Ask a physicist anything. (2)

[Cabal]

Dear Science Person,

Is this cool?

http://www.youtube.com/watch?v=SAnDeh9xNHU

Show me their radar signatures (or lack thereof) and we'll talk

Meh, it's not too bad. They seem to have a lot of filler footage and no actual shots of one of the planes actually completing a stunt that involves flying upside down at any point.

But, they seem to be nippy and turn well, so props.

If I had one, I would most assuredly scare children with it.

 

[Wiccan_Child]

Dear Science Person,

Is this cool?

http://www.youtube.com/watch?v=SAnDeh9xNHU

Dear Unwashed Masses,

Yes, indubidably. Also, I'm suing you on behalf of Her Majesty for infringing on the copywright of the DARPA Falcon Project.

 

[chilehed]

Dear physicists,

Please confirm whether or not my reasoning here is correct:

I think I understand the statistical entropy definition S = k[sub]B[/sub] lnΩ, where Ω is the number of microstates corresponding to the macrostate.

And because thermodynamic entropy S = -k[sub]B[/sub] ∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ), I gather that the information function –∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ) is equal to lnΩ.

Then since for any real process dS >/= 0, every real process results in a change in the information function greater than or equal to zero.

So that if someone insists that a process results in a reduction in information, they are insisting that the process violate the Second Law
​

Right?

 

[TerranceL]

AH, here's the one I originally meant to post.

http://www.youtube.com/watch?v=SDbQ5xvsrIU

I live nearby a airforce base, when I was a small kid we went there on a field trip. I saw one of these on display.

Loved them ever since. Still think it's the most beautiful plane ever designed.

 

[canehdianhotstuff]

Dear physicists,

Please confirm whether or not my reasoning here is correct:

I think I understand the statistical entropy definition S = k[sub]B[/sub] lnΩ, where Ω is the number of microstates corresponding to the macrostate.

And because thermodynamic entropy S = -k[sub]B[/sub] ∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ), I gather that the information function –∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ) is equal to lnΩ.

Then since for any real process dS >/= 0, every real process results in a change in the information function greater than or equal to zero.

So that if someone insists that a process results in a reduction in information, they are insisting that the process violate the Second Law
​

Right?

Yes, the ohm symbol is a natural number greater than 0, hence the entropy is always 0 or greater itself.

 

[chilehed]

Yes, the ohm symbol is a natural number greater than 0, hence the entropy is always 0 or greater itself.

Perhaps my question wasn't clear.

 

[Wiccan_Child]

Dear physicists,

Please confirm whether or not my reasoning here is correct:

I think I understand the statistical entropy definition S = k[sub]B[/sub] lnΩ, where Ω is the number of microstates corresponding to the macrostate.

And because thermodynamic entropy S = -k[sub]B[/sub] ∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ), I gather that the information function –∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ) is equal to lnΩ.

Then since for any real process dS >/= 0, every real process results in a change in the information function greater than or equal to zero.

So that if someone insists that a process results in a reduction in information, they are insisting that the process violate the Second Law
​

Right?

OK, I'll do this step by step.

"I think I understand the statistical entropy definition S = k[sub]B[/sub] lnΩ, where Ω is the number of microstates corresponding to the macrostate."

That's correct, but only for systems where Ω is meaningful (that is, the system in question actually has microstates and macrostates). The entropy of a system with a non-uniform density of states cannot be written as S = k[sup]B[/sub] ln(&#937.


"And because thermodynamic entropy S = -k[sub]B[/sub] ∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ), I gather that the information function –∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ) is equal to lnΩ."

Again, that's correct, but the logarithm is what the summation simplifies to in a specific case. Moreover, it is incorrect to call it the 'information' function: it is only loosely linked to the ill-defined concept of 'information' (and it's not even a function, but that's just being picky).


"Then since for any real process dS >/= 0, ..."
In a mechanically simple, thermodynamically closed system, yes. For any other system, entropy can easily increase: all you need is an energy source, or a local disturbance (e.g., Brownian motion, quantum fluctuations).


