e^pi vs pi^e using physics!?!
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- Опубліковано 2 чер 2024
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Finally, a real-world application of thermodynamics.
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"
A fun thing about this proof is you can replace π with any number greater than e and get the same result since the only feature of π it relied on was that π > e. So really this is a proof that eˣ > xᵉ for all x > e .
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.
Coincidentally, the temperature of the *Cosmic Microwave Background (CMB)* is *2.726 K* - approximately *_e_* kelvins! So deep space can act as a heat reservoir for the experiment described in this video. 🤯
good one!
It was lovely to see the application of Physics concepts
We get the strict inequality in the wrap-up only because we are dealing with an _irreversible_ exchange of heat; that is an exchange of heat through a finite temperature difference*. This is a physics idea smuggled into the Maths.
*The temp change could be brought about _reversibly_ by bringing up to the hot body an array of ideal heat reservoirs differing only infinitesimally in temperature.
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
Antimatter and black holes were first discovered with a bit of maths, being solutions to differential equations. It's good to see something like this helping to redress the balance.
To be rigorous we have to specify when calculating delta T1 that we consider a reversible process bringing the subsystem to the same final temperature as the actual process, and the fact that entropy is a state function.
Also the infinitesimal heat change is usually not a differential form so we write lowercase delta Q instead of "dQ".
The total entropy change is strictly greater than 0 because the process is not reversible, and since this is a closed system the calculation of Delta S is actually the total entropy created by the process.
Fun fact, we dont even know whether π^e is rational or not!
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
Combine e, pi, and sqrt[163], Exp[Pi*Sqrt[163]] an integer :) Martin Gardner
That's outrageous
and unfair
Take a seat.
Superb explanation
just a thought; but I wonder if this inequality is the reason that time/entropy has a direction...?
Kind of. When he asserted that delta S > 0, he used the 2nd law of thermodynamics.
We have the best universe
Beautiful!!!
Amazing! Impressive! Wonderful!
Wow... that'a incredible
Exquisite ✨
I disagree with the fact that this nice calculation is a real demonstration. In fact, we do not prove anything but the fact that within the model the total entropy (entropy of universe) variation is positive. It just means that following thermodynamics principles, this transformation is possible, considering the proposed model.
If it would have been negative, it would just mean that this transformation is not possible. So in that case you have to go bac to the principles by miximasing the entropy of the universe (or inquivalently by minimising the free energy of the system) to predict the evolution. It can be done and it is the case for more complicated models. Generally it means that a phase separation occurs.
but both integrals can be used to prove it without second law, just by comparing them ( 1/T
*NT problem*
Are there integers a and b such that a^5*b+3 and ab^5 +3 are both perfect cubes of integers?
That’s crazy! Given that entropy increases is a statistical thing, so statistics requires that final result!!??
1:48 replay button for the dad joke.
The blue region's temperature doesn't change. Why does that not mean that ΔS2 = 0?
Clever!
Cool!!
pi log base e(e) vs e log base e(pi)
Thanks sir for the video !!
Physics and applied math students-yes, you're right, we say "reversible _process,"_ not "reversible system." But I think we can cut the good professor some slack-this isn't really a physics lesson after all.
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!)
Incompressible solid vs. ideal heat reservoir is clearly frictionless elephants whose masses may be ignored city. I love physics...
You misunderstand the purpose of approximations in physics. You can get arbitrarily close to both of these limits by choosing an appropriate system.
I would say that since we know mathematically that one is bigger then the other, then , assuming the equations we have on heat transfer and capacity are correct, we have just proven the second law of thermodynamics
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 ◾
Assuming adiabatic system of course.
The 2nd thermodynamics law is a law (meaning it can only be observed but not proved). You can't use a physics law to prove a math theorem. It only shows that the law does not violate math rules in this particular case, not the other way around.
2nd law of thermodynamics is a mathematical theorem about shannon entropy in dynamical systems described by microcanonical ensamble
shouldn’t the final temp be pi since pi is greater than e ?
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.
Weird enough, one of the authors, Italo Bove, was my professor back in Universidad de la Republica in Montevideo, Uruguay, in the late 1990s.
I think I don't know what an "ideal thermal reservoir" is. I presume the volume of that reservoir increases as it absorbs heat, this way keeping the temperature (and pressure) constant.
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.
nice. But what I'd really like to see by this channel is a course on thermodynamics using a mathematically modern and (reasonably) rigorous formalism. I hate all the talk about infinitesimal and exact differentials...
physics is a part of math that the axioms. Appears of observation
Entropy!
How a mathematician thinks a physicist solves the problem: ^ How a physicist actually solves the problem: 3^3=3^3, come on guys smh
S=C*lnT
if to take S-digit e-ary number system
then S is a dimension(digit)
T+1 is a carry(infinity,motion) towards S+1 dimension(digit).
You teach physics better than physicist.
As a physics student i disagree, i cringed inside as it was apparent that they have no physics background
Not that I expected more, but the physics used here are ... rather unclear. I mean, so little was explained, why ΔS is what is it, why ΔS_2 is not 0 (as the formula would imply), you know, rather important things. Or that ΔS is always positive, which I guess the 2nd law of thermodynamics does say that, but it's more complicated than that, it's not FORCED to be positive, it just is all reasonable cases.
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 UA-cams 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.
:)
Well in engineering, delta S > 0; in physics delta S >= 0. Suppose you live in a curved universe where pi = e? 🙂
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.
Hang on. At which point was any property of Pi used?
The proof just relies on pi>e, as that's what determines the heat flow.
Significant caveat here is the fact that the proof holds only if the laws of thermodynamics is true which we can't be sure a hundred percent
More to say, it is not uncommon for small systems to spontaneously decrease their entropy for a while, I guess in this brief moments of misbehave π^e becomes greater than e^π
I proved it using a calculator
Some people have seen the video they can't even click like and they want calculate WHAT A PARADOX !!!!!
First principles: Math is to physics as grammar is to poetry.
The vehicle, not the content.