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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?

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  • $\begingroup$ 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? $\endgroup$ – TotallyRhombus Feb 4 '17 at 21:12
  • $\begingroup$ Isn't it $T(\mathbf x)\to T(r)$ s.t. $\partial_\phi T=\partial_\theta T=0$? $\endgroup$ – Kyle Kanos Feb 4 '17 at 21:12
  • $\begingroup$ 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. $\endgroup$ – Physther Feb 4 '17 at 21:26
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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: $$ T(\mathbf x)\equiv T(r,\,\phi,\,\theta)\to T(r) $$

Since the temperature no longer depends on the polar and azimuthal angles, then the derivatives with respect to those angles are zero and so the heat equation reduces to your second equation.

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