User talk:Gamebm

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Conductivity and Probability flow

Apologies for my ignorance. I understand in classical electrodynamics conductivity means flow of electrons, now since here conductivity is understood in term of solution of Schrodinger's equation, what is conductivity then, which differs conductor from insulator. Now if one looks at the Bloch wave, as one approximation mentioned in this article, I will intuitively assume conductivity, the flow of electrons, is connected with the probability flow of electrons, that is (from the conservation of probability implied by the Schrodinger's equation)

${\vec {J}}={\frac {\hbar }{2im}}(\Psi ^{*}\nabla \Psi -\Psi \nabla \Psi ^{*})$

substituting the expression of Bloch wave, I get ${\vec {J}}={\frac {\hbar {\vec {k}}}{m}}$. So no matter where the electron is, it is moving like as a flow. Now I kinda guess, the summation of all the reciprocal vector ${\vec {k}}$ inside the Brillouin zone cancels out when the valence band is filled up. Is it true? Electrons are actually move collectively in the valance band but do not present macroscopic flow due to cancellation?

Then another question is, how to understand this in the case superconductivity? Gamebm (talk) 17:13, 27 November 2012 (UTC)

I don't think your calculation is right. With $\Psi =ue^{ikr}$, I get

$\Psi ^{*}\nabla \Psi -\Psi \nabla \Psi ^{*}=2ik|u|^{2}+u^{*}\nabla u-u\nabla u^{*}$

Can you please double-check? --Steve (talk) 23:29, 27 November 2012 (UTC)

Also as a side note, this won't help you understand superconductivity at all, since it doesn't arise from the band structure...a13ean (talk) 23:47, 27 November 2012 (UTC)

Thanks for pointing that out. I was study the topic and for some reason incorrectly assumed that u must be real. Now it seems the first term ($|u|^{2}$) will not change the conclusion since if one sums up all the k at a given spatial point, it is just a constant. However if the second term does remain, it will give a non-zero contribution to the flow (locally). Then I may understand that a filled band does present some flow locally since the charge flow calculated from QM does not vanish(? - is it measurable?). Now globally, since the second term is surface like (or due to its periodic behavior), it still may not lead to macroscopic flow, (since the length scale in our question is of lattice size.) and for the very same argument, the first term reduces to a irrelevant normalization constant.

I have no idea what superconductivity is. I was just trying to think of it following this line of thought, if all those electron pairs are in the same state (no idea what the wave function shall look like, will study it), and one is again legitimate to calculate its probability flow, but if everything is in the same state, and the resulting flow does not vanish (since it conducts not by excited to another state, but simply staying in this state?), then does this imply that there is some preferred direction in space? Gamebm (talk) 18:04, 28 November 2012 (UTC)

Electron motion is properly calculated using the formula for the group velocity of the electron wavepacket ... not the formula for probability current. They must be related ... I would not be surprised if they are identical ... but I'm not sure of the details off the top of my head.

As pointed out by A13ean, this line of thought will not lead you towards understanding superconductivity. Band structures are part of the "single-particle picture", where you by-and-large ignore the fact that electron quasiparticles interact with each other. But interactions between electron quasiparticles are the entire basis for superconductivity. --Steve (talk) 19:14, 28 November 2012 (UTC)

If you're interested in that bit, the last chapter of Ashcroft and Mermin gives an overview of how superconductivity arises from electron-electron interactions. a13ean (talk) 19:25, 28 November 2012 (UTC)

Few notes before and after some reading of Ziman.

(1) "group velocity" and "probability flow". Probability flow is just a kind of wave which flows, itself may have phase or group velocity. Group velocity is just something comes out when one summing up ${\vec {k}}$ when considering a dispersion relation.

(2) In the book of Ziman for instance, $u({\vec {r}})$ (in Bloch wave) is ignored when discussing electron flow, the arguments may come from when one discusses group velocity of wave package in QM. So somehow it turns out to be a somewhat (reasonable) approximation not to think of the contribution of $u({\vec {r}})$ in the first place, at least, when one considers group velocity.

(3) The fact that the system is non-perturbative has not really anything to say about whether one can write down the state in turn of wave function. See for example, Eq.(11.55) of Ziman when flow in superconductor was discussed, the notion of wave function is used. The wave function is not of one particle, but the important thing is its symmetry in reciprocal space, see below.

(4) In the book of Ziman, it explains what I was asking for in the first place. In fact, my argument was more based on spacial isotropy of the wave function in reciprocal space, so that when summing over all possible momenta, isotropy implies that there does not exist a preferred direction, which implies no flow in coordinate space. In the conductor/semiconductor case, electron has to be exited to a high-energy state to break this symmetry, which in turn gives macroscopic flow. In the case of superconductivity, the argument is based on the fact that the ground state and energy gap themselves are solutions in respect to Fermi surface in stead of some fixed coordinates in reciprocal space. In fact, an external field may distort Fermi surface and make it anisotropic, therefore even though all the electron pairs are in the ground state, one may still have macroscopic electron flow, simply because the ground state is not isotropic as in the case of conductor. Gamebm (talk) 20:10, 2 December 2012 (UTC)

(1b) In the book of Ashcroft, it was actually show in Ch.12 and Appendix E that the group velocity and expectation of velocity operator (the later is proportional to probability flow) are the same thing. It can be understood as following: the physics of velocity operator comes (mainly) from probability flow, since the physics of probability flow, by definition, is related to the measurement of detected matter flow, it is not surprising at all when it turns out to be the the same as the velocity of the propagation of the maximum of wave function -- group velocity of wave function.

(4b) This cancellation of flow is explicitly discussed in the book of Ashcroft, it seems only sparkles to who is actually thinking of this. Gamebm (talk) 14:08, 18 September 2014 (UTC)

Conservation of energy and momentum

I have a simple question concerning the conservation law of the energy-momentum tensor, and I cannot figure out where did I make the mistake in the following calculations. Many thanks in advance!

