EoM

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this page is about the Equations of Motion for viscous hydrodynamics in the SPH approach.

Equations of Motion[edit | edit source]

<math>

T^{\mu\nu}{}_{;\mu} = 0 </math>

<math>

\partial_\mu T^{\mu\nu} + \Gamma^{\mu}_{\mu\alpha} T^{\alpha\nu} + \Gamma^{\nu}_{\mu\alpha} T^{\mu\alpha} = 0 </math>

<math>

\frac{1}{\sqrt{-g}}\partial_\mu ( \sqrt{-g} T^{\mu\nu} ) + \Gamma^{\nu}_{\mu\alpha} T^{\mu\alpha} = 0 </math>

Hyperbolic coordinates[edit | edit source]

<math>

g_{\mu\nu} = diag( 1, -1, -1, -\tau^2 ) </math>

<math>

\sqrt{-g} = \tau \\ \Gamma^{\tau}_{\eta\eta} = \tau \\ \Gamma^{\eta}_{\tau\eta} = \Gamma^{\eta}_{\eta\tau} = \frac{1}{\tau} </math>

<math>

\frac{1}{\sqrt{-g}}\partial_\mu ( \sqrt{-g} T^{\mu\nu} ) + \Gamma^{\nu}_{\eta\eta} T^{\eta\eta} + \Gamma^{\nu}_{\tau\eta} T^{\tau\eta} + \Gamma^{\nu}_{\eta\tau} T^{\eta\tau} = 0 </math>

<math>

\frac{1}{\sqrt{-g}}\partial_\mu ( \sqrt{-g} T^{\mu\nu} ) + \Gamma^{\nu}_{\eta\eta} T^{\eta\eta} + 2\Gamma^{\nu}_{\tau\eta} T^{\tau\eta} = 0 </math>

<math>

\frac{1}{\tau}\partial_\mu ( \tau T^{\mu\nu} ) + \delta^{\nu\tau} \tau T^{\eta\eta} + \frac{2 \delta^{\nu\eta} }{\tau} T^{\tau\eta} = 0 </math>

Ideal hydrodynamics[edit | edit source]

The energy-momentum tensor of ideal hydrodynamics is expressed as

<math>

T^{\mu\nu} = \omega u^\mu u^\nu - Pg^{\mu\nu} </math>

First derivation[edit | edit source]

Now we apply the covariant derivative and open the metric derivative terms

<math>

\frac{1}{\tau}\partial_\mu ( \tau \omega u^\mu u^\nu ) - \frac{1}{\tau} \partial_\mu (\tau P g^{\mu\nu}) + \delta^{\nu\tau} \tau (\omega (u^\eta) ^2 + \frac{P}{\tau^2}) + \frac{2 \delta^{\nu\eta} }{\tau} ( \omega u^\tau u^\eta) = 0 </math> it is convenient to show the expressions for each component

<math>

\frac{1}{\tau}\partial_\mu ( \tau \omega u^\mu u^\tau ) = \frac{1}{\tau} \partial_\tau (\tau P) - \tau (\omega (u^\eta) ^2 + \frac{P}{\tau^2}) \\ \frac{1}{\tau}\partial_\mu ( \tau \omega u^\mu u^i ) = - \partial_i P \\ \frac{1}{\tau}\partial_\mu ( \tau \omega u^\mu u^\eta ) = - \frac{1}{\tau^2} \partial_\eta P - \frac{2}{\tau}( \omega u^\tau u^\eta) </math>

By introducing a conserved quantity <math>\sigma</math> one gets

<math>

(\sigma u^\mu)_{;\mu} = \frac{1}{\sqrt{-g}} \partial_\mu (\sqrt{-g} \sigma u^\mu) = 0 \\ \frac{1}{\tau} \partial_\mu (\tau \sigma u^\mu) = 0 </math>


<math>

\sigma u^\mu \partial_\mu ( \frac{\omega u^\tau }{\sigma} ) = \frac{1}{\tau} \partial_\tau (\tau P) - \tau \omega (u^\eta) ^2 - \frac{P}{\tau} = \partial_\tau P - \tau \omega (u^\eta) ^2 </math>

