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$$ EMiTTER & DETECTOR equations & ENTHALPY E typo +eV & added line.
 [SPEEDO.wpd] ; Re: the speed of electron; Mon 16 Jan 2006 20:14.
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 EQUATiONs show 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 + L - (me*v^2 / 2) ; ..Added line;
= eM + eK      ; ..This line, VERY much, fits 526 keV, ABOVE;
= eF + L + eK
= eG + eK - eV ; ..THiS LiNE and NEXT *are-MUTUALLY-exclusive*;
= eG + eK + (me*v^2 / 2); Typo +eV WAS on (missing) BOTTOM LiNE.

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.

> 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.
 ##########          ###########      ##########      ##########

$$         EMiTTER & DETECTOR equations & NOTEs, etc.
> 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
                                  |
  2*pi*(h + nA*hbar)*Ni           |  (PLANCK Temp Tp)*h
= -- -- -- --- -- -- -- = --- ---;|= -- -- --- -- -- -- = -- --- -- --
  (PLANCK Temp Tp)*sec            |  (2*pi + nA)*Ni*sec
                                  |
  (h + nA*hbar)*hbar              |      2*pi*Rx
= -- -- --- -- -- -- = -- - -- --;|= -- -- --- -- -- = -- -- --- -- --
        Rx*sec                    |  (2*pi + nA)*sec
                                  |
LEFT side is a RECiEVER, DETECTOR;| The RiGHT side is an h*fL 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.
[SPEEDo.wpd]; $$ $$    $$ $$    $$ $$    $$ $$  $$  End of POST.
Re: EMiTTER & DETECTOR equations & ENTHALPY E typo & added line.