"...every real process results in a change in the information function greater than or equal to zero."

See above for why this is more or less inaccurate.


"So that if someone insists that a process results in a reduction in information, they are insisting that the process violate the Second Law"

Non sequitur. The Second Law only applies to certain thermodynamic systems, and the formulae you've used above apply for more general scenarios.


Basically, an increase in entropy doesn't necessarily information must go down (or up, or sideways).

 

[chilehed]

OK, I'll do this step by step.

"I think I understand the statistical entropy definition S = k[sub]B[/sub] lnΩ, where Ω is the number of microstates corresponding to the macrostate."

That's correct, but only for systems where Ω is meaningful (that is, the system in question actually has microstates and macrostates). The entropy of a system with a non-uniform density of states cannot be written as S = k[sup]B[/sub] ln(&#937.

I'm thinking about thermodynamic systems, which indeed have microstates and macrostates.

"And because thermodynamic entropy S = -k[sub]B[/sub] ∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ), I gather that the information function –∑(ρ[sub]i[/sub] lnρ[sub]i[/sub] ) is equal to lnΩ."

Again, that's correct, but the logarithm is what the summation simplifies to in a specific case. Moreover, it is incorrect to call it the 'information' function: it is only loosely linked to the ill-defined concept of 'information' (and it's not even a function, but that's just being picky).

My understanding is that in information theory, it is referred to as the information function, and in that context rho indicates the probablility of each microstate i.

"Then since for any real process dS >/= 0, ..."
In a mechanically simple, thermodynamically closed system, yes. For any other system, entropy can easily increase: all you need is an energy source, or a local disturbance (e.g., Brownian motion, quantum fluctuations).

So that would be yes for simple closed systems, and yes for any other system.

As I said, it holds for any real process.

"...every real process results in a change in the information function greater than or equal to zero."

See above for why this is more or less inaccurate.

You seem to have already agreed that the information function is identical to lnΩ. If that's the case then I fail to see how my statement is incorrect.

"So that if someone insists that a process results in a reduction in information, they are insisting that the process violate the Second Law"

Non sequitur. The Second Law only applies to certain thermodynamic systems, and the formulae you've used above apply for more general scenarios.

There you're dead wrong - the Second Law applies to all thermodynamic systems, and any process which violates the Second Law is not a real process. Have you taken thermo in college yet?

Basically, an increase in entropy doesn't necessarily information must go down (or up, or sideways).

What is your background in statistical mechanics?

 

[Wiccan_Child]

I'm thinking about thermodynamic systems, which indeed have microstates and macrostates.

Only if they're composed of particles. Thermodynamic media don't have microstates and macrostates in any meaningful sense.

My understanding is that in information theory, it is referred to as the information function, and in that context rho indicates the probablility of each microstate i.

I daresay information theory would have parallels to entropic formulae, since the two are somewhat related.

So that would be yes for simple closed systems, and yes for any other system.

As I said, it holds for any real process.

My apologies, I meant to say: "or any other system, entropy can easily decrease: all you need is an energy source, or a local disturbance (e.g., Brownian motion, quantum fluctuations)."

So it's 'yes' for simple, closed systems, and 'not necessarily' for any general system. So your statement "Then since for any real process dS >/= 0, ..." is, in general, false, since dS < 0 is an entirely possible (and readily observable) phenomenon. Ever notice that puddles evaporate despite being cooler than 100°C?

You seem to have already agreed that the information function is identical to ln&#8486;. If that's the case then I fail to see how my statement is incorrect.

I agreed that the information function simplifies to ln&#937; in specific cases, not that they're identical. Moreover, I said your statement was inaccurate, not outright false.

There you're dead wrong - the Second Law applies to all thermodynamic systems, and any process which violates the Second Law is not a real process. Have you taken thermo in college yet?

The Second Law of Thermodynamics states that the entropy of a closed thermodynamic system will tend to a maximum. Thus, it applies only to the entropy of closed thermodynamic system; it does not apply to open systems.

What is your background in statistical mechanics?

Will that change the veracity of my words?