<math>

\sigma u^\mu \partial_\mu ( \frac{\omega u^i }{\sigma} ) = - \partial_i P </math>

<math>

\sigma u^\mu \partial_\mu ( \frac{\omega u^\eta }{\sigma} ) = - \frac{1}{\tau^2} \partial_\eta P - \frac{2}{\tau}( \omega u^\tau u^\eta) </math>

Second derivation[edit | edit source]

From the energy-momentum conservation

<math>

T^{\mu\nu}{}_{;\mu} = 0 </math> we can write

<math>

(\omega u^\mu u^\nu){}_{;\mu} - g^{\mu\nu} P{}_{;\mu} = 0 </math> using that <math>g^{\mu\nu}{}_{;\mu} = 0</math>. Now we introduce a conserved quantity <math>\sigma</math>

<math>

\sigma u^\mu (\frac{\omega u^\nu}{\sigma})_{;\mu} - g^{\mu\nu} \partial_\mu P = 0 </math>

<math>

\sigma u^\mu (\partial_\mu (\frac{\omega u^\nu}{\sigma}) + \Gamma^{\nu}_{\mu\alpha} \frac{\omega u^\alpha}{\sigma} ) - g^{\mu\nu} \partial_\mu P = 0 </math>

<math>

\sigma u^\mu \partial_\mu (\frac{\omega u^\nu}{\sigma}) = - \Gamma^{\nu}_{\mu\alpha} \omega u^\mu u^\alpha + g^{\mu\nu} \partial_\mu P </math>

the components are

<math>

\sigma u^\mu \partial_\mu (\frac{\omega u^\tau}{\sigma}) = - \Gamma^{\tau}_{\eta\eta} \omega (u^\eta)^2 + \partial_\tau P = - \tau \omega (u^\eta)^2 + \partial_\tau P \\ \sigma u^\mu \partial_\mu (\frac{\omega u^i}{\sigma}) = - \partial_i P \\ \sigma u^\mu \partial_\mu (\frac{\omega u^\eta}{\sigma}) = - 2\Gamma^{\eta}_{\eta\tau} \omega u^\eta u^\tau - \frac{1}{\tau^2} \partial_\eta P = - \frac{2}{\tau}\omega u^\eta u^\tau - \frac{1}{\tau^2} \partial_\eta P </math>

one can absorb the first term in the rhs of the <math>\eta</math>-component as such

<math>

\sigma u^\mu \partial_\mu (\frac{\omega \tau^2 u^\eta}{\sigma}) = - \partial_\eta P </math>

Hyperbolic coordinates[edit | edit source]

<math>

\partial_\mu (\tau \sigma u^\mu) = 0 \\ v^\mu \partial_\mu ( \tau \sigma u^\tau ) + \tau \sigma u^\tau \partial_\mu v^\mu = 0 </math>

we introduce the quantity <math>\bar\sigma = \tau \sigma u^\tau </math>

<math>

v^\mu \partial_\mu \bar\sigma + \bar\sigma \partial_\mu v^\mu = 0 </math>


<math>

\frac{ \sigma u^\mu }{\tau} \partial_\mu (\frac{\tau\omega u^\nu}{\sigma}) - g^{\mu\nu} \partial_\mu P = 0 \\ v^\mu \partial_\mu (\frac{\tau\omega u^\nu}{\sigma}) = \frac{ \tau }{ u^\tau \sigma } g^{\mu\nu} \partial_\mu P </math>

we introduce the quantity <math>q^\mu = \frac{\tau\omega u^\nu}{\sigma} </math>

<math>

\frac{ \sigma u^\mu }{\tau} \partial_\mu (\frac{\tau\omega u^\tau}{\sigma} ) - \partial_\tau P = 0 \\ \frac{ \sigma u^\mu }{\tau} \partial_\mu ( \frac{\tau\omega u^i }{\sigma} ) + \partial_i P = 0 \\ \frac{ \sigma u^\mu }{\tau} \partial_\mu ( \frac{\tau\omega u^\eta }{\sigma} ) + \frac{1}{\tau^2}\partial_\eta P = 0 </math>

