Why doesn't the Higgs produces a cosmological constant

BDOA
Messages
31
Reaction score
0
If the vacuum contains all these Higgs bosons, at expectation of <246> GeV in vacuum, why isn't there a cosmological constant, given the vacuum an energy density of 246 GeV^4 , instead we see dark energy at a few meV^4. Could just say that the graviton doesn't couple to the Higgs at all, but doesn't that violate the
equivalence principle. Do we need another tachyonic world, which has a tachyo-Higgs at -246 GeV to cancel
out the vacuum energy? Is there any other way to solve this mystery?
 
Physics news on Phys.org
It does, but nobody knows why there is a discrepancy between the predicted value of the comsological constant from the standard model of elementary particles and the experimentally observed value.

The discrepancy is much larger than what you indicated in your post, due to the nontrivial quantum field theory vacuum:

http://en.wikipedia.org/wiki/Vacuum_catastrophe
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top