Re: Is temperature relative?
- From: sal <pragmatist@xxxxxxxxxx>
- Date: Sun, 21 Aug 2005 23:12:30 -0400
By the way, a more entertaining restatement of the problem is in
<pan.2005.08.22.02.45.18.505480@xxxxxxxxxx>
On Sun, 21 Aug 2005 11:59:24 -0700, Ken S. Tucker wrote:
>
> 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.
Uh ... I hate to change horses in mid-stream, 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
>
> 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
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.
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.
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!
> 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...
> 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.
> 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...
> 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
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