Key Notes on Equation of Motion of Hydrodynamic Model

Key Notes on Equation of Motion of Hydrodynamic Model

参考文献列表

 * arXiV:arXiv:0901.4355 Early collective expansion: Relativistic hydrodynamics and the transport properties of QCD matter by Ulrich Heinz
 * arXiV:hep-ph/0407360 Concepts of Heavy-Ion Physics by Ulrich Heinz
 * A first course in General Relativity by Schutz
 * Fluid Mechanics by Landau
 * General Relativity and Cosmology by Steven Weinberg
 * Introduction to RHIC by L.P. Csernai

理想流体运动方程,能动张量守恒
一个比较好的方式是把流体力学方程表达成垂直于四速度与平行于四速度的分量形式. 首先,我们给出不考虑粘滞度情况下的流体力学方程(参考Heinz讲义),它们是


 * $$\begin{align}

&\partial_{\mu}N^{\mu}=0 \\ &N^{\mu}=nu^{\mu} \\ &\partial_{\mu}T^{\mu\nu}=0 \\ &T^{\mu\nu}=(\varepsilon+P)u^{\mu}u^{\nu}-Pg^{\mu\nu} \end{align}$$

对多组元体系,流守恒对每个独立的守恒荷分别成立,但是所有的守恒荷通过相同的粒子流承载故具有相同的四速度


 * $$\begin{align}

&\partial_{\mu}N_p^{\mu}=0 \\ &N_p^{\mu}=n_pu^{\mu} \end{align}$$

利用投影算符,利用关系 $$\partial^{\mu}=u^{\mu}D+\nabla^{\mu}$$ ,可以化为垂直于速度的分量 $$\nabla^{\mu} \equiv \Delta^{\mu\nu} \partial_{\nu}$$ (其中 $$\Delta^{\mu\nu}\equiv g^{\mu\nu}-u^{\mu}u^{\nu}$$ )和平行于速度的分量 $$D \equiv u^{\nu}\partial_{\nu}$$.引入记号


 * $$\begin{align}

&\theta \equiv \partial \cdot u\equiv \partial_{\mu}u^{\mu} \\ &\dot f =u^{\mu} \partial_{\mu}f \equiv u\cdot \partial f\equiv Df \end{align}$$

其中第一个记号相当于体系在随动系的膨胀速率.这是因为


 * $$\begin{align}

&\partial_{\mu}(\rho u^{\mu})=u^{\mu}\partial_{\mu}\rho+\rho\partial_{\mu}u^{\mu}=0 \\ &\partial_{\mu}u^{\mu}=-\frac{u^{\mu}\partial_{\mu}\rho}{\rho}=-\frac{\dot{\rho}}{\rho} \end{align}$$

在它不等于零的时候,它反映了体系的膨胀速率.注意到这里有个负号.而对于不可压缩的体系,存在简单的关系


 * $$\begin{align}

\partial_{\mu}(\rho u^{\mu})=\rho\partial_{\mu}u^{\mu}=0 \end{align}$$

第二个记号是任意物理量随固有时的变化率或者在随动系中对时间的变化率.另外需要特别注意到这与下面讨论SPH自由度时在Bjorken坐标系 $$\eta-\tau$$ 中引入的记号不同


 * $$\begin{align}

Df \ne d_{\tau'}f =\frac{1}{\gamma} u\cdot \partial f \end{align}$$

其中的 $$\tau' \equiv \sqrt{t^2-z^2}$$ 是一个时间坐标而非固有时. 我们利用(下面第一式因为是平直空间)


 * $$\begin{align}

&\partial_{\sigma}g^{\mu\nu}=0\\ &u_{\mu}u^{\mu}=1 \end{align}$$

可以证明


 * $$\begin{align}

&u^{\mu}\partial_{\sigma}u_{\mu}=0\\ &Dg^{\mu\nu}=\dot g^{\mu\nu}=0 \end{align}$$

从而可以化简
 * $$u_{\mu}u^{\sigma}\partial_{\sigma}(eu^{\mu}u^{\nu})=u^{\sigma}\partial_{\sigma}(eu^{\nu}) $$

又如


 * $$\begin{align}

&u_{\mu}Dq^{\mu}=u_{\mu}Dg^{\mu\nu}q_{\nu}=u_{\mu}g^{\mu\nu}Dq_{\nu}=u^{\mu}Dq_{\mu}\\ &u_{\mu}\nabla^{\sigma}q^{\mu}=u^{\mu}\nabla^{\sigma}q_{\mu} \\ &\Delta^{\mu\nu}D u_{\nu}=(g^{\mu\nu}-u^{\mu}u^{\nu})D u_{\nu}=D u^{\mu} \end{align}$$

另外利用投影算符的性质比如与速度垂直 $$u_{\mu}\Delta^{\mu\nu}=0$$ 以及 $${\Delta^{\mu}}_{\sigma}\Delta^{\sigma\nu}=\Delta^{\mu\nu}$$ .略去冗长但是不复杂的化简,可以证明


 * $$\begin{align}

\partial_{\mu}N^{\mu}=0 \end{align}$$ 等价于


 * $$\begin{align}

\dot n=-n\theta \end{align}$$ 而


 * $$\begin{align}

\partial_{\mu}T^{\mu\nu}=0 \end{align}$$ 垂直于和平行于速度的分量分别为


 * $$\begin{align}

&\Delta_{\sigma\nu}\partial_{\mu}T^{\mu\nu}=0 \\ &u_{\nu}\partial_{\mu}T^{\mu\nu}=0 \end{align}$$

它们分别等价于


 * $$\begin{align}

&\dot u^{\mu}=\frac{\nabla^{\mu}P}{P+\varepsilon}\\ &\dot \varepsilon=-(P+\varepsilon) \theta \end{align}$$

第一个方程与弯曲空间Bjorken坐标下的方程形式非常接近,如果把一般导数理解为协变导数,其形式完全一致.这是一个数学上的巧合,是Bjorken坐标的简单形式决定的,不具有一般性.在非相对论极限下为欧乐方程(见下) 第二个方程实际上等价于熵流守恒,证明如下 利用第一定律


 * $$\begin{align}

dE=TdS-PdV+\sum_p \mu_pdN_p \end{align}$$

由标度不变性得到(所有广延量放大 $$\lambda$$ 倍表达式同样成立)
 * $$E=TS-PV+\sum_p \mu_pN_p$$

上式用强度量表达为


 * $$\begin{align}

P+\varepsilon=Ts+\sum_p \mu_pn_p \end{align}$$

利用这个表达式,把第一定律中的所有广延量都写成强度量乘以体积,


 * $$\begin{align}

&E=\varepsilon V \\ &S=sV \\ &N_p=n_pV \end{align}$$

我们发现与压强有关的项自然的从等式两边消去,我们得到


 * $$\begin{align}

&d\varepsilon=Tds+\sum_p \mu_pdn_p \\ &u\cdot\partial \varepsilon=Tu\cdot\partial s+\sum_p \mu_pu\cdot\partial n_p \end{align}$$

或者


 * $$\begin{align}

D\varepsilon=TDs+\sum_p \mu_pDn_p \end{align}$$

需要指出热力学第一定律并不导致


 * $$\begin{align}

\partial^{\mu}\varepsilon=T\partial^{\mu}s+\sum_p \mu_p\partial^{\mu}n_p \end{align}$$

原因是热力学第一定律在随动系中成立,对应在随动系中对时间的导数,从而对应对固有时的全微分,而另一方面上述偏导数是一个很强的要求,一般情况下并不成立.换言之,热力学定律仅仅对应上述导数在速度方向上的投影. 由运动方程的第二式,以及上面的等式


 * $$\begin{align}

&u \cdot \partial \varepsilon + (P+\varepsilon)\partial \cdot u=0 \\ &u \cdot (T\partial s +\sum_p \mu_p dn_p) + (Ts+\sum_p\mu_p n_p)\partial \cdot u=0 \\ &T \partial \cdot (su) + \sum_p \mu_p \partial \cdot (n_p u)=0 \end{align}$$

由于粒子数流守恒 $$ \partial \cdot (n_p u)=0$$ ,我们最终得到


 * $$\begin{align}

\partial \cdot (su) =0 \end{align}$$

上面的证明中,实际上我们涉及到一个与体系演化方程无关的热力学第一定律与标度定律的推论,对于只有一个守恒荷的体系,它对固有时的全微分不变意味着每个粒子的熵不变,这是一个重要的热力学关系


 * $$\begin{align}

nTdS_{N}=d \varepsilon - \frac{(P+\varepsilon)}{n}dn \end{align}$$

其中 $$n=\frac{N}{V}$$ ,而 $$S_{N}=\frac{S}{N}$$ 为比熵,即单位粒子的熵,其余量都为密度量.对于多组元体系


 * $$\begin{align}

n_pTdS_{N_p}=d \varepsilon - \frac{(P+\varepsilon)}{n_p}dn_p \end{align}$$

其中 $$n_p=\frac{N_p}{V}$$ ,而 $$S_{N_p}=\frac{S}{N_p}$$ 为比熵,即单位守恒荷的熵.我们将在最后附上该公式的简单证明.利用此表达式,单位守恒荷的熵随固有时的演化为


 * $$\begin{align}

n_pT\frac{dS_{N_p}}{d\tau}=\frac{d \varepsilon}{d\tau} - \frac{(P+\varepsilon)}\frac{d n_p}{d\tau}=D \varepsilon - \frac{(P+\varepsilon)}{n_p}Dn_p \end{align}$$

注意到等式的右边正是体系的演化方程


 * $$\begin{align}

u\cdot \partial \varepsilon - (P+\varepsilon)\frac{u \cdot \partial n_p}{n_p}=u\cdot \partial \varepsilon - (P+\varepsilon)\partial \cdot u=0 \end{align}$$

这里我们证明这个关系式,其实只要注意到


 * $$\begin{align}

n_pTdS_{N_p}=n_pTu\cdot \partial(\frac{s}{n_p})=T \partial \cdot((\frac{s}{n_p})(n_pu))=T \partial \cdot(su) \end{align}$$

借助上面的推导过程即得所需表达式


 * $$\begin{align}

&n_pTdS_{N_p}=T \partial \cdot(su) \\ &=TDs+Ts\partial\cdot u \\ &=TDs+Ts\partial\cdot u+\sum_p\mu_p\partial\cdot (n_pu) \\ &=(TDs+\sum_p\mu_pDn_p)+(Ts+\sum_p\mu_pn_p)\partial\cdot u \\ &=D\varepsilon+(\varepsilon+P)\partial\cdot u=0 \end{align}$$

最后一步等号是运动方程的第二式.

理想流体运动方程的等价形式与讨论
动量方程实际上等价于相对论形式的欧乐(Euler)以及纳维－斯托克斯(Navier-Stokes)方程.参见Schutz A first course in GR(4.55-56)或者Hans-Thomas Elze,J.Phys.G25(1999)1935.

与欧乐方程的比较

 * 与非相对论欧乐方程的比较

上面得到的理想流体运动方程,即能动张量守恒,方程垂直于速度方程等价于


 * $$\begin{align}

\dot u^{\mu}=\frac{\nabla^{\mu}P}{P+\varepsilon} \end{align}$$

考虑方程的空间分量,在非相对论极限下做代换


 * $$\begin{align}

&P+\varepsilon \to \varepsilon=\rho \\ &\dot u=u^{\mu} \partial_{\mu}u \to \left( \frac{\partial}{\partial t}+{\bold u}\cdot\nabla \right){\bold u} \end{align}$$

自然的得到欧乐方程


 * $$\begin{align}

\rho\left( \frac{\partial}{\partial t}+{\bold u}\cdot\nabla \right){\bold u}=-\nabla p \end{align}$$

如果考虑外力和应力张量,上述欧乐方程直接可以推广为纳维－斯托克斯方程(wiki)


 * $$\begin{align}

\rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v} \right) = -\nabla p + \nabla \cdot\mathbb{T} + \mathbf{f} \end{align}$$ 其中$$ \nabla\cdot\mathbb{T}$$来自扣除压力部分(应力张量的迹的三分之一)后的应力,因为相应总应力
 * $$\begin{align}

{\mathbf F}=\int d{\mathbf S}\cdot\mathbb{T} =\oint dV (\nabla\cdot\mathbb{T} ) \end{align}$$ 而 $$\mathbf{f}$$ 为外源作用在单位体积上的力.


