---
url: /GR/lesson-13-action/index.md
---
作用量原理：我们的目的是测地线方程，
$$
\frac{\text{d}^2x^\mu}{\text{d}\tau^2}+\Gamma^\mu{}*{\nu\lambda}\frac{\text{d}x^\nu}{\text{d}\tau}\frac{\text{d}x^\lambda}{\text{d}\tau} = 0
$$
粒子的作用量表达为
$$
I\_M = -\sum\_nm\_n\int*{-\infty}^\infty\text{d}p\left\[-g\_{\mu\nu}(x\_n)\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\right]^{1/2}
$$
我们想通过其变分得到测地线方程，有
$$
\begin{aligned}
\delta I\_M &=-\sum\_nm\_n\int\_{-\infty}^\infty\text{d}p\cdot\frac{1}{2}\left\[-g\_{\mu\nu}(x\_n)\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\right]^{-1/2}\Big{-\frac{\partial g\_{\mu\nu}}{\partial x\_n^\lambda}\delta x\_n^\lambda\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\\\\
&\quad -g\_{\mu\nu}(x\_n)\left(\frac{\text{d}}{\text{d}p}\delta x\_n^\mu\frac{\text{d}}{\text{d}p}x\_n^\nu+\frac{\text{d}}{\text{d}p}x\_n^\mu\frac{\text{d}}{\text{d}p}\delta x\_n^\nu\right) \Big}\\\\
&= -\sum\_nm\_n\int\_{-\infty}^\infty\text{d}\tau\cdot\frac{1}{2}\Big{-\frac{\partial g\_{\mu\nu}}{\partial x\_n^\lambda}\delta x\_n^\lambda\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}-g\_{\mu\nu}(x\_n)\left(\frac{\text{d}}{\text{d}p}\delta x\_n^\mu\frac{\text{d}}{\text{d}p}x\_n^\nu+\frac{\text{d}}{\text{d}p}x\_n^\mu\frac{\text{d}}{\text{d}p}\delta x\_n^\nu\right) \Big}\\\\
&= \cdots\cdots
\end{aligned}
$$
后面过程略，但是最后会得到 Christoffel 符号的定义式.

对于一个带电的体系，上面的作用量加一项场的部分和一项电荷之间相互作用的部分. 其中场的部分是
$$
-\frac{1}{4}\int\text{d}^4x\cdot g^{1/2}(x)F\_{\mu\nu}(x)F^{\mu\nu}(x)
$$

> 这里的 $g^{1/2}$ 来源于张量密度，$\text{d}^4x$ 不是普通标量.

