Re: Is temperature relative?

On Sun, 21 Aug 2005 22:45:04 +0000, Tom Roberts wrote:

> shuba wrote:
>> I don't think your idea that the power P=P' is too far off the
>> mark, perhaps subject to certain reasonable assumptions. After
>> all, and ignoring Tom's caveat that power is not a scalar, power is
>> d(Energy)/dt, and both the energy and the time elements are dilated
>> by gamma when transformed to a frame moving with respect to the
>> resistor.
>
> There are two meanings of "energy" in relativity:
>
> A. for a SINGLE PARTICLE (i.e. pointlike object) the time
> component of its 4-momentum.
>
> B. energy density is the (time,time) component of the
> energy-momentum tensor.
>
> This confusion in terminology is rooted in ancient history and is
> unlikely to change.
>
> You seem to be using meaning A, but in this case you need to use
> meaning B (energy density) -- as it is a component of a rank-2
> tensor, for simple cases with all other components negligible in the
> object's rest frame, it transforms with a factor of \gamma^2. That
> dt in the "denominator" also has a \gamma in its transform which
> cancels one of them. Then remember that the energy density must be
> integrated over the volume of the resistor, and the other \gamma
> gets canceled (this is a SIMPLE situation in which "length
> contraction" can be applied). So for the SIMPLE SITUATION IN WHICH
> ALL OTHER ENERGY-MOMENTUM COMPONENTS ARE NEGLIGIBLE AND "LENGTH
> CONTRACTION" APPLIES, P'=P.

Thank you. That was helpful.

Need we add, this applies to heat-like forces only?


> You got the right answer via the wrong argument. The right
> argument also shows its limitations. For situations in which
> those limitations do not apply, power will transform in a more
> complicated manner -- it is not a scalar.
>
>
> But underlying all this is the problem: what does it mean to measure
> or compute an energy density (or power density, or power) in one
> frame for an object (resistor) at rest in another frame?

I'm glad you asked that! :-)

Here's the experiment that's been floating around at the back of my
mind. It's a little more complicated than the original post but is, I
think, precisely specified, at least; and it is specified entirely in
terms of times, voltages, and currents, measured in a hypothetical
laboratory using at least somewhat plausible apparatus.

* * *

We have two identical resistors, with the following properties:

R = 1 Megohm

Specific heat = 20 degrees/watt-second

Initial temperature = 20 degrees Celsius

They're thermally insulated, so they won't radiate or conduct away
heat; they'll just get warm.

Lay resistor A on the table.

Put resistor B into an ultra-ultra-ultra-centrifuge, and spin it up so
that

v = 0.1 C

Note that for resistor B, we have,

g = gamma = 1/sqrt(1-0.01) ~ 1.005

(I'll be using the approximate gamma and low-precision math
throughout to avoid the need for a calculator; hope that's OK.)

Connect both resistors, via low-resistance wires, to a _stationary_
power supply. Resistor B's wires must go through some fancy
slip-rings or swivel joints but we're not concerned with the details
of the connections; let's just assume they work.

Turn on the power, set to

100 volts

on the wires to each resistor. We measure the current going from the
power supply to each resistor. We find, much to our surprise,

I(A) = 0.1 mA (as expected)

I(B) = 0.099 mA (surprise)

[Do you agree so far?]

Power going to the two resistors is, therefore, different:

P(A) = 10 mW

P(B) = 9.9 mW

We now leave the power to resistor A turned on for 1000 seconds (16
minutes 40 seconds), and we leave the power to resistor B turned on
for 1010 seconds (16 minutes 50 seconds):

Time-on(A) = 1000 seconds

Time-on(B) = 1010 seconds

as a result of which the total ENERGY going from the power supply to
each resistor is

P = 10 watt-seconds

NOTE WELL: The same amount of total energy has gone to each resistor.

We then turn off the power, stop the centrifuge, and measure the
TEMPERATURE of each resistor. And we find:

temp(A) = 220 degrees

temp(B) = 219 degrees

[Do you agree?]

Resistor B's temperature didn't rise as much as resistor A's
temperature, by 1/2 percent.

Somewhere, 0.05 watt-seconds of energy were lost.

Where did the lost energy go? ;-)



> Note, however, that one could easily define a useful quantity
> that is "proper power" -- power dissipated in the resistor's
> rest frame. This is obviously an invariant, and it probably
> captures all interesting aspects of power dissipation (this is
> essentially what is done for temperature).
>
>
> Tom Roberts tjroberts@xxxxxxxxxx

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