 * 与相对论欧乐方程的比较

在平直时空下


 * $$\partial_{\mu}T^{\mu\nu}=0,T^{\mu\nu}=(\varepsilon+P)u^{\mu}u^{\nu}-Pg^{\mu\nu}$$
 * $$\partial_{\mu}T^{\mu\nu}=\partial_{\mu}[(\varepsilon+P)u^{\mu}u^{\nu}]-\partial_{\mu}(Pg^{\mu\nu})$$
 * $$\partial_{\mu}T^{\mu\nu}=\partial_{\mu}[su^{\mu}\frac{\varepsilon+P}{s}u^{\nu}]-g^{\mu\nu}\partial_{\mu}P$$
 * $$\partial_{\mu}T^{\mu\nu}=su^{\mu}\partial_{\mu}(\frac{\varepsilon+P}{s}u^{\nu})-g^{\mu\nu}\partial_{\mu}P$$
 * $$\partial_{\mu}T^{\mu\nu}=s\gamma\frac{d}{dt}(\frac{\varepsilon+P}{s}u^{\nu})-g^{\mu\nu}\partial_{\mu}P$$

考虑时间部分，


 * $$s\gamma\frac{d}{dt}(\frac{\varepsilon+P}{s}u^{0})-g^{\mu 0}\partial_{\mu}P=0$$
 * $$s\gamma\frac{d}{dt}(\frac{\varepsilon+P}{s}\gamma)-g^{00}\partial_{0}P=0$$
 * $$s\gamma\frac{d}{dt}(\frac{\varepsilon+P}{s}\gamma)=\frac{\partial P}{\partial t}$$

考虑空间部分，


 * $$s\gamma\frac{d}{dt}(\frac{\varepsilon+P}{s}\gamma v^{i})-g^{\mu i}\partial_{\mu}P=0$$
 * $$s\gamma\frac{d}{dt}(\frac{\varepsilon+P}{s}\gamma)v^{i}+s\gamma(\frac{\varepsilon+P}{s}\gamma)\frac{dv^{i}}{dt}-g^{\mu i}\partial_{\mu} P=0$$
 * $$\frac{\partial P}{\partial t}v^{i}+(\varepsilon+P)\gamma^{2}\frac{dv^{i}}{dt}+\partial_{i}P=0$$

其中最后一步利用了时间部分的结果.上式可以写成矢量形式：


 * $$\frac{d\vec{v}}{dt}=-\frac{1}{(\varepsilon+P)\gamma^{2}}(\vec{\nabla}P+\vec{v}\frac{\partial P}{\partial t})$$

或者


 * $$\frac{\partial\vec{v}}{\partial t}+(\vec{v}\cdot\vec{\nabla})\vec{v}=-\frac{1-v^{2}}{\varepsilon+P}(\vec{\nabla}P+\vec{v}\frac{\partial P}{\partial t})$$

上面推导过程中用到熵流守恒以及度规张量分量为常数的性质.

流体静力学和伯努力方程

 * 流体静力学和伯努力方程

我们可以进一步得到流体静力学方程和伯努力方程(参见朗道流体力学).下面的论述来自朗道的流体力学.对于流体静力学,流体的速度场恒为零,且流体在静止情况下不会产生剪切应力,故无迹的应力张量无贡献,方程可化简为


 * $$\begin{align}

\nabla p = \mathbf{f} \end{align}$$

在仅仅存在重力场时


 * $$\begin{align}

\nabla{p}=\bold{f}=\rho\bold{g} \end{align}$$

伯努力方程需要引入流线的概念,它的定义是


 * $$\begin{align}

\frac{dx}{v_x}=\frac{dy}{v_y}=\frac{dz}{v_z} \end{align}$$

利用关系


 * $$\begin{align}

\nabla(\frac{1}{2}v^2)=\vec{v}\times(\nabla\times\vec{v})+\vec{v}\cdot(\nabla{v}) \end{align}$$

可以把纳维－斯托克斯方程重新写为


 * $$\begin{align}

&\rho(\frac{\partial\vec{v}}{\partial{t}}+\frac{1}{2}\nabla{v^2}-\vec{v}\times(\nabla\times{v}))\\ &=-\nabla{p}+\nabla\mathbb{T}+\bold{f}\\ &=-\nabla{p}+\nabla\mathbb{T}+\rho\bold{g}\\ &=-\nabla{p}+\nabla\mathbb{T}-\rho\nabla\phi \end{align}$$

伯努力方程考虑稳定流体对着流线积分的形式,故速度场不是时间的函数且与速度垂直的项对线积分没有贡献,从而直接得到我们熟悉的形式


 * $$\begin{align}

\frac{1}{2}\rho{v^2}+p+\rho{gh}=const. \end{align}$$

与玻尔兹曼方程的联系

 * 与玻尔兹曼方程的联系

由于能动张量的具体形式为


 * $$\begin{align}

T^{\mu\nu}=\sum_np^{\mu}_n\frac{dx^{\nu}_n}{dt}\delta^3(x-x_n(t))=\frac{1}{(2\pi)^3}\int\frac{d^3p}{E}p^{\mu}p^{\nu}f(x,p) \end{align}$$

利用它的第一步等式中求和形式,可以直接验证对仅包含瞬间短程作用力的体系满足能动张量守恒$$\partial_{\mu}T^{\mu\nu}=0$$,具体参见温伯格的宇宙学和引力论. 我们这里重点讨论连续的积分形式和玻尔兹曼方程的联系. 玻尔兹曼方程描写粒子数密度函数 $$f \equiv f(\mathbf{x},\mathbf{p},t)$$ 所满足的方程.玻尔兹曼方程的一个简单的推导如下.首先考虑粒子流守恒的情况.因为穿透三维表面 $$d\sigma_{\mu}$$ 的世界线的数目为(参见L.P. Csernai Introduction to RHIC P.43)


 * $$\begin{align}

\Delta N=\int d\sigma_{\mu} \frac{d^3p}{p^0} p^{\mu}f(x',p) \end{align}$$

上述表达式可以理解如下.首先量纲是正确的,因为分布函数乘以三动量乘以空间等于粒子数无量纲,其次上述表达式洛仑兹协变的,最后它的物理意义是粒子流与面积元的内积. 如果表面是类时的,它等于(某坐标系内)确定时刻大小为 $$d\sigma_{\mu}$$ 的体积元内的粒子数,如果表面是类空的,它等于单位时间,穿过大小为 $$d\sigma_{\mu}$$ 表面的粒子数.如果粒子数守恒,粒子的世界线不会创生也不会消灭,(且世界线必须都处于因果光锥之内.)故世界线的面积分对封闭四维曲面为零,即


 * $$\begin{align}

\oint d\sigma_{\mu} \frac{d^3p}{p^0} p^{\mu}f(x',p) =0 \end{align}$$

利用高斯定理,并注意到动量和坐标独立,高斯定理仅仅对坐标作用,从而


 * $$\begin{align}

&\int d^4x\frac{d^3p}{p^0}p^{\mu}\frac{\partial f}{\partial x^{\mu}} =0 \\ &p^{\mu}\frac{\partial f}{\partial x^{\mu}} =0 \end{align}$$

如果考虑到粒子的产生湮灭


 * $$\begin{align}

p^{\mu}\frac{\partial f}{\partial x^{\mu}} =C(x,p) \end{align}$$ 其中


 * $$\begin{align}

&C(x,p)=\frac{1}{2}\int\frac{d^3p_1}{p_1^0}\frac{d^3p'}{{p'}^0}\frac{d^3{p'}_1}{{p'}_1^0}[f'f'_1W(p',{p'}_1|p,p_1)-ff_1W(p,p_1|p',{p'}_1)] \\ &f'\equiv f(x,p'), f'_1\equiv f(x,{p'}_1), f\equiv f(x,p), f_1\equiv f(x,p_1) \end{align}$$

$$W(p',p'_1|p,p_1)$$ 为过程 $$p',p'_1 \to p,p_1$$ 跃迁几率,总的跃速率与该几率与初态粒子数密度成正比,且对所有亚元求和.这里仅仅考虑了两元碰撞.

在非相对论近似下,世界线上粒子数密度的变化满足(参见维基百科)


 * $$\begin{align}

&\frac{\partial f}{\partial t} + \frac{d\mathbf{x}}{dt}\cdot \frac{\partial f}{\partial \mathbf{x}} + \frac{d\mathbf{p}}{dt}\cdot \frac{\partial f}{\partial \mathbf{p}} =C[\rho] \\ &\frac{\partial f}{\partial t} + \mathbf{v} \cdot \nabla_x f + \mathbf{f} \cdot \nabla_p f =C[f] \end{align}$$

其中 $$C[f]=C_{gain}-C_{loss}$$ 来源于碰撞.当没有外力时,直接可推广为上面它的相对论协变形式

在一定的假设下,可以由玻尔兹曼方程导出能动张量守恒,如下(参见L.P. Csernai Introduction to RHIC P.53) 我们仅仅考虑两元碰撞,并且假定微观碰撞四动量守恒,即对过程 $$p_k,p_l \to p_i,p_j, \Psi_k \equiv \Psi_k(x,p_k)= a_k(x)+b_{\mu}(x)p_k^{\mu}$$ 满足


 * $$\begin{align}

\Psi_k+\Psi_l=\Psi_i+\Psi_j \end{align}$$

我们定义


 * $$\begin{align}

C_{k}(x,p_k)=\frac{1}{2}\int\frac{d^3p_l}{p_l^0}\frac{d^3p_i}{p_i^0}\frac{d^3p_j}{p_j^0}[f_if_jW(p_i,p_j|p_k,p_l)-f_kf_lW(p_k,p_l|p_i,p_j)] \end{align}$$

可以证明


 * $$\begin{align}

\int\frac{d^3p_k}{p_k^0}\Psi_kC_{kl}(x,p_k)=0 \end{align}$$

证明主要是注意到上述表达式对四个动量的积分部分完全对称,差别在于剩余部分.并且我们注意到跃迁几率具有初始状态之间的对称性和末态之间的对称性.具有对称性.为了利用$$\Psi_k+\Psi_l=\Psi_i+\Psi_j$$,我们把指标$$k$$依次轮换为其他指标,并且把四项加在一起,由于所有指标都是亚元,结果应该为元表达式的四倍.但是根据前面提到的对称性和表达式的具体形式,发现结果积分号内可以提出因子$$\Psi_k+\Psi_l-\Psi_i-\Psi_j$$,从而问题得证.

对玻尔兹曼方程


 * $$\begin{align}

p_k^{\mu}\frac{\partial f}{\partial x^{\mu}} =C_k(x,p_k) \end{align}$$

两边乘以$$\Psi_k$$并对末态求和$$\int\frac{d^3p_k}{p_k^0}$$ 等式左边为零,


 * $$\begin{align}

\int\frac{d^3p_k}{p_k^0}\Psi_k^{\mu}p^{\mu}_k\frac{\partial f}{\partial x^{\mu}} =0 \end{align}$$ 其中常数部分导致守恒荷 $$q_k$$ 的粒子数守恒


 * $$\begin{align}

\int\frac{d^3p_k}{p_k^0}q_kp_k^{\mu}\frac{\partial f}{\partial x^{\mu}} ={Q^{\mu}}_{,\mu}=0 \end{align}$$

其中与动量正比部分得到能动张量守恒


 * $$\begin{align}

\int\frac{d^3p_k}{p_k^0}p^{\nu}_kp_k^{\mu}\frac{\partial f}{\partial x^{\mu}} ={T^{\mu\nu}}_{,\nu}=0 \end{align}$$

方程数目与变量数目
理想流体情况,不考虑粘滞系数,热流,化学流的情况下.假设为单守恒荷体系具有四速度(3)热力学参数(3),所以共有6个独立的自由度.而另一方面,我们有守恒流(1)能动张量守恒(4)状态方程(1),正好6个独立的方程. 非理想流体情况,由于粘滞系数(1+5)和通过热流或者化学流体现的热传导机制(3)具有共计9个独立自由度,所以需要9个新的方程.需要由模型决定.

考虑粘滞系数热流化学流情况下的流体力学方程
不管如何,我们总是可以把能动张量投影到平行于速度和垂直于速度的方向上.这样第一个方程仍旧和熵流有关,只是在一般情况下不再守恒.因为在相对论重离子碰撞的应用中,重子数密度很小,所以以下我们采用Landau的随动系的定义来讨论.我们暂时在形式上保留 $$q^{\mu}$$ ,只在最后时刻取为零. 对守恒流


 * $$\begin{align}

&N^{\mu}=nu^{\mu}+j_{\nu}\Delta^{\mu\nu}\equiv nu^{\mu}+V^{\mu} \\ &\partial_{\mu}N^{\mu}=0 \end{align}$$

利用前面引入的投影算符,利用关系 $$\partial^{\mu}=u^{\mu}D+\nabla^{\mu}$$ ,可以化为垂直于速度的分量 $$\nabla^{\mu} \equiv \Delta^{\mu\nu} \partial_{\nu}$$ 和平行于速度的分量 $$D \equiv u^{\nu}\partial_{\nu}$$.利用前面定义的记号


 * $$\begin{align}

&\theta \equiv \partial \cdot u\equiv \partial_{\mu}u^{\mu} \\ &\dot f =u^{\mu} \partial_{\mu}f \equiv u\cdot \partial f\equiv Df \end{align}$$

我们得到


 * $$\begin{align}

0=\partial_{\mu}N^{\mu}=u^{\mu}\partial_{\mu}n+n\partial_{\mu}u^{\mu}+u_{\mu}\dot V^{\mu}+\nabla_{\mu}V^{\mu} \end{align}$$

即


 * $$\begin{align}

&\dot n=-n\theta-[\dot{(u\cdot V)}-V\cdot\dot u]-\nabla\cdot V \\ &\dot n=-n\theta+V\cdot\dot u-\nabla\cdot V \end{align}$$

而对于能动张量守恒


 * $$\begin{align}

&\partial_{\mu}T^{\mu\nu}=0 \\ &T^{\mu\nu}={\varepsilon}u^{\mu}u^{\nu}-(P+\Pi)\Delta^{\mu\nu}+\pi^{\mu\nu}+2q_{\lambda}\Delta^{\lambda(\mu}u^{\nu)}={\varepsilon}u^{\mu}u^{\nu}-(P+\Pi)\Delta^{\mu\nu}+\pi^{\mu\nu}+q^{\mu}u^{\nu}+q^{\nu}u^{\mu} \end{align}$$

在上面第二步等号中,为了节省笔墨,我们采用默认约定 $$\pi^{\mu\nu}$$ 已经对称化与速度垂直并且无迹, $$j^{\mu}$$ 和 $$q^{\mu}$$ 已经与速度垂直,从而下面关系成立


