A decrease in entropy .. and magnets and solvation

Discussion in 'Chemistry' started by DRZion, Nov 26, 2009.

1. noodlerBannedBanned

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Entropy is invariant under Lorentz transforms, but the volume isn't.

3. TrippyALEA IACTA ESTStaff Member

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How is this relevant to the topic at hand?

How does this contradict or affect my 'resolution' to the original question?

5. noodlerBannedBanned

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It doesn't. Unless you accelerate a battery or a solution with dissolved 'particles' so a Lorentz formula becomes appropriate.

7. TrippyALEA IACTA ESTStaff Member

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But nobody mentioned that scenario, and even under a lorentz transformed system, the scenario I suggested still applies, and still gives a correct answer.

8. noodlerBannedBanned

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Yes, because entropy is invariant under the transform. The entropy volume is not invariant under the transform (the hard part).
At low velocities it doesn't really make sense to refer to a Lorentz frame, however.

Last edited: Nov 29, 2009
9. AlphaNumericFully ionizedRegistered Senior Member

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All physics is, by construction, invariant under Lorentz transformations.

Actually its piss easy. Given something of volume V in its rest frame if you then boost to a frame which has the object moving at speed v then its volume will be $\frac{V}{\gamma}$.

And its irrelevant because entropy per unit volume is not a relevant physical quantity.

Firstly, this sentence isn't even grammatically correct. Secondly going from one frame to another via a Lorentz boost is valid AT ALL velocities less than the speed of light. That's the point of Lorentz transforms, they are the full transformations you apply to go from one frame to another. Galilean transformations are only valid at low velocities precisely because they approximate the exact expressions obtained from Lorentz transformations. You'll struggle to tell the difference between the results obtained by doing a Galilean transformation compared to a Lorentz transformation for v = 0.000000000000000001m/s but that doesn't mean the Lorentz version isn't valid.

You were just throwing in something to try to look clever and you actually just displayed your ignorance.

10. noodlerBannedBanned

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"Energy per unit volume is not a relevant physical quantity",

Say what?

The sentence "at low velocities it doesn't really make sense to refer to a Lorentz frame" is relevant to a chemical reaction or charging a battery in a low-velocity frame. That is, although the last poster is able to gleefully point out that Lorentz means physics is invariant, or more generally covariant, this is "meaningless" in a frame where all clocks give the same time.

As he so righteously points out, "you'll struggle to tell the difference" between a full transform under covariance, and a first approximation in a Galilean frame.

So he's pointed out that in Galilean frame you have approximate results, but there's a "hard part" as I mentioned, which is that the results aren't exact. This "doesn't matter" as he also points out.

I think he's just trying to be clever.

... but note that I rewrote the sentence "entropy per unit volume" as "energy per unit volume", what's wrong with doing this?

What's wrong is that entropy is a measure of 'direction', entropy always increases, but energy can decrease or stay fixed. Entropy per unit volume is a measure of the time it took for a system to have a certain energy; entropy increasing is the "arrow" of thermodynamic time.

Last edited: Nov 29, 2009
11. AlphaNumericFully ionizedRegistered Senior Member

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There's no 'hard part' in doing a Lorentz or Galilean transformation at low velocities. You claimed the Lorentz case wasn't well defined at low velocity, which is wrong.

And I said entropy per unit volume. Don't quote me, edit the quote and then go "What?!". It's clear dishonesty.

Noodler, aren't you vkh.... whatever the name was. This is something you do all the time, just post crap, try to convince people you're not an ignorant fool and then continue digging yourself a hole.

12. noodlerBannedBanned

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So what?
Yea, but so what?

Who said a Lorentz transform isn't valid? Where did you pick this up from, considering I didn't state any such thing?
Why does the above state categorically that it's "hard" to tell the difference between a Galilean first-order and a Lorentz transform?

Here, for the record, is what I actually posted:
There's a not so "simple" answer too. Lorentz invariance and entropy-volume changes.
The entropy volume is not invariant under the transform (the hard part).

13. AlphaNumericFully ionizedRegistered Senior Member

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You said
You are saying that a frame obtained by a low velocity Lorentz transformation isn't well defined. Nonsense.

Here

Can't you do a Taylor expansion of gamma?

Except it isn't hard. A Lorentz boost to a frame with relative velocity v results in a gamma factor $\gamma$ and the new volume is $\frac{V}{\gamma}$. No need for pen and paper, no need for calculations, it's easy. I've already said this, you failed to address it.

I take it from the fact you ignored my comment that you used to have a different username and you were known for crank behaviour you can't defend yourself from such an accusation.

14. noodlerBannedBanned

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Right. when discussing a battery charging, or a fluid with particles suspended or dissolved in it, saying "it doesn't really make sense to refer to a Lorentz frame" must mean: "a Lorentz transform is invalid at low velocities"...

What would you have thought if it said "it isn't necessary to refer to a Lorentz frame" instead? Or "there's a perfectly good approximation, you don't need Lorentz"...

When you calculate the new volume over gamma, does the entropy-volume relation stay invariant, or just the total entropy?