Let us discuss a simple example. Considering a single free particle with 4-momentum $p^{\mu }$ in Cartesian coordinates. In the following, we will try to verify explicitly for a closed surface, one has

$I=\int d\sigma _{C\mu }T_{C}^{\mu 0}=0$

where the closed 3-surface $\sigma _{\mu }$ is defined by $\tau =\tau _{1},\tau _{2}$, $\eta =\eta _{1},\eta _{2}$, $x=x_{1},x_{2}$ and $y=y_{1},y_{2}$ with the following definitions

$\tau ={\sqrt {t^{2}-z^{2}}}$

$\eta ={\frac {1}{2}}\ln {\frac {t+z}{t-z}}$

Since the energy momentum tensor is evaluated in the Cartesian coordinates, the expression is not covariant, the vanishing contraction corresponds to the energy conservation in Cartesian coordinates. We use the following notations

$I=I_{\tau }+I_{\eta }+I_{x}+I_{y}$

As shown below, the 3-surface element at $\tau =const$ reads

$d\sigma _{C\mu }=dxdyd\eta \left({\begin{array}{c}\tau \cosh \eta \\0\\0\\-\tau \sinh \eta \end{array}}\right)$

and the integral on the 3-surface $\tau =\tau _{1},\tau _{2}$ yields

${I}_{\tau }=\int _{\tau _{1}}\tau dxdyd\eta (\cosh \eta T_{C}^{00}-\sinh \eta T_{C}^{30})-\int _{\tau _{2}}\tau dxdyd\eta (\cosh \eta T_{C}^{00}-\sinh \eta T_{C}^{30})$

Similarly, the 3-surface element at $\eta =const$ reads

$d\sigma _{C\mu }=dxdyd\eta \left({\begin{array}{c}\sinh \eta \\0\\0\\-\cosh \eta \end{array}}\right)$

and the integral on the 3-surface $\eta =\eta _{1},\eta _{2}$ yields

${I}_{\eta }=\int _{\eta _{1}}dxdyd\tau (\sinh \eta T_{C}^{00}-\cosh \eta T_{C}^{30})-\int _{\eta _{2}}dxdyd\tau (\sinh \eta T_{C}^{00}-\cosh \eta T_{C}^{30})$

Expressions for the other two 3-surfaces are more intuitive, one has

${I}_{x}=\int _{x_{1}}dtdydzT_{C}^{10}-\int _{x_{2}}dtdydzT_{C}^{10}$

and

${I}_{y}=\int _{y_{1}}dtdxdzT_{C}^{20}-\int _{y_{2}}dtdxdzT_{C}^{20}$

The energy momentum tensor of a point-like particle $q$ in Cartesian coordinates possesses the following form

$T_{C}^{\mu \nu }={\frac {p^{\mu }p^{\nu }}{p^{0}}}\delta ^{3}({\vec {x}}-{\vec {x}}_{q})={\frac {p^{\mu }p^{\nu }}{p^{0}}}\delta (x-x_{q})\delta (y-y_{q})\delta (z-z_{q})$

where

$\delta (z-z_{q})={\frac {\delta (\eta -\eta _{q})}{\tau \sinh \eta }}={\frac {\delta (\tau -\tau _{q})}{\cosh \eta }}$

depending on whether one deals with 3-surface parameterized in z, $\eta$ or $\tau$. Substituting into the expressions of ${I}$, one gets

${I}_{x}=\int _{x_{1}}dtdydzT_{C}^{10}-\int _{x_{2}}dtdydzT_{C}^{10}=\int dt(p^{1}-p^{1})=0$

${I}_{y}=\int _{y_{1}}dtdxdzT_{C}^{20}-\int _{y_{2}}dtdxdzT_{C}^{20}=\int dt(p^{2}-p^{2})=0$

${I}_{\tau }=\int _{\tau _{1}}d\eta (p^{0}{\coth \eta }-p^{3})\delta (\eta -\eta _{q})-\int _{\tau _{2}}d\eta (p^{0}{\coth \eta }-p^{3})\delta (\eta -\eta _{q})=\left\{{\begin{matrix}p^{0}(\coth \eta _{a}-\coth \eta _{b})&{\textrm {theworld-linegoesacrossbothsurfaces}}\\0&{\textrm {theworld-linegoesacrossneithersurface}}\\{\textrm {othervalues}}&{\textrm {othercases}}\end{matrix}}\right.$

${I}_{\eta }=\int _{\eta _{1}}d\tau (p^{0}\tanh \eta -p^{3})\delta (\tau -\tau _{q})-\int _{\eta _{2}}d\tau (p^{0}\tanh \eta -p^{3})\delta (\tau -\tau _{q})=\left\{{\begin{matrix}p^{0}(\tanh \eta _{1}-\tanh \eta _{2})&{\textrm {theworld-linegoesacrossbothsurfaces}}\\0&{\textrm {theworld-linegoesacrossneithersurface}}\\{\textrm {othervalues}}&{\textrm {othercases}}\end{matrix}}\right.$

where $\eta _{a},\eta _{b}$ are the values of $\eta$ coordinates when the particle does go across the surfaces $\tau _{1},\tau _{2}$. Among the four possible surfaces discussed above, one notes that the particle always goes across two of them. In practice, one may always choose a big enough area of $\tau _{1},\tau _{2}$ surface, so that the particle goes across $\tau _{1},\tau _{2}$ surfaces and, one ignores the integral on the $\eta _{1},\eta _{2}$ surfaces. In a more general case, if the system consists of many point-like particles, one can usually find big enough area of $\tau _{1},\tau _{2}$ so that all the particles go through the $\tau _{1},\tau _{2}$ area. One sees that the above expression does not vanish as expected from the conservation of energy.

• corrections: The expression of :$\delta$ function above is not correct, it should consider contain the information of the world-line of the particle, but the above expression does not. The correct expression is

$\delta (z-z_{q})={\frac {\delta (\eta -\eta _{q})}{\tau \cosh \eta -v_{z}\tau \sinh \eta }}$

when parameterized in term of :$\eta$. The final result is, of course, :$I=0$.

Surface element and initial energy

To evalute the surface element, one may transform (the orientation of) a surface element in hyperbolic coordinates $\tau =const.$, which is a unitary vector in hyperbolic coordinates, into Cartesian coordinates (here we use :$B$ to indicate hyperbolic coordinates and :$C$ to indicate Cartesian coordinates.)

