F_{21}(x) = Ax + a \ \ \ (1)[/tex]

    (2)speed of light is equal to 1. This led to the condition

    [tex]A^{t} \eta A = \alpha \eta \ \ \ (2)[/tex]
    which attaches a proportionality factor [itex]\alpha[/itex] to the above linear transformation.

    SET B:

    (1)space and time are homogeneous.
    (2)Space is isotropic.
    (3)Speed of light is 1;

    [tex]d\bar{s}^{2} = \alpha ds^{2} \ \ \ (3)[/tex]

    We will now show that both sets independently imply [itex]\alpha = +1[/itex] and that eq(1) represents the most general [itex]\alpha = 1[/itex] solution of eq(3).

    SET A:
    From eq(2) it follows that

    [tex](\det A)^{2} = \alpha \ \ \ (4)[/tex]

    Thus, [itex]\alpha[/itex] is strictly positive number.
    The inverse transformation [itex]F_{12}[/itex] is given by

    [tex] x = F_{12}(\bar{x}) = A^{-1} \bar{x} - A^{-1}a[/tex]

    Thus [itex]F_{12}[/itex] is characterized by the matrix [itex]A^{-1}[/itex]. A simple manipulation of eq(2) gives us

    [tex](A^{-1})^{t} \eta (A^{-1}) = \frac{1}{\alpha}\eta[/tex]

    So the proportionality factor corresponding to [itex]F_{12}[/itex] is [itex]\frac{1}{\alpha}[/itex]. As the two systems are equivalent, we conclude that

    [tex]\frac{1}{\alpha} = \alpha \ \ \Rightarrow \alpha^{2} = 1[/tex]

    and from eq(4) it follows that [itex]\alpha = +1[/itex].

    SET B:
    Notice that the relative velocity [itex]\vec{v}_{21}[/itex] and coordinate x are the only variables available for [itex]\alpha[/itex] to depend on. So, let us assume that [itex]\alpha = \alpha (x^{a}, \vec{v}_{21})[/itex].But because of the principles (1) and (2), [itex]\alpha[/itex] can only be a function of the absolute value of the relative velocity, i.e.,

    [tex]\alpha = \alpha ( |\vec{v}_{21}| ) = \alpha ( |\vec{v}_{12}| )[/tex]

    Since there is nothing special about the primed observer, it is also true that

    [tex]ds^{2} = \alpha ( |\vec{v}_{12}| ) d\bar{s}^{2}[/tex]

    This, together with eq(3) of principle (3), leads to

    [tex]\alpha ( |\vec{v}_{12}| ) = \pm 1 \ \ \forall |\vec{v}_{21}|[/tex]

    Before choosing one of the above values, observe that [itex]\alpha[/itex] can not be equal to (+1) for some [itex]|v|[/itex] and (-1) for other [itex]|v|[/itex], for this would means that there exists an [itex]|v|[/itex] for which [itex]-1< \alpha < +1[/itex] which is impossible.
    Since, for identical observers (the identity transformation) we have

    [tex]d\bar{s}^{2} = ds^{2} \ \ (5)[/tex]

    with [itex]\alpha = +1[/itex]. Therefore continuity implies that

    [tex]\alpha ( |\vec{v}| ) = +1 \ \ \forall |\vec{v}|[/tex]

    Ok, let us now solve eq(5). For this purpose we consider an arbitrary infinitesimal coordinates transformation

    [tex]\bar{x}^{a} = x^{a} + f^{a}(x)[/tex]

    Inserting this in eq(5) leads to the following homogeneous PDE

    [tex]\partial^{a}f^{b} + \partial^{b}f^{a} = 0[/tex]

    This has the following general solution

    [tex]f^{a}(x) = \omega^{a}{}_{c}x^{c} + \epsilon^{a}[/tex]

    where [itex]\omega^{ab} = - \omega^{ba}[/itex] and [itex]\epsilon^{a}[/itex] are the infinitesimal integration constants.
    Form this we find our finite linear transformation

    [tex]\bar{x} = \Lambda (\omega) x + a[/tex]


    [tex]\Lambda^{t}\eta \Lambda = \eta \ \ (6)[/tex]

    Before considering the conformal transformation, I would like to say one last thing about the meaning of the proportionality factor that can appear in the linear transformation law. Notice that eq(6) does not determine the matrix [itex]\Lambda[/itex] uniquely! Indeed, if [itex]\lambda(\Lambda)[/itex] is a function of [itex]\Lambda[/itex] with the property [itex]\lambda^{2} = 1[/itex], then [itex]\Lambda \rightarrow \lambda(\Lambda) \Lambda[/itex] leaves eq(6) invariant.
    So, what does it mean for the observers to be related by linear transformation of the form [itex]\bar{x}^{a} = \lambda \Lambda^{a}{}_{b}x^{b} + a^{a}[/itex]?
    It is a scale factor that arises because nothing we have said after eq(6) requires both observers to use the same scale of length. If we require that all observers choose the same standard of length, then we can argue that [itex]\lambda[/itex] must always be 1. Indeed, it is easy to show that [itex]\lambda(\Lambda) = +1[/itex] for all proper orthochronous [itex]\Lambda[/itex].