 

[chilehed]

Only if they're composed of particles.

I'm not aware of any thermodynamic systems that aren't composed of particles.

My apologies, I meant to say: "or any other system, entropy can easily decrease: all you need is an energy source, or a local disturbance (e.g., Brownian motion, quantum fluctuations)."

There's a distinction between entropy as an intensive state property, and the entropy generated during a process. Usually the context of the question provides a clue as to which is being discussed.

I'm talking about the entropy generated during a process.

So it's 'yes' for simple, closed systems, and 'not necessarily' for any general system. So your statement "Then since for any real process dS >/= 0, ..." is, in general, false, since dS < 0 is an entirely possible (and readily observable) phenomenon.

You're certainly entitled to your opinion, but in fact there has never been obeserved a process which resulted in a negative generation of entropy. That is quite simply an impossibility. All such processes are fictional, and if you said otherwise on your thermo exam you'd get that question marked off.

The Second Law of Thermodynamics states that the entropy of a closed thermodynamic system will tend to a maximum. Thus, it applies only to the entropy of closed thermodynamic system; it does not apply to open systems.

Again, you are certainly entitled to your opinion.

Will that change the veracity of my words?

I'm merely trying to understand what reason you have for thinking you know what you're talking about.

 

[canehdianhotstuff]

Incorrect, entropy has been shown to go negative, even if only for a short period of time.

Second law of thermodynamics "broken" - 19 July 2002 - New Scientist

Entropy is not exclusive of a closed system, in basic chemical reactions, which are not closed systems by any means, entropy is obviously observed and calculated.

 

[chilehed]

Incorrect, entropy has been shown to go negative, even if only for a short period of time.

Second law of thermodynamics "broken" - 19 July 2002 - New Scientist

Entropy is not exclusive of a closed system, in basic chemical reactions, which are not closed systems by any means, entropy is obviously observed and calculated.

What they observed is called Brownian motion. It's a well known result of the fact that statistical fluctuations become more and more noticable as the number of interacting particles decreases. It doesn't constitute a violation of the Second Law.

Notice that they are focused on the changes in the kinetic energy of the particle; it appears that they completely ignore that of the water molecules as well as any changes in the latent heat of the particle itself. They quite simply did not gather sufficient information to do a Second Law analysis, and the claim of a violation appears to have gone off half-cocked.

 

[Wiccan_Child]

I'm not aware of any thermodynamic systems that aren't composed of particles.

It was a theoretic aside. Though a case could be made for magnetic fields (though, quantum mechanically, they're just particles anyway).

There's a distinction between entropy as an intensive state property, and the entropy generated during a process. Usually the context of the question provides a clue as to which is being discussed.

Indeed, and since the context is the Second Law, one would presume that we were talking about the intensive state property.

You're certainly entitled to your opinion, but in fact there has never been obeserved a process which resulted in a negative generation of entropy. That is quite simply an impossibility. All such processes are fictional, and if you said otherwise on your thermo exam you'd get that question marked off.

Have you ever seen an ice-cube come out a refrigerator? Have you ever noticed how a puddle evaporates even though it's less than 100°C? These things happen because there has been a decrease in entropy.

The very concept of a heat engine performing work is based on the idea that heat can be pumped against a gradient. In other words, on the idea that entropy can be decreased.

The Second Law states that entropy must tend to a maximum... but only for certain thermodynamic systems. It most certainly does not apply to all systems. Specifically, it only applies to closed systems: the entropy of an open system can go up, down, or stay the same. Any input of external energy, or any perturbation of internal energy, can lower energy (for example, a heat engine pumping the heat out of an enclosed freezer unit, lowering the entropy of a small container of water, thus creating ice cubes).

Honestly, this is entropy 101. Next you'll be saying that the Second Law disproves atheism .

Again, you are certainly entitled to your opinion.

Please, tell me which parts were my 'opinion'. The part where I said the Second Law states that the entropy of a closed thermodynamic system will tend to a maximum? Or the part where I said it applies only to the entropy of closed thermodynamic system? The former is a mathematical proof, and the latter is simply a restating of a premise of said proof. I fail to see where 'opinion' enters into hard mathematics.

