Re: What if the higgs don't exist....
From: John C. Polasek (jpolasek_at_cfl.rr.com)
Date: 11/18/04
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Date: Thu, 18 Nov 2004 17:03:28 GMT
On Wed, 17 Nov 2004 23:32:52 -0800, "FrediFizzx"
<fredifizzx@hotmail.com> wrote:
>"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
>news:2crmp0dpu79d1st1iob74lnc6sr37g30b3@4ax.com...
>| On Tue, 16 Nov 2004 21:35:09 -0800, "FrediFizzx"
>| <fredifizzx@hotmail.com> wrote:
>|
trimmed for trimming's sake
>| >| >a
>| >| >| >valid approach where Q = sqrt(hbar*c) = vacuum charge.
>| >| >| >
>| >| >| >FrediFizzx
>| >| >| Fred: Please explain the units for vacuum charge.
>| >| >
>| >| >The units for vacuum charge is always charge. In natural units of
>hbar =
>| >c
>| >| >= 1, it is dimensionless. In cgs, it is (sqrt(mass*length^3))/time.
>In
>| >SI,
>| >| >it is amp*sec. In SI, vac charge = +,- sqrt(4pi*eps0*hbar*c).
>| >|
>| >| Fredi I agree the SI vac charge works out to 11.71 electron charges.
>| >| 11.71 = 1/sqrt(alpha). So if you add alpha to your SI expression e.g.
>| >| Qa = sqrt(4pi*eps0*hbar*c * alpha) = 1 electron = 1.6e-19Coul.
>| >| Do you have a place for alpha in your scheme? I like the expression
>| >| better; it's really full of bigshot coefficients now. :-)
>| >
>| >Vac charge is 11.71*e in any system of units. Of course alpha is in my
>| >scheme. I highly suspect it is a geometric factor of some kind other
>than
>| >simply being the ratio of unit electronic charge to vacuum charge.
>|
>| Fred take another look at my Qa (includes alpha) above. You will see
>| Qa, my revised version of vacuum charge, is simply the embodiment of
>| the definition for the FSC alpha, only instead of solving for alpha,
>| we are assuming alpha as given and solving for the charge on the
>| electron. Alpha is defined in SI as:
>| alpha = e^2/4pi*eps0*hbar*c or, solving for e
>| e^2 = alpha*4pi*eps0*hbar*c
>| It is hard to see why sqrt(alpha) would be a valid scale factor for a
>| new fundamental charge. JP
>
>??? Qa is just electronic charge not vacuum charge. The sqrt(alpha) is the
>basic probability for an electron and photon to interact. Of course the big
>question is why? My research leads me to geometrical factors. Like the
>ratio of volumes.
>
>| >| I don't argue that the CGS system of units is self-consistent, but the
>| >| units are hokey. They don't bear any logical relationship to real
>| >| life.
>| >
>| >Probably true but you get used to it after working with them.
>| >
>| >| In CGS they delete eps0 which is a legitimate unit of coulombs/volt
>| >| and actually characterizes the polarizability of space. And what
>| >| happens when you take out a coefficient of coulombs/volt? The products
>| >| get bollixed up.
>| >| Volts^2 are dynes. Vac charge is sqrt(erg cm)
>| >|
>| >| >| I get
>| >| >|
>| >| >| Q^2 = hbar*c = 3.1e-26 meter Joules.
>| >| >|
>| >| >| I don't even want to take the square root of that.
>| >| >
>| >| >Why not?
>| >| >
>| >| >| In cgs you really can't use units anymore after you "simplify" the
>| >| >| legitimate expression for E as:
>| >| >|
>| >| >| E = q'/r^2 volt/meter (OK)
>| >| >|
>| >| >| because now
>| >| >|
>| >| >| q' = q/4pieps0 = 1.44e-9 volt meters (for electron)
>| >| >|
>| >| >| This funny electron is now volt meters and coulombs are no longer
>| >| >| involved. Very seldom will you see units affixed to cgs equations
>and
>| >| >| little wonder.
>| >| >|
>| >| >| Now it becomes impossible to find force of any kind in the following
>| >| >| where you bring another electron into the field:
>| >| >|
>| >| >| F = q'q'/r^2 = 2.07e-18 V^2 m^2/r^2 = volts^2
>| >| >| Volts^2 is never a force.
>| >| >
>| >| >In cgs, volt^2 has the same dimensions as force. There is no problem
>and
>| >| >nothing needs to be fixed.
>| >| >
>| >| >| To fix that you can (and must, and cgs'ers do) fix it by multiplying
>| >| >| by Coulomb's constant
>| >| >| k = 4pi*eps0
>| >| >| F = k*q'q'/r^2 = 2.3e-28 Newton m^2/m^2 = Newtons
>| >| >|
>| >| >| The large point is that while we must use k to repair the force
>| >| >| equation to get real Newtons, now every time you want to use a real
>| >| >| coulombic" electron you must put a k in front of it or the electron
>| >| >| remains as volt meters as a consequence of the original sin.
