Say you are modelling heat diffusion from hot air into a sphere. The thermal conductivity is the same at all positions on the sphere.

Using the 1D diffusion equation, the temperature of the sphere is only a function of its radius and time. If I model the system using spherical coordinates, the concentration is still only a function of its radius and time in every direction.

I imagine that the 1D profile of heat should be equivalent to a profile through sphere as a function of its radius.

If that is the case, what is the point of being able to model diffusion in a sphere, when it can be modeled more simply with a 1D equation when there is no variation of the thermal conductivity in any direction?

Or am I totally missing something?

  • 1
    $\begingroup$ You seem to be missing that fact that "radially symmetrical" is not the same as "one-dimensional in a straight line". Sure, you can write the radially symmetrical diffusion as a first-order differential equation where the only variables are the radius and time, but it is a different equation from the genuinely 1-D diffusion equation. $\endgroup$
    – alephzero
    Apr 28 '21 at 21:47
  • $\begingroup$ radial diffusion is a 3D process where you can ignore two-dimensions due to symmetry. True 1D diffusion...there are no other two-dimensions to account for. Perhaps think about it like this. in the radial case you can estimate a rate through a patch of surface area of the sphere. In a 1D system, this has no meaning. The underlying results therefore are very different. $\endgroup$ Apr 29 '21 at 1:57
  • $\begingroup$ @mathstackuser12 But isn't 1D diffusion essentially equivalent to just diffusion through an infinitesimally small surface area? $\endgroup$ Apr 29 '21 at 6:50
  • $\begingroup$ @DavidPierson yes apologies i should have been more clear. 1D heat diffusion derives from flux through an infinitesimal area dydz say (assuming diffusion in the x-direction). if there is no transfer in any other direction, then in the limit this becomes the one-dimensional heat equation. There is no symmetry here and there is effectively no other dimension regarding the transfer. In the radially symmetric case, the diffusion is 3-dimensional. The radially symmetric variable 'points' in all directions in space. Perhaps i should move from comments to an answer. $\endgroup$ Apr 29 '21 at 9:22

Let's work it out. Let $\Phi(x,y,z,t)$ be the temperature profile of your sphere, which by assumption is a solution to the heat equation:

$$\Phi_t = \Delta\Phi$$

(setting the constant diffusion coefficient equal to 1). By your assumptions the initial conditions are spherically symmetric, and $\Phi$ will be spherically symmetric at all times.

Let $\Psi$ be the function $\Phi$ written in spherical coordinates, so $\Psi(r, \phi, \theta, t) = \Phi(x,y,z,t)$. The Laplacian in spherical coordinates applied to $\Psi$ is

\begin{align*}\Delta\Psi &= \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \Psi}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial \Psi}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 \Psi}{\partial \varphi^2} \\ &= \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \Psi}{\partial r}\right),\end{align*}

the second equality by your assumptions which imply that $\Phi$ is spherically constant, so $\Psi$ doesn't depend on $\theta$ and $\phi$.

We see now that the heat equation in spherical coordinates in this spherically symmetric situation becomes

$$\frac{\partial\Psi}{\partial t} = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \Psi}{\partial r} \right),$$

not quite the same form as the heat equation in 1D.


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