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I have a very limited knowledge of tensor calculus, and I've never had proper exposure to general relativity, but I hope that the people reading this forum are able to help out.
So I'm doing some stat. mech. and a part of a system's free energy is
\mathcal{F} = \int V(\rho)\nabla^2\rho dx
I'd like to derive the stress/pressure tensor. According to a reference, taking linear combinations of all symmetric tensors s.t. the equilibrium condition is satisfied (div P = 0), I should get the form
P_{\alpha\beta} = ((\mu \rho - f) + (\nabla V(\rho))\cdot (\nabla\rho) + V(\rho)\nabla^2\rho)\delta_{\alpha\beta} - \partial_\alpha V(\rho)\partial_\beta\rho - \partial_\alpha \rho\partial_\beta V(\rho)
where \mu = \frac{\delta\mathcal{F}}{\delta \rho}, f = V(\rho)\nabla^2\rho
Ok, I thought, I want to test out a rather obscure reference from 30 years ago, where a form for the pressure tensor was derived as
\frac{1}{2}\sqrt{g}P_{\alpha\beta} = \frac{\delta \mathcal{F}}{\delta g^{\alpha\beta}}
So writing the free energy in generalized coordinates, I suppose I'd get something like
\mathcal{F} = \int \sqrt{g}V(\rho) \frac{1}{\sqrt{g}} \partial_i(\sqrt{g}g^{ij}\rho_{,j}) dx
Taking the variation, I'm led to believe that I get EDIT: I believe this is wrong, as the variation depends on derivatives of gij as well. Using this fact and the Cristoffel symbols, for the final operator, I think I might get somewhere
P_{\alpha\beta} = (\mu \rho - f)g_{\alpha\beta} + 2\frac{\partial f}{\partial g^{\alpha\beta}}
where f = V(\rho) \frac{1}{\sqrt{g}} \partial_i(\sqrt{g}g^{ij}\rho_{,j})
This means I should take the derivative of the Laplace-Beltrami operator wrt the metric tensor. I tried using different identities for the derivatives of √g, but couldn't get where I wanted to. Any help would be greatly appreciated. I'm a bit worried that the derivatives on V might not appear on the final form, as they do in the reference (derived through a different route).
So I'm doing some stat. mech. and a part of a system's free energy is
\mathcal{F} = \int V(\rho)\nabla^2\rho dx
I'd like to derive the stress/pressure tensor. According to a reference, taking linear combinations of all symmetric tensors s.t. the equilibrium condition is satisfied (div P = 0), I should get the form
P_{\alpha\beta} = ((\mu \rho - f) + (\nabla V(\rho))\cdot (\nabla\rho) + V(\rho)\nabla^2\rho)\delta_{\alpha\beta} - \partial_\alpha V(\rho)\partial_\beta\rho - \partial_\alpha \rho\partial_\beta V(\rho)
where \mu = \frac{\delta\mathcal{F}}{\delta \rho}, f = V(\rho)\nabla^2\rho
Ok, I thought, I want to test out a rather obscure reference from 30 years ago, where a form for the pressure tensor was derived as
\frac{1}{2}\sqrt{g}P_{\alpha\beta} = \frac{\delta \mathcal{F}}{\delta g^{\alpha\beta}}
So writing the free energy in generalized coordinates, I suppose I'd get something like
\mathcal{F} = \int \sqrt{g}V(\rho) \frac{1}{\sqrt{g}} \partial_i(\sqrt{g}g^{ij}\rho_{,j}) dx
Taking the variation, I'm led to believe that I get EDIT: I believe this is wrong, as the variation depends on derivatives of gij as well. Using this fact and the Cristoffel symbols, for the final operator, I think I might get somewhere
P_{\alpha\beta} = (\mu \rho - f)g_{\alpha\beta} + 2\frac{\partial f}{\partial g^{\alpha\beta}}
where f = V(\rho) \frac{1}{\sqrt{g}} \partial_i(\sqrt{g}g^{ij}\rho_{,j})
This means I should take the derivative of the Laplace-Beltrami operator wrt the metric tensor. I tried using different identities for the derivatives of √g, but couldn't get where I wanted to. Any help would be greatly appreciated. I'm a bit worried that the derivatives on V might not appear on the final form, as they do in the reference (derived through a different route).
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