为求电荷相互作用的项，首先写出狭义相对论下的流：
$$
J^\mu(x) = \sum\_ne\_n\frac{\text{d}x\_n^\mu}{\text{d}t}\delta^3(x-x\_n(t)) = \sum\_ne\_n\frac{\text{d}x\_n^\mu}{\text{d}\tau}\frac{\text{d}\tau}{\text{d}t}\delta^3(x-x\_n(t))
$$
简单地把 $3$ 改成 $4$ 就是四维形式，但是并不广义相对论协变，因为 $\delta^4$ 并不是标量，还需要乘一个张量密度的系数，
$$
J^\mu(x) = \sum\_ne\_n\int\text{d}\tau\cdot g^{1/2}(x\_n(\tau))\frac{\text{d}x\_n^\mu}{\text{d}\tau}\delta^4(x-x\_n(\tau))
$$
由流，得到粒子之间的电磁相互作用，
$$
\begin{aligned}
&\int\text{d}^4x\cdot g^{1/2}(x)A\_\mu(x)J^\mu(x)\\\\
\=&\int\text{d}\tau\int\text{d}^4x\sum\_n\delta^4(x-x\_n(\tau))A\_\mu(x)e\_n\frac{\text{d}x\_n^\mu}{\text{d}\tau}\\\\
\=&\sum\_ne\_n\int\_{-\infty}^\infty\text{d}p\frac{\text{d}x\_n^\mu(p)}{\text{d}p}A\_\mu(x\_n(p))
\end{aligned}
$$
相对于之前无电荷的作用量，这里多了两项：
$$
\begin{aligned}
I\_M &=-\sum\_nm\_n\int\_{-\infty}^\infty\text{d}p\left\[-g\_{\mu\nu}(x\_n)\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\right]^{1/2}-\frac{1}{4}\int\text{d}^4x\cdot g^{1/2}(x)F\_{\mu\nu}(x)F^{\mu\nu}(x)\\\\
&\quad -\sum\_ne\_n\int\_{-\infty}^\infty\text{d}p\frac{\text{d}x\_n^\mu(p)}{\text{d}p}A\_\mu(x\_n(p))
\end{aligned}
$$
变分后，第一项和第三项都有 $x\_n^\mu$，对于第一项变分之后得到测地线方程，下面考虑对第三项变分，
$$
\begin{aligned}
\delta I\_3 &= \sum\_n\int\_{-\infty}^\infty\text{d}p\cdot\left\[\frac{\text{d}\delta x\_n^\mu}{\text{d}p} A\_\mu(x\_n(p))+\frac{\text{d}x\_n^\mu}{\text{d}p}\left.\frac{\partial A\_\mu}{\partial x^\nu}\right|*{x=x\_n(p)}\delta x\_n^\mu(p)\right]\\\\
&= \sum\_ne\_n\int*{-\infty}^\infty\text{d}p\left(-F\_{\mu\nu}\frac{\text{d}x\_n^\mu}{\text{d}p}\right)\delta x^\nu+\text{boundary terms}
\end{aligned}
$$

> 这里的第一项可以通过 integral by part，
> $$
> \frac{\text{d}\delta x\_n^\mu}{\text{d}p}A\_\mu(x\_n(p))\longrightarrow -\delta x\_n^\mu\left.\frac{\partial A\_\mu}{\partial x^\nu}\right|\_{x=x\_n(p)}\frac{\text{d}x\_n^\nu(p)}{\text{d}p}
> $$

电磁场部分，
$$
\begin{aligned}
\delta I\_A &= \int\text{d}^4x\sqrt{g}\left(-\frac{1}{4}F\_{\mu\nu}F^{\mu\nu}\right)
\end{aligned}
$$
其中后面 $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$，是反对称的，所以可以写成 $2\partial^\mu A^\nu$. 之后对这个 $A^\nu$ 做变分，
$$
\delta I\_A = \int\text{d}^4x\sqrt{g}(-1)(\partial\_\mu A\_\nu-\partial\_\nu A\_\mu)\partial^\mu\delta A^\nu = \int\text{d}^4x\sqrt{g} F\_{\mu\nu}{}^{;\nu}
$$

***

下面用作用量原理推导场方程.

场的相互作用是 local 的，因为没有长程关联，也就没有双重的积分. 取一个标量的作用量，
$$
I\_G = -\frac{1}{16\pi G}\int\text{d}^4x\cdot g^{1/2} R
$$
变分，
$$
\begin{aligned}
\delta\left(g^{1/2}R\right) &= \delta\left(g^{1/2}g^{\mu\nu}R\_{\mu\nu}\right)\\\\
&= \delta g^{\mu\nu}g^{1/2}R\_{\mu\nu}+\delta g^{1/2}R+g^{1/2}g^{\mu\nu}\delta R\_{\mu\nu}
\end{aligned}
$$
最难处理的是最后这一项，因为根本不知道 $R$ 是什么. 但是我们知道 $R\_{\mu\nu}$ 没缩并之前是由 $\Gamma$ 构成的，而且 $\delta\Gamma$ 是 tensor (因为使得 $\Gamma$ 不是 tensor 的那一项变分的时候没有了). 因此
$$
\delta R\_{\mu\nu} = \text{something's derivative}
$$
这一项的贡献是
$$
\int\text{d}^4x\sqrt{g}\cdot D\_\mu \tilde J^\mu
$$
由 Stokes 可以化为面积分，将面积趋于无穷大，此式得零. 因此这一项对运动方程没有贡献.

