Re: 2nd law of thermodynamics in question

Paul wrote:
Sorcerer wrote:
"Radium" <glucegen1@xxxxxxxxxx> wrote in message
news:1163829026.834570.61770@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|I hate the 2nd law of thermodynamics. I want it to stop existing. I
| don't like the fact that the overall amount of disorder in the universe
| cannot decrease. Is there anything that can be done about the sadistic
| second law? Any escape?

No.


I don't adhere to the 2nd law of thermodynamics, but I know physicists
that acknowledge I found error in such a law.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c1
Quote, "Second Law of Thermodynamics: It is not possible for heat to
flow from a colder body to a warmer body without any work having been
done to accomplish this flow. Energy will not flow spontaneously from a
low temperature object to a higher temperature object."

Please analyze the following experiment for error.

Consider a carbon resistor in series with a metal film resistor.

Metal film resistor:
Resistance: R
Voltage noise: V

Carbon resistor:
Resistance: R
Voltage noise: V + |Vc|

|Vc| is the extra noise from carbon resistor due to flicker noise.

The total voltage noise: (V + (V + |Vc|))^0.5 = (V*2 + |Vc|)^0.5

Current: (V*2 + |Vc|)^0.5 / (R*2)

Both resistors receive the same amount of power.

Power from carbon resistor noise: V * I = (V + |Vc|) * ((V*2 +
|Vc|)^0.5 / (R*2))

Power from metal film resistor noise: V * I = V * ((V*2 + |Vc|)^0.5 /
(R*2))

Power ratio: ((V + |Vc|) * ((V*2 + |Vc|)^0.5 / (R*2))) / (V * ((V*2 +
|Vc|)^0.5 / (R*2))) = 1 + |Vc| / V

Both resistors receive the same amount of power, but the carbon
resistor generates more power than the metal film resistor. The carbon
resistor will be colder than the metal film resistor. Therefore heat
will flow from carbon resistor to metal film resistor.



Regards,
Paul




I would like to offer another experiment for analysis. I'll try not to
be overly defensive.


Lets analyze real observed white noise within a bandwidth range. We can
split such noise into as many frequencies as required given long enough
sampling time. First lets consider 1000 frequencies. The average peak
voltage when considering all such frequencies is X volts given the
sample period. Theoretically all 1000 frequencies could momentarily
sync in such a manner so as to sum to a total of X*1000 volts peak. If
we sample 4 times longer then we can distinguish four times as many
frequencies. The bandwidth of each frequency is now one fourth.
Therefore the rms voltage of each frequency segment is half. V = (4 K T
R B)^0.5. B is bandwidth. We can distinguish four times as many
frequencies, each consisting of half the voltage. The following
equation applies when summing multiple voltage _noise_ sources Vt =
(V1 + V2 + V3 + V4 ...)^0.5 Therefore, four voltage noise sources,
each half, sum to Vt = (0.5 * 4)^0.5 = sqrt(2). Therefore, given four
times the sampling period, it is theoretically possible for all
frequencies to momentarily sync for a total of X * 1000 * (0.5 * 4)^0.5
= X * 1000 * sqrt(2) volts peak.

If the above is true, then longer sampling periods increase the
probability of obtaining higher noise crests. This seems in agreement
with my observation of noise. Therefore, consider the following
circuit.

Consider one real resistor in series with one LED diode contained
within an isolated system. LED power dissipation greatly increases as
the voltage approaches the manufactures specified data*** value.
LED's do indeed dissipate power far below the data*** threshold.
Furthermore, the voltage noise from the resistor will periodically
reach crest values over the LED's specific data*** value. Therefore,
we have one LED that occasionally emits a photon. Such photons could be
aimed at any part or area within the isolated system. First lets
analyze an experiment where the photons are aimed at the real resistor,
given the resistor is coated with paint or some surface material of
high emissivity with respect to the photons wavelength. In such a case,
the resistor absorbs its own energy; i.e., the voltage noise from the
resistor periodically emits a photon through the LED, which the
resistor absorbs. Now lets analyze another flavor of the experiment
where the only difference being the LED absorbs the photon. An opaque
surface could be coated over the LED in such a way the LED absorbs its
own photon. Lets now compare the two experiments, first analyzing just
the resistor. The resistor in the later experiment will, on average, be
slightly colder as compared to the first experiment. The opposite is
true when analyzing the LED; i.e., the LED in the later experiment
will, on average, be slightly hotter as compared to the first
experiment.



Regards,
Paul

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