Timeline for The derivation of the Planck distribution
Current License: CC BY-SA 3.0
10 events
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| Jan 31, 2019 at 2:15 | comment | added | Hans | As I suspected, point 3. is a hard question, not answerable by your hand-waving but hard thinking by outstanding mathematicians. See my edit to point 3. of my question. | |
| Dec 21, 2018 at 11:06 | comment | added | Hans | @OktayDoğangün: I did not see your comment until now. Question 2. needs the quantum electrodynamics theory. My question 3 enquires about the affect of the geometry of the box on the distribution. Do you have any thoughts on its answer? | |
| Jul 13, 2018 at 7:06 | comment | added | Oktay Doğangün | The rigorous way for a photon in a box does not necessarily require QFT. But it surely needs relativistic quantum mechanics since photon is a relativistic particle, and Schrödinger equation is strictly non-relativistic and requires $m \neq 0$. | |
| Jul 1, 2017 at 22:41 | comment | added | Hans | 3) The note you link to has the exact same expression for $n$ and is premised on it being a cube --- even a rectangular box will make the expression different, not to mention a sphere which requires spherical harmonics or a cylinder which requires Bessel functions. By what do you conclude "photons inside some cavity with an arbitrary shape, and such shape doesn't really accounts when summing states in the thermodynamic limit"? That is a pretty big statement. Again, I have no use for hand waving arguments which come a dime a dozen, well actually come for free. Please derive it rigorously. | |
| Jul 1, 2017 at 22:21 | comment | added | rsaavedra | 2) I think you are right, the rigorous way to do it is QFT 3) If you calculate the number of particles $N$ you would see that the energy has dependency $N/L$ which is constant in the thermodynamic limit. I don't know what is a fractal geometry, but the point is that you have photons inside some cavity with an arbitrary shape, and such shape doesn't really accounts when summing states in the thermodynamic limit. I recommend reading the notes of D. Tong about quantum gases damtp.cam.ac.uk/user/tong/statphys/three.pdf | |
| Jul 1, 2017 at 21:58 | comment | added | Hans | 2) What is the "similar way"? Please do not give me a hand waving argument, which I can manufacture myself easily. What is the equation, as it is not Schrodinger? It must be QFT then? 3) I understand it is for $N\rightarrow\infty$. The $L$ is divided away in the unit energy form. However, it is under the assumption that the cube is not changing as $N\rightarrow\infty$ and it depends on the square root of sum form of $n$ in the article. A different geometry may not give the same result. Will you be able to prove the same thing for, say, a fractal geometry? | |
| Jul 1, 2017 at 21:35 | comment | added | rsaavedra | 2) Ok the problem is not exactly the same, but can be done in a similar way remembering that the momentum of a photon is $p=h\nu/c$. 3) What you usually mean by distribution is "particle number distribution", which is explicitly independent of $L$... but it is true it may depend of it via the energy per particle $\epsilon$. Of course the energy function will have a different form if you consider a different geometry of the system, e.g. photons in cylinder, sphere, but the thermodynamic limit, $N\rightarrow\infty$, $L\rightarrow\infty$ will always get rid of any dependency on the geometry | |
| Jul 1, 2017 at 21:29 | comment | added | Hans | 1) You are right. Thank you. 2) You are wrong. The link you provided is for a positive mass particle. A photon is massless, thus does not obey the Schrodinger equation. 3) You are wrong. Look at Equation (1). It has explicitly $L$ the length of the box in it, and the form of the square root of the sum is implicitly obtained by the boundary being rectangular. The equation above Equation (3) is an approximation when $n$ is large and when $n$ assumes the square root sum form which is constrained by the geometry. | |
| Jul 1, 2017 at 21:01 | history | edited | rsaavedra | CC BY-SA 3.0 |
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| Jul 1, 2017 at 20:32 | history | answered | rsaavedra | CC BY-SA 3.0 |