Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Finsler Geometry and Doubly Special Relativity

0 views
Skip to first unread message

eric alan forgy

unread,
Jul 24, 2002, 1:28:12 AM7/24/02
to

Hello,

Not that I am very knowledgeable in either of the two areas of the subject
line, but I can't help but to think these two things are related.

http://www.arxiv.org/abs/hep-ph/0009305
Can Neutrinos and High-Energy Particles Test Finsler Metric of Space-Time?
Authors: G.S.Asanov

http://arxiv.org/abs/gr-qc/0207085
Generalized Lorentz invariance with an invariant energy scale
Authors: Joao Magueijo, Lee Smolin

It is difficult to deny the eery similarities that are obvious even to me.
Hmm... does this mean that I now should go out and learn Finsler geometry?
:)

In Magueijo and Smolin, they have this dispersion relation

E^2 = p^2 + m^2 + lambda*E^3

which is motivated from results in LQG from

http://arxiv.org/abs/hep-th/0108061
Loop quantum gravity and light propagation
Authors: Jorge Alfaro, Hugo A. Morales-Tcotl, Luis F. Urrutia

It seems to me that a natural way to interpret the above dispersion
relation is via a modification of the inner product so that for a
4-momentum P = (p,E), instead of having

P.P
= p^2 - E^2
= -m^2

as usual, you would have

P.P
= p^2 - E^2 + lambda*E^3
= -m^2.

Therefore, the inner product is still a relativistic invariant however, it
is no longer quadratic. That is where Finsler geometry comes in. Finsler
geometry is (from my rudimentary understanding) a generalization of
Riemannian geometry to the case where the inner product is not a bilinear
form, but is rather a function F: TM x TM -> R satisfying some intuitive
relations very reminiscent of the usual quadratic metric tensor.

I think my brain is going to explode...

Eric

PS: The paper on Finsler geometry seems to suggest that the sign of lambda
depends on whether the particle is a particle or antiparticle (I think).

0 new messages