$e_{B}^{\mu }=\left({\begin{array}{c}1\\0\\0\\0\end{array}}\right)$

$e_{C\mu }=\eta _{\lambda \mu }e_{C}^{\lambda }=\eta _{\mu \lambda }{\frac {\partial x_{C}^{\lambda }}{\partial x_{B}^{\nu }}}e_{B}^{\nu }=\left({\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}}\right)\left({\begin{array}{cccc}\cosh \eta &0&0&\tau \sinh \eta \\0&1&0&0\\0&0&1&0\\\sinh \eta &0&0&\tau \cosh \eta \end{array}}\right)\left({\begin{array}{c}1\\0\\0\\0\end{array}}\right)=\left({\begin{array}{c}\cosh \eta \\0\\0\\-\sinh \eta \end{array}}\right)$

It is straightforward to verify at this point $|e_{C\mu }|=|e_{B\mu }|=1$. The area of the surface element can be calculated as following

$d{\vec {r}}_{1}\equiv {\frac {\partial x_{C}^{\mu }}{\partial \eta }}d\eta =\left({\begin{array}{c}\tau \sinh \eta d\eta \\0\\0\\\tau \cosh \eta d\eta \end{array}}\right)$

$d{\vec {r}}_{2}\equiv {\frac {\partial x_{C}^{\mu }}{\partial x}}dx=\left({\begin{array}{c}0\\dx\\0\\0\end{array}}\right)$

and

$d{\vec {r}}_{3}\equiv {\frac {\partial x_{C}^{\mu }}{\partial y}}dy=\left({\begin{array}{c}0\\0\\dy\\0\end{array}}\right)$

for completeness, we also write down here

$d{\vec {r}}_{0}\equiv {\frac {\partial x_{C}^{\mu }}{\partial \tau }}d\tau =\left({\begin{array}{c}\cosh \eta d\tau \\0\\0\\\sinh \eta d\tau \end{array}}\right)$

$|d\sigma _{C\mu }|_{(\tau =const.)}\equiv |d\sigma _{C\mu }|=|d\sigma _{C}^{\mu }|=|d{\vec {r}}_{1}||d{\vec {r}}_{2}||d{\vec {r}}_{3}|=\tau dxdyd\eta$

which had used the fact that the three vectors are mutually orthogonal. In fact, this also implies

$|d\sigma _{B\mu }|=\tau dxdyd\eta \rightarrow |d\sigma _{C\mu }|=\tau dxdyd\eta$

where $d{\vec {r}}$ are the 4-vectors in Cartesian coordinates. One sees that the size of the surface element defined by $dx,dy,d\eta$ does not change.

As a side note, in the above, the surface element was calculated by (1) assuming that it s a vector, therefore its components in Cartesian coordinates was obtained by coordinate transformation (2) the module of the surface element was obtained by using the fact the surface $\tau =const.$ is parameterized in terms of orthogonal coordinates, therefore all displacements corresponding to each variables ($\tau ,\eta ,x,y$) are orthogonal, the total area of the 3-surface element is simply a product of the module of three displacements.

On the other hand, the surface element can be calculated by using some more general formulae, which reads

$d\sigma _{\mu }=\epsilon _{\mu \alpha \beta \gamma }e_{1}^{\alpha }e_{2}^{\beta }e_{3}^{\gamma }d^{3}y$

where $\epsilon _{\mu \alpha \beta \gamma }={\sqrt {-g}}[\mu ,\alpha ,\beta ,\gamma ]$ is the Levi-Civita tensor, $y_{1},y_{2},y_{3}$ parameterize the 3-surface, and $e_{1}^{\alpha }=\partial x^{\alpha }/\partial y_{1},e_{2},e_{3}$ describe the connection between the coordinates and its parameters. For instance, see the discussion near (3.2.2) of . In our case of the Cartesian coordinates, ${\sqrt {-g}}=1$, therefore $d\sigma _{C\mu }=[\mu ,\alpha ,\beta ,\gamma ]e_{1}^{\alpha }e_{2}^{\beta }e_{3}^{\gamma }d^{3}y$. It is straightforward to verify that one obtains the same result, namely,

$d\sigma _{C\mu }=dxdyd\eta \left({\begin{array}{c}\tau \cosh \eta \\0\\0\\-\tau \sinh \eta \end{array}}\right)$

The energy momentum tensor

Up to this point, we have not discussed the explicit form of energy momentum tensor. According to ,

${\begin{aligned}T_{C}^{\mu \nu }=\sum _{i}{\frac {p_{i}^{\mu }p_{i}^{\nu }}{p_{i}^{0}}}\delta ^{3}({\vec {x}}-{\vec {x}}_{i})\end{aligned}}$

Integrating on the $t=t_{0}$ surface, one obtains

${\begin{aligned}E_{C}({\text{IC}})=\int _{\sigma _{\mu }(t=const.)}T_{C}^{\mu 0}=\int dxdydzT_{C}^{00}=\sum _{i}p_{i}^{0}\end{aligned}}$

which is as expected. --Gamebm (talk) 19:05, 28 April 2014 (UTC)

Your question is far from simple (at least for someone like me who does not do these calculations often). I have not checked everything you did above, so I cannot testify to its correctness. But I do see one possible source for your problem — your division by $p^{0}$ in the expression for the stress-energy tensor. The article says "In special relativity, ...". This implies that you are using the kind of Cartesian coordinates in which the Lorentz transformation is expressed. But you are not using such a system of coordinates. Accordingly, you should adjust the formula to make the factor of p by which you are dividing orthogonal to the surface in terms of which the delta-functions are expressed, for example, the $\tau =\tau _{1}$ surface. JRSpriggs (talk) 05:17, 2 May 2014 (UTC)

Hi JRSpringgs, I found my mistake and corrected it, it came from the $\delta$ function, as a result, the integral cancels out perfectly. I would not say the problem is not simple, it is the energy conservation problem of a free particle, the geodesic of a particle in flat space-time. I was careful about the energy momentum tensor, that part was all right. Again, thanks for the help! --Gamebm (talk) 12:09, 5 May 2014 (UTC)

WikiProject Investment

Hey there! Id like to invite you to the WikiProject Investment

I'd like to invite you to join the Investment WikiProject. There are a lot of Investment related articles on Wikipedia that could use a little attention, and I hope this project can help organize an effort to improve them. So please, take a look and if you like what you see, help get this project off the ground and a few Investment pages into the front ranks of Wikipedia articles. Thanks!

Cheers! WikiEditCrunch (talk) 19:01, 24 August 2017 (UTC)

Anisotropy as a signature of transverse collective flow, Phys. Rev. D46, 229 (1992), by Jean-Yves Ollitrault

本文是第一篇讨论偏心率和对应的集体流的文章,文章提出了独立点粒子源模型.