 * $$\begin{align}

&\pi^{\mu\nu}a_{\mu\nu} =\pi^{\mu\nu}\frac{1}{2}(a_{\mu\nu}+a_{\nu\mu}) \\ &\pi^{\mu\nu}{\Delta_{\nu}}^{\sigma} ={\pi^{\mu\sigma}} \\ &\pi^{\mu\nu}{u_{\nu}} =0 \\ & j^{\mu}{\Delta_{\mu\nu}} =j_{\nu} \\ & j^{\mu}{u_{\mu}} =0 \\ &q^{\mu}{\Delta_{\mu\nu}} =q_{\nu} \\ &q^{\mu}{u_{\mu}} =0 \end{align}$$

严格的,上面关系我们可以通过投影算符来验证


 * $$\begin{align}

&\Pi^{\mu\nu}u_{\nu}=0 \\ &\pi^{\mu\nu}=\Delta^{\mu\nu\alpha\beta}\Pi_{\alpha\beta}=\Pi^{\mu\nu}-\Delta^{\mu\nu}\frac{{\Pi^{\alpha}}_{\alpha}}{3}\\ &\Pi\equiv-\frac{{\Pi^{\alpha}}_{\alpha}}{3}\\ &q^{\mu}\equiv\Delta^{\mu\alpha}T_{\alpha\beta}u^{\beta}=\Delta^{\mu\alpha}(T_{\alpha\beta}-T^0_{\alpha\beta}-\Pi_{\alpha\beta})u^{\beta} \end{align}$$

其中引入了得到张量对称无迹部分的投影算符


 * $$\begin{align}

\Delta^{\mu\nu\alpha\beta} \equiv \frac{1}{2}\left[\Delta^{\mu\alpha}\Delta^{\nu\beta}+\Delta^{\mu\beta}\Delta^{\nu\alpha}-\frac{2}{3}\Delta^{\mu\nu}\Delta^{\alpha\beta}\right] \end{align}$$

带入能动张量的一般形式后得到


 * $$\begin{align}

&\partial_{\mu}T^{\mu\nu}\\ &=\partial_{\mu}(\varepsilon u^{\mu}u^{\nu}-(P+\Pi)\Delta^{\mu\nu}+\pi^{\mu\nu}+q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= u^{\mu}u^{\nu}\partial_{\mu}\varepsilon+\varepsilon u^{\nu} \partial_{\mu}u^{\mu}+\varepsilon u^{\mu} \partial_{\mu}u^{\nu} -\Delta^{\mu\nu}\partial_{\mu}(P+\Pi)-(P+\Pi)\partial_{\mu}\Delta^{\mu\nu} +\partial_{\mu}\pi^{\mu\nu}+\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= u^{\nu}\dot\varepsilon+\varepsilon u^{\nu} \theta+\varepsilon \dot u^{\nu} -\nabla^{\nu}(P+\Pi)+(P+\Pi)\dot u^{\nu}+u^{\nu}(P+\Pi)\theta +\partial_{\mu}\pi^{\mu\nu}+\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= u^{\nu}\dot\varepsilon+(\varepsilon+P+\Pi) u^{\nu} \theta+(\varepsilon+P+\Pi) \dot u^{\nu} -\nabla^{\nu}(P+\Pi)+\partial_{\mu}\pi^{\mu\nu}+\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \end{align}$$

平行于和垂直于速度的分量分别为


 * $$\begin{align}

&u_{\nu}\partial_{\mu}T^{\mu\nu}=0 \\ &\Delta_{\sigma\nu}\partial_{\mu}T^{\mu\nu}=0 \end{align}$$

对平行四速度部分,引入定义


 * $$\begin{align}

\sigma^{\mu\nu}\equiv\frac{1}{2}(\nabla^{\mu}u^{\nu}+\nabla^{\nu}u^{\mu}) \end{align}$$


 * $$\begin{align}

& 0=u_{\nu}\partial_{\mu}T^{\mu\nu}= \dot\varepsilon+(\varepsilon+P+\Pi) \theta+u_{\nu}(\varepsilon+P+\Pi) \dot u^{\nu} -u_{\nu}\nabla^{\nu}(P+\Pi) +u_{\nu}\partial_{\mu}\pi^{\mu\nu}+u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta+u_{\nu}(\varepsilon+P+\Pi) \dot u^{\nu} -u_{\nu}\nabla^{\nu}(P+\Pi) +u_{\nu}\partial_{\mu}\pi^{\mu\nu} +u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta +u_{\nu}\partial_{\mu}\pi^{\mu\nu}+u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta +u_{\nu}\partial_{\mu}\pi^{\mu\nu}-\partial_{\mu}(u_{\nu}\pi^{\mu\nu}) +u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi^{\mu\nu}(\partial_{\mu}u_{\nu}) +u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}(\partial^{\mu}u^{\nu}) +u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\Delta^{\nu\sigma}(\partial_{\sigma}u^{\nu}) +u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\nabla^{\mu}u^{\nu}+u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\frac{1}{2}(\nabla^{\mu}u^{\nu}+\nabla^{\nu}u^{\mu}) +u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+u_{\nu}\partial_{\mu}(q^{\mu}u^{\nu})+u_{\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+\partial_{\mu}(u_{\nu}q^{\mu}u^{\nu})-q^{\mu}u^{\nu}\partial_{\mu}u_{\nu}+u_{\nu}u^{\mu}\partial_{\mu}q^{\nu} +u_{\nu}q^{\nu}\partial_{\mu}u^{\mu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+\partial_{\mu}(u_{\nu}q^{\mu}u^{\nu})+u_{\nu}u^{\mu}\partial_{\mu}q^{\nu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+\partial_{\mu}q^{\mu}+u_{\nu}\dot q^{\nu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+u_{\mu}\dot q^{\mu}+\nabla_{\mu}q^{\mu}+u_{\nu}\dot q^{\nu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+2u_{\mu}\dot q^{\mu}+\nabla_{\mu}q^{\mu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu}+\dot{(2q^{\mu} u_{\mu})}-2q^{\mu}\dot u_{\mu}+\nabla_{\mu}q^{\mu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu} -2q^{\mu}\dot u_{\mu}+\nabla_{\mu}q^{\mu} \\ &= \dot\varepsilon+(\varepsilon+P+\Pi) \theta -\pi_{\mu\nu}\sigma^{\mu\nu} -2q\cdot \dot u+\nabla\cdot q \end{align}$$

即


 * $$\begin{align}

\dot\varepsilon= -(\varepsilon+P+\Pi) \theta+\pi_{\mu\nu}\sigma^{\mu\nu}+2q\cdot \dot u-\nabla\cdot q \end{align}$$

对于垂直四速度部分


 * $$\begin{align}

& 0=\Delta_{\sigma\nu}\partial_{\mu}T^{\mu\nu}\\ &= \Delta_{\sigma\nu}(\varepsilon+P+\Pi) \dot u^{\nu} -\Delta_{\sigma\nu}\nabla^{\nu}(P+\Pi)+\Delta_{\sigma\nu}\partial_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= \Delta_{\sigma\nu}(\varepsilon+P+\Pi) \dot u^{\nu} -\nabla_{\sigma}(P+\Pi)+\Delta_{\sigma\nu}\partial_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) +\Delta_{\sigma\nu}\partial_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) +\Delta_{\sigma\nu}u_{\mu}\dot\pi^{\mu\nu}+\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) +D(\Delta_{\sigma\nu}u_{\mu}\pi^{\mu\nu})-\pi^{\mu\nu}D(u_{\mu}\Delta_{\sigma\nu}) +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu}+\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi^{\mu\nu}D(u_{\mu}\Delta_{\sigma\nu}) +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi^{\mu\nu}\Delta_{\sigma\nu}D(u_{\mu}) +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}D(u^{\mu}) +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu}+q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu}+\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}\partial_{\mu}(q^{\mu}u^{\nu})+\Delta_{\sigma\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +\Delta_{\sigma\nu}q^{\mu}\partial_{\mu}u^{\nu}+\Delta_{\sigma\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +g_{\sigma\nu}q^{\mu}\partial_{\mu}u^{\nu}+\Delta_{\sigma\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +q^{\mu}\partial_{\mu}u_{\sigma}+\Delta_{\sigma\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +q_{\rho}\Delta^{\rho\mu}\partial_{\mu}u_{\sigma}+\Delta_{\sigma\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +q_{\rho}\nabla^{\rho}u_{\sigma}+\Delta_{\sigma\nu}\partial_{\mu}(q^{\nu}u^{\mu}) \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +q_{\rho}\nabla^{\rho}u_{\sigma}+\Delta_{\sigma\nu}q^{\nu}\partial_{\mu}u^{\mu}+\Delta_{\sigma\nu}u^{\mu}\partial_{\mu}q^{\nu} \\ &= (\varepsilon+P+\Pi) \dot u_{\sigma} -\nabla_{\sigma}(P+\Pi) -\pi_{\mu\sigma}\dot u^{\mu} +\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +q_{\rho}\nabla^{\rho}u_{\sigma}+q_{\sigma}\partial_{\mu}u^{\mu}+\Delta_{\sigma\nu}\dot q^{\nu} \end{align}$$

即


 * $$\begin{align}

(\varepsilon+P+\Pi) \dot u_{\sigma} = \nabla_{\sigma}(P+\Pi) +\pi_{\mu\sigma}\dot u^{\mu} -\Delta_{\sigma\nu}\nabla_{\mu}\pi^{\mu\nu} +(q\cdot \nabla) u_{\sigma}+q_{\sigma}\theta+\Delta_{\sigma\nu}\dot q^{\nu} \end{align}$$

上面推导方程的精神如下,对于粘滞系数,我们不保留它的导数,而总是通过部分积分转换为体系整体膨胀速率$$\theta$$和速度梯度$$\sigma^{\mu\nu}$$的函数.

在上面的推导中,我们实际上涉及到一个有用的表达式,它是对速度导数的写法


 * $$\begin{align}

&\partial_{\mu}u_{\nu}=u_{\mu}Du_{\nu}+\nabla_{\mu}u_{\nu} \\ &=u_{\mu}Du_{\nu}+\frac{1}{2}(\nabla_{\mu}u_{\nu}+\nabla_{\nu}u_{\mu})+\frac{1}{2}(\nabla_{\mu}u_{\nu}-\nabla_{\nu}u_{\mu})\\ &=u_{\mu}Du_{\nu}+\sigma_{\mu\nu}+\frac{1}{3}\Delta_{\mu\nu}\theta+\Omega_{\mu\nu} \end{align}$$

其中引入定义(包括上面已经涉及到的这里重复书写)


 * $$\begin{align}

&\sigma_{\mu\nu}\equiv\frac{1}{2}(\nabla_{\mu}u_{\nu}+\nabla_{\nu}u_{\mu}-\frac{2}{3}\Delta_{\mu\nu}\theta)\\ &\theta\equiv\partial_{\mu}u^{\mu}\\ &\Omega_{\mu\nu}\equiv\frac{1}{2}(\nabla_{\mu}u_{\nu}-\nabla_{\nu}u_{\mu}) \end{align}$$

一般在文献中引入如下记号


 * $$\begin{align}

A^{<\mu\nu>}\equiv\Delta^{\mu\nu\alpha\beta}A_{\alpha\beta} \end{align}$$

和下面的定义


 * $$\begin{align}

&\nabla^{<\mu}u^{\nu>}\equiv 2\sigma^{\mu\nu}=2\nabla^{(\mu}u^{\nu)}-\frac{2}{3}\Delta^{\mu\nu}\theta\\ &\nabla^{(\mu}u^{\nu)}\equiv\nabla^{\mu}u^{\nu}+\nabla^{\nu}u^{\mu} \end{align}$$

上面第一个定义是无迹的,第二个有迹.但是如果是作用在$$\pi_{\mu\nu}$$上,两者没有区别,因为这是迹的部分没有任何贡献.