>| >| >
>| >| >While I agree that charge has funny dimensions in cgs, it is not a
>| >problem
>| >| >for me. SI is a bit "richer" in certain physics problems but cgs is
>| >| >sometime easier to use. With the concept of vacuum charge, I can deal
>| >with
>| >| >any system of units.
>| >| >
>| >| >FrediFizzx
>| >|
>| >| You might look again at that alpha thing I mentioned above.
>| >
>| >For me in any system of units, alpha is simply the ratio between unit
>| >electronic charge and vacuum charge. Always.
>|
>| Well, no, you mean sqrt(alpha).
>
>Whoops! Yes, the sqrt(alpha).
>
>| >Now the "real" main question is what should the "real" coulomb constant
>be?
>| >Is it 1 or is it 1/4pi*eps0? I personally don't think it is exactly
>either
>| >one of them. But I would have to say that 1/4pi*eps0 is probably closer
>to
>| >the "truth".
>|
>| No, k = 4pi*eps0. Remember CGS makes a new charge q' = q/k as detailed
>| above. Now when you have charge squared, you have q^2/k^2, so you now
>| have 4pi*eps0 squared in the denominator where only one is allowed.
>| Thus the fix is multiply by k = 4pi*eps0. That is, to accommodate the
>| original apostate electron q', every other electron we bring up to it
>| has to be exorcised of the k in the denominator by multiplying by
>| Coulomb's constant.
>
>John, you keep getting it upside down. In SI, k_e = 1/4pi*eps0. Your
>problem here is with q' = q/k. In cgs, it is q' = q/sqrt(k). See the
>Appendix on units in Jackson's "Classical Electrodynamics". Now of course
>if you set k_e = 1, then it doesn't much matter.
Well, Fredi you got me there! You're right about k, it is the
inverse, and I was all wrong, and that leaves the electrons honest ie
q' = q. What a relief!
But this Jackson interpretation looks fishy. It looks too "fixy".
In the end if you stick to SI these discussions would not be taking
place, but then you would have to believe in permittivity.
And you can't do that if you endorse the bogus equation:
D = E + 4piP
(in preference to D = E*eps0)
that is of no earthly use and denies permittivity. I invite you to
attach units to this equation.
>
>| > We know one thing for sure from experiments and that is that
>| >k_e/k_m = c^2. The electric constant divided by the magnetic constant is
>| >equal to the speed of light squared. With vacuum charge, I get c^2 =
>| >(lambda/2pi)^2(1/(Cvac*Lvac)), where Cvac is vacuum capacitance and Lvac
>is
>| >vacuum inductance. These hold true in any system of units. So let's
>take a
>| >shot at k_e = lambda/(2pi*Cvac) and k_m = 2pi*Lvac/lambda.
>| >
>| >Now, what is Cvac and Lvac in the different systems. In hbar = c = 1,
>| >
>| >Cvac = Lvac = lambda/2pi = c/w
>| >
>| >Which we can see that k_e = k_m = 1.
>| >
>| >In gaussian cgs,
>| >
>| >Cvac = lambda/2pi and Lvac = lambda/(2pi*c^2)
>| >
>| >Which we can see that k_e = 1 and k_m = 1/c^2
>| >
>| >In SI,
>| >
>| >Cvac = 2*eps0*lambda and Lvac = mu0*lambda/8pi^2
>| >
>| >We can see k_e = lambda/(2pi*2*eps0*lambda) = 1/4pi*eps0,
>| >and k_m = 2pi*mu0*lambda/(8pi^2*lambda) = mu0/4pi.
>| >
>| >Everything is perfectly consistent with the simple postulation of vacuum
>| >charge. Without vacuum charge, Cvac and Lvac can't exist. Therefore, I
>| >contend that the real coulomb constant and real magnetic constant are,
>| >
>| >k_e = lambda/(2pi*Cvac) and k_m = 2pi*Lvac/lambda,
>| >
>| >and are true in any system of units. And it is easy to see from these
>| >expressions that SI's coulomb constant is a closer match to the real k_e
>| >where we have length/capacitance. Ditto for the magnetic constant. QED.
>| >
>| >FrediFizzx
>| I am not sure what lambda is but I still say you can avoid a lot of
>| heresy by using God's units, also known as SI.
>
>All units are God's units. They all have their own unique purposes in
>describing nature. Lambda is wavelength. But a good question. How does
>wavelength relate to this? Via hbar*c. It only works if hbar can be due to
>a vacuum process but if we are to seriously consider that a "free" photon
>has energy of hbar*w, then hbar has to be a vacuum process. We already know
>that c is a vacuum process.
>
>hbar*c = hbar*w*lambda/2pi
or c = f*L
>So lamda/2pi = hbar*c/hbar*w
or L = c/f
>>Which is charge^2/energy. Another way to express capacitance.
or C = q/v
>FrediFizzx
>
John Polasek
If you have something to say, write an equation.
If you have nothing to say, write an essay
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