New answers tagged

0

After a long discussion with "curiousone" I would like to like to share the relevant points of our discussion (hopefully I will do them justice) and some extra bits I added after thinking it over First Law of thermodynamics While the equation $$TdS = k_b T \ln N dN + dU -PdV $$ is quite general to any system where particle number is not conserved. We ...


0

John G. Cramer discussed G4V in a recent Analog Alternate View Column (Mar. 2016), and how Advanced LIGO data could possibly falsify G4V, General Relativity or even both of them (Their predicted gravity wave signatures signatures differ.) Cramer also stated that there would be no dark energy since G4v explains distant receding Type IIa supernova dimming as ...


0

I don't really have the background to understand such matters, but this presentation by Alan Guth gave me at least some idea of what it means to say the energy of a gravitational field is negative (which Guth calls a "miracle of physics"). Starting at 0:52:00 in that video clip, Guth presents this thought experiment... ...where it's taken as a given that ...


4

1) In view of the fact that we know how to measure the speed of light, it follows that a change in the speed of light would be detectable. 2) Any change in the speed of light would have to be accompanied by either a change in $\mu_0$, a change in $\epsilon_0$, or (far more drastically) a failure of Maxwell's equations, any of which would be easy to ...


5

I believe it can be useful to define the following concepts (I won't be very formal here for pedagogical reasons): Any event can be described through four real numbers, which we take to be: the moment in time it happens, and the position in space where it takes place. We call this four numbers the coordinates of the event. We collect these numbers in a ...


4

You're right. $\eta_{\mu\nu}\rightarrow \eta_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\,\mu^{'}}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\,\nu^{'}}$ just says that the metric transforms as a tensor as you would expect from its indices; there's nothing special about that. Being invariant means that when you make the transformation you get back the same matrix: ...


0

Trying to answer my old question myself: The cited piece is talking about an infrared cutoff, which basically amounts to giving the propagating particle a mass which appears squared in a propagator. So I think, what is meant is considering $$\lim_{\Lambda\to 0}\int \frac{d^4p}{p^2+\Lambda}$$ and this $\Lambda$ of course has a mass dimension of 2.



Top 50 recent answers are included