熵流定义的推广
我们首先顺着Philipe硕士论文的思路利用能动张量的明显形式来推导熵流方程,引入熵流的定义的推广,然后我们顺着Heinz一文的思路从另一角度来理解这一推广的定义.最后我们给出热力学第二定律的形式. 我们知道能动张量守恒平行四速度在理想流体的情况下对应熵流守恒,在存在粘滞系数热流化学流的情况下,该方程的物理内涵应该不变,同时由于耗散的存在,熵流不会守恒,由热力学第二定律,这时体系满足不可逆过程熵增加.我们重复上面的过程,运动方程平行四速度分量为


 * $$\begin{align}

&0=u_{\mu}\partial_{\nu}T^{\mu\nu}\\ &=\partial_{\nu}(u_{\mu}T^{\mu\nu})-T^{\mu\nu}\partial_{\nu}u_{\mu}\\ &=\partial_{\nu}(\varepsilon u^{\nu}+q_{\lambda}\Delta^{\lambda\nu})-T^{\mu\nu}\partial_{\nu}u_{\mu}\\ &=\partial_{\nu}(\varepsilon u^{\nu}+q_{\lambda}\Delta^{\lambda\nu}) -(\varepsilon u^{\mu}u^{\nu}-(P+\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\Delta^{\mu\nu} +\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}+2q_{\lambda}\Delta^{\lambda(\mu}u^{\nu)})\partial_{\nu}u_{\mu}\\ &=\partial_{\nu}(\varepsilon u^{\nu}+q_{\lambda}\Delta^{\lambda\nu}) +((P+\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\Delta^{\mu\nu} -\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}-q_{\lambda}\Delta^{\lambda\mu}u^{\nu})\partial_{\nu}u_{\mu}\\ &=\partial_{\nu}(\varepsilon u^{\nu}+q_{\lambda}\Delta^{\lambda\nu}) +((P+\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\Delta^{\mu\nu}-\Delta^{(\mu\rho}\Delta^{\nu)\sigma} \pi_{\sigma\rho})\partial_{\nu}u_{\mu}-q\cdot \dot u \\ &=\partial_{\nu}(\varepsilon u^{\nu}+q_{\lambda}\Delta^{\lambda\nu}) +(P+\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta-\Delta^{(\mu\rho}\Delta^{\nu)\sigma} \pi_{\sigma\rho}\partial_{\nu}u_{\mu}-q\cdot \dot u \end{align}$$

因为在随动系中(Eckart或者Landau)满足热力学第一定律


 * $$\begin{align}

u^{\mu}\partial_{\mu}\varepsilon=Tu^{\mu}\partial_{\mu}s+\mu u^{\mu}\partial_{\mu}n \end{align}$$

以及热力学关系


 * $$\begin{align}

&\varepsilon=-P+Ts+\mu n \\ &(\varepsilon+P)\partial_{\mu}u^{\mu}=(Ts+\mu n)\partial_{\mu}u^{\mu} \end{align}$$

从而


 * $$\begin{align}

\partial_{\mu}(\varepsilon u^{\mu})=-P\partial_{\mu}u^{\mu}+T\partial_{\mu}(su^{\mu})+\mu \partial_{\mu}(nu^{\mu}) \end{align}$$

代入上式消去压强部分得


 * $$\begin{align}

T\partial_{\mu}(su^{\mu})+\mu \partial_{\mu}(nu^{\mu})=-\partial_{\nu}(q_{\lambda}\Delta^{\lambda\nu}) -(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u \end{align}$$

上述推导中中间一个步骤压强项在等号两边被消去,这是很自然的,因为理想流体相当于把所有粘滞系数和热流取为零,我们已经知道这时等式左边等于零,所以等式右边不可能含有与压强有关的非平庸项.把这个结果再代入流守恒关系


 * $$\begin{align}

0=\partial_{\mu}N^{\mu}=\partial_{\mu}(nu^{\mu})+\partial_{\mu}(j_{\nu}\Delta^{\mu\nu}) \end{align}$$ 引入定义 $$\alpha\equiv\frac{\mu}{T}, \beta\equiv\frac{1}{T}$$ ,得到


 * $$\begin{align}

&T\partial_{\mu}(su^{\mu})-\mu \partial_{\mu}(j_{\nu}\Delta^{\mu\nu})=-\partial_{\nu}(q_{\lambda}\Delta^{\lambda\nu}) -(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u \\ &T\partial_{\mu}(su^{\mu})-\mu \partial_{\mu}(j_{\nu}\Delta^{\mu\nu})+\partial_{\mu}(q_{\lambda}\Delta^{\lambda\mu}) =-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u \\ &\partial_{\mu}(su^{\mu})-\frac{\mu}{T} \partial_{\mu}(j_{\nu}\Delta^{\mu\nu})+\frac{1}{T}\partial_{\mu}(q_{\lambda}\Delta^{\lambda\mu}) =\frac{1}{T}\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right]\\ &\partial_{\mu}(su^{\mu})- \partial_{\mu}(\frac{\mu}{T}j_{\nu}\Delta^{\mu\nu})+\partial_{\mu}(\frac{1}{T}q_{\lambda}\Delta^{\lambda\mu}) =- j_{\nu}\Delta^{\mu\nu} \partial_{\mu}(\frac{\mu}{T})+q_{\lambda}\Delta^{\lambda\mu}\partial_{\mu}(\frac{1}{T}) +\frac{1}{T}\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right] \\ &\partial_{\mu}(su^{\mu})-\partial_{\mu}(\alpha j_{\nu}\Delta^{\mu\nu})+\partial_{\mu}(\beta q_{\lambda}\Delta^{\lambda\mu}) =- j_{\nu}\Delta^{\mu\nu} \partial_{\mu}\alpha+q_{\lambda}\Delta^{\lambda\mu}\partial_{\mu}\beta +\frac{1}{T}\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right] \\ &\partial_{\mu}(su^{\mu}-\alpha j_{\nu}\Delta^{\mu\nu}+\beta q_{\lambda}\Delta^{\lambda\mu}) =- j_{\nu}\Delta^{\mu\nu} \partial_{\mu}\alpha+q_{\lambda}\Delta^{\lambda\mu}\partial_{\mu}\beta +\frac{1}{T}\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right] \\ &\partial_{\mu}(su^{\mu}-\alpha j_{\nu}\Delta^{\mu\nu}+\beta q_{\lambda}\Delta^{\lambda\mu}) =- j_{\nu}\Delta^{\mu\nu} \partial_{\mu}\alpha+q_{\lambda}\Delta^{\lambda\mu}\partial_{\mu}\beta +\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right] \\ &\partial_{\mu}S^{\mu} =- j_{\nu}\Delta^{\mu\nu} \partial_{\mu}\alpha+q_{\lambda}\Delta^{\lambda\mu}\partial_{\mu}\beta +\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right] \end{align}$$

其中最后引入广义的熵流的定义
 * $$\begin{align}

S^{\mu}=su^{\mu}-\alpha j_{\lambda}\Delta^{\mu\lambda}+\beta q_{\lambda}\Delta^{\mu\lambda} \end{align}$$ (注意到Philipe论文的推导中少了一项).但是上述对广义的熵流的引入似乎有些随意,即把可以写为散度乘以温度的部分归入熵流之中,而把剩余部分认为是熵流的变化的源.从Heinz一文的推导中,上面做法的物理意义更为清晰.(虽然Heinz的推导中也漏了一项,其文章中的(23)式并不正确.)基本思路如下.我们把存在粘滞和热流的体系的守恒流和能动张量都写成微扰形式,即平衡理想体系的相应物理量加上微扰项.换言之


 * $$\begin{align}

& T^{\mu\nu}=T_{EQ}^{\mu\nu}+\delta T^{\mu\nu} \\ & T_{EQ}^{\mu\nu}=\varepsilon u^{\mu}u^{\nu}-P\Delta^{\mu\nu}   \\ &\delta T^{\mu\nu}=-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\Delta^{\mu\nu}+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}+2q_{\lambda}\Delta^{\lambda(\mu}u^{\nu)} \\ &N^{\mu}=N_{EQ}^{\mu}+\delta N^{\mu} N_{EQ}^{\mu}=nu^{\mu} \\ &\delta N^{\mu}=j_{\nu}\Delta^{\mu\nu} \end{align}$$

对于理想平衡的体系,


 * $$\begin{align}

&\varepsilon=-P+Ts+\mu n \\ &S_{EQ}^{\mu}\equiv su^{\mu}=\beta\varepsilon u^{\mu}+P\beta u^{\mu}-\alpha n u^{\mu} =P\beta u^{\mu}-\alpha N_{EQ}^{\mu}+\beta T_{EQ}^{\mu\nu} u_{\nu} \end{align}$$

其中引入了与前面一致的定义 $$\alpha\equiv\frac{\mu}{T}, \beta\equiv\frac{1}{T} $$ 将 $$\varepsilon=-P+Ts+\mu n$$ 对时间的全微分与 $$D\varepsilon=TDs+\mu Dn$$ 取差,我们得到


 * $$\begin{align}

&DP=sDT+ nD\mu \\ &u_{\mu}\partial^{\mu}P=su_{\mu}\partial^{\mu}T+ nu_{\mu}\partial^{\mu}\mu \end{align}$$

此即Gibbs-Duhem关系.利用这个关系我们计算我们演算下面的表达式


 * $$\begin{align}

&u^{\mu}\partial_{\mu}(P\beta)=\beta u^{\mu}\partial_{\mu}P+Pu^{\mu}\partial_{\mu}\beta \\ &=\beta (su^{\mu}\partial_{\mu}T+ nu^{\mu}\partial_{\mu}\mu)+Pu^{\mu}\partial_{\mu}\beta \\ &=\beta su^{\mu}\partial_{\mu}T+ \beta nu^{\mu}\partial_{\mu}\mu+Pu^{\mu}\partial_{\mu}\beta \\ &=\beta su^{\mu}\partial_{\mu}T+ \beta nu^{\mu}\partial_{\mu}\mu+( n\mu u^{\mu}\partial_{\mu}\beta-n\mu u^{\mu}\partial_{\mu}\beta)+Pu^{\mu}\partial_{\mu}\beta \\ &=\beta su^{\mu}\partial_{\mu}T+ nu^{\mu}\partial_{\mu}\alpha-n\mu u^{\mu}\partial_{\mu}\beta+Pu^{\mu}\partial_{\mu}\beta \\ &=-\frac{s}{\beta} u^{\mu}\partial_{\mu}\beta+ nu^{\mu}\partial_{\mu}\alpha-n\mu u^{\mu}\partial_{\mu}\beta+Pu^{\mu}\partial_{\mu}\beta \\ &=nu^{\mu}\partial_{\mu}\alpha+(P-\frac{s}{\beta} -n\mu)u^{\mu}\partial_{\mu}\beta \\ &=nu^{\mu}\partial_{\mu}\alpha-\varepsilon u^{\mu}\partial_{\mu}\beta \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}u_{\nu}\partial_{\mu}\beta \end{align}$$

所以


 * $$\begin{align}

&\partial_{\mu}(P\beta u^{\mu})\\ &=u^{\mu}\partial_{\mu}(P\beta)+P\beta\partial_{\mu} u^{\mu} \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}u_{\nu}\partial_{\mu}\beta+P\beta\partial_{\mu} u^{\mu} \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}u_{\nu}\partial_{\mu}\beta+P\beta\partial_{\mu} u^{\mu} \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}\partial_{\mu}(\beta u_{\nu})+T_{EQ}^{\mu\nu}\beta\partial_{\mu}(u_{\nu})+P\beta\partial_{\mu} u^{\mu} \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}\partial_{\mu}(\beta u_{\nu})+((\varepsilon+P)u^{\mu}u^{\nu}-Pg^{\mu\nu})\beta\partial_{\mu}(u_{\nu})+P\beta\partial_{\mu} u^{\mu} \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}\partial_{\mu}(\beta u_{\nu})-Pg^{\mu\nu}\beta\partial_{\mu}(u_{\nu})+P\beta\partial_{\mu} u^{\mu} \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}\partial_{\mu}(\beta u_{\nu}) \end{align}$$

这是一个有用的结果,即Heinz文中(23)式.因为


 * $$\begin{align}

&\partial_{\mu}S_{EQ}^{\mu}=\partial_{\mu}(P\beta u^{\mu}-\alpha N_{EQ}^{\mu}+\beta T_{EQ}^{\mu\nu} u_{\nu}) \\ &=\partial_{\mu}(P\beta u^{\mu})-\partial_{\mu}(\alpha N_{EQ}^{\mu})+\partial_{\mu}(\beta T_{EQ}^{\mu\nu} u_{\nu}) \\ &=N_{EQ}^{\mu}\partial_{\mu}\alpha-T_{EQ}^{\mu\nu}\partial_{\mu}(\beta u_{\nu})-\partial_{\mu}(\alpha N_{EQ}^{\mu})+\partial_{\mu}(\beta T_{EQ}^{\mu\nu} u_{\nu}) \\ &=-\alpha \partial_{\mu}N_{EQ}^{\mu}+\beta u_{\nu} \partial_{\mu}T_{EQ}^{\mu\nu} \end{align}$$

如果我们假定推广的熵流可以写成下面的形式


 * $$\begin{align}

&S^{\mu}\equiv S_{EQ}^{\mu}+\Phi^{\mu}\\ &=P\beta u^{\mu}-\alpha N^{\mu}+\beta T^{\mu\nu} u_{\nu}+Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) \\ &= S_{EQ}^{\mu}-\alpha \delta N^{\mu}+\beta \delta (T^{\mu\nu} u_{\nu})+Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) \end{align}$$

其导数为


 * $$\begin{align}

&\partial_{\mu}S^{\mu}\\ &=\partial_{\mu}(P\beta u^{\mu})-\partial_{\mu}(\alpha N^{\mu})+\partial_{\mu}(\beta T^{\mu\nu} u_{\nu})+\partial_{\mu}Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) \\ &=-\delta N^{\mu}\partial_{\mu}\alpha+\delta T^{\mu\nu}\partial_{\mu}(\beta u_{\nu})+\partial_{\mu}Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) \end{align}$$

这就是文中的(24)式.其中利用 $$\partial_{\mu}N^{\mu}=0$$ 和 $$\partial_{\mu}T^{\mu\nu}=0$$ 这个结果与Philipe的结果类似,但是其实很不同.因为出发点非常不同.Philipe的推导的出发点是能动张量守恒的时间分量应该与熵流有关,这里的推导是熵流应该可以在理想平衡体系的熵流的形式上做推广的基础上再引入附加项 $$Q^{\mu}$$ .但是,不难注意到,在最后推广的熵流的定义两者完全一致的(!).为了看清两个结果的联系,把上式该写为


 * $$\begin{align}

&\partial_{\mu}S^{\mu}\\ &=-\delta N^{\mu}\partial_{\mu}\alpha+\delta T^{\mu\nu}\partial_{\mu}(\beta u_{\nu})+\partial_{\mu}Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) \\ &=-\delta N^{\mu}\partial_{\mu}\alpha+\delta (T^{\mu\nu}u_{\nu})\partial_{\mu}\beta+\beta\delta T^{\mu\nu}\partial_{\mu}u_{\nu}+\partial_{\mu}Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) \end{align}$$

注意到


 * $$\begin{align}

&\delta N^{\mu}=j_{\nu}\Delta^{\mu\nu} \\ &\delta T^{\mu\nu}=-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\Delta^{\mu\nu}+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}+2q_{\lambda}\Delta^{\lambda(\mu}u^{\nu)} \\ &\delta (T^{\mu\nu}n_{\nu})=\delta T^{\mu\nu}n_{\nu}=2q_{\lambda}\Delta^{\lambda(\mu}u^{\nu)}u_{\nu}=q_{\lambda}\Delta^{\lambda\mu}u^{\nu}u_{\nu} =q_{\lambda}\Delta^{\lambda\mu}\\ &\beta\delta T^{\mu\nu}\partial_{\mu}u_{\nu}=\beta (q^{\mu}u^{\nu}+q^{\nu}u^{\mu})\partial_{\mu}u_{\nu} +\beta (-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\Delta^{\mu\nu}+ \Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho})\partial_{\nu}u_{\mu}\\ &=\beta q^{\nu}u^{\mu}\partial_{\mu}u_{\nu}+\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}\right]\\ &=\beta q\cdot \dot u+\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}\right] \end{align}$$

两者完全一致,所以实际上比较两个等式的右边


 * $$\begin{align}

&\partial_{\mu}Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu}) +\beta q\cdot \dot u+\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}\right] \\ &=\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right] \partial_{\mu}Q^{\mu}(\delta N^{\mu},\delta T^{\mu\nu})\\ &=0 \end{align}$$

这意味着 $$Q^{\mu}$$ 是守恒的.

