Meaning of Beta

The Lagrange multiplier, β, showed up when maximizing the entropy with energy constraints. It turns out that β is not an arbitrary constant, but has an important physical meaning.

[@lookthesky9632]

Hi Professor Steven, may I ask why as T → inf, chair state and boat state are approximately equal? In my understanding, should it all turns to boat state as T → inf? because boat state is upper energy level, and as you said if you increasing temperature, the particles in state EA will level up to state EB, so follow this logic, finally all particle in state EA will level up to state EB, this will result to 100% boat state only. Or because conservation of energy, when particle gains temperature to level up to EB state, it needs to be balanced so that a particle from EB state will drop to EA? But this also seems not right because you said increasing temperature will increase the probability of particles in EB state to cause more chance for boat state. If one level up from EA and another one drop from EB immediately, then it can never change the probability of both chair state and boat state. I am confused. Could you please help me? Thank you very much.

[@PhysicalChemistry]

Good question. This is a very common source of confusion. The key is that temperature is not the same thing as potential energy. Having a higher temperature *allows* you to reach higher PE states, but doubling the T doesn't require that you have twice as much PE. Here's an analogy that might be helpful. Instead of cyclohexane occuping boat / chair states, consider instead your probability of buying your favorite sneakers at full price, or when they are on sale. The equivalent of temperature here is your salary. If you are poor, then you will almost always wait for a sale to buy the sneakers. If you are rich, then it doesn't require you to pay full price. It just means you don't care too much about the price. It gives you the ability not to care about their cost. So you may buy them on sale, or may not. You don't *prefer* to buy them when they are more expensive. It's just that you can't be bothered to put much effort into searching out a sale. Similarly for cyclohexane. If the system is hot, it doesn't much care whether the molecule is in the chair or boat state. It has enough energy that can access both easily, but that doesn't make it prefer the higher-energy state over the lower-energy one.

[@lookthesky9632]

@@PhysicalChemistry I really appreciate it! So this means in very hot situation, the particles got enough energies that are able to choose any energy state they want(in this example, they can choose EA or EB energy state as they wish). Just wondering in this situation, how those particles make "decision" to choose any of state since they don't have to level up to EB state or stay in EA state? Or they don't make decisions just randomly back and forth between EA state and EB state in this case? Thank you very much!

[@PhysicalChemistry]

@@lookthesky9632 Yes, that's right. And the "decision" is just random for each particle.

[@lookthesky9632]

@@PhysicalChemistry Thank you!

[@mortezakhoshbin]

Although i got your example so clear but Can we see this from entropy perspective? As T goes to infinity we obtain maximum entropy by pA=pB Not by PB=1. And if T is equal to zero minimum entropy can be achieved by pA=1. Thanks alot

[@PhysicalChemistry]

Yes, you're absolutely correct. And I certainly approve of always trying to think about thermodynamic topics from an entropic point of view. You're right that entropy is large when beta is small (and T is large). But an even more fundamental connection between beta and entropy is that beta is the amount that the entropy increases when you add energy to a system

[@mortezakhoshbin]

@@PhysicalChemistry So nice point thank you alot

[@Eta_Carinae__]

The expression for beta has a factor of the log of the ratios of the probabilities. In information theory, the sum of the log of the ratios of probabilities multiplied by the numerating probability is the KL divergence. It's not precisely the same, but I'm wondering if there is an analogue for the divergence in thermo./stat. mech.

[@PhysicalChemistry]

Very interesting question, thanks! I wasn't familiar with the KL divergence, so your comment sent me down an information theoretical rabbit hole. In this video, we're just comparing two discrete probabilities, but of course you could sum these over a distribution, as you are aware, i.e. when calculating entropy. I'm not aware of any thermodynamic equivalent to the KL divergence, though. The KL divergence is a means of comparing two different distributions. There are occasions we might want to do this in thermo, but probably not with a non-symmetric distance, as it wouldn't correspond to a state function.

[@Eta_Carinae__]

@@PhysicalChemistry Ah, that's alright. I suspected as much. The reason I ask is that in the free-energy principle, you essentially have the difference between an entropy and a divergence term, and this is to be taken as a free-energy analogue, so I'm just trying to see whether the divergence itself has a thermo. analogue, that allows it to be cast as a component of a free-energy.

[@PhysicalChemistry]

@@Eta_Carinae__ Perhaps. But if so, the other component would also have to be a non-state function. Like U = q + w

[@juv7026]

Do you have some reference to prove that beta is inversely proportional to T? (As in, with just the info that you got we could've very well said beta=1/kT^3)

[@conormcdaniel2066]

en.m.wikipedia.org/wiki/Thermodynamic_beta This page does a pretty good job of showing the complete derivation of Beta. It is in fact inversely proportional to T and not a power of T, it just takes more calculus legwork to get to a definitive proof than the intuitive explanation provided here.

[@mortezakhoshbin]

why dont we say if the temprature goes to infinity the probability of being in B state is equal to one . i mean there is alot of energy and all of the A state turns to B state. like the water and stream if the energy goes up we dont say probability of being stream and water will be equal and all the water will be turn to stream. thanks alot

[@PhysicalChemistry]

This is a great question. You're right that the probability of having steam is 100% above the boiling point, not 50%. But that is because there are more than 2 different microstates. A gas has higher entropy than the liquid. It has **many** more microstates than the liquid does. At high temperatures, all of the microstates are equally likely, but it is **much** more likely that the substance will be in one of the gas-like states than a liquid-like state. (Here's a link to a later video on phase transitions, but it comes after we develop the idea of free energy: kzbin.info/www/bejne/Z3alkoCortKshKM) The example in this video describes a simpler system, with only 2 possible states. For an even simpler example, consider an electron with 2 spin states: spin up or spin down. One of those has a greater energy than the other when it is in an electric field. But at high temperatures, we expect (and find) that the electron will spend 50% of its time in the spin up state, and 50% spin down.

[@mortezakhoshbin]

@@PhysicalChemistry THANKS...