Was energy quantization implicit even in classical mechanics?

[DavidAngelMX]

Let's analyze the case of a rigid body, say a paralelepiped (a x b x c), which presents a certain temperature distribution U(x,y,z) in the stationery state (the distribution doesn't change over time). We can keep the lower face to a certain temperature u, and the other 5 faces at zero degrees, in whatever temp units you like. Then, Fourier's law of heat transfer says:

 

dU/dt (1/c^2) = div(grad U) ; which can be simplified as div(grad U) = 0, since ?U/?t = 0

 

By solving (using separation of variables) this equation, we obtain:

 

U(x,y,z) = (16u/pi^2)Sum[ Sum[ (1/mn)(sinh(q*(c - z))/sinh(q*c)) sin(m?x/a) sin(n?y/b) ] ] ; You can see "Mathematical Physics, by Eugene Butkov if u don't believe me.

 

-The sums are over m and n respectively.

-The coefficient q is actually q sub mn, since it reads:

 

q = sqrt( (m/a)^2 + (n/b)^2)

 

As you can see, this is a series of eigenfunctions, summing over a discrete spectrum depending on m and n. Don't you think this is a very strong evidence of the quantization of energy, since heat here behaves as a discrete spectrum of permitted values (wave numbers)? Actually the eigenvalues m and n come out of the solutions of the ordinary DE in the process of solution, so it has to do with the physical meaning of this model.

 

Opinions, please.

[Tom Mattson]

As you can see' date=' this is a series of eigenfunctions, summing over a discrete spectrum depending on m and n. Don't you think this is a very strong evidence of the quantization of energy, since heat here behaves as a discrete spectrum of permitted values (wave numbers)?

[/quote']

 

Sure, there are quantized energies in classical mechanics. It happens with standing waves on strings, too. But that's not to say that QM is derivable from classical mechanics or that the so-called first and second quantizations are appropriate in classical mechanics. In QM, we get energy quantization where it is classically unexpected (as in, say, the central force problem), which is the real break from the old theory.

[DavidAngelMX]

I agree with you, it was not my intention to suggest that QM was derivable from Classical Mechanics, I just imply that the idea of quantizised energy shoulden't be such a head-braker when this kind of situations are often seen in classical physics.

[Tom Mattson]

I very much agree.

[DavidAngelMX]

I guess many people was scared away with those equations in my first post, but everyone, it's actually pretty simple (at least the physical purpose of it, the mathematical demonstration of the principles behind the solution method are a bit more tricky though).

 

Either way, if Math is the most universal Language, then Physics is the most beautiful kind of Poetry.

[Johnny5]

I guess many people was scared away with those equations in my first post' date='

 

[/quote']

 

I wasn't scared off at all, (in fact this was one of the few threads that interested me) but Latex isn't working.

 

You have a parallelipiped, say a cube, defined by three vectors A,B,C. Then i think only one side of it had a nonzero temperature, so that you could apply a mathematical model, which had wave solutions. Then you were just saying that in the solutions (which are way complicated to follow without more explanation on your part), natural numbers appear, so that quantization of variables appears in classical mechanics as well as quantum mechanics. I followed what you said.

 

In order to really follow you though, I would need to see and do more math, and as I said Latex isn't working.

 

PS: Ok let me have a look at it. You gave this:

 

dU/dt (1/c^2) = div(grad U)

 

RHS first:

Divergence of a gradient. Divergence of gradient of U. In general, U(x,y,z,t), where x,y,z are spatial coordinates in some 3D reference frame, and t is the time coordinate in that frame (measured by a stationary clock).

 

Then you stipulated that the parallelipiped is in a stationary state, which means that dU/dt=0, actually partial of U with respect to t=0. Following you so far. So that U is just a function of the spatial variables x,y,z. U is the temperature of the parallelipiped.

 

So U is a scalar function, not a vector function, so it has a gradient. OK.

 

so RHS is:

 

[math] \nabla \bullet \nabla U [/math]

 

Then you state without proof, that the divergence of U is zero. I don't see why right away. The proof is purely mathematical, can you provide it?

 

Ok LHS now:

 

dU/dt (1/c^2)

 

My next question is, why does the model equate this to the RHS?

 

This is a wave equation, so that the temperature U is waving in space, but it's not waving in time, since you have it in a stationary state. I don't fully get this model. Standing wave pattern for sure, but of temperature?