I'm merely trying to understand what reason you have for thinking you know what you're talking about.

Logic and reason. I also keep the civic highground, which gives me no end of pleasure.

 

[chilehed]

..Indeed, and since the context is the Second Law, one would presume that we were talking about the intensive state property.

Then it appears that you really don't understand the Second Law, because the Second Law forbids a negative generation of entropy during a real process. It certainly does NOT forbid increases in the entropy of one of the thermal masses during the process. Go take a closer look at a refrigeration cycle.

..Have you ever seen an ice-cube come out a refrigerator? Have you ever noticed how a puddle evaporates even though it's less than 100°C? These things happen because there has been a decrease in entropy.

Actually, you have it backwards. Water at 0C has a higher entropy than ice at 0C, and water vapor at a temperature has a higher entropy than liquid water at the same temperature. Check your tables. The heat required to melt the ice and vaporise the water comes from another body, which body experiences a reduction in its entropy. And the process as a whole results in a positive generation of entropy, as required by the Second Law.

..The very concept of a heat engine performing work is based on the idea that heat can be pumped against a gradient. In other words, on the idea that entropy can be decreased.

At no time during during any real heat cycle does heat flow from a cold object to a hot object. The fact that you claim otherwise demonstrates that you've never studied thermodynamics.

..The Second Law states that entropy must tend to a maximum... but only for certain thermodynamic systems. It most certainly does not apply to all systems. Specifically, it only applies to closed systems: the entropy of an open system can go up, down, or stay the same. Any input of external energy, or any perturbation of internal energy, can lower energy (for example, a heat engine pumping the heat out of an enclosed freezer unit, lowering the entropy of a small container of water, thus creating ice cubes).

Honestly, this is entropy 101.

So says the one who has never actually studied the subject.

I find it interesting that you feel you can take the high ground merely because I asked you what your qualifications are. It's a reasonable question. For my part, I hold a BSME degree from a highly respected school, graduated with honors (got a 98.7 in thermo), and have been working in the design of power generation and fluid thermal systems for nearly 20 years. It's all very well for you to wave your hands and claim that evaporating water and refrigerators involve Second Law violations, but if you want to convince those who've really studied the subject you need to do a bit of math.

 

[Wiccan_Child]

Then it appears that you really don't understand the Second Law, because the Second Law forbids a negative generation of entropy during a real process.

Sure... in closed systems. In open systems, the Second Law is irrelevant.

Actually, you have it backwards. Water at 0C has a higher entropy than ice at 0C, and water vapor at a temperature has a higher entropy than liquid water at the same temperature. Check your tables.

Which completely misses both my points.

First, I wasn't talking about ice cubes melting, I was talking about ice cubes themselves: how do you think they formed in the first place, if not by a local decrease in entropy? As you yourself state, water has a higher entropy than ice at the same temperature. The blob of water has been cooled (thus lowering its entropy) and subsequently frozen (thus lowering its entropy further).

Second, I wasn't talking about the puddle as a whole. I was talking about the fact that the puddle is at less than 100°C - the temperature required for a water molecule to evaporate off the surface. The reason water nonetheless evaporates is because thermal motion endows the odd molecule with enough energy to escape: the entropy in that region is temporarily lowered, since we have gone from a uniform heat distribution to a temperature differential. Thus, entropy has decreased.

How can you say entropy cannot decrease, when it quite clearly can? Cracking an egg or melting an ice cube are crude examples of increasing entropy, but how do you think chickens make eggs in the first place? By decreasing entropy. Some scientists define life as negative entropy machines.

At no time during during any real heat cycle does heat flow from a cold object to a hot object. The fact that you claim otherwise demonstrates that you've never studied thermodynamics.

