🚀 go-pugleaf

$$ [MORE h*fL EMiTTER & RECiEVER equations @ item 2.]; SPEEDO.wpd.
       Re: the speed of electron; Mon, 16 Jan 2006 20:14:10
Tom Roberts wrote: > double precision wrote: > big magician wrote:
> >>> does anyone knows the speed of the accelerated deviated
> >>> electrons in a crt?
> >> The accelerating voltage of a typical CRT is about 15 kV, so the
> >> kinetic nergy of an electron when it hits the screen is about 15
> >> keV; it therefore has v/c ~ 0.03. -=- I see I made a mistake.
> >> Sorry.  I use units with c=1: keV for energy, keV/c^2 for mass,
> >> and keV/c for momentum (since c=1 these distinctions are
> >> notational only).
>
> KE = 15 keV ; m = 511 keV/c^2 ; E = m + KE = 526 keV

Those equation shows no LaGrangian L; And, no Angular momentum pA:
$$                      GR equation corrected.
GUESS iSS GR G_uv = G_absolute / G = 1 / (n - 1) = G_relative = Gr. 

ENTHALPY E = m*c^2 + LaGrangian L + Volt*Amp*sec energy eV
           = eM    +     L        +     eV
           = m*c^2 +    pL*c      +    pA*fA
           = m*c^2 +    pL*c      - (me*v^2 / 2)
           = m*c^2 +   m1*c^2     - (me*v^2 / 2)
           = m*c^2 +    h*fL      +    pA*fA
           = m*c^2 +    h*fL      +  nA*hbar*fA
           = m*c^2 + nL*h*c / wl  +  nA*hbar*fA
= iNTRiNSiC energy + LaGrangian L + Volt*CHARGE energy eV
= eM + L + eV
= eM + eK     ..note, this line very much fits 526 keV, above;
= eF + L + eK
= eG + eK - eV
= eG + eK + (me*v^2 / 2)
= eM + eK + (me*v^2 / 2) + eV.

Hamiltonian ANALYSiS confused POTENTiAL energy especially, which
is WHY it only fits with LaGrangian ANALYSiS in PARTiCULAR cases.
You CAN only understand this if REST mass*c^2 = iNTRiNSiC energy.
Note c constant is NOT equal to 1 in GUESS iSS units, but arrives
at the SAME ANSWER ..and also, with the SAME form as noted above.
                                                        ```Brian.
> P = sqrt(E^2 - m^2) = 125 keV/c ; > beta = v/c = P/E = 0.24
> So this is more relativistic than I originally said, but not
> enormously so. gamma-1 is a typical measure of how large
> relativistic effects are, and that is 0.029 or ~3%. To obtain
> good crisp images across the face of a CRT (especially position
> being linearly related to the signal voltage)
> I suspect they must be taken into account in the design.
>
> As others have pointed out, bigger screens use larger voltages.
> Increasing to 25 kV only increases v/c to 0.30 and gamma-1 to ~5%.
>
> Tom Roberts     tjoberts@lucent.com
Re: the speed of electron.
 ##########          ###########      ##########      ##########

$$ [MORE h*fL EMiTTER & RECiEVER equations @ item 2.]; PD wrote:
> Length is not an intrinsic physical property of the spaceship,
> so asking whether it "PHYSICALLY increases or decreases" is a
> nonsequiter. The same is true of its clock rate. > PD

Especially in GR, WHERE the GR TEST-mass is a GR POiNT-mass ..ONLY.!!
But PUT GR aside:
1. h/2*pi*c = me*wls = {mph}*ls = Moo*loo/4*pi = eR*loo/2*pi*c^2.

2. h*fL = pL*c = nL*h*c / wl = m1*c^2 = nA*{mph}*ls*c^2 / rA, ENERGY;
     = pi*(h + nA*hbar) / (pi + 1)*sec ..The ENERGY for at a RECiEVER;
    = 2*pi*(h + 2*hbar) / (2*pi + nA)*sec ..The ENERGY for an EMiTTER;
           = (m1*v1^2/2)*{(m1/M1) + 1), The GUESS STANDARD LaGrangian.
SAME...
LEFT side is a RECiEVER DETECTOR; AND|OR EMiTTER (below) ENERGY, h*fL;
                                     |
  pi*(h + nA*hbar)                   |  2*pi*(h + 2*hbar)
= -- -- ---  -- -- = -- -- --- -- --;|= -- -- -- -- -- --; = -- ---
    (pi + 1)*sec                     |   (2*pi + nA)*sec
                                  |
 2*pi*(h + nA*hbar)               |       Ra*h         2*pi*Ra*hbar
= -- -- -- -- -- -- = -- -- -- --;|= -- -- -- -- -- = -- -- -- -- --
       Ra*sec                     |  (2*pi +nA)*sec   (2*pi +nA)*sec
                                  |
LEFT side is the RECiEVER DETECTOR| ; The RiGHT side is the EMiTTER. 
                                  |
SAME, but for the AMBiENT media ..or, iN-TRANSiT space ENERGY, h*fL;

  nA*{mph}*ls*c^2     nL*h*c   | m1*v1^2   m1     |
= -- -- -- -- -- -- = --- -- = |(-- - --)*(-- + 1)| = pL*c = m1*c^2.
         rA             wl     |    2      M1     |
                               STANDARD LaGrangian.

