The entropy of systems

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17 years 9 months ago #19146 by GD
Replied by GD on topic Reply from
I have come up with an experiment to test the definition of equilibrium & entropy:

the (a)microscope versus the (b)telescope
subject of observation: a “wad of stuff” (to use Larry’s expression)

In both observations you notice there is motion, but this motion tends to bring molecules together (for (a) and galaxies together for (b). Would you say that:

1) -(a) & (b) are in equilibrium
2) -(a) & (b) are not in equilibrium
3) - there is no relationship between observations (a) & (b)


4)- entropy increase in (a) & (b)
5)- entropy decrease in (a) & (b)
6)- entropy has nothing to do with these observations.

Larry, if I assume right, your answers would be 3 & 6

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17 years 9 months ago #19047 by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, again I'm not with you. The term entropy is from the ancient Greek and it means, roughly, the road in or out. That heat flows "downhill" is a real observation, that someone decided to call entropy. That seems fair enough to me [:)]

GD, a very good question. The nature of your experiment needs work though I think. A passive microscope/telescope will alter the entropy of the system, think of them as heat engines or thermometers. Active instruments could dramatically alter the entropy of the system. Of course we're getting into the realm of the observer and observed of quantum experiments.

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17 years 9 months ago #19148 by GD
Replied by GD on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
<br />Hi Jim, again I'm not with you. The term entropy is from the ancient Greek and it means, roughly, the road in or out. That heat flows "downhill" is a real observation, that someone decided to call entropy. That seems fair enough to me [:)]

<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

In order for this theory to be true, there would need to be more energy out than in. Is this a possibility?
The atom's electrons would have to slow down continuously (although very slowly for atoms at the surface of the earth, but more rapidly at the center of the Earth).

Varying energy in a mass would also mean varying electron speed of the atom in a mass. Is this logical?

- the continuous variation in energy of the atom would mean that you would never be able to verify its energy state at a set moment in time.

This is why the previous question was meant as an observation that would not take the instruments into account.

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17 years 9 months ago #19050 by Stoat
Replied by Stoat on topic Reply from Robert Turner
If we use a simple kinetic model, with no cohesive forces involved, then an atom would simply radiate away in about a trillionth of a second. That's obviously not the case, so we need to look at he cohesive forces of real things.

Now I know GD that you are a little reluctant to look at what's been said on these matters but you simply have to read up on the QED and the SED accounts of the atom. Once again you have hit upon an important aspect of the issue, in that you have asked about boundaries. Any problems you encounter, from a basic physics book, will be different from those that people who can just soak it all in emperically but you will not lose track of where you're at.

QED is Quantum electro dynamics and SED is Stochastic electo dynamics. Read how the two views look at the atom, then drink tea and watch football for a few hours, then think about how it fits in with your viewpoint.

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17 years 9 months ago #19149 by Larry Burford
GD,

You have not provided enough information. I can think of microscopic and macroscopic observations that would fall into zero or more of all six of the categories you listed. And I can think of several categories that you did not list.

Lets start at the begining - I need to know:

===

1) The definition of equilibrium. (This is the one from the dictionary, the one that everyone else uses.)

2) The definition of GD-equilibrium. (This is the definition that you use.)

3) A discussion of the differences between equilibrium and GD-equilibrium. IOW, the things that you see as the important distinctions.

4) The definition of entropy. (This is the one from the dictionary, the one that everyone else uses.)

5) The definition of GD-entropy. (This the definition that you use.)

6) A discussion of the differences between entropy and GD-entropy. IOW, the things that you see as the important distinctions.

LB

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17 years 9 months ago #19054 by GD
Replied by GD on topic Reply from
O.K. lets start with equilibrium:

The dictionary version: a force which counteracts another so the resultant is zero. in other words: acceleration of a body is present but it is counteracted by an opposite force from another body.
Resultant of opposing forces =0


My version: acceleration is not present, therefore a body is in equilibrium. Forces =0

How twisted is that?

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