热力学第二定律
热力学第二定律意味着


 * $$\begin{align}

\frac{dS^{0}}{dt}+\nabla\vec S=\partial_{\mu}S^{\mu}\ge 0 \end{align}$$

为此,我们把前面得到的熵流更详细的写出


 * $$\begin{align}

&\partial_{\mu}S^{\mu} =- j_{\nu}\Delta^{\mu\nu} \partial_{\mu}\alpha+q_{\lambda}\Delta^{\lambda\mu}\partial_{\mu}\beta +\beta\left[-(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u\right]+\partial_{\mu}Q^{\mu}\\ &T\partial_{\mu}S^{\mu} =- j_{\nu}\Delta^{\mu\nu} T\partial_{\mu}\alpha+q_{\lambda}\Delta^{\lambda\mu}T\partial_{\mu}\beta -(\Pi+\frac{\Delta^{\rho\sigma}\pi_{\sigma\rho}}{\Delta})\theta+\Delta^{(\mu\rho}\Delta^{\nu)\sigma}\pi_{\sigma\rho}\partial_{\nu}u_{\mu}+q\cdot \dot u+T\partial_{\mu}Q^{\mu}\\ &=- j^{\mu} T\partial_{\mu}\alpha+q^{\mu}T\partial_{\mu}\beta-\Pi\theta+\pi^{\mu\nu}\partial_{\nu}u_{\mu}+q\cdot \dot u+T\partial_{\mu}Q^{\mu}\\ &=- j^{\mu} T\partial_{\mu}\alpha-q^{\mu}\frac{\partial_{\mu}T}{T}-\Pi\theta+\pi^{\mu\nu}\partial_{\nu}u_{\mu}+q\cdot \dot u+T\partial_{\mu}Q^{\mu}\\ &=- j^{\mu} T\partial_{\mu}\alpha-q^{\mu}\frac{\partial_{\mu}T}{T}+q^{\mu} \dot u_{\mu}-\Pi\theta+\pi^{\mu\nu}\partial_{\nu}u_{\mu}+T\partial_{\mu}Q^{\mu}\\ &=- j^{\mu} T\partial_{\mu}\alpha-q^{\mu}\frac{\nabla_{\mu}T}{T}+q^{\mu} \dot u_{\mu}-\Pi\theta+\pi^{\mu\nu}\nabla_{\nu}u_{\mu}+T\partial_{\mu}Q^{\mu}\\ &=- j^{\mu} T\partial_{\mu}\alpha-q^{\mu}(\frac{\nabla_{\mu}T}{T}- \dot u_{\mu})-\Pi\theta+\pi^{\mu\nu}\nabla_{\nu}u_{\mu}+T\partial_{\mu}Q^{\mu}\\ &=- j^{\mu} T\partial_{\mu}\alpha-q^{\mu}(\frac{\nabla_{\mu}T}{T}- \dot u_{\mu})-\Pi\theta+\pi^{\mu\nu}\nabla_{(\mu}u_{\nu)}+T\partial_{\mu}Q^{\mu} \end{align}$$

其中我们已经采用默认约定 $$\pi^{\mu\nu}$$ 已经对称化与速度垂直并且无迹, $$j^{\mu}$$ 与 $$q^{\mu}$$ 已经与速度垂直,如果引入记号


 * $$\begin{align}

&X\equiv -\theta =-\nabla\cdot u \\ &X_{\mu}\equiv (\frac{\nabla_{\mu}T}{T}- \dot u_{\mu})\\ &X_{\mu\nu}\equiv \nabla_{(\mu}u_{\nu)} \end{align}$$

并且注意到,在实际物理问题中,我们不可能同时 $$j^{\mu}$$ 与 $$q^{\mu}$$ ,只能取其一,如果取采用Eckart随动参照系 $$j^{\mu}=0 $$与 $$q^{\mu}\ne 0$$ 我们得到Heinz一文的(25)式的结果


 * $$\begin{align}

T\partial_{\mu}S^{\mu}=\Pi X-q^{\mu} X_{\mu}+\pi^{\mu\nu}X_{\mu\nu}+T\partial_{\mu}Q^{\mu} \ge 0 \end{align}$$

如果取Landau随动参照系 $$j^{\mu}\ne 0$$ 与 $$q^{\mu}= 0$$ ,利用Heinz文中给出的关系 $$j^{\mu}=-\frac{n}{\varepsilon+P}q^{\mu}$$ 以及


 * $$\begin{align}

X_{\mu}\equiv (\frac{\nabla_{\mu}T}{T}- \dot u_{\mu})=-\frac{nT}{\varepsilon+P}\nabla_{\mu}\left(\frac{\mu}{T}\right) \end{align}$$

则两个结果完全等价,因为这时我们有


 * $$\begin{align}

- j^{\mu} T\partial_{\mu}\alpha=-q^{\mu}(\frac{\nabla_{\mu}T}{T}- \dot u_{\mu}) \end{align}$$

一阶粘滞理论(Navier-Stokes理论)
我们需要通过模型来得到新的独立变量,粘滞系数,所满足的方程.但是通过上述热力学第二定律的形式,直观而方便的可以假设粘滞系数和热流满足下述线性关系,称为代数假定(Algebraic Ansatz)或者Fick定律


 * $$\begin{align}

&\Pi=-\zeta \theta\\ &q^{\mu}=-\lambda\frac{nT^2}{\varepsilon+P}\nabla^{\mu}\left(\frac{\mu}{T}\right) \\ &\pi^{\mu\nu}=2\eta \sigma^{\mu\nu} \end{align}$$

其中的系数$$\zeta, \lambda, \eta$$正定.这样保证了两点,第一由于这时熵流具有两次项形式,熵增加原理得以保证


 * $$\begin{align}

T\partial_{\mu}S^{\mu}=\frac{\Pi^2}{\zeta}-\frac{q^{\mu}q_{\mu}}{2\lambda T}+\frac{\pi^{\mu\nu}\pi_{\mu\nu}}{2\eta} \ge 0 \end{align}$$

其中第二项为负因为热流 $$q^{\mu}$$ 是类空的,其内积为负数.第二是因为自然的给予了粘滞与热流其物理意义.例如,体粘滞是对体系整体膨胀 $$\theta$$ 的反应,剪切粘滞是对速度导数 $$\sigma^{\mu\nu}$$ 的反应,而热流是对温度和化学势梯度 $$\nabla^{\mu}\left(\frac{\mu}{T}\right)$$ 的反应.前面三个体系的动力学量按Fick定律的思路被视为热力学力,分布称为标量,张量,矢量热力学力.而满足热力学第二定律的最简单的一阶理论,假定这些反应都是线性的.可惜这种瞬时的反应机制导致流体的演化违反因果律,即可以导致超光速信号的传播.我们必须进而寻求两阶或者高阶理论.一个简单的关于这类方程的数学上和物理上的讨论可以参见上述Tomoi的文章.

两阶理论和高阶粘滞理论
从数学的角度出发,一阶粘滞理论相当于热传导的抛物线(Parabolic)形偏导方程,.数学上需要引入弛豫项,从而改变为双曲线形(Hyperbolic)偏导方程抛物线.物理上,需要讨论对应于非平衡态的推广的热力学,这相当于引入非零的 $$Q^{\mu}$$ 项即非平衡流,或者明显的引入弛豫项对应的格林函数改写Fick定律.

这里给出两个例子,第一个例子是Israel-Stewart理论.该模型引入两个假设, $$Q^{\mu}$$ 的具体形式,熵流仍然可以写成正定的两次型.在不考虑热流的情况下,假定 $$Q^{\mu}$$ 的形式为


 * $$\begin{align}

Q^{\mu}=-(\beta_0\Pi^2+\beta_2\pi_{\nu\lambda}\pi^{\nu\lambda})\frac{u^{\mu}}{2T} \end{align}$$

由此得到熵流的形式为


 * $$\begin{align}

T\partial_{\mu}S^{\mu}=\Pi \left[-\theta-\beta_0\dot\Pi-\Pi T\partial_{\mu}\left(\frac{\beta_0u^{\mu}}{2T}\right)\right] +\pi^{\alpha\beta}\left[\sigma_{\alpha\beta}-\beta_2\dot\pi_{\alpha\beta}-\pi_{\alpha\beta}T\partial_{\mu}\left(\frac{\beta_2u^{\mu}}{2T}\right)\right] \end{align}$$

如果熵流仍然可以写成


 * $$\begin{align}

T\partial_{\mu}S^{\mu}=\frac{\Pi^2}{\zeta}+\frac{\pi^{\mu\nu}\pi_{\mu\nu}}{2\eta} \ge 0 \end{align}$$

则


 * $$\begin{align}

&\Pi= \zeta\left[-\theta-\beta_0\dot\Pi-\Pi T\partial_{\mu}\left(\frac{\beta_0u^{\mu}}{2T}\right)\right] \\ &\pi_{\alpha\beta}=2\eta\left[\sigma_{\alpha\beta}-\beta_2\dot\pi_{\alpha\beta}-\pi_{\alpha\beta}T\partial_{\mu}\left(\frac{\beta_2u^{\mu}}{2T}\right)\right] \end{align}$$

比较前面的线性理论,我们很自然的看到这里新出现的非线性项.原则上所有新的系数都应该由微观的动力学理论得到,从流体力学层次,这些系数可以看作模型参数.为了讨论新引入参数的物理意义,我们把上面的方程改写成


 * $$\begin{align}

&\dot\Pi= -\frac{1}{\tau_{\Pi}}\left[\Pi+\zeta\theta+\Pi\zeta T\partial_{\mu}\left(\frac{\tau_{\Pi}u^{\mu}}{2\zeta T}\right)\right] \approx -\frac{1}{\tau_{\Pi}}\left[\Pi+\zeta\theta\right]   \\ &\dot \pi_{\alpha\beta}=-\frac{1}{\tau_{\pi}}\left[\pi_{\alpha\beta}-2\eta \sigma_{\alpha\beta}+\pi_{\alpha\beta}\eta T\partial_{\mu}\left(\frac{\tau_{\pi}u^{\mu}}{2\eta T}\right)\right]  \approx -\frac{1}{\tau_{\pi}}\left[\pi_{\alpha\beta}-2\eta \sigma_{\alpha\beta}\right] \end{align}$$

其中 $$\tau_{\Pi} = \zeta\beta_0,  \tau_{\pi} = 2\eta\beta_2$$ 的物理意义为弛豫时间.理由如下,考虑体粘滞系数的情况,因为如果某种原因运动学上体系没有整体膨胀 $$\theta =0$$ ,则关于体粘滞系数方程退化为


 * $$\begin{align}

\dot\Pi= -\frac{1}{\tau_{\Pi}}\Pi \end{align}$$ 他的解为


 * $$\begin{align}

\Pi = \Pi_0 e^{-\frac{\tau}{\tau_{\Pi}}} \end{align}$$

第二个例子是记忆模型.该模型直接假设粘滞系数和热流对体系动力学参数的依赖关系由历史决定,而非瞬时决定.例如取代


 * $$\begin{align}

\Pi=-\zeta \theta \end{align}$$

模型假设


 * $$\begin{align}

\Pi=-\int_{-\infty}^{\tau} d\tau'G(\tau,\tau')\zeta' \theta' \end{align}$$

其中引入记号的规定 $$f' \equiv f(\tau')$$ ,如果格林函数为 $$\delta$$ 函数,则模型退化为代数假定.类似的我们可以写下


 * $$\begin{align}

\dot\Pi=D\Pi=-\int_{-\infty}^{\tau} d\tau'DG(\tau,\tau')\zeta' \theta'-G(\tau,\tau)\zeta \theta \end{align}$$

