e^pi vs pi^e using physics!?! 

I can smell the salt through this comment, and that's totally reasonable.

Ha!

I teach thermo and found this comment hilarious.

Repeat 10 times "My ignorance is not proof"

​@@DeJay7Repeat 10 times "My ignorance is not proof"

Nice

Now that I think about it, that's very intuitive. For small x≥0, e^x>x^e. Just test x=0. They'd meet only at x=e, and by derivatives they should be tangent to each other. For x>e, the derivative of x^e is (e/x) x^e, and 0

It doesn't even use the fact that x>e, we just need x>0. So in fact it proves it for all x>0

@@watchnarutoshippuden3228 Also true for x = 0. 1 > 0.

@@rainerzufall42 that's right. I just said we used x>0 because it was a physics proof and no body can be at absolute 0 temperature.

good one!

If we assume that the exchange of heat is brought about _reversibly_ then we can "prove" e^pi = pi^e. What we can't "prove" is e^pi < pi^e, but then we don't want to prove that

E^pi is irrational. If x is algebraic but not 0 or 1 and C is an imaginary algebraic number, x^C is transedental. Let a be algebraic but not 0. So (-1)^(-ia)=e^(a pi) is transedental.

@@user-yz3he2jm4o We know e^π is, but we don't know much about π^e

and unfair

Take a seat.

Kind of. When he asserted that delta S > 0, he used the 2nd law of thermodynamics.

it‘s fun because he states the definition of ΔS for reversible „systems?!?“ then procedes to apply it for the thermalization process of a hot body to a lower temperature, wich is clearly irreversible.

the fun part is that he prooved also that he isn’t a physicist.

​@@tomtomspa Well, Prof. Penn never claimed to be a physicist, so that's not exactly a revelation. As for the process of heat transfer itself-you're right, the process described in the video is irreversible. Which is why the entropy change of the complete system ΔS is greater than zero-a fact that was used to prove that e^π > π^e. Note that entropy is a function of *state,* meaning, the change in entropy depends only on the initial and final states of the system. It does NOT depend on whether the process was reversible or irreversible. So for calculating the change in entropy, you can use a reversible process, because that usually makes calculations easier. For a reversible heat transfer, the object loses heat infinitesimally slowly to an infinite series of reservoirs, each at an infinitesimally lower temperature than the previous. That's how we get ΔS₁ = ∫dQ/T. That equation is NOT valid if heat transfer takes place irreversibly, for example, across a finite (non-infinitesimal) temperature difference, as shown in this video. But, like I said before, ΔS₁ itself is independent of the process. So you can calculate ΔS₁ for a reversible process, and it'll be the same for an irreversible process as well. Likewise with ΔS₂ and ΔS. Prof. Penn did not explain all this because, of course, this isn't a physics lesson. However, I can see this same issue has confused other viewers as well. In any case, if you're interested, go ahead and read the original paper by Vallejo and Bove in the May 1, 2024 issue of the American Journal of Physics. I can't put the link here for fear of getting blocked, but you can google it. The article is free. They also give an alternative derivation involving a perfect gas.

@@nHans This is the fun part: seeing Michael naively get elementary physics concepts wrong, when he usually shows enormous competence.

@@tomtomspa He's not wrong! Entropy is a state function, so you just need to pick any process connecting the initial and the final thermodynamic state and the variation ΔS will stay the same. Of course the most convenient is a reversible process whose infinitesimal heat exchange is δQ = mcdT. "Reversible system" doesn't really make sense to me, though... xD (Sorry in case I misunderstood your point!)

It is a cycle. Doesn't matter where you start, the ending temperature is gonna be the same as the starting one. See here en.wikipedia.org/wiki/Carnot_cycle

​@@zafiroshinthe process used in the video is not a cycle

The cold body is assumed to be so much larger that the temperature changes negligibly.

​@@zafiroshinThanks. Carno's cycle happens on dof-lte line(electric). Absolutely same cycle happens on space-time line(magnetic). D-A - iso-time decompression ,for example from 1d to 2d in 0 second, A-B- iso dimensional decompression, time begin to flow in same dimension. and likewise B-C iso-time compression C-D iso-dimensional compression and of ,course, space-dimension S=C*lnT; where T-time.

