Re: Is temperature relative?
- From: "Androcles" <Androcles@ MyPlace.org>
- Date: Sun, 21 Aug 2005 19:33:45 GMT
"Ken S. Tucker" <dynamics@xxxxxxxxxxxx> wrote in message
news:1124650764.013859.83610@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|
| sal wrote:
| > On Sat, 20 Aug 2005 23:03:50 -0700, Ken S. Tucker wrote:
| >
| > >
| > > Excellent post Tom!
| > > Sure hope Sal reads it.
| >
| > I _always_ read Tom's posts when he replies to something I wrote.
| >
|
|
| > I may disagree with his pedantry but I typically learn things from
his
| > posts, even when he's driving me nuts picking nits. :-)
|
| Well Sal, it get's nitty...
|
| I think your transformation of power using V'=gV and I'=gV
| is ok, but there is a splendid complication.
|
| Power is usually ignored by relativistist's let's see why.
|
| Recall that a current flowing threw a resistor creates heat
| by quanta's. The heat and power output is quantized just
| as the infared radiation from a warm resistor is, by the
| emission of infared photons.
|
| If that makes sense, then what you're really doing is measuring
| and transforming the radiation emitted by the resistor to various
| FoR's, and using a timed rate of emission to get POWER.
|
| That's hot stuff!
|
| Recall Lorentz force in GR is,
|
| f_u = q*F_uv U^v , U^v = dx^v/ds .
|
| It follows POWER=force*velocity,
|
| f = f_u U^u = 0 ,
|
| which is perfectly in accord with GR as invariant forces and
| accelerations vanish.
|
| Let's do some easy math...
|
| By association using the metric g_uv, I can set,
|
| f = g_uv f^v U^u =0.
|
| Watch this, the covariant derivative of g_uv is zero, i.e.
|
| g_uv,w = 0 ,
|
| so I will treat g_uv as a constant when I integrate,
|
| $ f ds = g_uv $ f^v dx^u = invariant scalar constant "P".
|
| Set the integral
|
| $ f^v dx^u = P^vu
|
| where P^vu is the POWER tensor, and
|
| P = g_uv P^uv ,
|
| where P is quantized, because it's a constant from the
| integration above.
|
| Ignoring Doppler Effects (DE) we can set,
|
| P = g_00 P^00
|
| and therefore, using g^00 = 1/g_00,
|
| P^00 = P*g^00 = P/ (1-2m/r) .
|
| Again ignoring DE set the 2m/r == v^2 and so,
|
| P^00 = P/ (1-v^2) = P*g^2 ,
|
| in agreement with Sal's thinking, using
|
| P'^00 = P*g'^2 .
|
| P^00 = P*g^2
|
| Of course there is a DE applied to the measure of the
| emitted photon radiation and thus the POWER but I
| figure that's wasn't necessary to include as everybody
| probably would know that.
|
| What's neat about Sal's problem is the convergence
| of three somewhat distinct theories, GR, EM and QT,
| and they all need to get along together when doing
| POWER calculations, as is outlined above.
|
| Regards
| Ken S. Tucker
| kxsxt
It is known that Maxwell's electrodynamics--as usually understood at the
present time--when applied to moving bodies, leads to asymmetries which
do not appear to be inherent in the phenomena. Take, for example, the
reciprocal optical action of a Sagnac ring and a observer. The
observable phenomenon here depends only on the relative motion of the
observer and the Sagnac ring, whereas the customary view draws a sharp
distinction between the two cases in which either the one or the other
of these bodies is in motion. For if the Sagnac ring is in motion and
the observer at rest, there arises in the neighbourhood of the Sagnac
ring a beat frequency with a certain definite period, producing a
fringe at the places where parts of the observer are situated.
sal is a phuckwit anyway, he cannot comprehend the reciprocal view of
Sagnac.
Androcles.
.
- References:
- Is temperature relative?
- From: sal
- Re: Is temperature relative?
- From: Tom Roberts
- Re: Is temperature relative?
- From: Ken S. Tucker
- Re: Is temperature relative?
- From: sal
- Re: Is temperature relative?
- From: Ken S. Tucker
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