By Claude Daviau, Jacques Bertrand

We expand to gravitation our past research of a quantum wave for all debris and antiparticles of every new release (electron + neutrino + u and d quarks for instance). This wave equation is shape invariant below Cl3*, then relativistic invariant. it's gauge invariant below the gauge staff of the traditional version, with a mass time period: this used to be very unlikely sooner than, and the end result was once an impossibility to hyperlink gauge interactions and gravitation.

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**Extra info for The Standard Model of Quantum Physics in Clifford Algebra**

**Example text**

Therefore Eq. 50) ∇φp σ1 σ21 + qAφp σ1 + m(−e−iβp )(−φp σ1 ) = 0. 51) −∇φp σ21 + qAφp + me−iβp φp = 0. 52) −φp ∇φp σ21 + qφp Aφp + mρp = 0. This means that only the diﬀerential part of the wave equation is changed. Instead of the system Eq. 10) to Eq. 55) 0=v +V , 0 3 0 = −w + V , 3 2 0 = −w , 0=w , 0 0=v . 60) page 46 August 28, 2015 13:33 Limit Theorem for Nonlinear-9586 The homogeneous nonlinear wave equation livre_CD_JB 47 So the charge conjugation does not really change the sign of the charge nor the sign of the mass, only the sign 3 of the diﬀerential term of the wave equation.

A reduced mass m = m0 c/ is proportional to the inverse of a space-time length, which is a frequency. This is exactly what E = hν says. Otherwise, the existence of Planck’s constant is linked to the fact that m and ρ are not separately invariant, but only their product is. Or again, the existence of Planck’s constant is linked to the invariance under the Cl3∗ group, greater than the invariance group of the restricted relativity. Somewhere we can say: the existence of Planck’s constant was not fully understood from the physical point of view.

The two Cliﬀord algebras are not identical. And yet there was until this work no known physical reason to prefer one to the other algebra. Remark 2: The even sub-algebra Cl1,3 + , formed by all the even elements N = s + b + ps, is 8-dimensional and is isomorphic to Cl3 . We shall see this in the next subsection by using the Dirac matrices. The privileged diﬀerential operator, in Cl1,3 , is: ∂ = γ μ ∂μ ; γ 0 = γ0 ; γ j = −γj , j = 1, 2, 3. 73) It satisﬁes: ∂∂ = = (∂0 )2 − (∂1 )2 − (∂2 )2 − (∂3 )2 .