由于这里格林函数的物理意义是体系对过去的依赖程度,所以数学上至少引入一个参数 $$\tau_R$$ 描写体系的弛豫时间
 * $$\begin{align}

G(\tau,\tau')\equiv G(\tau,\tau';\tau_R) \end{align}$$

比较IS模型中含有弛豫时间参数的体粘滞系数的方程(或者独立于IS模型,我们考虑上述方程应该含有一项描写一阶IS模型,另一项当独立存在时给出指数衰减弛豫形式解),我们得到


 * $$\begin{align}

&G(\tau,\tau;\tau_R) \equiv \frac{1}{\tau_R} \\ &DG(\tau,\tau;\tau_R) \equiv -\frac{1}{\tau_R}G(\tau,\tau;\tau_R) \end{align}$$

上面第二个方程显然有指数形式的解,把第一个方程作为它的边界条件,我们得到


 * $$\begin{align}

G(\tau,\tau';\tau_R) \equiv \frac{1}{\tau_R}e^{-(\tau-\tau')/\tau_{R}} \end{align}$$

代回体粘滞的导数的表达式,我们得到


 * $$\begin{align}

\dot{\Pi}=-\frac{\Pi}{\tau_R}-\frac{\zeta\theta}{\tau_R} \end{align}$$

如果积分有限,不难验证体粘滞系数可以表达为


 * $$\begin{align}

\widetilde{\Pi}=-\int_{\tau_0}^{\tau} d\tau'G(\tau,\tau')\frac{\zeta'\theta'}{\sigma'}+e^{-(\tau-\tau_0)/\tau_R}\widetilde{\Pi}_0 \end{align}$$

我们可以类似的写下其余方程,这里总结如下


 * $$\begin{align}

&\Pi=-\int_{-\infty}^{\tau} d\tau'G(\tau,\tau')\zeta' \theta'\\ &q^{\mu}=-\int_{-\infty}^{\tau} d\tau'G(\tau,\tau')\lambda'\frac{n'T'^2}{\varepsilon'+P'}\nabla_{\mu}\left(\frac{\mu'}{T'}\right) \\ &\pi^{\mu\nu}=\int_{-\infty}^{\tau} d\tau'G(\tau,\tau')2\eta' \sigma'^{\mu\nu} \end{align}$$

双曲线Bjorken坐标系及其度规
Bjorken坐标系的变量定义为


 * $$\begin{align}

&\tau=\sqrt{t^2-z^2} \\ &\eta = \frac{1}{2} \ln \left(\frac{t+z}{t-z}\right) \end{align}$$

利用一些有名的关于协变导数的恒等式,如


 * $$\begin{align}

&\Gamma^i_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right) = {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m}) \\ &\Gamma^i_{ki}=\frac{1}{2}g^{im}\frac{\partial g_{im}}{\partial x^k}={1 \over 2g}\frac{\partial g}{\partial x^k}=\frac{\partial \log \sqrt{|g|}}{\partial x^k}\\ &D_{\mu}\phi\equiv(\phi)_{;\mu}=\partial_{\mu}\phi\\ &D_{\mu}V_{\nu}\equiv(V_{\nu})_{;\mu}=\partial_{\mu}V_{\nu}-\Gamma^{\alpha}_{\mu\nu}V_{\alpha}\\ &D_{\mu}V^{\nu}\equiv(V^{\nu})_{;\mu}=\partial_{\mu}V^{\nu}+\Gamma^{\nu}_{\mu\alpha}V^{\alpha}\\ &D_{\mu}V_{\alpha\beta}\equiv(V_{\alpha\beta})_{;\mu}=\partial_{\mu}V_{\alpha\beta}-\Gamma^{\lambda}_{\alpha\mu}V_{\lambda\beta}-\Gamma^{\lambda}_{\beta\mu}V_{\lambda\alpha}\\ &D_{\mu}V^{\mu}\equiv(V^{\mu})_{;\mu}=\frac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}V^{\mu})\\ &D_{\mu}V^{\mu\nu}\equiv(V^{\mu\nu})_{;\mu}=\frac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}V^{\mu\nu})+\Gamma^{\nu}_{\alpha\mu}V^{\alpha\mu} \end{align}$$

以外,一些等式涉及Bjorken坐标下度规的具体形式
 * $$\begin{align}

g_{\mu\nu}=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -{\tau^2} \end{pmatrix} \end{align}$$


 * $$\begin{align}

\sqrt{-g}=\tau \end{align}$$

显然度规的逆为


 * $$\begin{align}

g^{\mu\nu}=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -\frac{1}{\tau^2} \end{pmatrix} \end{align}$$

利用导致克里斯多夫符号的定义


 * $$\begin{align}

\Gamma^i_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right) \end{align}$$

我们知道,对应该度规,克里斯多夫符号仅仅三项不为零,它们是


 * $$\begin{align}

\Gamma^{0}_{33}=\tau, \Gamma^{3}_{30}=\Gamma^{3}_{03}=1/\tau \end{align}$$

一些公式的具体推导参见Key Notes on Viscous Hyperbolic SPH Equation of Motion of Hydrodynamic Model

SPH自由度
这里仅仅列出主要结果,一些公式的具体推导参见Key Notes on Viscous Hyperbolic SPH Equation of Motion of Hydrodynamic Model

记号约定.注意在用SPH自由度表达时,所有在实验室坐标系中的量用上标$$^*$$标记,比如这里的记号$$\sigma^*$$ 表示在实验室坐标系中的强度量,而在本小节以外,上述记号原则上不出现,任何量比如$$\sigma,P,\Pi,\epsilon$$ 表示在随动坐标系中的强度量.比如能动张量就是用四速度和随动系中的能量密度,压强,体粘滞系数来表达的.这个自然的做法对于剪切粘滞系数存在歧义,因为剪切粘滞系数在能动张量中是在任意坐标系中的形式,但是我们没有使用星号.这点需要特别注意.

实际应用中,我们需要把方程表达为SPH自由度.按SPH的物理涵义,对任何强度物理量$$A$$在空间任何点的实验室坐标系的值由该物理量在SPH粒子$$i$$处实验室坐标系的取值(依赖于状态方程)以及守恒荷$$\nu$$(对应密度$$\sigma$$)确定下来


 * $$\begin{align}

A_j\equiv \sum_i \left(\frac{A}{\sigma^*}\right)_i\nu_iW_{ji}=\sum_i \left(\frac{A}{\sigma \tau\gamma}\right)_i\nu_iW_{ji} \end{align}$$

注意到一个重要的约定,如果$$A$$本身也是一个守恒荷$$B$$在实验室坐标系中的密度$$b^*$$,那么由于


 * $$\begin{align}

&A=b^*\\ &\left(\frac{A}{\sigma^*}\right)_i=\left(\frac{b^*}{\sigma^*}\right)_i=\left(\frac{b}{\sigma}\right)_i\\ &A_j=\sum_i \left(\frac{A}{\sigma^*}\right)_i\nu_iW_{ji}=\sum_i B_iW_{ji} \end{align}$$

而对于一般的物理量,如压强,则没有上面的简化形式,运动方程中涉及对物理量的空间偏导,从而也必须表达为SPH自由度


 * $$\begin{align}

(\partial A)_j\equiv\sum_i \left(\frac{A}{\sigma^*}\right)_i\nu_i\nabla_jW_{ji} \end{align}$$

出于懒惰关系,其中位置矢量并没有加矢量箭头,其定义准确的写开为


 * $$\begin{align}

&W_{ji}\equiv W_{ji}(|r_j-r_i|)\equiv  W_{ji}(|\vec r_j-\vec r_i|) \\ &\nabla_jW_{ji}\equiv \nabla_jW_{ji}(|r_j-r_i|)\equiv \frac{\partial W_{ji}(|\vec r_j-\vec r_i|)}{\partial \vec r_j}   \\ &\nabla_iW_{ji}\equiv \nabla_iW_{ji}(|r_j-r_i|)\equiv \frac{\partial W_{ji}(|\vec r_j-\vec r_i|)}{\partial \vec r_i} \end{align}$$

它自然的满足


 * $$\begin{align}

\nabla_iW_{ji}(|r_j-r_i|)=-\nabla_jW_{ji}(|r_j-r_i|) \end{align}$$

在实际运用中,我们需要计算四速度的空间导数,这时我们似乎没有相互作用力的对称性的限制,所以我们直接写下


 * $$\begin{align}

(\partial u_k)_i=\sum_j\nu_j\left(\frac{u_k}{\sigma^*}\right)_j\nabla_i W_{ij} \end{align}$$

对于本模型,有两项需要写为SPH自由度时,需要特别注意.第一项是压强的空间偏导,
 * $$\partial_i {\tilde P}$$

其中


 * $$\begin{align}

\widetilde{P}\equiv P+\Pi \end{align}$$

我们不采取上面的简化形式.理由是出于对称性的考虑,我们必须引入一个貌似多余步骤.这实际上从一个侧面说明了SPH方法不是第一性原理,一些物理量的表达没有唯一性,这时必须引入一些准则,比如表达式必须是对称的(物理意义为两点之间的相互作用满足牛顿第三定律,在不进行求和的情况下,表达式关于源点和目标点对称).不难看到,直接利用SPH核函数的导数求和不满足这个对称性.另外该形式可以比较自然的由变分原理得到.


 * $$\begin{align}

(\partial {\tilde P})_i = \sum_j \nu_j \sigma_i^* \left(\frac{{\tilde P}_i}{\sigma_i^{*2}}+\frac{{\tilde P}_j}{\sigma_j^{*2}}\right) \bigtriangledown_i W_{ij} \end{align}$$

另外一个运动方程中必须表达为SPH自由度的项是
 * $$\partial_i v^i $$

它涉及流体速度从而是对时间的导数(之后再对空间求偏导),与重子数密度或者压强不同,我们并不直接计算,而是利用守恒荷密度对应的守恒流来得到它的表达式,我们最后得到


 * $$\begin{align}

\left(\partial_i v^i\right)_j = - \frac{1}{\sigma^*}\sum_k \nu_k (\dot r_j - \dot r_k) \cdot  \nabla_j W_{jk} \end{align}$$

作为一个例子,这里导出理想流体在双曲线坐标度规空间的SPH自由度的运动方程.我们知道,在理想流体的情况下,能动张量守恒的时间部分等价于熵流守恒并不带来新的方程,而能动张量守恒的空间部分对应决定了SPH粒子"速度"对时间导数的三个运动方程.注意到这里我们并不把能动张量守恒投影到垂直和平行于四速度的方向上,而是直接取能动张量守恒的空间部分.在度规下$$\sqrt{-g}=\tau$$,我们有


 * $$\partial^\mu(\tau T_{\mu i})=0$$
 * $$\partial^\mu(\tau (\epsilon+P)u_\mu u_i-\tau g_{\mu i}P)=0$$
 * $$\tau s u_\mu\partial^\mu\left(\frac{(\epsilon+P)}{s}u_i\right)-\partial^\mu(\tau g_{\mu i}P)=0$$
 * $$\tau s \gamma \frac{d}{d\tau}\left(\frac{(\epsilon+P)}{s}u_i\right)-\partial^\mu(\tau g_{\mu i}P)=0$$
 * $$\tau s \gamma \frac{d}{d\tau}\left(\frac{(\epsilon+P)}{s}u_i\right)-g_{\mu i}\partial^\mu (\tau P)-\tau P\partial^\mu( g_{\mu i})=0$$
 * $$\tau s \gamma \frac{d}{d\tau}\left(\frac{(\epsilon+P)}{s}u_i\right)-\partial_i (\tau P)=0$$
 * $$\frac{d}{d\tau}\left(\frac{(\epsilon+P)}{s}u_i\right)-\frac{1}{s\gamma}\partial_i P=0$$
 * $$\frac{d}{d\tau}\left(\frac{(\epsilon+P)}{s}\gamma g_{ij}v^j\right)-\frac{1}{s\gamma}\partial_i P=0$$

这就是Yogiro的综述文章中给出的运动方程.上述推导中用到熵流守恒以及度规张量导数为零的性质.

考虑仅仅体积粘度在Bjorken坐标系中的流体的相对论流体力学方程
这里我们具体给出在Bjorken坐标系中考试体粘滞系数情况下的流体动力学方程.

这一情况下一共有5个方程.其中3个来自于能动张量守恒的空间部分(在随动坐标系中等价于垂直流体四速度的部分),而能动张量守恒的时间分量(在随动坐标系中等价于平行流体四速度的部分)在考虑粘度的情况下不再平庸,给出熵密度的演化方程,最后我们还需要一个与模型有关的关于体积粘度的方程,我们选用Tomoi和Kodama提出的记忆模型.对应上述5个方程,我们选取5个变量如下,逆变四速度的三个空间分量 $$u^i (i=1,2,3)$$ ,熵密度除以守恒荷密度 $$s/\sigma$$ ,体积粘度除以守恒荷密度 $$\Pi/\sigma$$ ,其中 $$s$$ 是熵密度, $$\Pi$$ 是体积粘度,由于熵流并不守恒,我们额外的引入一个守恒荷 $$\sigma$$ 满足


 * $$\begin{align}

\left(\sigma u^{\mu}\right)_{;\mu}=0 \end{align}$$

我们得到对任意流 $$f u^{\mu}$$  ,有


 * $$\partial \cdot (fu)

=\left( f u^{\mu}\right)_{;\mu} = \sigma u^{\mu}\left(\frac{f}{\sigma}\right)_{;\mu} = \sigma u^{\mu}\left(\frac{f}{\sigma}\right)_{,\mu} =\sigma\gamma\frac{d}{d\tau}\left(\frac{f}{\sigma}\right) =\frac{\sigma^*}{\tau}\frac{d}{d\tau}\left(\frac{f}{\sigma}\right) \equiv\frac{\sigma^*}{\tau}d_{\tau}\left(\frac{f}{\sigma}\right) $$

其中倒数第二步中注意到 $$\tau$$ 并非固有时,从而产生一个 $$\gamma=u^0$$ 因子.而实验室坐标系中的对此守恒荷的观测密度与正则随动系中的密度的关系为 $$\sigma^*=\tau \gamma \sigma$$

另外有一点需要特别注意,在这里的推导中记号中 是Bjorken坐标系的类时坐标,而非固有时.所以与上面推导流体动力学方程中引入的记号完全不同,对 的导数并不是对固有时求导,而是对一个类时坐标的导数.


