Re: Is temperature relative?


sal wrote:
> By the way, a more entertaining restatement of the problem is in
>
> <pan.2005.08.22.02.45.18.505480@xxxxxxxxxx>

Thanks Sal, I try to read everybodys posts. I studied your post
before replying.

> On Sun, 21 Aug 2005 11:59:24 -0700, Ken S. Tucker wrote:

> > 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.
>
> Uh ... I hate to change horses in mid-stream,

I do it all the time :-).

> but the first
> complication is that my transformation was indeed wrong. The
> current's fine but I got the voltage wrong; it's not V'=gV. Here's
> how I blew it, in detail.
>
> First, orient the resistor perpendicular to the line of travel --
> otherwise we end up in Simultaneity Swamp. So, let's say resistor is
> along the Y axis, motion is along the X axis.
>
> Given measurements in one frame, there are two equally workable ways
> to get the voltage in the other frame. We can find the E field in the
> other frame, and then path-integrate it to get the voltage difference
> between two points. Or, we can start with the vector potential, and
> transform it directly. They both give the same answer, of course. I
> used the E field approach, so let's stick with that for now.
>
> I'll just lift the E field transform from MTW and skip the messy
> matrices. P. 78, eq. 3.23:
>
> Ey' = g*(Ey + v x B(perpendicular))
>
> (That "x" is the cross product.) Though I didn't actually state this,
> I actually assumed that I could use a large resistor, reduce the
> current to something very small, so the B field was negligible, and
> then ignore the vxB term. In that case we obtain
>
> Ey' = g*Ey
>
> WRONG -- In the stationary frame the motion of the resistor creates a
> B field which cannot be ignored.
>
> SO ... We need to go into the resistor's frame. In that FoR, if we
> use a large resistor and small current, then the B field really is
> small -- as small as we like. So we get something like this
> (flipping the sign of v to invert the transform -- hope that's right)
>
> Ey = g*(Ey' - v x B'(perpendicular))
>
> but if B' is very small (which, as I said, we can assure) we get
> something very close to
>
> Ey = g*Ey'
>
> And of course that implies
>
> Ey' = Ey/g .... oops
>
> Multiply by the length of the resistor and we get
>
> V' = V/g
>
> Alternatively, if we again assume B is very small in the resistor's
> frame, we can transform the 4-vector potential; in the resistor's
> frame, that'll look like this:
>
> A = (V',A^1,A^2,A^3) ~ (V',0,0,0)
>
> and it transforms as
>
> Lorentz(A)= (gV', gvV', 0, 0)
>
> Again we see
>
> V = gV', or V' = V/g

Delta energy is charge*potential = q*V = mass, so Volts
transforms like mass in ordinary SR since q is invariant.

> > Power is usually ignored by relativistist's let's see why.
>
> Fnord. I've seen a number of power discussions. Tom's talked about
> it in another post:
>
> <Qt7Oe.246$rS4.130@xxxxxxxxxxxxxxxxxxxxxxxxxx>
>
> and pointed out that a resistor is an extended body so the approach I
> used to find the power isn't really kosher, but I'll repeat it here
> anyway. (I've assumed the resistor is a "point particle" but so it
> goes.)
>
> Thing is, energy of a point particle in relativity is g*m. One can
> then designate as power the frame-dependent derivative
>
> P = d(g*m)/dt
>
> That's the rate at which energy is going into an object of rest mass
> m, measured in the frame that uses time coordinate "t". If we keep
> that in mind, then things work out.
>
> From this, we can also see how the power must transform in my little
> resistor problem, from first principles. In the fixed frame,
>
> P = d(g*m)/dt = g*dm/dt
>
> since g is constant. In the moving frame,
>
> P' = dm/dtau = (dm/dt)*(dt/dtau) = g*dm/dt

No, in simplistic terms, you're transforming m/t. I would use,

m'=m*g , t' = t/g and find

P' = m'/t' = g^2 m/t = g^2 P .


> and they are indeed _equal_. P = P'.
>
> This appears to be a somewhat general result, applicable to
> non-extended bodies and heat-like forces.
>
> It is pleasing to note that with the (corrected) correspondence
> between V, V', I, and I', we have
>
> P'= I' * V' = (I*g) * (V/g) = I*V = P
>
> and the books balance (if we assume I*V is a reasonable measure of
> power).
>
>
> > Recall that a current flowing through 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!
>
> Actually I'm just collecting it as heat, a temperature rise in the
> resistor. But if you insist we can look at radiation.

Shuba explained the difference between Power (heat output) and
temperature, please keep them distinct.

> Fire one photon out the front _and_ one photon out the back,
> simultaneously, every K seconds, so that the moving body doesn't
> accelerate as it radiates, and so we don't have anything difficult we
> need to do with angles to figure out the red/blue shift. Now, look at
> it from the point of view of a frame moving at velocity v along the
> line of travel.

Thanks good point.

> First, note that the frequency of emission drops by 1/gamma.
>
> Then, note that energy of a photon goes as frequency. The front
> photon is blueshifted by
>
> (1/gamma)*(1/(1-v)) = gamma*(1+v)
>
> and the back photon is redshifted by
>
> (1/gamma)*(1/1+v)) = gamma*(1-v)
>
> and the average energy shift of the two photons is
>
> gamma*(1+v+1-v)/2 = gamma
>
> Overall, moving to another frame, we multiply by the change in photon
> emission rate _and_ the average change in photon power to get the
> change in power level. That will be
>
> (1/gamma) * gamma = 1
>
> So the frame-to-frame mapping of the power is, once again
>
> P' = P
>
> unless, of course, I did that wrong!

I think you made a good point but your conclusion is wrong.

Consider a light-bulb emitting 40 watts.
Call it's output

N photons/sec * (n cycles/second == frequency)

IOWs,

Power = (N/t)*(n/t) = g^2 P

where "P" is calculated in the rest frame.

Boost either the rate of photons emitted or the frequency (energy),
and each contributes to the Power output. The local FoR clock
gives "t".

> > Recall Lorentz force in GR is,
>
> Sorry I didn't read the next section very carefully, in large part
> 'cause I'm sure something's wrong, since the conclusion is off a
> bit...

That's ok, I hope the light-bulb above is easier anyway.

> > f_u = q*F_uv U^v , U^v = dx^v/ds .
> >
> > It follows POWER=force*velocity,
>
> And in this case the force looks something like this:
>
> (F^0, 0, 0, 0)
>
> It's a heat-like force.

Sorry Sal I thought you were operational with tensors,
so yes disregard what I wrote using tensors.

> > 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
>
> But the trouble is, this formula is wrong.
>
> So there _must_ be something wrong with the derivation...

I think it's ok using the light-bulb above.
Regards
Ken

.

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