You misunderstand the purpose of approximations in physics. You can get arbitrarily close to both of these limits by choosing an appropriate system.

i think dof-lte line(electricity) works on aplitudes(dof from below lte from above) while space-time line (magnetism) works on frequences (space from below time from above). ideal thermal reservoir is huge enough to increase local termal equilibrium by 1 degree, so it remains on previuose lte(carry not happens), despite absorbing some energy. This extra energy causes electricity( degree of freedom(dof,amplitude)curves). According to Heron's formulae mass^2=s(s-a)(s-b)(s-c);a,b.c - amplitudes; s=(a+b+c)/2; c

Ideal thermal reservoir just means that the temperature of it is left unchanged after interacting with the object. The first way you could do that is with something like melting ice or boiling water. A second way would be to consider an object so much bigger than the other one that the absorbed/released heat will be negligible for the thermal reservoir, and by all practical porpuses, you could consider it to be constant temperature.

Note that it does have to be a closed system though, otherwise the entropy increasing equation wouldn't necessarily be true. So boiling water is out.

@@marioveca6033 At first, I thought it may be an ideal gas in a container with a lid of very tiny mass that can move freely, but ice in a melting phase seems to be a better example. Thank you.

@@GeorgeFoot you can have boiling water with a closed system, provided you put it in a container at constant pressure that doesn't let the steam go out.

you proven it for one specific configuration of body and temperatures, which is not a thing mathematician would be proud about

This is something called a backward proof, and it's an easy to trap to fall into. Just because we can use an assumption to prove something we know to be true doesn't mean our assumption was correct. For instance, let's assume a * b = a + b. Using this assumption, we see that 2 * 2 = 2 + 2 = 4, which is true! But this doesn't mean our assumption is always correct: it also tells us 2 * 3 = 2 + 3 = 5, which is incorrect. In the same way, you can use the laws of thermodynamics to get correct results, but that doesn't imply the laws themselves are always correct. It's possible they _only_ work on the specific case we tested.

My car is black, therefore all cars are black QED ◾

2nd law of thermodynamics is a mathematical theorem about shannon entropy in dynamical systems described by microcanonical ensamble

As a physics student i disagree, i cringed inside as it was apparent that they have no physics background

The formula does not suggest that delta S_2 is 0. Delta S_2 is the integral of dQ/T Here T is a constant since the temperature of the reservoir is constant. So we simply get Q/T when we integrate that. We already know Q because whatever heat leaves the second system is what enters the second system. And since the heat is leaving the reservoir (this is just an assumption and if it is the other way around, we will simply get the opposite sign of Q), we take it to be negative of the 'Q' for the first system.

@@watchnarutoshippuden3228 But the integral is defined to have bounds from T_i to T_f, and the initial and final temperature of the reservoir is e, so is it NOT an integral from e to e, therefore 0?

The supposed "proof" is rubbish. Of course it proved nothing, not in the sense of an actual "proof". It wasn't even physics, no actual observations were made. Some integral dropped from the heavens, as a given. Then some hand-waving stuff around ΔS_2, exactly as you have observed. And all of a sudden something-something entropy-blah must be greater than zero, because "everyone knows". You could hear it in Dr. Penn's unsure voice at exactly those steps, and how he had to double check and consult his scripted notes, how uncomfortable he was with all that lah-di-dah hand-wavey stuff. But he went through with it anyway, I suppose for the sake of the RU-vids entertainment. But this video was neither math nor physics, it was pure entertainment.

​@@DeJay7 indeed, things are a little more complicated than that. In general, you can't assume that dQ is an exact differential, namely, it is not true in general that the integral is just a 1 variable integral from a starting temperature to a final temperature. Rather, it is a path integral between initial and final point in a space of more variables, where you have to specify the path from the starting to final point. In general, such an integral would depend on the path. However, the entropy variation is defined as the integral of dQ/T, and that is an interesting quantity because it is one of the few cases where the result is independent on the path (it can be shown). Anyway, the entropy integral for the object 1 is meant to be from T1 to T2 with other variables constant; the entropy integral for the thermal bath, on the contrary, is an integral along a path of constant temperature while other variables are changing (there is no need to specify what variables because heat absorbed and released by 1 and 2 is the same, making the integral's result trivial). I hope this clarifies things a little bit.

@@marioveca6033 This does clarify things, thanks a lot. I think you'll agree that this isn't at all "intuitive", although I have seen path dependent/independent integrals before, if I remember correctly Michael didn't anywhere mention that it's a special kind of integral, in that dQ is more than just the change in one variable.

curvature is quantity dimension is a quality. if we look at curvature then we don't see dimesnsion. if we look at dimension then we don't see curvature.

The proof just relies on pi>e, as that's what determines the heat flow.