 * $$\begin{align}

d_{\tau}f =\frac{1}{\gamma} u\cdot \partial f \ne Df=u\cdot \partial f=\dot f \end{align}$$

下面我们将逐项讨论流体力学的运动方程,其中一些公式的具体证明细节参见Key Notes on Viscous Hyperbolic SPH Equation of Motion of Hydrodynamic Model.

能动张量守恒方程的空间部分

 * 能动张量守恒方程的空间部分

逆变四速度空间分量 $$u^i (i=1,2,3)$$ 满足的方程对应能动张量守恒
 * $$\begin{align}

\left(T^{\mu\nu}\right)_{;\mu} = 0 \end{align}$$ 的空间部分.理论上,我们可以写下能动张量守恒与四速度垂直的部分,在此意义上得到的表达式应该更为对称,但是除了数学上更繁琐外,最后的结果形式上其实完全相同,参见平直空间无粘滞的情况(Key Notes on Hydrodynamics).利用能动张量的具体形式,


 * $$\begin{align}

T^{\mu\nu}&=&{\tilde\omega}u^{\mu}u^{\nu}-{\tilde P}g^{\mu\nu}=\left(\varepsilon+P+{\Pi}\right)u^{\mu}u^{\nu}-\left(P+{\Pi}\right)g^{\mu\nu} \end{align}$$

其中 $${\tilde\omega}= \varepsilon + {\tilde P}$$ 和 $${\tilde P}= P + \Pi$$ .取空间部分 $$\nu=i=1,2,3$$ 可表达为


 * $$\begin{align}

&{\left(\tilde \omega u_{\mu}u_{i}-{{\tilde P}g_{\mu i}}\right)_{;}}^{\mu} = 0 \\ &{\tilde \omega}u_{i}\partial \cdot u+{\tilde \omega}u\cdot \partial u_{i}+u_{i} u \cdot \partial {\tilde \omega}-\partial_{i}{\tilde P} = 0  \\ &{\tilde \omega}u_{i}\partial \cdot u+{\tilde \omega}u\cdot \partial u_{i}+(1+c_s^2) u_{i} u \cdot \partial {\varepsilon}+u_{i} u \cdot \partial {\Pi}-\partial_{i}{\tilde P} = 0 \end{align}$$

其中 $$c_s$$ 是声速,
 * $$\begin{align}

c_s^2 \equiv \left(\partial P/\partial \varepsilon\right)_{ad} \end{align}$$

下面我们逐一讨论上面方程涉及到的各项.首先是记号约定的问题.

$$\partial \cdot u, u\cdot \partial u^{i}, u \cdot \partial {\tilde \omega}$$ 中不含上下指标的偏导符号 $$\partial$$ 默认都是协变偏导,而非(前面推导运动方程过程中一直默认使用的)一般偏导.我们有


 * $$\begin{align}

&\partial \cdot u \equiv \left(u^{\mu}\right)_{;\mu} \\ &= \frac{1}{\tau} \partial_{\mu} \left(\tau u^{\mu}\right)  \\ &= \frac{1}{\tau} \left(\partial_{\tau} \left(\tau \gamma\right)+\partial_i\left(\tau\gamma v^i \right)\right)   \\ &= \frac{1}{\tau} \left(\partial_{\tau} \left(\tau \gamma\right)+v^i \partial_i\left(\tau\gamma  \right)\right)+\frac{1}{\tau}(\tau\gamma)\partial_i v^i   \\ &= \frac{1}{\tau} d_{\tau} (\tau\gamma) + \gamma \partial_i v^i  \\ &= \frac{\gamma}{\tau} + d_{\tau}\gamma+\gamma \partial_i v^{i} \end{align}$$

我们可以进一步把方程右边逐项表达成上面选定的变量


 * $$\begin{align}

&d_{\tau}\gamma = \frac{d}{d\tau}\left(1-g_{ij}u^iu^j\right)^{1/2}  \\ &= -\frac{g_{ij}u^i}{\gamma}\frac{du^j}{d\tau}-\frac{u^iu^j}{2\gamma}\frac{dg_{ij}}{d\tau}  \\ &= -\frac{u_j}{\gamma}\frac{du^j}{d\tau}+\frac{\left(u^3\right)^2\tau}{\gamma} \end{align}$$

同样的


 * $$\begin{align}

&u\cdot \partial u_{i} = \gamma \frac{du_i}{d\tau} - u^{\mu} u_{\lambda} \Gamma^{\lambda}_{i\mu}  \\ &= \gamma \frac{du_i}{d\tau}   \\ &= - \gamma\frac{du^1}{d\tau}\delta_{i1}- \gamma\frac{du^2}{d\tau}\delta_{i2}- \left(\gamma\tau^2\frac{du^3}{d\tau}+2\tau\gamma u^3 \right)\delta_{i3} \end{align}$$

其中第二步等式是利用了度规和斯托克斯符号的明显形式


 * $$\begin{align}

- u^{\mu} u_{\lambda} \Gamma^{\lambda}_{i\mu} = \left(-\tau\gamma u^3 - \frac{1}{\tau}\gamma u_3 \right)\delta_{i3}= 0\times\delta_{i3} = 0 \end{align}$$ 上面能动张量守恒空间部分方程中涉及对体粘滞的导数的一项为


 * $$\begin{align}

u\cdot\partial \Pi=\gamma{d_{\tau}\Pi}=-\frac{1}{\tau_R}\left(\Pi+{\left(\zeta+{\tau_R\Pi} \right )\partial\cdot u}\right ) \end{align}$$

推导过程中最后利用了记忆模型.这个表达式同时是体积粘度满足的方程,在下面会继续讨论.四速度方程的最后一项类似的可以表达为


 * $$\begin{align}

&u \cdot \partial {\varepsilon} = u \cdot \partial {\varepsilon}+\omega \partial \cdot u - \omega \partial \cdot u  \\ &= T u \cdot \partial s + T s \partial \cdot u + \sum_p \mu_p u \cdot \partial n_p + \sum_p \mu_p n_p \partial \cdot u - \omega \partial\cdot u \\ &= T \partial \cdot (su) + \sum_p \mu_p \partial \cdot (n_pu) - \omega \partial \cdot u  \\ &= T \partial \cdot (su) - \omega \partial \cdot u   \\ &= \frac{T \sigma^*}{\tau}d_{\tau}\left(\frac{s}{\sigma}\right) - \omega \partial \cdot u \end{align}$$

其中用到守恒流满足的关系


 * $$\begin{align}

\sum_p \mu_p \partial \cdot (n_p u)=0 \end{align}$$

还涉及热力学第一定律


 * $$\begin{align}

&\partial \varepsilon=\partial(\frac{E}{V})=\frac{V\partial E-E\partial V}{V^2} \\ &\partial E =T\partial S -P \partial V +\sum_p \mu \partial N_p\\ &\partial S = \partial (sV)=V\partial s+s\partial V\\ &\sum_p \mu \partial N_p=\sum_p \mu V\partial n_p+\sum_p \mu n_p \partial V\\ &\partial \varepsilon =T\partial s+\sum_p\mu_p \partial n_p \end{align}$$

其中关于压强的项自然的被抵消.最后有用到广延量的性质


 * $$\begin{align}

\omega=\varepsilon+p=Ts-p+\sum_p \mu \partial n_p+p=Ts+\sum_p \mu \partial n_p \end{align}$$

能动张量守恒方程的时间部分,熵流方程

 * 能动张量守恒方程的时间部分,熵流方程

由于我们选取 $$s/\sigma$$ 作为变量,需要一个关于它的方程,这个方程实际上可由能动张量平行于四速度的分量得到


 * $$\begin{align}

&u_{\nu}({\tilde \omega}u^{\mu}u^{\nu})_{;\mu}-u \cdot \partial {\tilde P} = 0 \\ &({\tilde \omega}u^{\mu})_{;\mu}-u \cdot \partial {\tilde P} = 0  \\ &u \cdot \partial {\tilde \omega} +{\tilde \omega} \partial \cdot u -u \cdot \partial {\tilde P} = 0  \\ &u \cdot \partial \varepsilon +{\tilde \omega} \partial \cdot u = 0  \\ &T u \cdot \partial s + \sum_p \mu_p u \cdot \partial n_p + (Ts + \sum_p \mu_p n_p + \Pi) \partial \cdot u = 0 \\ &T (su^{\mu})_{;\mu} + \Pi \partial \cdot u  = 0   \\ &\frac{T\sigma^*}{\tau} d_{\tau} \left(\frac{s}{\sigma}\right) + \Pi \partial \cdot u = 0 \end{align}$$

推导过程用到的守恒流关系与广延量性质上面已经都有论述,故略去.容易看到,对无粘滞性的情况, $$\Pi\equiv 0$$ ,上式等价为


 * $$\begin{align}

(su^{\mu})_{;\mu} = 0 \end{align}$$

即熵流守恒.

粘滞性方程

 * 粘滞性方程

运动方程涉及的另一个变量是 $$\Pi/\sigma$$,关于体粘滞系数的方程在前面实际上已经有所涉及,这里给出其在Bjorken坐标系中的方程的具体形式.我们实际上需要把体粘滞系数$$\Pi$$满足的方程


 * $$\begin{align}

\dot{\widetilde{\Pi}}=-\frac{\widetilde{\Pi}}{\tau_R}-\frac{\zeta\theta}{\sigma\tau_R} \end{align}$$

在现有的坐标系中表达出来.按Gabriel的笔记(一种模型),我们将对故有时的导数表达为四速度与协变导数的缩并,


 * $$\begin{align}

\dot{\widetilde{\Pi}}\equiv\frac{D\widetilde{\Pi}}{D\tau} \end{align}$$

而将体系的膨胀速率记为协变散度


 * $$\begin{align}

\theta\equiv\partial\cdot u \end{align}$$ 由此我们得到我们需要的关于体粘滞系数的方程.一方面,我们有,


 * $$\begin{align}

\dot{\widetilde{\Pi}}\equiv\frac{D\widetilde{\Pi}}{D\tau}=u^\mu D_\mu\widetilde{\Pi}=u^\mu \partial_\mu\widetilde{\Pi}=\gamma d_\tau\left(\frac{\Pi}{\sigma} \right ) \end{align}$$

另一方面
 * $$\begin{align}

d_{\tau}\left(\frac{\Pi}{\sigma}\right) = \frac{d_{\tau}\Pi}{\sigma}-\frac{\Pi}{\sigma^2}d_{\tau}\sigma = \frac{d_{\tau}\Pi}{\sigma} + \frac{\Pi}{\sigma^2}\left(\frac{\sigma}{\gamma}\partial \cdot u \right) = \frac{d_{\tau}\Pi}{\sigma} + \frac{\Pi}{\sigma}\left(\frac{1}{\gamma}\partial \cdot u \right) \end{align}$$

其中利用关系


 * $$\begin{align}

0 = \left(\sigma u^{\mu}\right)_{;\mu} = u \cdot \partial \sigma + \sigma \partial \cdot u \end{align}$$

从而


 * $$\begin{align}

d_\tau\sigma=\frac{d\sigma}{d\tau}=-\frac{\sigma}{\gamma}(\partial\cdot u) \end{align}$$

联合两者,我们得到需要的方程


 * $$\begin{align}

d_{\tau}\left(\frac{\Pi}{\sigma}\right) = \frac{d_{\tau}\Pi}{\sigma} + \frac{\Pi}{\sigma^2}\left(\frac{\sigma}{\gamma}\partial \cdot u\right) =\frac{1}{\gamma}\left(-\frac{\widetilde{\Pi}}{\tau_R}-\frac{\zeta\theta}{\sigma\tau_R} \right ) =-\frac{1}{\gamma\tau_R\sigma}\left(\Pi+{\zeta\partial\cdot u}\right ) \end{align}$$

从中,我们可以得到另一个被用到的关系


 * $$\begin{align}

{d_{\tau}\Pi}=-\frac{1}{\gamma\tau_R}\left(\Pi+{\left(\zeta+{\tau_R\Pi} \right )\partial\cdot u}\right ) \end{align}$$

与无粘度情况下运动方程的比较

 * 与无粘度情况下运动方程的比较

无粘度情况下的方程显然是这里讨论的一个特例.如上所述,能动张量守恒的的时间分量部分,退化为熵流守恒等式.对于空间部分,注意到,面我们利用度规的具体形式可以证明


 * $$\begin{align}

u\cdot \partial u_{i} = u_{\mu}{\left(u_i\right)_{;}}^{\mu} = u_{\mu}{\left(u_i\right)_{,}}^{\mu} \end{align}$$

为了证明普通偏导和协变导数一致,需要证明对于现有的度规,两者的差别等于零,相应的需要利用了前面已经用到的一个和克里斯多夫符号有关的关系


 * $$\begin{align}

u_{\mu} \Gamma^{\mu}_{i\lambda}u^{\lambda} =\left(u_0 \Gamma^{0}_{33}u^{3}+u_3\Gamma^3_{30}u^0\right)\delta_{i3}= \left(\gamma\tau u^3   + u_3\frac{1}{\tau}\gamma  \right)\delta_{i3} = 0 \end{align}$$

从而能动张量守恒的空间部分可化简为


 * $$\begin{align}

&{\left(\tilde \omega u_{\mu}u_{i}-{{\tilde P}g_{\mu i}}\right)_{;}}^{\mu} = 0 \\ &\sigma u_{\mu} {\left(\frac{{\tilde \omega}u_i}{\sigma}\right)_{;}}^{\mu} - \partial_{i}{\tilde P} = 0\\ &\sigma u_{\mu} {\left(\frac{{\tilde \omega}u_i}{\sigma}\right)_{,}}^{\mu} - \partial_{i}{\tilde P} =0\\ &\sigma\gamma\frac{d}{d\tau}\left(\frac{{\tilde \omega}\gamma g_{ij}u^j}{\sigma}\right) - \partial_{i}{\tilde P}= 0\\ &\sigma\gamma\frac{d}{d\tau}\left(\frac{{\tilde \omega}\gamma u^i}{\sigma}\right) + \partial_{i}{\tilde P}= 0 \end{align}$$

把$$\tilde{P}$$替换为$$P$$即得无粘滞度系统的运动方程.在直角坐标系中,利用度规的具体形式 $$g_{ij}=-\delta_{ij}$$,即得到更简单的形式.