(2.1)

这里定义的$S$矩阵是一个3X3的对称矩阵.含有6个独立参数. 文中指出,其实只有3个参数包含物理意义,其他3个参数可以通过选择合适的坐标系来确定下来.

首先,反应平面(事件平面)的定义是碰撞轴和碰撞参数所决定的平面,而系统对反应平面具有镜像对称. 具体的,对于反应平面的镜像操作$P$,我们有$P^{-1}=P$以及$PSP=S$. 同时对于垂直与反应平面的法向量${\hat {n}}$,我们有$P{\hat {n}}=-{\hat {n}}$. 由此

$PS{\hat {n}}=PSP^{-1}P{\hat {n}}=S(-{\hat {n}})=-S{\hat {n}}$

所以$S{\hat {n}}$必然正比于反应平面的法向量$S{\hat {n}}\propto {\hat {n}}$,记为

$S{\hat {n}}=f_{2}{\hat {n}}$

换言之,法向量${\hat {n}}$是矩阵$S$的一个本征矢量. 这个方向必然垂直与碰撞轴. 我们把碰撞轴取为${\hat {z}}$方向,取${\hat {y}}\equiv {\hat {n}}$. 剩下的方向取为${\hat {x}}$,反应平面就是$x-z$平面,${\hat {x}}$正是碰撞参数的方向.

因为$S$是对称矩阵,它的三个本征矢量是互相垂直的. 根据上述约定,其中一个本征矢量沿着${\hat {y}}$方向,垂直于反应平面和垂直于碰撞轴,但另外两个本征矢量并不一定沿着坐标轴${\hat {x}},{\hat {z}}$的方向. 我们注意到,除了对反应平面的镜像对称性以外,虽然碰撞后的体系在反应平面上对${\hat {y}}$轴具反演对称性,这并不保证剩余两个本征矢量的方向. 但是利用本征矢量相互垂直的性质,剩余的两个本证矢量必然躺在反应平面上. 按文中的约定,我们把更大的本征值$f_{3}$与碰撞轴的夹角$\theta _{F}$记为"流角". 这是因为在极端情况下,粒子全部沿着上述本征矢量方向(正负方向)出射,对应的本征矢量就是粒子流的方向.

文中指出,在中低能核碰撞中,对边缘碰撞,$\theta _{F}\sim 0$,对中心碰撞,$\theta _{F}\sim \pi /2$. 在极端相对论碰撞中,我们知道在中心快度区域,粒子流主要处于反应平面上,且在$\theta _{F}\sim \pi /2$附近角度区间基本满足Bjorken分布. 但是,作者指出,因为核子互相穿越(nuclear transparency),大多数能量集中在纵向的旁观者,这时$\theta _{F}\sim 0$,并不是一个好的观测量. 在此基础上,作者提出了椭圆流(偏心率)的概念.

(2.2)

由于矩阵$S={\begin{pmatrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33}\end{pmatrix}}$的一个本征值与$3$轴垂直, 他必然具有形式 ${\begin{pmatrix}A\\B\\0\end{pmatrix}}$ 以满足

${\begin{pmatrix}0&0&1\end{pmatrix}}{\begin{pmatrix}A\\B\\0\end{pmatrix}}=0$

同时由于它是本征值,我们有

${\begin{pmatrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33}\end{pmatrix}}{\begin{pmatrix}A\\B\\0\end{pmatrix}}=C{\begin{pmatrix}A\\B\\0\end{pmatrix}}$

这样

$S_{31}A+S_{32}B=0$

${\frac {S_{11}A+S_{12}B}{A}}={\frac {S{21}A+S_{22}B}{B}}$

由上面第一式得到的关系消去第二式中的${\frac {A}{B}}$,并注意到$S$矩阵是对称矩阵,即得(2.2)的结果.

(2.3)

我们给出(2.3)的推导. 首先(2.3)必须是一个对称矩阵,三个本征值是$f_{1},f_{2},f_{3}$,其中$f_{2}$对应的本征矢量是,在之前(2.2)的计算中取$A=0,B=1$

${\hat {n}}={\hat {y}}={\begin{pmatrix}0\\1\\0\end{pmatrix}}$.

代入本证方程得到

${\begin{pmatrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33}\end{pmatrix}}{\begin{pmatrix}0\\1\\0\end{pmatrix}}=f_{2}{\begin{pmatrix}0\\1\\0\end{pmatrix}}$.

由此得到$S_{12}=S_{32}=0,S_{22}=f_{2}$.

由于$f_{3}$对应的本征矢量与${\hat {y}}$垂直,与${\hat {z}}={\begin{pmatrix}0\\0\\1\end{pmatrix}}$夹角为$\theta _{F}$,归一后它的形式是

${\hat {n}}_{3}={\begin{pmatrix}\sin \theta _{F}\\0\\\cos \theta _{F}\end{pmatrix}}$

代入本证方程得到

${\begin{pmatrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33}\end{pmatrix}}{\begin{pmatrix}\sin \theta _{F}\\0\\\cos \theta _{F}\end{pmatrix}}=f_{3}{\begin{pmatrix}\sin \theta _{F}\\0\\\cos \theta _{F}\end{pmatrix}}$

由此得到

$S_{11}\sin \theta _{F}+S_{13}\cos \theta _{F}=f_{3}\sin \theta _{F}$

$S_{31}\sin \theta _{F}+S_{33}\cos \theta _{F}=f_{3}\cos \theta _{F}$

$S_{21}\sin \theta _{F}+S_{23}\cos \theta _{F}=0$.

类似的,最后一个对应$f_{1}$本征矢量是

${\hat {n}}_{1}={\begin{pmatrix}\cos \theta _{F}\\0\\-\sin \theta _{F}\end{pmatrix}}$

代入本证方程得到

${\begin{pmatrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33}\end{pmatrix}}{\begin{pmatrix}\cos \theta _{F}\\0\\-\sin \theta _{F}\end{pmatrix}}=f_{1}{\begin{pmatrix}\cos \theta _{F}\\0\\-\sin \theta _{F}\end{pmatrix}}$

由此得到

$S_{11}\cos \theta _{F}-S_{13}\sin \theta _{F}=f_{1}\cos \theta _{F}$

$S_{31}\cos \theta _{F}-S_{33}\sin \theta _{F}=-f_{1}\sin \theta _{F}$

$S_{21}\cos \theta _{F}-S_{23}\sin \theta _{F}=0$.

