• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

203 submissions , 161 unreviewed
5,007 questions , 2,163 unanswered
5,341 answers , 22,655 comments
1,470 users with positive rep
815 active unimported users
More ...

  Under which circumstances is the thermalization time proportional to the square of the length?

+ 1 like - 0 dislike

I had heard that the thermalization time (i.e. the time required for an object to reach its final temperature, or say 99% of it) was proportional to its length squared. I hadn't given it much thoughts until I found out this statement in Onsager's famous 1931 free access paper, on page 419:

We may also recall that the time needed for equalization of temperature in a body is proportional to the square of its linear dimensions

But now that I think about it, I do not think this holds under any circumstances, and I would like to know when this holds, and when this doesn't.

Here are my thoughts:

  • By length, I assume they mean the length in the direction of heat propagation.
  • Intuitively, I think the statement holds for objects to which Dirichlet boundary conditions are applied. For example, let's take a cylinder of height L (let's say it's a metallic rod), whose ends are kept at different constant temperatures, and initially it has a uniform temperature. In that case, if I solve the PDE (which is simplified to a 1 dimensional PDE since there's no radial nor angular dependence of the temperature) $\kappa \frac{\partial T}{\partial z} = C_p\frac{\partial T}{\partial t}$ and if I cheat by considering a temperature as being equal to absolute $0$ (else I do not get a nice quantization of eigenvalues), I am almost able to reach that the solution is an infinite series of terms containing $\exp (\alpha_n t/ L^2)$. Note that the radius and cross section area of the cylinder does not enter into play. Thus the statement holds, in agreement with intuition.
  • But also intuitively, if instead of keeping the cylinder ends at fixed temperature, if one applies a constant heat flux, which would correspond to Neumann boundary conditions, then the statement should not hold anymore. Why? Because heating an end with, say a $10 W$ power source, will certainly not have the same effect on a small radius cylinder compared to a bigger radius cylinder. Clearly, a huge metal cylinder is going to heat up more slowly if its radius is much bigger than a cylinder with a tiny radius, assuming they have equal length. In fact, since the volume of the cylinder goes as the radius squared, I expect a cylinder with a radius twice as big as the one of a smaller cylinder to take 4 times more time to thermalize. However, I see no way whatsoever for the cross sectional area, or the cylinder of the radius, to appear in the heat equation (and its boundary conditions), even with Neumann b.c., which would be something of the sort: $\frac{\partial T}{\partial x}_{\text{one end}}=\text{constant}$. I am therefore unable to formally work out what my intuition claims. Can someone clear this stuff up? Am I right in claiming that for constant power source applied on the end(s) of an object, Onsager statement does not hold? If so, how can I see it mathematically? How come I am unable to make the radius of the cylinder appear for the solution to the heat equation under these boundary conditions? How does the radius appear?
  • Clearly, radiation effects are neglected. It is clear to me that the statement would hold when radiation effects are not taken into account. These are nonlinear effects that are proportional to the surface area.
asked Sep 7, 2018 in General Physics by comuldlens_*%tester (5 points) [ no revision ]

indeed, it is a rough law in general, but its application is approximative and only in the enumerated conditions. Further, other variables are assumed to be random, excluding specific symmetric configurations and boundaries are considered.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights