# Spherical Coordinates Heat Equation

Can anyone help me to understand the difference between spherical coordinates and spherical coordinates with symmetry? I found two formulations for the heat equation:

$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(kr^2\frac{\partial T}{\partial r}\right) + \frac{1}{r^2 \sin^2\phi}\frac{\partial}{\partial \phi }\left(k\frac{\partial T}{\partial \phi}\right)+\frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta }\left(k \sin\theta \frac{\partial T}{\partial \theta}\right) = \rho C_p \frac{\partial T}{\partial t} \tag{1}$$ and: $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(kr^2\frac{\partial T}{\partial r}\right) = \rho C_p \frac{\partial T}{\partial t} \tag{2}$$

And I don't really understand why the last one is reduced. I searched in different places and I find sometimes that there can be some differences because of... symmetry of the sphere. Can anyone help me to understand why a symmetrical condition would cancel the sin terms and what is a symmetrical sphere? Aren't all the spheres symmetrical?

• A general function on $\mathbb R^3$ depends on $r$ as well as $\theta$ and $\phi$. If the function is isotropic around some point however, it depends only on $r$. Does this help? – TotallyRhombus Feb 4 '17 at 21:12
• Isn't it $T(\mathbf x)\to T(r)$ s.t. $\partial_\phi T=\partial_\theta T=0$? – Kyle Kanos Feb 4 '17 at 21:12
• Thank you, guys. @fs137, It might help, I just need to read about isotropy. Kyle, can you please explain more? The second part of the formulation makes sense. I don't understand the T(x) -> T(r) part. – Physther Feb 4 '17 at 21:26

Spherical symmetry basically indicates that the dependence of the function, the temperature in this case, is only in the radial direction, $r$ (i.e., it does not depend on $\phi$ and $\theta$). In this case, the temperature at $T(100,\,30^\circ,\,60^\circ)$ is the same at $T(100,\,20^\circ,\,10^\circ)$ but different from $T(90,\,30^\circ,\,60^\circ)$. Since it only depends on the radial direction, we can just drop the $\theta$ and $\phi$ arguments of the function: $$T(\mathbf x)\equiv T(r,\,\phi,\,\theta)\to T(r)$$