Maintenant, on arrive à elle, j'étais sûr que ce nouveau développement de la technologie serait une catastrophe écologique d'une certaine sorte, combien de colibri de poussée de 80 tonnes à Saturne? Quant à la (soi-disant faux ...) proposition plus tard, vous pensez que les licornes ne sont pas suffisamment rare pour prendre un péage sur la diversité de leur population?
Lei capisce ?
Sounds a mite gangsta.
Yes, but are his papers publicly available and if so where ?Sorry, didn't mean to sound picky.OK. So there's a good chance I still have the '64 Sciama here somewhere, at least.
Interesting, note the 50KHz vs couple Gig. Capacitive effect of dielectric surface to microwaves, check.Virtual proton/positron creation is a strong function of photon energy (Hz).
The Mach effect... was derived from a simplified form of general relativity
1) Woodward's derivation uses a flat Minkowski space. In that sense he does not use Einstein's General Relativity. He uses Special Relativity (and says so in a number of places). There is no curvature of space in Woodward's derivation. There is no covariant, contravariant or mixed tensors. No Riemannian geometry.Gravity is given "ab initio" (unlike Einstein's General Relativity where gravity is a result of curving of space by massive objects), yet goes on to postulate transient mass effects due in most part to most distant objects. The justification appears to be isotropy of spacetime and local flatness of spacetime.2) He uses the [rest energy/volume] relationship to [rest density] Eo=rho c^2 sometimes here and sometimes there.3) I have not found Woodward's transient mass effect equations in any paper by Sciama. He apparently uses some results from Sciama's 1953 paper and goes on from there.
"Pair production" can only occur if the photons have an energy exceeding twice the rest energy of an electron (0.511 MeV rest energy, which doubled is --->1.022 MeV).
Virtual pairs are never produced. The question is can they carry momentum ?
Quote from: Notsosureofit on 09/22/2014 02:25 am"Pair production" can only occur if the photons have an energy exceeding twice the rest energy of an electron (0.511 MeV rest energy, which doubled is --->1.022 MeV). As I understand, that is correct but I also understand that this pair is a real electron and a real antielectron (positron). And when they self annihilate they produce real energy. Not what we are seeing. QuoteVirtual pairs are never produced. The question is can they carry momentum ?Not produced in any way that we know of. They appear from the quantum vacuum and disappear into it. I believe they do so leaving no trace of their passing. The fly in that ointment is that the electron is real as I understand it while the positron is not but to reach that conclusion I go back to the 1930's as the theory was developed. Point is, virtual electron/positron pairs do not leave an energy trace when they annihilate. Hence virtual, but the electron is real else why bother developing the theory in the first place?
Identifying a positron as a backwards in time electron, is an elegant interpretation that exhibits in the Feynman diagrams the CPT symmetry they must obey.What I am saying is : the statement: "positrons are backward going electrons" is a convenient and accurate mathematical representation for calculation purposes. "As if". There has not been an indication, not even a tiny one, that in nature as we study it experimentally anything goes backwards in time, as we define time in the laboratory .
...JohnFornato...
Quote from: Rodal on 09/22/2014 01:10 pm...JohnFornato...You say Fornato. I say Fornaro.But hey.
Quote from: JohnFornaro on 09/21/2014 07:37 pmQuote from: Rodal on 09/21/2014 06:23 pmThere is no distinction between the cavity's interior and the (copper ?) metal wall in the COMSOL finite element analysis display of the Electric Field, so my reading of this is that the metal wall was modeled as a Boundary Condition for the field.2. If the [metal walls are modelled as a] boundary condition as Rodal suggests, then what difference does it make what you make the truncated conical frusturm thingy out of?....6. The usual rejoinder.It makes just as much difference as for example when you model the end supports of a beam made with material modulus Ea supported inside another material with modulus Eb at both ends:blah, blah, blah...Lei capisce ?It's all in how one models the Boundary Conditions. Maxwell's equations are differential equations, and to solve them one needs to satisfy Boundary Conditions, just as when one solves a beam equation.
Quote from: Rodal on 09/21/2014 06:23 pmThere is no distinction between the cavity's interior and the (copper ?) metal wall in the COMSOL finite element analysis display of the Electric Field, so my reading of this is that the metal wall was modeled as a Boundary Condition for the field.2. If the [metal walls are modelled as a] boundary condition as Rodal suggests, then what difference does it make what you make the truncated conical frusturm thingy out of?....6. The usual rejoinder.
There is no distinction between the cavity's interior and the (copper ?) metal wall in the COMSOL finite element analysis display of the Electric Field, so my reading of this is that the metal wall was modeled as a Boundary Condition for the field.
It's fixed now
Quote from: Rodal on 09/21/2014 07:49 pmQuote from: JohnFornaro on 09/21/2014 07:37 pmQuote from: Rodal on 09/21/2014 06:23 pmThere is no distinction between the cavity's interior and the (copper ?) metal wall in the COMSOL finite element analysis display of the Electric Field, so my reading of this is that the metal wall was modeled as a Boundary Condition for the field.2. If the [metal walls are modelled as a] boundary condition as Rodal suggests, then what difference does it make what you make the truncated conical frusturm thingy out of?....6. The usual rejoinder.It makes just as much difference as for example when you model the end supports of a beam made with material modulus Ea supported inside another material with modulus Eb at both ends:blah, blah, blah...Lei capisce ?It's all in how one models the Boundary Conditions. Maxwell's equations are differential equations, and to solve them one needs to satisfy Boundary Conditions, just as when one solves a beam equation.That's what I'm getting at, my gangsta godfadda. What I took the above conversation to mean was that the results (all those Roy G. Biv color schemes) cannot be well understood without knowing the boundary conditions. Since there was "no distinction" between the cavity interor and the metal wall, you assume (which is what your "reading" of that is) that there must be a boundary condition defined somewhere. That boundary condition is not specified, and would differ significantly on whether the boundary's various modulii were based on copper or compressed hummingbird wings.Ergo, (ipso fatso being my preferred translation, BTW) there are limited conclusions which can be drawn from the FEA display of the electric field.Non?
Quote from: Rodal on 09/22/2014 01:58 pm It's fixed nowWhat??? You changed the Akashic record?
I would just recommend the book. If Dr. Rodal wants to be placed on Woodward's general reading list where he can have a dialog on this issue, have him send me a note to this effect with a couple sentences of his background and interest and I'll forward this to Jim.