    In Minkowski space-time, conformal transformation is the only answer to the following question;
    What is the most general coordinates transformation that preserves the light-cone structure in the sense of

    [tex]d\bar{s}^{2} = S(x) ds^{2}[/tex]

    i.e. locally scaling the metric by a positive scalar field S(x)?
    We can show, see post #1 in
    that these transformation have the form

    \bar{x}^{a} = \lambda \Lambda^{a}{}_{b}\left( \frac{x^{b} - c^{b}x^{2}}{1 - 2c.x + c^{2}x^{2}}\right) + a^{a}
    The non-linear nature of these transformations represents the fact that space-time need not be necessarily homogeneous or isotropic. The absence of these properties is presented by the fixed vector [itex]c^{a}[/itex] (the conformal parameters). Notice that the transformation relating two inertial observers must have [itex]c^{a} = 0[/itex] because otherwise (1) the transformation becomes singular along the surface [itex]1 - 2c.x + c^{2}x^{2} = 0[/itex], and (2) inertial motions in the [itex]x[/itex]-system appear to be accelerated in the [itex]\bar{x}[/itex]-system.

    Good luck

  11. Aug 30, 2010 #10
    Man! Don't worry! After such a comprehensive explanation!

    I'm familiar with special and general relativity, the latter not being presented in a "too elegant" way, i.e. indices are welcome, but not equation involving [tex]\phi^*[/tex] maps, duals and biduals spaces, fibre bundle spaces etc, because when I encounter these objects I always choose a coordinate system and write them with indices anyway, so I'm much happier when the writer has done this work for me! Your notation was perfect.

    Regarding group theory, same story. I'm familiar with the "usual groups" (galilean, poincaré, SU(n),...) as they are typically applied to QFT, so no problem with representations, structure constants and central charges, but please don't come to me with para-compact spaces and theorems with weird mathematicians names!

    In quick words.


    Conformal group = set of invertible transformations T: unprimed -> primed that preserve speed of light:

    [tex]ds=0\Rightarrow ds'=0[/tex]

    which implies



    Poincaré group = set of T that preserve the interval:


    so it is the subgroup of the conformal group with f = 1.

    This implies that the jacobian of T is a Lorentz matrix independent of the point, so in particular it's a linear affine transformation that sends straight lines to straight lines.

    My question was how isotropy and homogeneity (physical requirements) mathematically select the f = 1 elements of the conformal group. I don't like "T sends lines to lines" because then why not "T sends parabolas to parabolas"? This seems to rely on newtonian mechanics of uniform motion.
    Last edited: Aug 30, 2010
  12. Aug 31, 2010 #11
    What do you call "conformal group? The 11-parameter one or 15-parameter one?
  13. Aug 31, 2010 #12
    The one such that ds = 0 => ds' = 0: 15 parameters.
  14. Aug 31, 2010 #13
    Then, you should rule out those that give rise to singularities. You do not want to have singularities, right? And denominators of special conformal transformations have zeros. What remains is 11-parameter group. On the other hand you can abandon [tex]R^4[/tex] as the arena and move to a homogeneous space of the full 15-parameter conformal group, where it acts without singularities. Your choice.
    Last edited: Aug 31, 2010
  15. Aug 31, 2010 #14


    User Avatar
    Science Advisor

    But this is exactly what it is! Parabola implies (and is implied by) the existence of force field. The presence of force field 1) defines a preferred direction,( i.e., space is no longer isotropic) and 2) assigns different physical properties to different points, (i.e., spacetime is no longer homogeneous).
    It is the absence of forces (i.e., the presence of linear motions) that gives spacetime its homogeneous and isotropic structures.
    So, as I said in the last post, to go from conformal transformation to Poincare’ transformation (by imposing homogeneity and isotropy), you need to get rid of accelerations in all coordinate systems. This means, you need to make your transformation linear by setting the (special) conformal parameters [itex]c^{a}[/tex] equal to zero.
    For the sake of completeness, I will now prove what I have stated (without proof) in the last post. That is, when we require all observers to use the same length of scale, then [itex]\lembda (\Lambda ) = 1[/itex] for all [itex]\Lambda[/itex].

    Putting [itex]c^{a}=0[/itex] in the conformal transformation gives

    [tex]\bar{x}^{a} = \lambda \Lambda^{a}{}_{b}x^{b} + a^{a}[/tex]

    Now consider the class of observers who agree on a single event to be the origin of their coordinates systems. Any two such observers are related by

    [tex]\bar{x} = \lambda \Lambda x[/tex]

    Observe that there cannot be two observers related to the x-observer by the same matrix [itex]\Lambda[/itex] but different [itex]\lambda[/itex]. Indeed, if

    [tex]\bar{x}=\lambda_{1}\Lambda x, \ \ \bar{y}=\lambda_{2}\Lambda x,[/tex]

    these two observers would be related to each other by

    [tex]\bar{x} = \frac{\lambda_{1}}{\lambda_{2}}\bar{y}[/tex]

    and would not agree on the standard of length. Thus we must have

    [tex]\lambda_{1} = \lambda_{2} = \lambda(\Lambda)[/tex]

    Furthermore, any other observer must obtain the same function [itex]\lambda(\Lambda)[/itex], since there is now nothing special about the x-observer.
    Next, by comparing the transformations

    \bar{x}= \lambda(\Lambda_{1}) \Lambda_{1}y, \ \ y = \lambda(\Lambda_{2})\Lambda_{2} x


    [tex]\bar{x} = \lambda(\Lambda_{1}\Lambda_{2}) \Lambda_{1}\Lambda_{2}x[/tex]
    we find that the function [itex]\lambda(\Lambda)[/itex] forms a representation of Lorentz group;

    [tex]\lambda(\Lambda_{1})\lambda(\Lambda_{2}) = \lambda(\Lambda_{1}\Lambda_{2})[/tex]

    But the only one dimensional representation of the proper orthochronous Lorentz group is the identity representation. Thus, as I claimed in the last post,

    [tex]\lambda(\Lambda) = 1, \ \ \ \forall \Lambda[/tex]

    I think I have covered almost everything.


Share this great discussion with others via Reddit, Google+, Twitter, or Facebook