这里第一个方程与$f_{3}$的第一个方程联立,得到$S_{11}=f_{1}\cos ^{2}\theta _{F}+f_{3}\sin ^{2}\theta _{F}$和$S_{13}=(f_{3}-f_{1})\sin \theta _{F}\cos \theta _{F}$.

这里第二个方程与$f_{3}$的第二个方程联立,得到$S_{31}=(f_{3}-f_{1})\sin \theta _{F}\cos \theta _{F}$和$S_{33}=f_{1}\sin ^{2}\theta _{F}+f_{3}\cos ^{2}\theta _{F}$.

这里第三个方程与$f_{3}$的第三个方程联立,得到$S_{21}=S_{23}=0$.

注意到上面部分系数与文中结果差了因子$2$. 上述结果确认了$S$矩阵是对称矩阵. 另外,$S$为零的矩阵元使得(2.2)自动得以满足.

(2.4)

这里取极限$\theta _{F}\to 0$和$f_{3}\to \infty$,只考虑$S$矩阵在横向$(x,y)$平面上的投影. 这样就得到(2.4). 进一步,作者考虑中间快度的粒子,以忽略(不定式)$f_{3}\theta _{F}^{2}$的贡献.

(2.5-6)

这里(2.5)就是对椭圆流(文中假设了初始条件和集体流的线性相应,并不区分它和偏心率的差别)的定义. 比较arXiv:1003.0194一文的笔记,注意到这里已经假设了${\hat {y}}$方向是和事件平面垂直的方向,故粒子分布沿着${\hat {x}}$方向是最大值.

实际上,不难证明,(2.6)就是偏心率更为一般的定义.注意到

$\mathrm {tr} S^{\perp }=M\left(\sigma _{x}^{2}+\sigma _{y}^{2}\right)$

$\mathrm {det} S^{\perp }=M^{2}\left(\sigma _{x}^{2}\sigma _{y}^{2}-\sigma _{xy}^{2}\right)$

其中

$\sigma _{x}^{2}={\frac {1}{M}}\sum _{\nu =1}^{M}w(\nu )p_{x}(\nu )^{2}$

$\sigma _{y}^{2}={\frac {1}{M}}\sum _{\nu =1}^{M}w(\nu )p_{y}(\nu )^{2}$

$\sigma _{xy}={\frac {1}{M}}\sum _{\nu =1}^{M}w(\nu )p_{x}(\nu )p_{y}(\nu )$

文中(2.6)正是arXiv:1003.0194一文的(3)式,参见相关笔记.

(2.7)

这是由单粒子横向球张量(2.1)来定义由$M$个粒子构成的体系的横向球张量.

(3.1)

这个式子的直接证明并不容易,但是按定义,从$(S_{11},S_{12},S_{22})$到$(\alpha ,\theta ,{\mathcal {E}})$的变量变换总是可以完成的. 我们从另一个角度,我们验证(3.1)给出的形式分别满足三个变量$(\alpha ,\theta ,{\mathcal {E}})$的定义.

首先由迹,容易证明$\mathrm {tr} S^{\perp }={\mathcal {E}}$.

其次,由(2.6),通过直接的代数运算,容易证明(3.1)满足定义(2.6).

最后,我们要证明$\theta$是本征矢量与${\hat {x}}$的夹角. 按类似证明(2.3)的思路,我们知道归一后本证矢量是

${\hat {n}}_{1}={\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}$

代入矩阵(3.1),不难发现上式的确是本征矢量,且本征值为${\frac {\mathcal {E}}{2}}(1+\alpha )$

$S^{\perp }{\hat {n}}_{1}={\frac {\mathcal {E}}{2}}{\begin{pmatrix}1+\alpha \cos 2\theta &\alpha \sin 2\theta \\\alpha \sin 2\theta &1-\alpha \cos 2\theta \end{pmatrix}}{\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}={\frac {\mathcal {E}}{2}}(1+\alpha ){\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}$

类似的,另一个本征矢量

${\hat {n}}_{2}={\begin{pmatrix}\sin \theta \\-\cos \theta \end{pmatrix}}$

的本征值是${\frac {\mathcal {E}}{2}}(1-\alpha )$.

上述结果与(3.1)完全自洽.

(3.2)

利用下面mathematica代码,我们得到

$J(\alpha ,\theta ,{\mathcal {E}})={\frac {\partial (S_{11},S_{12},S_{22})}{\partial (\alpha ,\theta ,{\mathcal {E}})}}={\frac {1}{2}}\alpha {\mathcal {E}}^{2}$

与文中结果差了一个${\frac {1}{2}}$因子.

u[x_, y_, z_] := z/2 (1 + x Cos[2 y]);

v[x_, y_, z_] := z/2 x Sin[ 2 y];

w[x_, y_, z_] := z/2 (1 - x Cos[2 y]);

jacobian = D[{u[x, y, z], v[x, y, z], w[x, y, z]}, { { x, y, z } } ]

Simplify[Det@jacobian]

(3.5-6)

这里表达式的左边是由$M$个粒子构成的体系的概率分布函数,注意到这里的自变量是所有组成粒子的横动量. 等式左边一共涉及$2M$个随机变量. 而表达式的右边是由(2.7)定义的球张量的分量. 等式右边仅涉及$3$个随机变量. 按下面(3.7)的讨论,剩余的变量都是角度部分的变量,概率分布函数(3.5)不是这些变量的显函,故可以直接对他们积分得到一个平庸的常数因子.

按数理统计,对概率分布函数做换元时,只需要在原有的分布函数上乘以雅克比即可. 如果(3.5)成立,那么我们就可以利用换元及其雅克比(3.2)把概率分布函数改写为偏心率的函数. 这样,偏心率的概率分布函数只需要对多余参数(nuisance parameter)积分即可,这正是下面(3.7-3.11)给出的重要结果.

因此,问题的中心回到(3.5)的讨论. 本质上,(3.5)也涉及到换元,而且,表面上看,换元前后的随机变量数不同. 这一点是可以理解的,因为这仅仅要求概率分布函数并不是换元后的$2M-3$个多余参数的函数即可. 我们需要讨论的,是换元前概率分布函数的物理意义,以及换元过程涉及的且在(3.5)右侧并未具体体现雅克比的形式.