 

Last question:

Why would anyone model the temperature of a solid, as a standing wave?

[DavidAngelMX]

Look, the model I planted is not a wave equation, it is called Heat Equation, and it was proposed by Fourier, and it just differs from the wave equation in one derivative respecto to time. So, in the stationery state, U doesn't depend of time at all, so dU/dt = 0, leading to an equation which is analogous to a Potential Field. I think I wrote the equation wrong, the div is not zero, the Laplacian is, div(grad U) = 0, sorry .[div(grad U) = d2U/dx2 + d2U/dy2 +d2U/dz2]

 

Well, for a start, we propose a solution of the form U=X(x)Y(y)Z(z) [we forget about time, since U does no longer depend on it]. Thes, we substitute on the equation and simplify, yielding:

 

X''/X + Z''/Z = -Y''/Y ; It's easy to see that, for this equation to be valid, since X depends only on x and so on, every term must be a constant [X''/X + Z''/Z = -Y''/Y = k], allowing us to separate in three ordinary DE, say:

 

i)Y'' + Yk = 0

ii)Z'' - (v-k)Z = 0

iii)X'' - vX = 0 ; k,v are real constants

 

Solving all of these equations applying some boudary conditions, and discarding all trivial solutions, we can deduce that k is quadratic and negative (k = -r^2), and v is quadratic and positive (v = s^2); this is to avoid all trivial solutions, and it could be slightly different depending on the way the separation is done, leading to equivalent eigenfunction expansions as solutions for U, but let's do it this way so the solution comes in the form I proposed before.

 

Now, the obtained functions are:

 

i)X(x) = ASin(n*pi*x/a) ; A is A sub n

ii)Y(y) = BSin(m*pi*y/a); B is B sub m

iii)Z(z) = C(e^rz - e^-rz); C is C sub m,n

 

So, since U=XYZ we have: U = Sum{Sum[C(e^rz - e^-rz)*BSin(m*pi*y/a)*ASin(n*pi*x/a)]}; where the sums are over n and m.

 

Look, I probably made some mistakes so long since I have been solving this directly with my keyboard, and the resulting form will not be exactly the same I proposed before, but it will be equivalent, I can assure you. But now you have the steps to do it yourself more thoroughly. In the end, you can join together the constants A,B,C, into a constant D which depends on m and n. The next step is to apply the initial condition (that one of the faces in mantained at the temp. u; you choose the face), and substitute to the solution U, and so now you have only two functions inside, and a big constant with exponentials in it; this two functions are called Eigenfunctions, and the constant, as the form of the solution suggests, is the Fourier Sine Series Coefficient, in two dimensions, whis is obtained using a double integral of the eigenfunctions multyplied by the condition u and a constant which I think to remember is 4/ab. Solving this, and giving the exponential functions and constants in the resulting "thing" the shape of a Sinh, since it is a linear combination of exponentials, and so it generates the same "space", just substitute in the solution for U (the one with the sums), and Voila, it's done.

 

Look, this might not have been very instructive, so I recommend you to check out Mathematics for Phycisists, Arfken; or Physical Mathematics,Eugene Butkov. Or you can also search the web for "the method of separation of variables", obviously for partial differential equations. You might be interested in investigating a little bit about Fourier Series and Transforms, and "The method of eigenfunctions expansion", which can help you interpret the result in terms of the Fourier discrete spectrum of the coefficients in this solution, and the continous spectrum in the case of the solutions using Eigenfunction Expansions.

 

See ya

[Johnny5]

See ya

 

Like I said, without latex fixed it was a bit hard to read.

[Johnny5]

Or you can also search the web for "the method of separation of variables"' date=' obviously for partial differential equations. You might be interested in investigating a little bit about Fourier Series and Transforms, and "The method of eigenfunctions expansion", which can help you interpret the result in terms of the Fourier discrete spectrum of the coefficients in this solution, and the continous spectrum in the case of the solutions using Eigenfunction Expansions.

[/quote']

 

I know separation of variables. I also am familiar with fouries series, and transforms.

 

I have never heard of method of eigenfunction expansion though, sounds like something you made up.

 

At any rate, I would have paid closer attention had you used Latex. I really wish they would get that fixed. Additionally, I did understand your point, which was that quantization shows up in classical physics, and I vaguely remembered the heat equation.

