There has been a popular physics problem floating around regarding how one might cook a turkey by slapping it. Supposedly the heat generated by one slap making contact could cook it, leading to some exaggerated quantities necessary for the thermodynamics involved. As it turns out, this problem is easy enough to solve (and has been done before), so I will demonstrate here the process of solving the partial differential equation (PDE) analytically.<\/p>\n\n\n\n
As a first approximation, we will assume the chicken is in the shape of a sphere. The results of more appropriate shapes such as prolate spheroidal, oblate spheroidal, or elliptical geometry for the chicken will result in essentially the same overall behavior but with different separable Laplacians, orthogonal coordinate systems, characteristic orthogonal polynomial series approximations, etc. Given the spherical geometry we have the following form of the heat equation to solve:<\/p>\n\n\n
$$
\n\\frac{\\partial T}{\\partial t} = k \\nabla^2 T + f(\\rho, \\theta, \\phi, t), \\quad \\mathbf{x} \\in \\Omega,\\ t \\in [0, \\infty)
\n$$<\/p>\n
$$
\n\\Rightarrow \\frac{\\partial T}{\\partial t} = k \\left[
\n\\frac{1}{\\rho^2} \\frac{\\partial}{\\partial \\rho} \\left( \\rho^2 \\frac{\\partial T}{\\partial \\rho} \\right)
\n+ \\frac{1}{\\rho^2 \\sin \\theta} \\frac{\\partial}{\\partial \\theta} \\left( \\sin \\theta \\frac{\\partial T}{\\partial \\theta} \\right)
\n+ \\frac{1}{\\rho^2 \\sin^2 \\theta} \\frac{\\partial^2 T}{\\partial \\phi^2}
\n\\right] + f(\\rho, \\theta, \\phi, t)
\n$$<\/p>\n
$$
\nT(\\rho, \\theta, \\phi, t)\\big|_{\\mathbf{x} \\in \\partial \\Omega} = T_a
\n$$<\/p>\n\n\n\n
Now we need to standardize the heat at boundaries to look more like Dirichlet boundary conditions, so then:<\/p>\n\n\n
$$
\nu(\\rho, \\theta, \\phi, t) = T(\\rho, \\theta, \\phi, t) \\mminus T_a
\n$$<\/p>\n
$$
\n\\Rightarrow \\frac{\\partial u}{\\partial t} = k \\nabla^2 u + f(\\rho, \\theta, \\phi, t)
\n$$<\/p>\n
$$
\nu(\\rho, \\theta, \\phi, t)\\big|_{\\mathbf{x} \\in \\partial \\Omega} = 0
\n$$<\/p>\n\n\n\n
In this particular problem we have azimuthal symmetry so we can get rid of the \\(\\phi\\) dependence:<\/p>\n\n\n
$$
\n\\frac{\\partial u}{\\partial t} = k \\left[
\n\\frac{1}{\\rho^2} \\frac{\\partial}{\\partial \\rho} \\left( \\rho^2 \\frac{\\partial u}{\\partial \\rho} \\right)
\n+ \\frac{1}{\\rho^2 \\sin \\theta} \\frac{\\partial}{\\partial \\theta} \\left( \\sin \\theta \\frac{\\partial u}{\\partial \\theta} \\right)
\n\\right] + f(\\rho, \\theta, t)
\n$$<\/p>\n
$$
\nu(\\rho, \\theta, t)\\big|_{\\mathbf{x} \\in \\partial \\Omega} = 0
\n$$<\/p>\n\n\n\n
By inspection we can see the left hand side only depends on time, and one term of the right hand side depends only on radial position. This allows us to form a solution by separation of variables. We form the product of separable functions:<\/p>\n\n\n
$$u(\\rho, \\theta, t) = R(\\rho)\\Theta(\\theta)T(t)$$<\/p>\n\n\n\n
Substituting this into the general homogeneous equation gives us:<\/p>\n\n\n
$$
\nR(\\rho)\\Theta(\\theta)T'(t) =
\nk \\left[
\n\\frac{1}{\\rho^2} \\frac{\\partial}{\\partial \\rho} \\left( \\rho^2 R'(\\rho)\\Theta(\\theta)T(t) \\right)
\n+ \\frac{1}{\\rho^2 \\sin \\theta} \\frac{\\partial}{\\partial \\theta} \\left( \\sin \\theta R(\\rho)\\Theta'(\\theta)T(t) \\right)