方程清单

 * 方程清单

至此我们已经把所有运动方程表达为5个变量


 * $$\begin{align}

u^1,u^2,u^3,\frac{s}{\sigma},\frac{\Pi}{\sigma} \end{align}$$

之间的关系,上面的方程的作用既是确定下面的系数矩阵和系数列矢量.


 * $$\begin{align}

\left( \begin{array}{c} r_1 \\ r_2 \\ r_3 \\ r_4 \\ r_5 \end{array}\right) =\left( \begin{array}{ccccc} m_{11} & m_{12} & m_{13} & m_{14} & m_{15}\\ m_{21} & \ddots &  &  & \\ m_{31} &  &  &  & \\ m_{41} &  &  &  & \\ m_{51} &  &  &  & \end{array}\right) \left( \begin{array}{c} d_{\tau}u^1 \\ d_{\tau}u^2 \\ d_{\tau}u^3 \\ d_{\tau}\left(s/\sigma \right ) \\ d_{\tau}\left(\Pi/\sigma \right ) \end{array}\right) \end{align}$$

下面中我们逐一复述这五个方程


 * $$\begin{align}

&u\cdot \partial u_{i} = -\frac{1}\left({\widetilde \omega}u_{i}\partial \cdot u+(1+c_s^2) u_{i} u \cdot \partial {\varepsilon}+u_{i} u \cdot \partial {\Pi}-\partial_{i}{\widetilde P} \right ) \\ &d_{\tau} \left(\frac{s}{\sigma}\right) =-\frac{\tau}{T\sigma^*} \Pi \partial \cdot u \\ &d_{\tau}\left(\frac{\Pi}{\sigma}\right) =-\frac{1}{\gamma\tau_R\sigma}\left(\Pi+{\zeta\partial\cdot u}\right ) \end{align}$$

其中所有的量都有定义,这里再次给出如下


 * $$\begin{align}

&\partial \cdot u = \left(u^{\mu}\right)_{;\mu} = \frac{1}{\tau} \partial_{\mu} \left(\tau u^{\mu}\right) = \frac{1}{\tau} \left(\partial_{\tau} \left(\tau \gamma\right)+\partial_i\left(\tau\gamma v^i \right)\right) = \frac{1}{\tau} d_{\tau} (\tau\gamma) + \gamma \partial_i v^i = \frac{\gamma}{\tau} + d_{\tau}\gamma+\gamma \partial_i v^{i} \\ &d_{\tau}\gamma = \frac{d}{d\tau}\left(1-g_{ij}u^iu^j\right)^{1/2} = -\frac{g_{ij}u^i}{\gamma}\frac{du^j}{d\tau}-\frac{u^iu^j}{2\gamma}\frac{dg_{ij}}{d\tau} = -\frac{u_j}{\gamma}\frac{du^j}{d\tau}+\frac{\left(u^3\right)^2\tau}{\gamma}\\ &u\cdot \partial u_{i} = \gamma \frac{du_i}{d\tau} - u^{\mu} u_{\lambda} \Gamma^{\lambda}_{i\mu} = \gamma \frac{du_i}{d\tau} = - \gamma\frac{du^1}{d\tau}\delta_{i1}- \gamma\frac{du^2}{d\tau}\delta_{i2}- \left(\gamma\tau^2\frac{du^3}{d\tau}+2\tau\gamma u^3 \right)\delta_{i3} \\ &u\cdot\partial \Pi=\gamma{d_{\tau}\Pi}=-\frac{1}{\tau_R}\left(\Pi+{\left(\zeta+{\tau_R\Pi} \right )\partial\cdot u}\right )\\ \end{align}$$

考虑体积粘滞系数和剪切粘滞系数在Bjorken坐标系中的流体的相对论流体力学方程
具体推导参见参见Key Notes on Viscous Hyperbolic SPH Equation of Motion of Hydrodynamic Model

方程清单

 * 方程清单

我们在这里把所有的运动方程和待求解变量总结如下.体系的待求解变量为


 * $$\begin{align}

u_i,\frac{s}{\sigma},\frac{\Pi}{\sigma},\frac{\pi_{\mu\nu}}{\sigma} \end{align}$$

其中剪切系数的5个独立变量选取为除去33指标的纯空间指标系数,其余系数都可以用这5个系数表达如下


 * $$\begin{align}

&\pi^{33}=\left(1-\frac{u_3^2}{\gamma^2}\right)^{-1}\frac{1}{\tau^2}\left[\sum_{i\ne j}\frac{u_iu_j}{\gamma^2}\pi^{ij}-\left(1-\frac{u_1^2}{\gamma^2}\right)\pi^{11}-\left(1-\frac{u_2^2}{\gamma^2}\right)\pi^{22} \right]\\ &\pi^{0i}=-\frac{u_j}{\gamma}\pi^{ji}\\ &\pi^{00}=\pi^{11}+\pi^{22}+\tau^2\pi^{33}=\frac{u_iu_j}{\gamma^2}\pi^{ij} \end{align}$$

上述待求解变量满足的运动方程为
 * $$\begin{align}

d_{\tau}\left(\frac{\Pi}{\sigma}\right) = \frac{d_{\tau}\Pi}{\sigma} + \frac{\Pi}{\sigma^2}\left(\frac{\sigma}{\gamma}\partial \cdot u\right) =\frac{1}{\gamma}\left(-\frac{\widetilde{\Pi}}{\tau_R}-\frac{\zeta\theta}{\sigma\tau_R} \right) =-\frac{1}{\gamma\tau_R\sigma}\left(\Pi+{\zeta\partial\cdot u}\right) \end{align}$$
 * $$\begin{align}

&\frac{\tau_R}{\sigma}(\gamma u_\mu\pi^j_\nu+\gamma u_\nu\pi^j_\mu-u_\mu\pi^0_\nu u^j-u_\nu\pi^0_\mu u^j)\frac{du_j}{d\tau} +\gamma\tau_R\frac{d\widetilde{\pi}_{\mu\nu}}{d\tau}+\frac{\tau_R}{\sigma\tau^3}u_3(\delta_{\mu 0}\pi_{\nu 3}+\delta_{\nu 0}\pi_{\mu 3})\\ &+\frac{\tau_R}{\sigma\tau}u_3(\delta_{\mu 3}\pi_{\nu 0}+\delta_{\nu 3}\pi_{\mu 0}) -\frac{\gamma\tau_R}{\sigma\tau}(\delta_{\mu 3}\pi_{\nu 3}+\delta_{\nu 3}\pi_{\mu 3})+\frac{\pi_{\mu\nu}}{\sigma}\\ &=\frac{\eta}{2\sigma}[\partial_\mu u_\nu+\partial_\nu u_\mu]-\frac{\eta\gamma}{2\sigma}\left[u_\mu\frac{du_\nu}{d\tau}+u_\nu\frac{du_\mu}{d\tau}\right]-\frac{\eta}{3\sigma}\Delta_{\mu\nu}(\partial\cdot u)\\ &-\frac{\eta\gamma\tau}{\sigma}\delta_{\nu 3}\delta_{\mu 3}-\eta\frac{u_3}{\sigma\tau}[\delta_{\mu 0}\delta_{\nu 3}+\delta_{\mu 3}\delta_{\nu 0}]-\frac{\eta(u_3)^2}{2\sigma\tau^3}(u_\mu\delta_{\nu 0}+u_\nu\delta_{\mu 0}) \end{align}$$
 * $$\begin{align}

&\partial_i(P+\Pi)= \gamma(\varepsilon+P+\Pi)\frac{du_i}{d\tau}+u_i(\varepsilon+P)(\partial\cdot u)+u_i{\gamma}\left(1+\frac{dP}{d\varepsilon}\right)\frac{1}{\gamma}\left({T\sigma\gamma}d_\tau\left(\frac{s}{\sigma}\right)-(\varepsilon+P)(\partial\cdot u)\right)\\ &+u_i\gamma\frac{d\widetilde{\Pi}}{d\tau}+\sigma\frac{d\widetilde{\pi}_{0i}}{d\tau}-v_j\partial^j\pi_{0i}+\partial^j\pi_{ji}+\frac{\pi_{0i}}{\tau}-\frac{{\pi}_{0i}}{\gamma}(\partial\cdot u) \end{align}$$
 * $$\begin{align}

\\d_\tau(\frac{s}{\sigma}) =\frac{1}{T\sigma\gamma}\Pi(-(\partial\cdot u))+\frac{1}{T\sigma\gamma} \left[\left(\pi_0^j-\pi_{00}\frac{u^j}{\gamma}\right)\frac{du_j}{d\tau}+(\pi^{ij}+\pi^{00}v^iv^j-\pi^{i0}v^j-\pi^{0j}v^i)\partial_iu_j-\frac{\gamma}{\tau^3}\pi_{33}+\frac{2u_3}{\tau^3}\pi_{03}-\frac{(u_3)^2}{\gamma\tau^3}\pi_{00}\right] \end{align}$$

方程中的记号涉及到下面关系
 * $$\begin{align}

&\theta\equiv Du\equiv\partial \cdot u = \left(u^{\mu}\right)_{;\mu} = \frac{1}{\tau} \partial_{\mu} \left(\tau u^{\mu}\right) =\frac{\gamma}{\tau}+\partial_{\mu}u^{\mu} =\frac{\gamma}{\tau} + d_{\tau}\gamma+\gamma \partial_i v^{i} \\ &d_\tau\sigma=\frac{d\sigma}{d\tau}=-\frac{\sigma}{\gamma}(\partial\cdot u) \frac{du_\mu}{d\tau}=\frac{d\gamma}{d\tau}\delta_{\mu 0}+\frac{du_i}{d\tau}\delta_{\mu i}\\ &\frac{du_\mu}{d\tau}=\frac{d\gamma}{d\tau}\delta_{\mu 0}+\frac{du_i}{d\tau}\delta_{\mu i}\\ &d_{\tau}\gamma = \frac{d}{d\tau}\left(1-g_{ij}u^iu^j\right)^{1/2} = -\frac{g_{ij}u^i}{\gamma}\frac{du^j}{d\tau}-\frac{u^iu^j}{2\gamma}\frac{dg_{ij}}{d\tau} = -\frac{u_j}{\gamma}\frac{du^j}{d\tau}+\frac{\left(u^3\right)^2\tau}{\gamma}\\ &\partial_k\gamma=-v^i\partial_ku_i\\ &u\cdot \partial u_{i} = \gamma \frac{du_i}{d\tau} - u^{\mu} u_{\lambda} \Gamma^{\lambda}_{i\mu} = \gamma \frac{du_i}{d\tau} = - \gamma\frac{du^1}{d\tau}\delta_{i1}- \gamma\frac{du^2}{d\tau}\delta_{i2}- \left(\gamma\tau^2\frac{du^3}{d\tau}+2\tau\gamma u^3 \right)\delta_{i3} \\ &\gamma\frac{du_i}{d\tau}=\gamma\frac{du_1}{d\tau}\delta_{i1}+\gamma\frac{du_2}{d\tau}\delta_{i2}+\gamma\frac{du_3}{d\tau}\delta_{i3}\\ \end{align}$$

如需要,我们有下面的相互转化关系
 * $$\begin{align}

{d_{\tau}\Pi}=-\frac{1}{\gamma\tau_R}\left(\Pi+{\left(\zeta+{\tau_R\Pi} \right)\partial\cdot u}\right) \end{align}$$
 * $$\begin{align}

\frac{d{\pi}_{\mu\nu}}{d\tau}=\sigma\frac{d\widetilde{\pi}_{\mu\nu}}{d\tau}-\frac{{\pi}_{\mu\nu}}{\gamma}(\partial\cdot u) \end{align}$$
 * $$\begin{align}

\frac{ds}{d\tau}=\sigma{d_\tau(\frac{s}{\sigma})}-\frac{s}{\gamma}(\partial\cdot u) \end{align}$$