首先,我们指出,对于更为简单的由$M$个iid球对称正态分布点源

${\frac {dP}{d^{2}p_{T}}}={\frac {1}{2\pi }}{\frac {dP}{p_{T}dp_{T}}}={\frac {1}{\pi }}{\frac {dP}{d\left(p_{T}^{2}\right)}}={\frac {dP}{dp_{T1}dp_{T2}}}={\frac {1}{2\pi \sigma ^{2}}}\exp \left[-{\frac {1}{2\sigma ^{2}}}\left(p_{T1}^{2}+p_{T2}^{2}\right)\right]={\frac {1}{2\pi \sigma ^{2}}}\exp \left[-{\frac {p_{T}^{2}}{2\sigma ^{2}}}\right]$

构成的点源模型(即arXiv:1604.07230v2一文(5)上方的表达式),其概率分布函数正是(3.5)的特例. 证明如下. 由$M$个正态分布iid点源得到的概率分布函数为

${\frac {dP}{d^{2}p_{T}(1)\cdots d^{2}p_{T}(M)}}\propto \exp \left[-{\frac {1}{2\sigma ^{2}}}\sum _{\nu =1}^{M}\left(p_{T1}^{2}(\nu )+p_{T2}^{2}(\nu )\right)\right]=\exp \left[-{\frac {1}{2\sigma ^{2}}}\left(S_{11}^{\perp }+S_{22}^{\perp }\right)\right]=\exp \left[-\left({\frac {S_{11}^{\perp }}{{\frac {2}{M}}{\overline {S}}_{11}}}+{\frac {S_{22}^{\perp }}{{\frac {2}{M}}{\overline {S}}_{22}}}\right)\right]=\exp \left[-{\frac {1}{2}}M\left({\frac {S_{11}^{\perp }}{{\overline {S}}_{11}}}+{\frac {S_{22}^{\perp }}{{\overline {S}}_{22}}}\right)\right]=\exp \left[-{\frac {M}{2}}\mathrm {tr} \left(S^{\perp }{\overline {S}}^{-1}\right)\right]$

其中注意到,两个矩阵$S^{\perp },{\overline {S}}$这时都是对角的. 后者来自事件平均,它的定义除了事件平均还要注意到${\overline {S}}_{ij}=M\langle p_{i}p_{j}\rangle$,其中$\langle p_{i}p_{j}\rangle$是单个点源的协方差矩阵分量. 对于各向同性的点源模型,$\langle p_{T1}^{2}\rangle =\langle p_{T2}^{2}\rangle =\sigma ^{2}$,$\langle p_{T1}p_{T2}\rangle =0$. 角度$\theta$的积分是各向同性的,那么$\langle \alpha \sin 2\theta \rangle =\langle \alpha \cos 2\theta \rangle =0$,注意到上述平均值是事件平均值,我们有

${\overline {S}}={\frac {\overline {\mathcal {E}}}{2}}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}$

换言之,(3.1)的事件平均值正比于单位对角矩阵在转动下不变,且${\overline {\alpha }}=0$. 但注意到$M$有限故对于某特定的事件,我们一般有$\alpha \neq 0$.

现在我们考虑更为一般的情况,当偏心率不为零,由(3.1)出发,我们可以通过事件平均$\langle \cdots \rangle$来定义球张量的平均值${\overline {S}}$

${\overline {S}}=M\langle S^{\perp }\rangle ={\frac {\overline {\mathcal {E}}}{2}}{\begin{pmatrix}1+\langle \alpha \cos 2\theta \rangle &\langle \alpha \sin 2\theta \rangle \\\langle \alpha \sin 2\theta \rangle &1-\langle \alpha \cos 2\theta \rangle \end{pmatrix}}$

其中

${\overline {\mathcal {E}}}=\langle \mathrm {tr} S^{\perp }\rangle =\left\langle \sum _{\nu =1}^{M}\left(p_{T1}^{2}(\nu )+p_{T2}^{2}(\nu )\right)\right\rangle =M\langle p_{T}^{2}\rangle$

我们指出,上述做法中的因子化

${\overline {S}}=\left\langle {\frac {\mathcal {E}}{2}}{\begin{pmatrix}1+\alpha \cos 2\theta &\alpha \sin 2\theta \\\alpha \sin 2\theta &1-\alpha \cos 2\theta \end{pmatrix}}\right\rangle =\left\langle {\frac {\mathcal {E}}{2}}\right\rangle \left\langle {\begin{pmatrix}1+\alpha \cos 2\theta &\alpha \sin 2\theta \\\alpha \sin 2\theta &1-\alpha \cos 2\theta \end{pmatrix}}\right\rangle ={\frac {\overline {\mathcal {E}}}{2}}{\begin{pmatrix}1+\langle \alpha \cos 2\theta \rangle &\langle \alpha \sin 2\theta \rangle \\\langle \alpha \sin 2\theta \rangle &1-\langle \alpha \cos 2\theta \rangle \end{pmatrix}}$

的数学基础是基于概率分布函数能够被因子化${\frac {dP({\mathcal {E}},\theta ,\alpha )}{d{\mathcal {E}}d\theta d\alpha }}={\frac {dP({\mathcal {E}})}{d{\mathcal {E}}}}{\frac {dP(\theta ,\alpha )}{d\theta d\alpha }}$,但是分布函数(3.8)由于其指数部分的具体形式,其实并不满足上述性质.