\n\\right]
\n$$<\/p>\n\n\n\n
So then dividing through by the function we have:<\/p>\n\n\n
$$
\n\\frac{T'(t)}{T(t)} =
\nk \\left[
\n\\frac{1}{\\rho^2 R(\\rho)} \\frac{d}{d \\rho} \\left( \\rho^2 \\frac{d R}{d \\rho} \\right)
\n+ \\frac{1}{\\rho^2 \\sin \\theta \\Theta(\\theta)} \\frac{d}{d \\theta} \\left( \\sin \\theta \\frac{d \\Theta}{d \\theta} \\right)
\n\\right]
\n$$<\/p>\n\n\n\n
This results in a term that depends only on \\( t \\), so we are permitted to set this invariant as an unknown constant, \\( C_1 \\). Upon algebraic rearrangement of variables:<\/p>\n\n\n
$$
\n\\frac{C_1 \\rho^2}{k} =
\n\\frac{1}{R(\\rho)} \\frac{d}{d \\rho} \\left( \\rho^2 \\frac{d R}{d \\rho} \\right)
\n+ \\frac{1}{\\sin \\theta \\Theta(\\theta)} \\frac{d}{d \\theta} \\left( \\sin \\theta \\frac{d \\Theta}{d \\theta} \\right)
\n$$<\/p>\n
$$
\n\\Rightarrow
\n\\frac{1}{R(\\rho)} \\frac{d}{d \\rho} \\left( \\rho^2 \\frac{d R}{d \\rho} \\right)
\n\\mminus \\frac{C_1 \\rho^2}{k} = \\mminus C_2
\n$$<\/p>\n\n\n\n
So now we have:<\/p>\n\n\n
$$
\nT'(t) = C_1 T(t)
\n$$<\/p>\n
$$
\n\\rho^2 R^{\u201d}(\\rho) + 2 \\rho R'(\\rho) + \\left( C_2 \\mminus \\frac{C_1 \\rho^2}{k} \\right) R(\\rho) = 0
\n$$<\/p>\n
$$
\n\\Theta^{\u201d}(\\theta) + \\cot \\theta \\Theta'(\\theta) \\mminus C_2 \\Theta(\\theta) = 0
\n$$<\/p>\n\n\n\n
The spherical Laplacian<\/strong> acts on a scalar function \\( u(r, \\theta, \\phi) \\) and is given by:<\/p>\n\n\n $$ $$ $$ where \\( \\nabla_{\\Omega}^2 \\) is the angular Laplacian<\/strong>.<\/p>\n\n\n\n $$ $$ $$ $$ Now we can narrow the solution down better.<\/p>\n\n","protected":false},"excerpt":{"rendered":" There has been a popular physics problem floating around regarding how one might cook a turkey by slapping it. Supposedly the heat<\/p>\n
\n\\nabla^2 u =
\n\\frac{1}{r^2} \\frac{\\partial}{\\partial r} \\left( r^2 \\frac{\\partial u}{\\partial r} \\right)
\n+ \\frac{1}{r^2 \\sin \\theta} \\frac{\\partial}{\\partial \\theta} \\left( \\sin \\theta \\frac{\\partial u}{\\partial \\theta} \\right)
\n+ \\frac{1}{r^2 \\sin^2 \\theta} \\frac{\\partial^2 u}{\\partial \\phi^2}
\n$$<\/p>\n\n\n\n
\n\n\n\nSolution via Separation of Variables<\/strong><\/h3>\n\n\n
\nu(r, \\theta, \\phi) = R(r) Y(\\theta, \\phi)
\n$$<\/p>\n
\n\\nabla^2 u =
\nY(\\theta, \\phi) \\left[ \\frac{1}{r^2} \\frac{d}{dr} \\left( r^2 \\frac{dR}{dr} \\right) \\right]
\n+ R(r) \\left[ \\frac{1}{r^2} \\nabla_{\\Omega}^2 Y(\\theta, \\phi) \\right]
\n$$<\/p>\n\n\n\n
\n\n\n\nAngular Part: Spherical Harmonics<\/h3>\n\n\n
\n\\nabla_{\\Omega}^2 Y_{\\ell m}(\\theta, \\phi) = \\mminus \\ell(\\ell + 1) Y_{\\ell m}(\\theta, \\phi)
\n$$<\/p>\n\n\n\n
\n\n\n\nRadial Part<\/h3>\n\n\n
\n\\frac{1}{r^2} \\frac{d}{dr} \\left( r^2 \\frac{dR}{dr} \\right)
\n\\mminus \\frac{\\ell(\\ell + 1)}{r^2} R = 0
\n$$<\/p>\n
\n\\Rightarrow \\frac{d}{dr} \\left( r^2 \\frac{dR}{dr} \\right)
\n\\mminus \\ell(\\ell + 1) R = 0
\n$$<\/p>\n\n\n\n
\n\n\n\nGeneral Solution<\/h3>\n\n\n
\n\\nabla^2 u = 0 \\quad \\Rightarrow \\quad
\nu(r, \\theta, \\phi) = \\sum_{\\ell=0}^\\infty \\sum_{m=\\mminus\\ell}^{\\ell} \\left(
\nA_{\\ell m} r^\\ell + \\frac{B_{\\ell m}}{r^{\\ell + 1}}
\n\\right) Y_{\\ell m}(\\theta, \\phi)
\n$$<\/p>\n\n\n\n\n
\n\n\n\n