Research Paper Notes on Viscous Hydrodynamics
Research Paper Notes on Viscous Hydrodynamics
本文档除了包括推导,疑惑外,做读书重点的记录
参考文献
- arXiv:1004.2023v2 [nucl-th] Constraining the viscous freeze-out distribution function with data obtained at the BNL RHIC by Matthew Luzum and Jean-Yves Ollitrault
- arXiv:0909.0754 [nucl-th] Radiative energy loss and v(2) spectra for viscous hydrodynamics by Kevin Dusling, Guy D. Moore , Derek Teaney
- arXiv:hep-ph/0602249 Dissipative Hydrodynamics and Heavy Ion Collisions, by Rudolf Baier, Paul Romatschke, and Urs Achim Wiedemann
arXiv:1004.2023v2 [nucl-th] Constraining the viscous freeze-out distribution function with data obtained at the BNL RHIC
本文考察在强子化过程中在平衡态附近强子气体的能动张量的表达式.
Eq.(1) 下面的剪切粘滞系数定义有误,应该为<math>\pi^{\mu\nu}\equiv (\Pi^{\mu\nu}-\frac{1}{3}{\Pi^{\alpha}}_{\alpha}\Delta^{\mu\nu})</math>
Eq.(9) 这个表达式应为张量且在粒子动量为零时能动张量应该趋近于零.然后在下面决定<math>\delta f</math>的形式.
Eq.(10) 标量<math>\delta f</math>的最低级近似,表达为能动张量偏离平衡态的线性函数.换言之<math>\Pi^{\mu\nu}</math>与另一个张量的内积,该张量是强子动量的函数,第一项无迹,第二项为迹.代入<math>\Pi^{\mu\nu}=\pi^{\mu\nu}+\Pi\Delta^{\mu\nu}</math>即可由Eq.(10)的第一步得到第二步等式.其中第一步等式是在Landau随动系中写出的,似乎没有必要.在Landau随动系中<math>\Pi^{\mu\nu}</math>的第一行和第一列都为零,故它与四动量内积的时候,仅仅是四动量空间分量有贡献,与<math>g^{\mu\nu}</math>,仅仅其空间部分<math>\delta</math>函数有贡献.
Eq.(11-12) Eq.(12)比较容易得到,将Eq.(10)代回Eq.(9),考虑与体粘滞系数相关的项为
- <math>\begin{align}
\Delta^{\mu\nu}\Pi = \int\frac{d^3p}{(2\pi)^3E}p^{\mu}p^{\nu}B(p)\Pi \end{align}</math> 两边用<math>g_{\mu\nu}</math>内积,即得Eq.(12)
Eq.(11)采用类似的方法,这时考虑与剪切粘滞系数相关的项为
- <math>\begin{align}
\pi^{\mu\nu}=\int\frac{d^3p}{(2\pi)^3E}p^{\mu}p^{\nu}A(p)\frac{1}{p^2}p^{\mu'}p^{\nu'}\pi_{\mu'\nu'}\equiv \mathcal{O}^{\mu\nu\mu'\nu'}\pi_{\mu'\nu'} \end{align}</math>
利用算符作用在<math>\pi_{\mu'\nu'}</math>上的性质,我们得到
- <math>\begin{align}
\mathcal{O}^{\mu\nu\mu'\nu'}=\int\frac{d^3p}{(2\pi)^3E}A(p)\frac{1}{p^2}p^{\mu}p^{\nu}p^{\mu'}p^{\nu'}=\alpha g^{\mu\mu'}g^{\nu\nu'}+(1-\alpha) g^{\mu\nu'}g^{\nu\mu'}+\beta g^{\mu\nu}g^{\mu'\nu'}+\Gamma^{\mu\nu}g^{\mu'\nu'} \end{align}</math>
其中<math>\Gamma^{\mu\nu}</math>代表任意张量,比如<math>\Gamma^{\mu\nu}=\gamma u^{\mu}u^{\nu}</math>.但是我们注意到<math>\mathcal{O}^{\mu\nu\mu'\nu'}</math>关于其四个指标是完全对称的.这导致
- <math>\begin{align}
&\alpha=\beta=\frac{1}{2}\\ &\Gamma^{\mu\nu}=0 \end{align}</math>
换言之,
- <math>\begin{align}
\mathcal{O}^{\mu\nu\mu'\nu'}=\int\frac{d^3p}{(2\pi)^3E}A(p)\frac{1}{p^2}p^{\mu}p^{\nu}p^{\mu'}p^{\nu'}=\frac{1}{2}( g^{\mu\mu'}g^{\nu\nu'}+g^{\mu\nu'}g^{\nu\mu'}+g^{\mu\nu}g^{\mu'\nu'}) \end{align}</math>
两边对<math>g_{\mu\nu}g_{\mu'\nu'}</math>做内积,我们得到
- <math>\begin{align}
g_{\mu\nu}g_{\mu'\nu'}\mathcal{O}^{\mu\nu\mu'\nu'}=\int\frac{d^3p}{(2\pi)^3E}A(p){p^2}=\frac{1}{2}( 4\times 4+4+4)=\frac{24}{2} \end{align}</math>
这与书上的结果不同,因为上述结果并不是一般的,实际上如果我们仅仅要求上述表达式在随动系中有相同的结果,我们可以有
- <math>\begin{align}
\tilde{\mathcal{O}}^{\mu\nu\mu'\nu'}=\int\frac{d^3p}{(2\pi)^3E}A(p)\frac{1}{p^2}p^{\mu}p^{\nu}p^{\mu'}p^{\nu'}=\frac{1}{2}( \Delta^{\mu\mu'}\Delta^{\nu\nu'}+\Delta^{\mu\nu'}\Delta^{\nu\mu'}+\Delta^{\mu\nu}\Delta^{\mu'\nu'}) \end{align}</math>
而速度投影算符<math>\Delta</math>的迹为<math>3</math>,我们得到文中的结果
- <math>\begin{align}
g_{\mu\nu}g_{\mu'\nu'}\tilde{\mathcal{O}}^{\mu\nu\mu'\nu'}=\int\frac{d^3p}{(2\pi)^3E}A(p){p^2}=\frac{1}{2}( 3\times 3+3+3)=\frac{15}{2} \end{align}</math>
两者的区别可以通过比如内积<math>g_{\mu\nu}u_{\mu'}u_{\nu'}</math>来得到,文中并未进行讨论,这似乎与模型有关,对此并不明确.
Eq.(13) 利用<math>E=p^{\mu}u_{\mu}</math>即得.
arXiv:0909.0754 [nucl-th] Radiative energy loss and v(2) spectra for viscous hydrodynamics
arXiv:hep-ph/0602249 Dissipative Hydrodynamics and Heavy Ion Collisions, by Rudolf Baier, Paul Romatschke, and Urs Achim Wiedemann
这篇文章的附录A给出了如果计算在化学冻结(freeze-out)表明上分布函数偏离平衡态的部分<math>\delta f</math>.