I see that you're the type who enjoys presuming things about what others say; it looks a lot like some sort of inadequacy you think you need to overcome, by slagging off others (and what they say).

Last edited: Nov 30, 2009
15. NasorValued Senior Member

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But you would have to expend energy to concentrate the salt. The useful energy that you were able to extract from the resulting temperature difference would be less than the energy needed to create the temperature difference in the first place.

16. DRZionTheoretical ExperimentalistValued Senior Member

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Or this EM energy could be used with a photovoltaic. Entropy of solution drops and entropy of batter drops.

Actually, entropy can spontaneously decrease as defined by fluctuation theorem.

Yes, even if instead of salt you would use a magnet and a magnetic solute the magnet will be doing work, and I don't suppose the magnetic potential could displace any layers of solvation that store energy.

You have a beaker with a partition in it, with x grams of magnetic solute solvated on one side of the partition. You have a magnet that is able to pull all the solute out.

Now you lift the partition, which increases entropy. The same magnet should still be able to pull all the solute out of the beaker, however extra heat should be generated if entropy is still going to be offset. There is no real limit to how large the beaker is, because eventually all the magnetic particles will get trapped by the magnet. IE the decrease in entropy is nearly limitless using just one small magnet.

17. TrippyALEA IACTA ESTStaff Member

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Photovoltaic?

Who said anything about the visible range?

18. AlphaNumericFully ionizedRegistered Senior Member

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Saying "it doesn't make sense" implies it's inaccurate or ill defined. Lorentz transforms are defined at all velocities below light. You said something inaccurate, just bloody accept it.

The entropy is left unchanged. Why are you even asking this? Several time in this thread people have said that the entropy is unchanged but the volume is, so something like entropy divided by volume is going to change. Rocket science this is not.

You said 'it doesn't make sense' when it does make sense to talk about Lorentz transforms. Since then you've both denied you got that wrong and you've put your foot in it several times. I'm not the one having to lie or try to concoct bull to dig myself out of a hole, you are.

If you say something incorrect, it isn't 'slagging off' to have someone correct you. The fact you're getting so defensive speaks for itself.

19. noodlerBannedBanned

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Ok I'll accept your "judgement" o wise and informed one, and issue the following correction:

"It makes sense to refer to a Lorentz frame, when charging a battery, even though you will struggle to see any difference from a Galilean one. This isn't hard to do (struggle, that is)"

Are we clear, now?

Oh, p.s. when you say "the entropy is unchanged" is that because a Lorentz transform is assumed to be continuous, or because entropy is assumed to be continuous (in, you know, time).

20. DRZionTheoretical ExperimentalistValued Senior Member

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Why would it have to be in the visible range ?? Its possible to make photovoltaics in the uv and in the mid-infrared as well.

Last edited: Nov 30, 2009

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22. TrippyALEA IACTA ESTStaff Member

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Who said anything about it being near the visible range?

Remember what I said about comparing it to NMR Spectroscopy?

NMR signals are measured as frequency shifts, those frequency shifts are measured in the 10-100 Hz range using 20 tesla magnetic fields.

Any EM radiation emitted by your experiment is probably going to be at the resonance frequency of the nucelii involved. The resonance frequency depends on the strength of the magnetic field. In the earths magnetic field (yes, you can conduct portable and cheap NMR spectroscopy using the earths intrinsic magnetic field) it's in the audio range. For a 21.2 Tesla magnet it is 900 MHz.
1Ghz requires a 23.5 Tesla magnet, and 800 MHz requires an 18.8 Tesla Magnet. Clearly then, each increment of 100 MHz requires an increase of about 2.4 Tesla. I'm not sure if the relationship is precisely linear, but doing a simple ratio calculation gives a resonant frequency of about 1.9 kHz, which is in the right ball park for what we were expecting.

So then the visible spectrum is in the range 790-400 THz, even the long wavelength edge of the Far Infrared if by 'Mid Infra red' you mean the 3-8 micron band, then even at 8 micron's, you're talking about a resonant frequency of 37.5 THz, which would require a magnetic field strength of 880,000 Tesla.

To put this in perspective, the strongest artificial magnetic field ever attained artificially is 2,800 Tesla, and that was acheived with an explosion. The strongest nondestructive pulsed magnetic field is 100 Tesla.

880,000 Tesla is closer to the realm of Neutron Stars (1,000,000 - 100,000,000 Tesla) than anything we can produce on earth.

Near UV would require a 750 THz resonance, which inturn requires a magnetic field strength 17,000,000 Tesla.

23. DRZionTheoretical ExperimentalistValued Senior Member

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I don't suppose its a good way to produce electricity by any means, I'm just trying to show that an exergonic solute-concentration 'reaction' can be coupled with a heat engine to lower the entropy of the universe.

I would suppose some kind of vibrational excitation may occur as the solute strikes the magnet, which may lead to emission of em radiation or simply relax as heat energy.

A permanent magnet is composed of magnetic domains that are aligned together - when a magnet binds metals, its attractive power decreases. Does this scatter the magnetic domains? This would certainly lead to an increase in entropy.

We can agree that the magnetic field decreases in strength, but does this lead to greater entropy of the magnet?

Last edited: Dec 1, 2009