Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 0 down vote favorite

The motivation is the following:

For each particle of mass m, we could write :

$- (E/m)^2 + (p_x/m)^2 + (p_y/m)^2 + (p_z/m)^2 = -1$,

which is nothing than a equation for a point in the hyperbolic space $\mathbb H3$, which can be seen, with Poincaré representation (Poincaré ball), as a (full) sphere (with hyperbolic distance)

For photons, we may give us a very tiny mass, so photons could be considered, in the limit where this tiny mass is zero, as living at the boundary of $\mathbb H3$, which is $\mathbb CP1$ (Riemann sphere $S^2$)

With these prescriptions, we cannot distinguish, photons with proportionnal momenta, but if the theory is momentum-scale invariant, it seems OK.

So could we use Hyperbolic Space $H3$ as representation space for momenta, in momentum-scale invariant theories?

share|improve this question

1 Answer

up vote 1 down vote

In English, unlike French, "impulsions" are known as "momenta".

Yes, massive particles' momentum 4-vectors belong to a $H^3$. On the other hand, massless particles' momentum 4-vectors don't belong to a boundary of $H^3$. Instead, they belong to a degenerate, singular version of an $H^3$ which is the same thing as $CP^1$ (semidirectly or fibered) times the scaling $R^+$. This statement means that the overall normalization of the photon's energy or momentum always matters. If two photons are moving in the same direction, to the same point of the $CP^1$, it clearly doesn't mean that they have the same momentum or energy, does it? An X-ray flying from Paris to Lyon is something else than a radio wave photon flying from Paris to Lyon.

In scale-invariant theories, particles have to be massless and physics of particles with one momentum 4-vector and its multiple is "physically equivalent" - related by a symmetry transformation. But it doesn't mean it's the same photon. The photons with differently scaled energies are still different and the ratio of energies of two photons, at least overlapping or nearby ones, may always differ from one and may be determined.

share|improve this answer
Thanks for language correction... But there is a subtelty between "physically equivalent" and "the photons are still different" that I don't understand. For instance, in calculating a scattering amplitude, if it is not scale-invariant in momenta, what does mean "physically equivalent" ? – Trimok Oct 9 '12 at 13:14
Dear Trimok, the scattering amplitude is scale-invariant, symmetric, which means that if you rescale all momenta $k$ times in it, you will get the same amplitude (up to some possible simple power-law rescaling of the amplitude by $k^\lambda$ where $\lambda$ is a simple exponent). But that doesn't mean that the actual photons don't care about the scaling of the energy. For example, if you rescale the (several) photons' momenta by different factors, you get a totally different amplitude. The symmetry only holds if you rescale everything. – Luboš Motl Oct 10 '12 at 5:06
Even if you do rescale everything - momenta of all photons - and you get the same amplitude up to a simple scaling of the amplitude, the states of the photon before and after rescaling are just not equal, much like an iPhone in Boston isn't equal to an iPhone as the same model iPhone in Marseille. – Luboš Motl Oct 10 '12 at 5:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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