... | ... | @@ -41,7 +41,7 @@ D _{R\theta} & = & D _{\theta R} & = &{1\over 2} (R{\partial \over \partial R}{ |
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We can now compute the viscous heating (see Eq. 27.30 pag. 92 of Mihalas and Mihalas 1984):
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$`Q^+ = 2\rho \nu (D_{RR}^2+D_{\varphi \varphi}^2+D_{\theta \theta}^2
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+(D_{R\varphi}^2+D_{\theta\varphi}^2+D_{R\theta}^2)-
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+2((D_{R\varphi}^2+D_{\theta\varphi}^2+D_{R\theta}^2))-
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{1\over 3}(\nabla \cdot \vec v)^2)`$
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We will consider in the following the heating from the central star.
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... | ... | @@ -62,13 +62,13 @@ $`\begin{array}{lll} |
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where $`Q^+`$ is the viscous heating and $Q^-$ is the radiative cooling.
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the stress tensor in the 2 dimensional case is:
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$`Q^+ = 2\Sigma \nu (D_{RR}^2+D_{\theta \theta}^2+D_{R\theta}^2-
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$`Q^+ = 2\Sigma \nu (D_{RR}^2+D_{\theta \theta}^2+2D_{R\theta}^2-
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{1\over 3}(\nabla \cdot \vec v)^2)`$
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in the hypothesis the the gas has only Keplerian speed (no radial velocity and
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$`v_{\theta} = \Omega_K R`$ then:
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$`Q^+ = \Sigma \nu D_{R\theta}^2 = {9\over 8}\Sigma \nu \Omega_K^2`$
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$`Q^+ = \Sigma \nu D_{R\theta}^2 = {9\over 4}\Sigma \nu \Omega_K^2`$
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this simple prescription can also be implemented in a 2 dimensional disk.
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