smk22 Posted August 25, 2014 Share Posted August 25, 2014 my question is that if graviton is not yet discovered then how physicist define their spin? Link to comment Share on other sites More sharing options...
ajb Posted August 25, 2014 Share Posted August 25, 2014 my question is that if graviton is not yet discovered then how physicist define their spin? It comes from the basic nature of the field that describes the graviton and the fact that the source of gravity of the energy-momentum tensor which is symmetric second rank tensor. Link to comment Share on other sites More sharing options...
JonathanApps Posted September 11, 2014 Share Posted September 11, 2014 (edited) Spin is always derived from the tensor rank of the field, as ajb says. If it's a scalar field (rank 0), the spin is zero. If it's a spinor (electrons) then the spin is 1/2; I guess a spinor is kind of a 1/2-rank tensor. Electromagnetic fields are vectors (1st rank tensors) so the spin is 1. The derivation comes from the field Lagrangian being a scalar. There is a good rundown of the derivation in Mandl & Shaw "Quantum Field Theory", Chapter 2 I think. ...although the gravity field is the metric tensor (not the stress - energy tensor; the latter is the _source_ of the field, not the field itself). But anyway, the metric tensor is 2nd rank, so - Spin =2. Edited September 11, 2014 by JonathanApps Link to comment Share on other sites More sharing options...
Sensei Posted September 17, 2014 Share Posted September 17, 2014 Spin is always derived from the tensor rank of the field, as ajb says. You're kinda turning everything upside down. Spins are measured by device, for particles we are able to. http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment or Zeeman effect http://en.wikipedia.org/wiki/Zeeman_effect From this experimental knowledge there has been created theory and conservation rules. Basing on conservation/formulas we can predict what spin will have to have particle that we have not detected yet. I don't have to search database of isotopes to check particular one if it has even quantity of protons and even quantity of neutrons - it will have nuclear spin 0 for sure. 2 Link to comment Share on other sites More sharing options...
JonathanApps Posted September 17, 2014 Share Posted September 17, 2014 (edited) "You're kinda turning everything upside down." Ah well, that's (ex) theoretical physicists for you. Never overestimate an ex theoretical physicist. :-/ Certainly it ultimately comes from experiment - for electrons we can measure the spin and yes, I guess that lead Dirac to hypothesise a spinor field. And after googling it you can measure photon spin by EM interaction of atoms. However for gravity I don't think we've measured anything and all we can do is theorise a spin of 2 based on the 2nd rank field. Could be wrong I suppose. I don't know if I even believe gravity has quanta; I think it's totally different from the other "forces". But most people would disagree I suspect. Edited September 17, 2014 by JonathanApps Link to comment Share on other sites More sharing options...
ajb Posted September 18, 2014 Share Posted September 18, 2014 However for gravity I don't think we've measured anything and all we can do is theorise a spin of 2 based on the 2nd rank field. I should also add that from very general arguments we know that any spin-2, electrically neutral, massless particle we find must be the gravition. That is by just assuming these properties just about the only consistent action is the equivalent to the Einstein-Hilbert action. Thus, if something were to 'pop-up' in a collider experiment we can be confident it is the gravition. Link to comment Share on other sites More sharing options...
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