[Tom Mattson]

I have never heard of method of eigenfunction expansion though' date=' sounds like something you made up.[/quote']

 

It's in any textbook on partial differential equations.

 

It is also very much related to the spin eigenstate expansion I posted in the "Math of QM" thread.

[DavidAngelMX]

Look again, it exists. Method f eigenfunction expansion

[Johnny5]

Look again, it exists. Method f eigenfunction expansion

 

Not calling you a liar, I just want to know if there is another name for it that I might know. Doesn't look like it though.

[DavidAngelMX]

I don't know how to write any math expressions here, but if you help me out or gimme an e-mail address, you'll get a full explanation, so you can trust me.

[DavidAngelMX]

Look, first I'll explain a problem like the one I posted first, then the Method of Eigenfunction Expansions and Finite Transform. I know my fingers will hurt afterwards, though.

Problem.pdf

Eigen.pdf

[Johnny5]

Look, first I'll explain a problem like the one I posted first, then the Method of Eigenfunction Expansions and Finite Transform. I know my fingers will hurt afterwards, though.

 

You typed all that up for this conversation?

 

Latex is easy to explain, I'll show you how later.

[DavidAngelMX]

You typed all that up for this conversation?

 

Of course I did!!! You wanted a clear solution, there it is. I hope you believe me now and give us some opinions. Either way, it seems like the subject´s been missing for a while, but anyway.

[Johnny5]

Of course I did!!! You wanted a clear solution, there it is. I hope you believe me now and give us some opinions. Either way, it seems like the subject´s been missing for a while, but anyway.

 

 

Ok excellent. I'm gonna go through it carefully, I could use the review, I'm sure I'll have a question or two.

 

I looked it over yesterday, yep it's very clear. Give me a day or two.

[DavidAngelMX]

Okay, just lemme know.

[Johnny5]

Alright, I am starting to run through this myself.

 

We begin with Fourier's heat equation which is this:

 

[math] \alpha^2 \nabla^2 U = \frac{\partial U}{\partial t} [/math]

 

In the equation above, U denotes temperature.

So in general, the temperature of the parallelepiped can change in time, the body could heat up, or cool down. But here we are going to discuss stationary states, meaning that the temperature cannot change in time, which means this:

 

[math] \frac{\partial U}{\partial t} = 0 [/math]

 

Which means that the value of U only depends upon the spatial variables, x,y,z. That is:

 

U(x,y,z)

 

You then give the boundary conditions of the parallelepiped, as well as a neat little picture.

 

In your opening post you say this:

 

We can keep the lower face to a certain temperature u' date=' and the other 5 faces at zero degrees, in whatever temp units you like. Then, Fourier's law of heat transfer says:

[/quote']

 

So now, we have to position the parallelepiped in some reference frame.

I'm trying to understand the boundary conditions right now, so I know where to put the origin of the frame. I want it to be consistent with your diagram, and the boundary conditions you wrote.

 

U(0,y,z) = U(L1,y,z) = U(x,0,z) = U(x,L2,z) = U(x,y,0) = 0

 

U(x,y,d) = f(x,y)

 

Ok, we can place the origin of the frame in the back left (in your picture).

 

So for the bottom face, z=0, and x can range from 0 to L1, and y can range from 0 to L2. Thats the bottom of the parallelepided you have drawn.

 

Then, for the grey shaded face, the y coordinate is L2, and x can range from 0 to L1, and z can range from 0 to d.

 

So this is actually a right-handed frame.

 

The other faces are similarly understood.

 

Ok, so I am this far along.

 

Ok, then you solve using separation of variables, which I know. So we have this:

 

[math] \alpha^2 \nabla^2 U = \frac{\partial U}{\partial t} = 0 [/math]

 

Now, we can divide through by alpha (so long as alpha isn't zero) to obtain:

 

[math] \nabla^2 U = 0 [/math]

[DavidAngelMX]

Yeah, you got it right. Just don't pay too much attention to the gray face, it's just the standard design for a parallellepiped in ms word

[DavidAngelMX]

Well, I guess this thread is not interesting to anyone anymore, so, Hasta la Vista

[Johnny5]

No it is still interesting. When they get the Latex fixed, I'll pick it up. If you aren't interested in it anymore fine... but I really still am, so I will finish it up. I need to work a problem using separation of variables again anyway, its been awhile since I've done one. So I am gonna finish this problem.

 

Regards