4
$\begingroup$

In a popular talk by Roger Penrose about spacetime geometry, when introducing his conformal cyclic cosmology starting at 17:15 I think he says that as soon as there are no massive particles left in the universe, the scaling of the spacetime metric loses its meaning, and we are left with a conformal geometry, which is scale invariant.

Did I interpret this correctly? If so, is it that all physics not involving massive particles is scale invariant? Is that so for all known physics, or for all possible physics (I suspect the latter, because right at 17:15 he says that scaling is essentially equivalent to how clocks measure time, and that without massive particles, there are no clocks).

Improve this question
$\endgroup$
  • 1
    $\begingroup$ This isn't a proper answer because I don't know a lot, but I believe that this is true in the classical theory but not in the quantum theory. I've heard that the Yang-Mills Lagrangian is scale invariant but the quantum theory is not. $\endgroup$ – Javier Jun 18 '17 at 23:49
  • 1
    $\begingroup$ @Javier Indeed, a classical field theory is scale invariant if there is no dimension-full (with dimension of mass) parameter. In a quantum theory however it is also necessary that the beta function vanishes. $\endgroup$ – Diracology Jul 15 '17 at 0:10
1
$\begingroup$

A general spacetime with no massive particles in it is certainly not conformally (or scale) invariant. For example, the Kerr black hole has no particles at all, and is not scale-invariant.

However, the specific CCC spacetime constructed by Penrose, which he is referring to, is conformally invariant in this case; see the Wikipedia article.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Certainly the black hole itself is considered to contain a non-zero mass, which is one of the fundamental properties of blackholes $\endgroup$ – R. Rankin Jun 19 '17 at 4:24
  • $\begingroup$ The spacetime has a mass scale. It's not a massive particle. The stress energy tensor is zero. And there are many, many other vacuum solutions which are not conformally invariant if you don't like that one. $\endgroup$ – AGML Jun 19 '17 at 4:25
  • $\begingroup$ It is not scale invariant, but maybe scale becomes unphysical, a gauge degree of freedom. For example null geodesics don't care about scale $\endgroup$ – doetoe Jun 19 '17 at 7:10
  • $\begingroup$ Null geodesics never care about scale. If you're referring to the Kerr black hole, scale is a physical asymmetry. A zoomed out black hole looks different from a zoomed in one. $\endgroup$ – AGML Jun 19 '17 at 13:30
  • $\begingroup$ No, I mean to say, what experiment can be done that doesn't involve massive particles to determine the scale? For example, the trajectories of massless particles are null geodesics, so that won't help. Without massive particles, there are no clocks. Does a zoomed out black hole really look different when there are no massive particles? Does scale have any physical consequence in a universe without massive particles? $\endgroup$ – doetoe Jun 19 '17 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.