在如果角度$\theta$的积分是各向同性的,那么上述平均值自动回到平均偏心率为零的结果. 如果实际的事件平均偏心率不为零,显然事件平均对应着的角度积分不再是平庸的结果,定义

$\langle \alpha \sin 2\theta \rangle ={\overline {\alpha }}\sin 2\Psi _{2}(\neq 0)$

$\langle \alpha \cos 2\theta \rangle ={\overline {\alpha }}\cos 2\Psi _{2}$

并把坐标值反向转动角度

$\Psi _{2}={\frac {1}{2}}\mathrm {arctan} {\frac {\langle \alpha \sin 2\theta \rangle }{\langle \alpha \cos 2\theta \rangle }}$

我们得到文中的(3.6)

${\overline {S}}={\frac {\overline {\mathcal {E}}}{2}}{\begin{pmatrix}1+{\overline {\alpha }}&0\\0&1-{\overline {\alpha }}\end{pmatrix}}$

一般的,从数学上,我们可以证明(3.5)给出的形式是一个多元高斯概率分布函数. 具体的,基础上面的考虑,我们暂时不考虑${\overline {S}}$矩阵元对应的物理意义,仅仅看成一个常数矩阵. 我们仍然可以把矩阵对角化,表达为(3.6)的形式. 按(3.5),我们做矩阵乘法并求迹,我们有

$S^{\perp }{\overline {S}}^{-1}={\frac {2}{\left(1-{\overline {\alpha }}^{2}\right){\overline {\mathcal {E}}}}}{\begin{pmatrix}S_{11}^{\perp }(1-{\overline {\alpha }})&S_{12}^{\perp }(1+{\overline {\alpha }})\\S_{21}^{\perp }(1-{\overline {\alpha }})&S_{22}^{\perp }(1+{\overline {\alpha }})\end{pmatrix}}$

$\mathrm {tr} \left(S^{\perp }{\overline {S}}^{-1}\right)={\frac {\mathcal {2}}{\left(1-{\overline {\alpha }}^{2}\right){\overline {\mathcal {E}}}}}\left(S_{11}^{\perp }(1-{\overline {\alpha }})+S_{22}^{\perp }(1+{\overline {\alpha }})\right)$

因为$S_{11}^{\perp }=\sum _{\nu =1}^{M}p_{T1}^{2}(\nu )$和$S_{22}^{\perp }=\sum _{\nu =1}^{M}p_{T2}^{2}(\nu )$. 这正是一个多元高斯概率分布函数,而且其中每个点源的每个分量的概率分布都是独立的.

(3.7)

这里指数函数前的因子$\left(\mathrm {det} S^{\perp }\right)^{(M-3)/2}$是椭圆极坐标的雅克比.

我们从量纲的视角先给出一个粗略的分析. 如果各向同性,我们可以把上述积分类比到高维求极坐标中球对称情况下的积分换元. 首先,$S_{11}^{\perp },S_{22}^{\perp },S_{12}^{\perp }$的量纲都是$p_{T}^{2}$. 故$\mathrm {det} S^{\perp }$的量纲是$p_{T}^{4}$,$\left(\mathrm {det} S^{\perp }\right)^{(M-3)/2}$的量纲是$p_{T}^{2(M-3)}$. 因此$dS_{11}^{\perp }dS_{22}^{\perp }dS_{12}^{\perp }\left(\mathrm {det} S^{\perp }\right)^{(M-3)/2}$的量纲是$p_{T}^{2M}$,与$d^{2}p_{T}(1)\cdots d^{2}p_{T}(M)$的量纲一致. 这意味着剩下的(那些不在指数中为显函的)变量都是角度部分的变量.

退而求其次,我们从球对称点源模型的角度来考虑(3.7)指数前的因子. 我们有$\mathrm {det} S^{\perp }=S_{11}^{\perp }S_{22}^{\perp }$ 对于概率分布函数的自变量,这里考虑两个独立的球对称子空间的直积$1\oplus 2$. 对于每个子空间,我们考虑推广的求极坐标

$r,\phi _{1},\cdots ,\phi _{M-1}$满足$r=p_{T1}>0$,$\phi _{M-1}\in [0,2\pi )$,以及对于其他角度坐标$\phi _{i}\in [0,\pi ]$.

和对应的坐标变换

$p_{T1}(1)=r\cos \phi _{1}$

$p_{T1}(2)=r\sin \phi _{1}\cos \phi _{2}$

$\cdots$

$p_{T1}(M-2)=r\sin \phi _{1}\sin \phi _{2}\cdots \cos \phi _{M-2}$

$p_{T1}(M-1)=r\sin \phi _{1}\sin \phi _{2}\cdots \sin \phi _{M-2}\cos \phi _{M-1}$

$p_{T1}(M)=r\sin \phi _{1}\sin \phi _{2}\cdots \sin \phi _{M-2}\sin \phi _{M-1}$

这个坐标变换的雅克比为$p_{T}^{M-1}\sin ^{n-2}(\phi _{1})\sin ^{n-3}(\phi _{2})\cdots \sin(\phi _{n-1}))$. 同时我们注意到

$\sum _{\nu =1}^{M}p_{T1}^{2}(\mu )=p_{T1}^{2}(1)+p_{T1}^{2}(2)+\cdots +p_{T1}^{2}(M)=p_{T1}^{2}=S_{11}^{\perp }$

因此,去除无关的角度部分,两个直积空间的雅克比的径向部分为

$dS_{11}^{\perp }\left(S_{11}^{\perp }\right)^{(M-2)/2}dS_{22}^{\perp }\left(S_{22}^{\perp }\right)^{(M-2)/2}$

但是上面结果并不能写成

$dS_{11}^{\perp }dS_{22}^{\perp }\left(\mathrm {det} S^{\perp }\right)^{(M-2)/2}$

甚至是

$dS_{11}^{\perp }dS_{22}^{\perp }dS_{12}^{\perp }\left(\mathrm {det} S^{\perp }\right)^{(M-3)/2}$

这里第一个表达式假设了行列式为对角,第个表达式进一步假设了引入$S_{12}^{\perp }$后并不改变行列式,这两个假设都不是正确的.

在球对称的情况下,概率分布函数的指数部分并不是$S_{12}^{\perp }$的显函,故原则上似乎不必引入随机变量$S_{12}^{\perp }$. 但是因为本文的论证逻辑利用了雅克比(3.2),为此我们必须得到(3.7)这个形式. JY提示说,这个结果可以由(3.5)严格导出,具体不清楚,需要查询相关数学书籍.

原则上,(3.7)相当于从所有$M$个粒子的横动量分量构成的$2M$个随机变量到$S_{11}^{\perp },S_{22}^{\perp },S_{12}^{\perp },\cdots$的坐标变换,证明雅克比中与 $S_{11}^{\perp },S_{22}^{\perp },S_{12}^{\perp }$相关的部分可以被因子化,因为概率分布函数与其他变量无关,故可以对其他变量积分. 要证明最后留下的因子正是$\left(\mathrm {det} S^{\perp }\right)^{(M-3)/2}$,且与坐标变换的具体形式无关. 后者是因为这个因子的物理意义是剩余变量的相空间的大小,因为这些变量的数值并不决定概率分布函数,在相空间中它们与$S_{11}^{\perp },S_{22}^{\perp },S_{12}^{\perp }$是正交的. 所以由雅克比的定义,它们对应的相空间的大小不依赖于变换的具体形式.