<math>

v^\mu \partial_\mu (\frac{\tau\omega u^\tau}{\sigma} ) = \frac{ \tau }{ u^\tau \sigma}\partial_\tau P \\ v^\mu \partial_\mu ( \frac{\tau\omega u^i }{\sigma} ) = - \frac{ \tau }{u^\tau \sigma} \partial_i P \\ v^\mu \partial_\mu ( \frac{\tau\omega u^\eta }{\sigma} ) = -\frac{ 1}{\tau u^\tau \sigma}\partial_\eta P </math>

so the evolution equations are

<math>

v^\mu \partial_\mu ( \frac{\tau\omega u^i }{\sigma} ) = - \frac{ \tau }{u^\tau \sigma} \partial_i P \\ v^\mu \partial_\mu ( \frac{\tau\omega u^\eta }{\sigma} ) = -\frac{ 1}{\tau u^\tau \sigma}\partial_\eta P \\ v^\mu \partial_\mu \bar\sigma + \bar\sigma \partial_\mu v^\mu = 0 \\ \dot x^i = v^i \\ \dot \eta = v^\eta </math>

SPH[edit | edit source]

Conserved density[edit | edit source]

<math>

j^\mu(x) = \sum_a \dot x_a^\mu W_a(x) </math>

<math>

j^\tau = \sum_a W_a(x) </math> since <math>\dot \tau_a = 1</math>

<math>

j^\mu(x) = \tau \sigma u^\mu = j^\tau v^\mu </math>

<math>

v^\mu = \frac{j^\mu}{j^\tau} = \frac{\sum_a \dot x_a^\mu W_a(x)}{\sum_a W_a(x)} </math>

<math>

\partial_\mu j^\mu(x) = 0 \\ \partial_\tau j^\tau + \partial_i j^i + \partial_\eta j^\eta = 0 </math>

<math>

\sum_a \dot \tau_a \partial_\tau W_a(x) + \sum_a \dot x_a^i \partial_i W_a(x) + \sum_a \dot \eta_a \partial_\eta W_a(x) = 0 </math>


<math>

\partial_\tau W_a(x) = - \dot x_a^i \partial_i W_a(x) - \dot \eta_a \partial_\eta W_a(x) </math>


Lagrangean[edit | edit source]

<math>

I = \int \tau dxdy d\eta d\tau ( \varepsilon(\sigma) + A \partial_\mu (\tau \sigma u^\mu) + B(g_{\mu\nu} u^\mu u^\nu - 1) ) </math>

Case 1[edit | edit source]
<math>

\varepsilon = \sum_a \frac{\varepsilon(\sigma_a)}{\rho_a} W_a(x) </math>

<math>

I = \int d\tau \sum_a \frac{\tau \varepsilon(\sigma_a)}{\rho_a} + \int \tau dxdy d\eta (-\tau \sigma u^\mu \partial_\mu A) + B(g_{\mu\nu} u^\mu u^\nu - 1) </math>

by using

<math>

j^\mu(x) = \sum_a \dot x_a^\mu W(x-x_a) </math> the conservation is satistied and by setting <math>v^\mu = \dot x^\mu</math> the normalization is also automatically satisfied.

