Why don't we experience dilatations in Minkowski spacetime?

The question

This question follows on from the use of projective coords for spacetime in Notation for Translation Group Generators . Under Felix Klein's Erlangen Program, Minkowski spacetime starts as a 4-d projective space. It is then made into an affine space (which has a notion of parallelism) by picking an invariant hyperplane at infinity; lines in the projective space which meet at infinity are parallel. An affine space admits dilatations which are expansions or contractions of space about the eigenpoint of the dilatation. When introduced projectively[1], dilatations are seen to be generalized translations because a translation is a dilatation in which the eigenpoint is on the hyperplane at infinity. Since one can smoothly change a dilatation into a translation by moving the eigenpoint onto the hyperplane at infinity, it is hard to avoid the conclusion that a dilatation is an active transformation on the same footing as translations, rotations and boosts. Yet we don't experience dilatations in everyday life. Why is this?

Some background observations

The general feeling in physics seems to be that dilatations are changes in the units of measurement, but this view is only possible because dilatations are introduced in an ad hoc way as scale transformations and the link with translations is missed.

In order to see that dilations are active transformations like translations, the translation generators of Notation for Translation Group Generators become, $$[P_{\Gamma}]^{\Phi}_{\ \Psi}=\delta^{\Phi}_{\Gamma}\delta^{4}_{\Psi}$$ where indices $\Phi,\Psi,\Gamma=0,1,2,3,4$. The translations are $P_{0},P_{1},P_{2},P_{3}$ and the dilatation generator is $P_{4}$.

The attempt to construct classical relativistic Hamiltonian particle mechanics in Minkowski spacetime with dilatations fails unless the particles are massless. Massive particles only exist by removing the dilatations by hand in order to restrict the group to the ten parameter Poincare group.

Edit: An attempt to derive dilatations without having to read Coxeter [1].

Start with Minkowski spacetime as a 4-d projective space; points are rays in a 5-d vector space $x^{\Phi}\in V_{5}$ with upper case Greek indices like $\Phi=0,1,2,3,4$. The spacetime group is GL(5,R). Introduce parallelism by picking an invariant hyperplane $\delta^{4}_{\Phi}\in \tilde{V}_{5}$. Invariance means, $$[D(g^{-T})]_{\Phi}^{\ \Psi}\delta^{4}_{\Psi}=\xi \delta^{4}_{\Phi}$$ where $\xi$ is a scale factor which is unimportant because the physical entities are rays. This implies that the group matrices are now have the block form, $$$$D(g)= \left[ \begin{array}{cc} \Lambda & a \\ 0 & 1+a^{4} \end{array} \right]$$$$ where $\Lambda^{\mu}_{\ \lambda}$ is a 4x4 matrix and $a^{\mu}$ is a 4x1 vector and lower case Greek indices run $\mu=0,1,2,3$. These affine group matrices factor as, $$$$D(g)= \left[ \begin{array}{cc} 1 & a \\ 0 & 1+a^{4} \end{array} \right] \left[ \begin{array}{cc} \Lambda & 0 \\ 0 & 1 \end{array} \right] \ .$$$$ The first matrix is a dilation. In tensor notation, the pure dilatations are, $$[D(g)]^{\Phi}_{\ \Psi}=\delta^{\Phi}_{\ \Psi}+a^{\Phi}\delta^{4}_{\Psi}$$ and a point transforms under a dilatation as, $$x'^{\Phi}=[D(g)]^{\Phi}_{\ \Psi}x^{\Psi}=x^{\Phi}+a^{\Phi}x^{4} \ .$$ This shows that a dilatation acts like the identity on points in the hyperplane at infinity (the points which we can write as $x^{\mu}$ because $x^{4}=0$) and the eigenpoint of the dilatation is $a^{\Phi}$ with eigenvalue $1+a^{4}$. The last equation implies that a dilatation is an expansion or contraction of spacetime about the eigenpoint because a dilatation is a linear combination of the point and the eigenpoint.

Consider two spacetime points $x^{\Phi}$ and $y^{\Phi}$. Transform them to $x'^{\Phi}$ and $y'^{\Phi}$ under a dilatation with eigenpoint $a^{\Phi}$. The dilatation moves the line joining $x$ and $y$ parallel to itself so that the line $xy$ (in an informal notation) is parallel to line $x'y'$. To see this, the point where the line $xy$ intersects the hyperplane at infinity is, $$b^{\Phi}=\delta^{\Phi \Psi}_{[\Gamma\Delta]}x^{\Gamma}y^{\Delta}\delta^{4}_{\Psi} =x^{\Phi}y^{4}-x^{4}y^{\Phi}$$ and the dilatation acts like the identity on $b$ which is in the hyperplane at infinity, so line $x'y'$ also intersects the hyperplane at infinity at $b$ and so $xy$ and $x'y'$ are parallel lines. Now consider the quadrilateral $x'y'yx$. We have just seen that the sides $xy$ and $x'y'$ are parallel. The quadrilateral can be made into a parallelogram if the sides $xx'$ and $yy'$ are also parallel. These lines are parallel if they intersect at a point in the hyperplane at infinity. But, these lines both go through the eigenpoint $a$ of the dilatation. Therefore, $xx'$ and $yy'$ are parallel if the eigenpoint $a$ is moved onto the hyperplane at infinity. With the eigenpoint $a$ on the hyperplane at infinity, we've made a parallelogram by starting with a line $xy$ and moving it parallel to itself to $x'y'$ with $xx'$ and $yy'$ also parallel. This is how to construct a parallelogram with a translation. In other words, a translation is a dilatation with its eigenpoint on the hyperplane at infinity. This construction is why I cannot see a reason to exclude dilatations from the spacetime group, it seems dilatations must be just like rotations and boosts.

[1] Coxeter, "Introduction to Geometry", chapter 5.

• In order to see that dilations are active transformations like translations, As far as I know, you seem to be using the terms "active" and "passive" in a nonstandard way. The standard usage that I've seen is this: en.wikipedia.org/wiki/Active_and_passive_transformation By this definition, dilatations, translations, and Lorentz transformations can all be described either actively or passively. – Ben Crowell Aug 8 '14 at 22:52
• Another issue is that your question is phrased in terms of the properties of Minkowski space as a geometry at the most general levels of the hierarchy of the Erlangen program -- as an affine or projective geometry. These geometries don't have a metric. Minkowski space does have a metric, so there is a special role for transformations that preserve that metric. – Ben Crowell Aug 8 '14 at 23:03
• @ Ben Crowell 1: I may be using the adjective active in a non-wikipedian way so here is what I mean. The translation generators $P_{\mu}$ for $\mu=0,1,2,3$ can be used to move Alice in spacetime relative to Bob. This should also be the case for the dilatation generator $P_{4}$; Bob should see Alice physically expand or contract but we don't experience dilatations. – Stephen Blake Aug 9 '14 at 6:52
• @ Ben Crowell 2 : Under the Erlangen Program, one adds the notion of distance between points by making the hyperplane at infinity into a hyperbolic space by giving it an invariant polarity which is an invariant tensor $\eta_{\mu\lambda}=diag(1,-1,-1,-1)$. The spacetime transformations then factor into a product of a dilatation and a collineation of the hyperplane at infinity into itself. The collineations are the homogeneous Lorentz transformations. The dilatations act like the identity on the hyperplane at infinity. I left out this part because it doesn't affect the dilatations. – Stephen Blake Aug 9 '14 at 7:05