It has been predicted by several background-independent approaches to Quantum Gravity (like LQG or spinfoams) that the physical dispersion relations in vacuum could take the following form:

$$ p^2 = E^2 \left( 1 + \xi \frac{E}{E_P} + {\cal O}\left( \frac{E^2}{E_P^2} \right) \right), $$

where $E$ is the observed energy of the massless particle, $p$ is the observed absolute value of the 3-momentum and $E_P = \text{const}$ is Planck's energy. The dimensionless coefficient $\xi$ is to be calculated by theory.

This anzatz is in apparent contradiction with Special Relativity, but actually this dispersion relations are a consequence of spacetime being discrete at the Planck scale. The idea is that it may be correct far beyond the domain of validity of SR. For observable energies $E \ll E_P$ we have $E = p$.

My question is: what are the most restrictive experimental bounds on the value of $\xi$? I've heard claims that tight bounds have been obtained from investigating the spectrum of distant astrophysical gamma-ray sources. Ideally, I want a reference to a peer-reviewed paper explicitly stating the experimental bound on $\xi$.

  • $\begingroup$ Could you elaborate why "this dispersion relations are a consequence of spacetime being discrete at the Planck scale"? Is it something analogous to the dispersion occurring for example in finite differences simulations of continuous equations? $\endgroup$ – user130529 Feb 24 '17 at 21:10
  • $\begingroup$ @claudechuber to my knowledge, the same anzatz for the dispersion relation arises in background-independent approaches to quantum gravity with discrete spacetime. Examples being canonical LQG with or without matter content, spinfoams with or without matter content. No, the discretness of spacetime isn't put by hand like in case of lattices. It is a prediction of the theory that spacetime is quantized (and thus behaves both continuously and discretely). It can no longer be approximated by Minkowski solution, thus the modification of dispersion relations. $\endgroup$ – Prof. Legolasov Feb 25 '17 at 13:18
  • $\begingroup$ @claudechuber The best analogy would be with quantum mechanics. Spacetime is hypothesized not to be discrete like a lattice, nor continuous like in GR, but it quantum, meaning completely different from what we encountered so far. The same goes for the photon, which as we know isn't a particle, nor is it a wave: it is described with a completely different language of quantum mechanics. However, this is probably unrelated to the question. I only answered because you asked me to. $\endgroup$ – Prof. Legolasov Feb 25 '17 at 13:21

The astrophysical gamma ray sources to which you are referring are gamma ray bursts, particularly the short bursts. These systems are ideal for testing Lorentz invariance because they can lead to the emission of high energy (tens of GeV) photons in a short interval of time, allowing for measurements of the potential delay in arrival of photons of different energies.

The strictest limit set on Lorentz invariance so far in the literature comes from Abdo et al 2009, in which they use the detection of a photon of approximately 30 GeV from the short GRB 090510. In your notation, they find

$$|\xi| < 0.82 \qquad $$

at about the 99% confidence limit. With less conservative assumptions, the limit drops even lower. To gain a more complete understanding of their analysis, you need to read the supplemental information, where they enumerate the various limits as they make their assumptions less and less conservative.

(Note that what they call $\xi_1$ in their paper is actually $1/\xi$ in your notation, and what they report is $\xi_1 > 1.22$).

  • $\begingroup$ Thanks! This really helps a lot! The bound of $|\xi| < 0.82$ doesn't appear very restrictive, however. $\endgroup$ – Prof. Legolasov Feb 25 '17 at 15:15
  • $\begingroup$ I'm glad it's helpful. I'm not qualified to answerhow restrictive this measurement is, though. My impression had been that the most popular variants of quantum gravity theory being tested had a for a while expected $\xi > 1$, but perhaps that is incorrect. $\endgroup$ – kleingordon Feb 25 '17 at 17:36
  • $\begingroup$ Hiya @kleingordon, can you please take a look at another related question of mine? physics.stackexchange.com/questions/412523/… - Cheers $\endgroup$ – Prof. Legolasov Jun 19 '18 at 13:16

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