从统计得知,其实

$T\equiv \left\{S_{11}^{\perp },S_{22}^{\perp },S_{12}^{\perp }\right\}=T(X)=T\left(p_{T1}(1),p_{T1}(2),\cdots ,p_{TM}(1),p_{TM}(2)\right)$

满足充分统计(sufficient statistical)的定义. 概率分布函数的参数$\theta =\left\{\theta _{1},\theta _{2},\cdots \right\}$就是矩阵${\overline {S}}$的分量. 记$X\equiv {X_{1},\cdots ,X_{2M}}=\left\{p_{T1}(1),p_{T1}(2),\cdots ,p_{TM}(1),p_{TM}(2)\right\}$,充分统计具有性质

$\mathbb {P} \left(X|T,\theta \right)=\mathbb {P} \left(X|T\right)$

利用条件概率的基本关系,我们有

$\mathbb {P} (T|\theta )={\frac {\mathbb {P} \left(X,T|\theta \right)}{\mathbb {P} \left(X|T,\theta \right)}}={\frac {\mathbb {P} \left(X,T|\theta \right)}{\mathbb {P} \left(X|T\right)}}={\frac {\mathbb {P} \left(X|\theta \right)}{\mathbb {P} \left(X|T\right)}}$

其中第二步利用了充分统计的性质,第三步利用了充分统计完全由随机变量决定的定义

$\mathbb {P} \left(X,T|\theta \right)=\mathbb {P} \left(T|X,\theta \right)\mathbb {P} \left(X|\theta \right)=\mathbb {P} \left(X|\theta \right)$

比较这个结果与之前的讨论,一方面分子就是$2M$个随机变量的分布函数,另一方面分母上的量与概率分布函数的参数$\theta$无关,仅仅与变量变换的雅克比有关. 这与之前的讨论完全一致. 换言之,分母其实对应了给定$T$情况下随机变量空间的大小,直观的,这个大小与除去统计充分以外的变量的具体形式无关.

向Alex求助.

(3.8)

利用(3.1)的具体形式,我们有$\mathrm {det} S^{\perp }={\frac {{\mathcal {E}}^{2}}{4}}(1-\alpha ^{2})$,与$\theta$无关. 代入(3.8)并考虑到(3.2)即得(3.8)指数函数前面的系数.

对于指数部分,考虑到(3.6)的逆阵也是对角的

${\overline {S}}^{-1}={\frac {2}{\left(1-{\overline {\alpha }}^{2}\right){\overline {\mathcal {E}}}}}{\begin{pmatrix}1-{\overline {\alpha }}&0\\0&1+{\overline {\alpha }}\end{pmatrix}}$

我们有

$S^{\perp }{\overline {S}}^{-1}={\frac {\mathcal {E}}{\left(1-{\overline {\alpha }}^{2}\right){\overline {\mathcal {E}}}}}{\begin{pmatrix}(1+\alpha \cos 2\theta )(1-{\overline {\alpha }})&\alpha \sin 2\theta (1+{\overline {\alpha }})\\\alpha \sin 2\theta (1-{\overline {\alpha }})&(1-\alpha \cos 2\theta )(1+{\overline {\alpha }})\end{pmatrix}}$

$\mathrm {tr} \left(S^{\perp }{\overline {S}}^{-1}\right)={\frac {\mathcal {E}}{\left(1-{\overline {\alpha }}^{2}\right){\overline {\mathcal {E}}}}}2\left(1-\alpha {\overline {\alpha }}\cos 2\theta \right)$

代入即得指数部分的结果.

(3.9)

这个积分可以用求导的方法得到.令 $f({\mathcal {E}})=e^{-\beta {\mathcal {E}}}$ 其中 $\beta ={\frac {M}{\overline {\mathcal {E}}}}{\frac {1-\alpha {\overline {\alpha }}\cos 2\theta }{1-{\overline {\alpha }}^{2}}}$

这样,积分可以改写为

$\int _{0}^{\infty }d{\mathcal {E}}{\mathcal {E}}^{M-1}f({\mathcal {E}})=\left(-{\frac {d}{d\beta }}\right)^{M-1}\int _{0}^{\infty }d{\mathcal {E}}f({\mathcal {E}})$

因为

$\int _{0}^{\infty }d{\mathcal {E}}f({\mathcal {E}})={\frac {1}{\beta }}$

我们有

$\int _{0}^{\infty }d{\mathcal {E}}{\mathcal {E}}^{M-1}f({\mathcal {E}})=\left(-{\frac {d}{d\beta }}\right)^{M-1}\left({\frac {1}{\beta }}\right)=(-1)^{M-1}(M-1)!\beta ^{-M}$

提取出$\beta$中与$\theta ,\alpha$相关的部分即得(3.9)的结果. 注意到我们可以忽略阶乘因为它会重新归一而并不真正反应概率分布对点源数目的依赖.

(3.10)

考虑定义(3.3)并在(3.9)中取${\overline {\alpha }}=0$即得. 这个结果在后续工作arXiv:1312.6555和arXiv:1604.07230中被使用.

(3.11)

利用(3.10),取极限$M\to \infty$并比较指数函数的展开即得.

为了将上述结果比较(3.4),我们考虑两维高斯分布. 一方面$\langle p_{T}^{2}\rangle =\langle p_{T1}^{2}+p_{T2}^{2}\rangle =\sigma ^{2}+\sigma ^{2}=2\sigma ^{2}$. 另一方面,$\langle p_{T}^{4}\rangle =\left\langle \left(p_{T}^{2}\right)^{2}\right\rangle =\langle p_{T1}^{4}+p_{T2}^{4}+2p_{T1}^{2}p_{T2}^{2}\rangle =3\sigma ^{4}+3\sigma ^{4}+2\sigma ^{2}\sigma ^{2}=8\sigma ^{4}$ 其中了利用高斯分布的动量矩的结果. 考虑到$\delta =\langle p_{T}^{4}\rangle /\langle p_{T}^{2}\rangle ^{2}$,即得$\delta =2$.

本文档除了包括推导,疑惑外,做读书重点的记录 ${\displaystyle }$

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本文档除了包括推导,疑惑外,做读书重点的记录 ${\displaystyle }$