\delta x_m[edit | edit source]
<math>

\sum_a \partial_m ( \frac{\tau_a \varepsilon_a}{\rho_a} ) \\ = \sum_a \frac{\tau_a}{\rho^2_a} ( \rho_a \partial_m \varepsilon_a - \varepsilon_a \partial_m \rho_a ) \\ = \sum_a \frac{\tau_a}{\rho^2_a} ( \rho_a \frac{\omega_a}{\sigma_a} \partial_m \sigma_a - \varepsilon_a \partial_m \rho_a ) \\ = \sum_a \frac{\tau_a P_a }{\rho^2_a} \partial_m \rho_a \\ = \sum_a \frac{\tau_a P_a }{\rho^2_a} \sum_b \partial_m W_{ba} \\ = \sum_a \frac{\tau_a P_a }{\rho^2_a} \sum_b W'_{ba} r_{ba} (\delta_{bm} - \delta_{am}) \\ = \sum_a \frac{\tau_a P_a }{\rho^2_a} \sum_b W'_{ba} r_{ba} \delta_{bm} - \sum_a \frac{\tau_a P_a }{\rho^2_a} \sum_b W'_{ba} r_{ba} \delta_{am} \\ = \sum_a \frac{\tau_a P_a }{\rho^2_a} W'_{ma} r_{ma} - \frac{\tau_m P_m }{\rho^2_m} \sum_a W'_{am} r_{am} \\ = - ( \frac{\tau_m P_m }{\rho^2_m} \sum_a W'_{am} r_{am} + \sum_a \frac{\tau_a P_a }{\rho^2_a} W'_{am} r_{am} ) \\ = - \sum_a (\frac{\tau_m P_m }{\rho^2_m} + \frac{\tau_a P_a }{\rho^2_a}) W'_{am} r_{am} </math>

<math>

- \sum_a \partial_m ( \frac{\tau_a \varepsilon_a}{\rho_a} ) = - (\frac{\tau_m P_m }{\rho^2_m} (\nabla\rho)_m + ( \nabla \frac{\tau P}{\rho})_m) </math>

\delta \dot x_m[edit | edit source]
<math>

\sum_a \partial_{\dot x_m} ( \frac{\tau_a \varepsilon_a}{\rho_a} ) \\ = \sum_a \frac{ \tau_a }{ \rho_a } \partial_{\dot x_m}\varepsilon_a \\ = \sum_a \frac{ \tau_a }{ \rho_a } \frac{\omega_a}{\sigma_a} \partial_{\dot x_m}\sigma_a \\ = \sum_a \frac{ \tau_a }{ \rho_a } \frac{\omega_a}{\sigma_a} \frac{\rho_a}{\tau_a} \partial_{\dot x_m} \gamma^{-1}_a \\ = - \frac{\omega_m}{\sigma_m} \gamma_a g_{ij} \dot x_a^j \delta_{am} \\ = - \frac{\omega_m}{\sigma_m} \gamma_m g_{ij} \dot x_m^j </math>

where we used that

<math>

\partial_{\dot x_m} \gamma^{-1}_a \\ = \partial_{\dot x_m} ( 1 + g_{ij} \dot x_a^i \dot x_a^j)^{1/2} \\ = \frac{1}{2}( 1 + g_{ij} \dot x_a^i \dot x_a^j)^{-1/2} g_{ij} \partial_{\dot x_m}( \dot x_a^i \dot x_a^j) \\ = \gamma_a g_{ij} \dot x_a^j \delta_{am} </math>

<math>

\sum_a \partial_{\dot x_m} ( \frac{\tau_a \varepsilon_a}{\rho_a} ) = - \frac{\omega_m}{\sigma_m} \gamma_m g_{ij} \dot x_m^j = q_j{}_m = - \frac{\tau_m \omega_m}{\rho_m} g_{ij} \gamma_m^2 \dot x_m^j = - g_{ij} \frac{\tau_m T^{\tau j}_m}{\rho_m} </math>

In order to extract the velocity out of the momentum one first computes

<math>

q_j = - \frac{\tau \gamma \omega}{\rho} g_{ij} u^j </math>

<math>

\sqrt{ -q^2 } = \sqrt{ -q^i q_i } = \sqrt{ -g^{ij}q_i q_j } = \frac{\tau \gamma \omega}{\rho} \sqrt{ -g_{ij} u^i u^j } = \frac{\tau \gamma \omega}{\rho} \sqrt{ -u^2 } \gt 0 </math> where we used that <math>\gamma^2 = 1 - g_{ij} \gamma^2 \dot x^i \dot x^j</math>. Therefore,

<math>

\frac{q_i}{ \sqrt{-q^2} } = - \frac{1}{ \sqrt{ \gamma^2 - 1 } } g_{ij}\gamma \dot x^j </math> or