Sun RADiATEs @ AMBiENT; AMBiENT RADiATEs 1400 Watts/m^2 @ 1*AU radius.

3. RESTmass*c^2 = iNTRiNSiC REST energy eM = EQUiVALENT energy e.



















 #######          ###########          #########

$$ iNTRiNSiC energy (Jim Carr), follows..


RePOST 1:            iNTRiNSiC energy (Jim Carr).
From: arcsign@my-deja.com; To: arcsign@my-deja.com 
Re: Do Photons Have Mass?; Date: Mon, 15 Jan 2001 20:05:56 GMT
--------------------------------------------------------------
In article  <93vike$u0a$1@nnrp1.deja.com>, srp@microtec.net wrote:
In article  <3a631377$1@pull.gecm.com>, richard.herring@baesystems.com
wrote: >  >  In article  <3A4646B6.E787BC16@microtec.net>,
srp (srp@microtec.net) wrote: >  >  >  Jim Carr a Ècrit :
In article  <3a362709.3795467@news.uq.edu.au> this@will.bounce writes:
>  >  >  >  > Mass is equivalent to energy and photons of course have
>  >  >  >  > energy.
>  >  >  >
>  >  >  >   Only partly true.  Mass is equivalent to (what I call)
>  >  >  >   _intrinsic_ energy, the energy a particle has that is
>  >  >  >   unrelated to its motion.
>  >  >  >   The relation is that m = sqrt{ E^2 - (pc)^2 }/c^2.
`````snip`````
>  >  Note the sign. He's *subtracting* the momentum contribution
>  >  from the total energy. What's left is the intrinsic energy.
>
>  I understand that. The point was that pc can only theoretically
>  be completely subtracted in that equation because in reality
>  no particle can be brought completely to rest with respect to
>  all other particles.
>
>  This means that when what Jim calls the intrinsic energy is to be
>  calculated, the simple original equation E=mc^2 has always been
>  and still is the real deal.

[arcsign to srp] YES, exactly, 'iNTRiNSiC energy' is just ANOTHER
simple SYNONYM for (Rest-mass)*c^2=(Rebel GR-invarint REST-mass)*c^2.

>  We know c, we can measure m, so we can very precisely calculate E.
>
>  On its part, p is relative, and varies depending with respect to
>  which other particle you consider the motion of the original
>  particle. The major point being that it can never equal zero for
>  any real particle with respect to all other particles.
>
>  --
>   André Michaud
>  Service de Recherche Pédagogique http://www.microtec.net/srp/
>  La nouvelle "Société Royale" http://www.microtec.net/srp/socroyfr.htm
>  "Les choses n'arrivent dans ce monde que lorsque quelqu'un les fait
>  arriver" - Anonyme.
>
>  Sent via Deja.com >  http://www.deja.com ; ® © ¿ ; End of rePOST 1.













RePOST 2:
From: alan (darkcloud@mac.SendMeSpam.not.com); sci.physics;
Date: 2000/06/19; Go-go Google: <
http://groups.google.com/groups?hl=en&lr=lang_en&ie=UTF-8&th=aae4c668cab37523&
seekm=8hm3vk%24j1c%241%40crib.corepower.com#link20 >:
> |  James A. Carr   <jac@scri.fsu.edu> > ================
> Dear Jim,
> Three times, you have quoted this error.
>
> The speed of light is NOT "the ratio of the electric permittivity
> of free space (vacuum)  to the magnetic permeability of free
> space.".
>
> It's the square root of the inverse of their product.
[jiM Carr to Brian (arcsign)]
Sorry for the error. Of course you are right, it is the "square root
of the inverse of their product".  Thank you for correcting me. I hate
being wrong, but now that you have corrected me I am no longer wrong.

> You numbskulls are pathetic.
>
> Sincerely, > brian a. m. stuckless
Re: Three questions from a layman.    ..end of rePOST 2.
$$ [MORE h*fL EMiTTER & RECiEVER equations @ item 2.]; End.