First, I said that heat can be pumped against a gradient. Second:

These are open systems (and thus the Second Law has no more relevance than, say, British law has in China), and, when switched on, their entropy drops: a temperature gradient is formed. Uniform heat distribution to an uneven one. Hot to cold. High entropy to low entropy. Entropy is lowered.
I don't know how else to say it.

So says the one who has never actually studied the subject.

Ah, ad hominems. Would you like to disparage my mother as well?

I find it interesting that you feel you can take the high ground merely because I asked you what your qualifications are. It's a reasonable question.

Why? What possible purpose could it serve?

For my part, I hold a BSME degree from a highly respected school, graduated with honors (got a 98.7 in thermo), and have been working in the design of power generation and fluid thermal systems for nearly 20 years.

Fascinating, yet utterly irrelevant.

It's all very well for you to wave your hands and claim that evaporating water and refrigerators involve Second Law violations, but if you want to convince those who've really studied the subject you need to do a bit of math.

Please, please show me where I claimed that the Second Law has ever been violated.

 

[chilehed]

Sure... in closed systems. In open systems, the Second Law is irrelevant....

As I said, you are entitled to your opinion.

Ah, ad hominems. Would you like to disparage my mother as well?

You're the one holding yourself out as an expert, so you ought to expect to have folks ask about your qualifications. You refuse to say what they are, and spout nonsense, so it's reasonable to observe that you probably haven't really studied the topic.

Now, if you'd like to do an actual Second Law analysis of a refrigeration cycle and show how the Law doesn't apply, have at it. Pick a refrigerant, working pressures, sink temperatures,...set it all up with real values, give pressure, temperature, enthalpy, entropy, entropy generation and heat flow values over the cycle, and show me how the Second Law doesn't apply.

I await with baited breath. You'll end up a very rich man if you can show a real process that violates the Second Law...if you can get the patent office to not throw your application in the can.

 

[chilehed]

AH, here's the one I originally meant to post.

http://www.youtube.com/watch?v=SDbQ5xvsrIU...

SWEEEET!

 

[Maxwell511]

So says the one who has never actually studied the subject.

For someone that is so condensing about his "alleged" lack of knowledge on the subject I am amazed that you didn't point out the obvious fact that he says closed system when he obviously means isolated system.Of course recognising the different types of systems in thermodynamic theory would force you to accept what he is saying is basically true and the fact that he is not suggesting any violation of the 2nd law, he really badly explained it, but he's not wrong.

Of course you need my credentials in order to accept my opinion...... so I assure you I have have read a book on thermodynamics.

Edit: Okay this is stupid "In open systems, the Second Law is irrelevant." However I still stand by my assertion that he knows (mostly) what he is talking about.

 

[dewaddict84]

Dude, Chilehed, a few points.

1. Make some sort of assertion already. So far you're asking if your reasoning is correct, being told no, and then getting whiny. Its looking like you came for a fight.

2. Even I know that the second law doesn't mean much in an open system. If this is opinion, then it's a widly held opinion by most everyone that isn't a crackpot.

3.

refrigeration cycle and show how the Law doesn't apply

There it is:

The energy that does the work of the compressor comes from somewhere. (about 50% from burning coal).
But that certainly doesn't violate the 2nd law, it means the law simply doesn't apply to this system (the fridge), because it's open. If you hooked it up to a generator and made some sort of closed system, then the 2nd law would tell us some things about it. Like the overall entropy of the system is going to... uh... decrease? Yeah, let's go with that.

 

[Maxwell511]

But that certainly doesn't violate the 2nd law,it means the law simply doesn't apply to this system (the fridge), because it's open.

So open systems don't have a temperature and there is no exchange of heat with its surroundings? Thus not involving them with the 2nd law.

Are you sure that you are not confusing magic systems with open systems? I sometimes get confused myself. The way I remember it is this; the second law applies to open systems, magic systems are just magic.

If you hooked it up to a generator and made some sort of closed system, then the 2nd law would tell us some things about it. Like the overall entropy of the system is going to... uh... decrease? Yeah, let's go with that.

And you could not tell that from the open system perspective? Also I am pretty sure you mean isolated when you say closed. What is with that today?