<math>

u^i = \frac{ \sqrt{ -u^2 } }{ \sqrt{-q^2} } ( - g^{ij} q_j ) </math>

equations of motion[edit | edit source]
<math>

\frac{d}{d\tau} (-\frac{\omega_m}{\sigma_m} \gamma_m g_{ij} \dot x_m^j) = -\sum_a (\frac{\tau_m P_m }{\rho^2_m} + \frac{\tau_a P_a }{\rho^2_a}) W'_{am} r_{am} </math>

<math>

\frac{d}{d\tau} q_m^j = - (\frac{\tau_m P_m }{\rho^2_m} (\nabla \rho)_m + (\nabla \frac{\tau P}{\rho})_m) </math>

where

<math>

\rho_m = \sum_a W_{am} \\ (\nabla \rho)_m = \sum_a W'_{am} r_{am} </math>

Furthermore

<math>

E = q^\tau = \sum_a v^i_a q_a{}_i - L \\ = \sum_a v^i_a q_a{}_i + \frac{\tau \varepsilon}{\rho} </math>

<math>

v^\mu \partial_\mu \frac{ \omega g_{ij} u^j }{ \sigma } \\ = \frac{\tau P}{\rho^2} \partial_i \rho + \partial_i \frac{\tau P}{\rho} \\ = \frac{1}{\rho} ( \frac{\tau P}{\rho} \partial_i \rho + \rho \partial_i \frac{\tau P}{\rho} ) \\ = \frac{1}{\rho} \partial_i( \tau P ) </math>

<math>

\sigma u^\mu \partial_\mu \frac{ \omega g_{ij} u^j }{ \sigma } = \partial_i P </math>

Case 2[edit | edit source]
<math>

j^\tau \varepsilon = \sum_a \varepsilon_a W_a(x) \\ \varepsilon = \frac{\sum_a \varepsilon_a W_a(x)}{\sum_a W_a(x)} </math>

<math>

\varepsilon = j^\tau E = \sum_a E_a W_a(x) </math>

SPH properties[edit | edit source]

We start we the identity (in n dimensions)

<math>

\phi(x) = \int d^d x \phi(x') \delta^d(x-x') </math>

Using the distribution W we rewrite this as

<math>

\phi(x) = \lim_{h\to 0} \int d^d x \phi(x') W^d(x-x') </math>

<math>

\phi(x) = \lim_{h\to 0}\lim_{N\to \infty} \sum_a^N V_a \phi(x_a) W^d(x-x_a) </math>


Given a conserved density

<math>

\rho(x) = \sum_a \nu_a W(x - r_a) </math> we can calculate its integral to be

<math>

\int d^3x \rho(x) = \sum_a \nu_a \int d^3x W(x - r_a) = \sum_a \nu_a </math>

We can relate another quantity with the conserved quantity by

<math>

A(x) = \sum_a \frac{\nu_a A_a}{\rho_a} W(x-r_a) </math> in this way we get

<math>

\int d^3x A(x) = \sum_a \frac{\nu_a A_a}{\rho_a} </math>

we first assume

<math>

\rho_a = \sum_b \nu_b W(r_a - r_b) = \nu_a W(0) + \sum_{b\neq a} \nu_b W(r_a - r_b) </math> and therefore,

<math>

A_a = \sum_b \frac{\nu_b A_b}{\rho_b} W(r_a - r_b) </math>

now if we use this is the previous definition of A(x) for self-consistency we get

<math>

\int d^3x A(x) = \sum_a \frac{\nu_a A_a}{\rho_a} = \sum_a \frac{\nu_a}{\rho_a} \sum_b \frac{\nu_b A_b}{\rho_b} W(r_a-r_b) \\ = \sum_b \frac{\nu_b A_b}{\rho_b} \sum_a \frac{\nu_a}{\rho_a} W(r_a-r_b) </math> so that,

<math>

\sum_a \frac{\nu_a}{\rho_a} W(r_a-r_b) = 1 </math>

<math>

\sum_a \frac{\nu_a}{\rho_a} \int d^3x W(r_a - x) = \int d^3x 1 </math>

<math>

\sum_a \frac{\nu_a}{\rho_a} = V = \sum_a V_a </math>