Petr Mugver said:
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?
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 F: \mathbb{R}^{4}\rightarrow M 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 x^{a} and \bar{x}^{a}, 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
\bar{x}^{a} = F^{a}(x) = A^{a}{}_{b}x^{b} + a^{a} \ \ (1)
(*) 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 P(x)\equiv P(\bar{x}) and Q(x + dx) \eqiv Q(\bar{x} + d\bar{x}) are two points lying on the world line of a photon. Then the requirement that two observers find the same speed of light, means
ds^{2} = \eta_{ab}dx^{a}dx^{b} = 0
if and only if
d\bar{s}^{2} = \eta_{ab}d\bar{x}^{a}d\bar{x}^{b} = 0
Or, in matrix form
ds^{2} = (dx)^{t} \eta (dx) = 0
if and only if
d\bar{s}^{2} = (dx)^{t} A^{t} \eta A (dx) = 0
Hence the two symmetric matrices \eta and A^{t}\eta A generate the same light cone
C = \{x^{a}| (x^{0})^{2} - (x^{1})^{2}- (x^{2})^{2} - (x^{3})^{2} = 0 \}
But then they must be proportional:
A^{t} \eta A = \alpha \eta
Or, if we do not want to use the linear transformation law, just
d\bar{s}^{2} = \alpha ds^{2}
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
\bar{x} = F_{21}(x) = Ax + a \ \ \ (1)
(2)speed of light is equal to 1. This led to the condition
A^{t} \eta A = \alpha \eta \ \ \ (2)
which attaches a proportionality factor \alpha to the above linear transformation.
SET B:
(1)space and time are homogeneous.
(2)Space is isotropic.
(3)Speed of light is 1;
d\bar{s}^{2} = \alpha ds^{2} \ \ \ (3)
We will now show that both sets independently imply \alpha = +1 and that eq(1) represents the most general \alpha = 1 solution of eq(3).
SET A:
From eq(2) it follows that
(\det A)^{2} = \alpha \ \ \ (4)
Thus, \alpha is strictly positive number.
The inverse transformation F_{12} is given by
x = F_{12}(\bar{x}) = A^{-1} \bar{x} - A^{-1}a
Thus F_{12} is characterized by the matrix A^{-1}. A simple manipulation of eq(2) gives us
(A^{-1})^{t} \eta (A^{-1}) = \frac{1}{\alpha}\eta
So the proportionality factor corresponding to F_{12} is \frac{1}{\alpha}. As the two systems are equivalent, we conclude that
\frac{1}{\alpha} = \alpha \ \ \Rightarrow \alpha^{2} = 1
and from eq(4) it follows that \alpha = +1.
SET B:
Notice that the relative velocity \vec{v}_{21} and coordinate x are the only variables available for \alpha to depend on. So, let us assume that \alpha = \alpha (x^{a}, \vec{v}_{21}).But because of the principles (1) and (2), \alpha can only be a function of the absolute value of the relative velocity, i.e.,
\alpha = \alpha ( |\vec{v}_{21}| ) = \alpha ( |\vec{v}_{12}| )
Since there is nothing special about the primed observer, it is also true that
ds^{2} = \alpha ( |\vec{v}_{12}| ) d\bar{s}^{2}
This, together with eq(3) of principle (3), leads to
\alpha ( |\vec{v}_{12}| ) = \pm 1 \ \ \forall |\vec{v}_{21}|
Before choosing one of the above values, observe that \alpha can not be equal to (+1) for some |v| and (-1) for other |v|, for this would means that there exists an |v| for which -1< \alpha < +1 which is impossible.
Since, for identical observers (the identity transformation) we have
d\bar{s}^{2} = ds^{2} \ \ (5)
with \alpha = +1. Therefore continuity implies that
\alpha ( |\vec{v}| ) = +1 \ \ \forall |\vec{v}|
Ok, let us now solve eq(5). For this purpose we consider an arbitrary infinitesimal coordinates transformation
\bar{x}^{a} = x^{a} + f^{a}(x)
Inserting this in eq(5) leads to the following homogeneous PDE
\partial^{a}f^{b} + \partial^{b}f^{a} = 0
This has the following general solution
f^{a}(x) = \omega^{a}{}_{c}x^{c} + \epsilon^{a}
where \omega^{ab} = - \omega^{ba} and \epsilon^{a} are the infinitesimal integration constants.
Form this we find our finite linear transformation
\bar{x} = \Lambda (\omega) x + a
where
\Lambda^{t}\eta \Lambda = \eta \ \ (6)
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 \Lambda uniquely! Indeed, if \lambda(\Lambda) is a function of \Lambda with the property \lambda^{2} = 1, then \Lambda \rightarrow \lambda(\Lambda) \Lambda leaves eq(6) invariant.
So, what does it mean for the observers to be related by linear transformation of the form \bar{x}^{a} = \lambda \Lambda^{a}{}_{b}x^{b} + a^{a}?
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 \lambda must always be 1. Indeed, it is easy to show that \lambda(\Lambda) = +1 for all proper orthochronous \Lambda.
**********
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
d\bar{s}^{2} = S(x) ds^{2}
i.e. locally scaling the metric by a positive scalar field S(x)?
We can show, see post #1 in
www.physicsforums.com/showthread.php?t=172461
that these transformation have the form
<br />
\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}<br />
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 c^{a} (the conformal parameters). Notice that the transformation relating two inertial observers must have c^{a} = 0 because otherwise (1) the transformation becomes singular along the surface 1 - 2c.x + c^{2}x^{2} = 0, and (2) inertial motions in the x-system appear to be accelerated in the \bar{x}-system.
Good luck
sam
