Re: 2nd law of thermodynamics in question

In message <1163518186.732439.209950@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul <softwarelabus@xxxxxxxxx> writes
Richard Herring wrote:
In message <1163444775.906688.99910@xxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul
<softwarelabus@xxxxxxxxx> writes
>Richard Herring wrote:
>> In message <1163436256.228189.76090@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul
>> <softwarelabus@xxxxxxxxx> writes
>> >jambaugh wrote:
>> >> Paul wrote:
>> >
>> >Which experiment generates more radiation? Answer: Experiment #1
>> >[previously corrected!]

That's very close to falsifying the record. The text I quoted looked like this:
===================================================
Which experiment generates more radiation? Answer: Experiment #2
===================================================

>> >
>> Sure about that? ;-)
>
>Yes, I am sure about that. Experiment #1. Did you see my typo
>correction within minutes after posting? Perhaps you want to be a j---.

No, it should be obvious that when I posted I had not seen your
correction.

It is courteous to view a person's entire picture / post before
commenting or criticizing.

<sigh>

I _had_ read your entire post. I'm not a mind-reader, so I don't see how you could expect me to know that you were about to post a correction.

[...]

>Doesn't take a genius, Richard, to understand experiment B radiates
>more power.

And where did I suggest otherwise?

Doesn't take a genius, Paul, to understand my experiment B has a greater
power flux than A.

>Experiment B will be cooler than experiment A.

Until it reaches thermal equilibrium.

Experiment B tends towards equilibrium faster than experiment A. So?

> Very simple
>stuff Richard. That is direct proof the 2nd Law of Thermodynamics is
>incorrect.

And that's what we call a non sequitur. Since, as you pointed out, there
are many formulations of the 2nd Law, perhaps you should clarify matters
by posting the one you are using, and then explain exactly which part of
it you think is contradicted.

Then you agree such an experiment contradicts some interpretations of
the 2nd law, right?

Not any of the correct ones.

Did you see my original quote from WikiPedia.com
where they stated the following?

Quote from WikiPedia,com:
"Since any thermodynamic engine requires such a temperature difference,
it follows that no useful work can be derived from an isolated system
in equilibrium; there must always be an external energy source and a
cold sink.

Do you know what "isolated system" means?

The second law is often invoked as the reason why perpetual
motion machines cannot exist."

In fact, it has been my experience that most physicists believe in the
above quote. I rest my case!

What case?

My cooling-water examples are exactly analogous to yours. Do you think
_they_ contradict the 2nd Law?

No answer?

>
>Lets simply this for you Richard. Neither experiment A or B have a
>power source except thermal noise.

Agreed.

>Experiment B radiates more power.

Agreed.

> If
>you disagree then show the error in the math. It is a very simple
>circuit. Over time, more energy is leaving experiment B than experiment
>A. Therefore experiment B will be colder than experiment A.

Until it reaches equilibrium.

So?

*!@#$%^&*()_+*!@#$%^&*()_+*!@#$%^&*()_+


So??? The object *was* at room temperature. Without adding any energy
to the system

System. That's an important word in thermodynamics.

What exactly is the "system" here?
Energy is not being added, but is any being taken out of the system?
What kind of system does your statement of the 2nd Law refer to?
Does _this_ system fit the definition?

we successfully lowered the temperature!

Of what? Everything in the isolated system?

Pull a switch
to disconnect the antenna and now it's not at equilibrium.

It wasn't at equilibrium before. Since energy is leaving the antenna, you have to consider where it's going to. You can't leave the energy sink out of an isolated system any more than you can have a battery with only one terminal.

You have a
source of so-called "free energy."

Not by any definition a physicist would use.


BTW, my definition of so-called "free energy"

Which one is that? Gibbs free energy? Helmholtz free energy?

merely includes the
process of extracting the energy, not the one time cost of buying the
parts and equipment. Therefore, I term solar cells as a source of
so-called "free energy."

>Do I have
>to draw you a picture?

Maybe you should draw yourself one. You need to include a heat source, a
heat sink, and a thermal resistance.

If the experiment is in a vacuum then the major factor is black body
radiation.

So add it to your equivalent circuit.

[...]

>> > Most loads, except a
>> >few such as radiation resistance, generate thermal noise.
>>
>> Anything that turns electrical energy into heat, in fact.
>> Fluctuation-dissipation theorem.
>
>So Richard Herring believes radiation resistance generates thermal
>noise?

Strawman. Since when did radiation resistance turn electrical energy
into *heat* ?

You should be the last person to retort to name-calling. It's not that
becoming of you.

ROFL. You still haven't looked up "strawman", have you?

Anyhow, you did not answer the question.

That's correct. When you attribute some position to me that I don't hold, I'm under no obligation to defend it.

You cannot suggest radiation
resistance generates thermal noise simply because electrical resistance
generates thermal noise.

I don't.

If you said, "Well, I detected thermal noise,
therefore radiation resistance must generate thermal noise."

I didn't.

I would
reply, "No, increase the antennas wire thickness and you'll see the
thermal noise decreases, but the radiation resistance remains
relatively the same."

But you can certainly treat the radiation resistance as a noise source,
if the antenna is coupled to other antennas being driven by thermal
sources. Does the name Jansky mean nothing to you?

Evidently not.

That's what I term "2D thinking." You should know better. I would
encourage you to break nature in to its elementary components.
Therefore, do not mix electrical resistance with radiation resistance
and you'll eliminate some of your limitations. It is theoretically
possible to decrease the antennas electrical resistance till it is
appreciably low. Example, the total thermal noise, less antennas
thermal noise, (such as from the power source, etc.) amounts to 1
uV/RtHz, but the antennas thermal noise could be 1 pV/RtHz, or at some
time in the future 1 fV/RtHz by means of future technology.

And what exactly do you think any of that rigmarole has to do with the antenna's ability to receive radio noise radiated by external thermal sources?

Of course, the noise is attenuated by the transmission loss between the
antennas, so its effective temperature is usually very low.
>
>> >I provided a
>> >broad example by stating "appropriate load." If both the source and
>> >load generate thermal noise then the power across the load is 4 K T dF,
>> >but if the load generates no noise then the power across the load is K
>> >T dF.
>>
>> I don't know what you think you mean by the power "across" a load. It's
>> either "into" or "out of".
>
>"Power across a load" means the consumed power as measured across a
>load.

Semantic confusion. You measure the voltage "across" the load, but
measure (or infer) the current "through" it, and the sign of their
product is significant.

Do you know a little computer programming? Then you should expand your
mental capacity to a concept called "Indirection."

It's called "indirect measurement" because you have to rely on additional assumptions. If those are incorrect there's a more important programming concept you'll have to reckon with: GIGO.

That's why dear
Richard an oscilloscope can display power across a load if you merely
pre-enter the loads resistance. :-)

So if I have a signal generator connected to a 50 ohm resistor via a coaxial cable, with a T-connector in the middle, and I let you use your oscilloscope to measure the voltage across it, do you think you can tell me which way the energy is travelling?

--
Richard Herring
.

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