接下来处理第一项，
$$
g\_{\nu\rho}g^{\rho\sigma} = \delta\_\mu{}^\sigma \Longrightarrow \delta g\_{\mu\rho}g^{\rho\sigma}+g\_{\mu\rho}\delta g^{\rho\sigma} = 0
$$
所以 $\delta g^{\rho\sigma}=-g^{\rho\mu}\delta g\_{\mu\nu}g^{\nu\sigma}$. 第二项是 $\delta g^{1/2}=\displaystyle{\frac{1}{2}}g^{1/2}g^{\mu\nu}\delta g\_{\mu\nu}$. 最终得到
$$
\delta I\_G = -\frac{1}{16\pi G}\int\text{d}^4x\left(-R^{\rho\sigma}+\frac{1}{2}g^{\rho\sigma}R\right)g^{1/2}\delta g\_{\rho\sigma}
$$
这个变分要和物质场的变分合在一起等于零. 如果场方程要正确，那么物质场对 $g\_{\rho\sigma}$ 变分必须得是能动张量. 这听起来很奇怪，不过这件事是正确的. 我们以一群粒子为例，
$$
I\_M = -\sum\_n m\_n\int\_{-\infty}^\infty\text{d}p\left(-g\_{\mu\nu}\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\right)^{1/2}
$$
对 $g\_{\mu\nu}$ 变分，
$$
\begin{aligned}
\delta I\_M &= -\sum\_nm\_n\int\_{-\infty}^\infty\text{d}p\cdot\frac{1}{2}\left(-g\_{\mu\nu}\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\right)^{-1/2}\left(-\frac{\text{d}x\_n^\mu}{\text{d}p}\frac{\text{d}x\_n^\nu}{\text{d}p}\delta g\_{\mu\nu}(x\_n)\right)\\\\
&=\sum\_nm\_n\int\_{-\infty}^\infty\text{d}p\cdot\frac{1}{2}\frac{\text{d}x\_n^\mu}{\text{d}\tau}\frac{\text{d}x\_n^\nu}{\text{d}\tau}\delta g\_{\mu\nu}(x\_n)\\\\
&= \int\text{d}^4x\sqrt{g(x)}\sum\_nm\_n\int\_{-\infty}^\infty\text{d}\tau\cdot\frac{1}{2}\frac{\text{d}x\_n^\mu}{\text{d}\tau}\frac{\text{d}x\_n^\nu}{\text{d}\tau}\cdot\sqrt{g(x)}\delta^4(x-x\_n(\tau))\delta g\_{\mu\nu}(x)
\end{aligned}
$$
而这个体系的能动张量是
$$
T^{\mu\nu}(x) = \int\text{d}\tau\sum\_nm\_n\frac{\text{d}x\_n^\mu}{\text{d}\tau}\frac{\text{d}x\_n^\nu}{\text{d}\tau} g^{1/2}(x)\delta^4(x-x\_n(\tau))
$$
验证了结果.

::: tip

现在来思考一下为什么这件事情是对的.
$$
S = \int\text{d}t\cdot L(q,\dot{q})
$$
的变分是
$$
\begin{aligned}
\delta S &= \int\_{t\_1}^{t\_2}\left(\frac{\delta L}{\delta q}\delta q+\frac{\delta L}{\delta \dot{q}}\delta\dot{q}\right)\text{d}t +\underset{\delta S\_2}{\underline{L(q(t\_2),\dot{q}(t\_2))\Delta t-L(q(t\_1),\dot{q}(t\_1))}}\\\\
&= \int\_{t\_1}^{t\_2}\left\[\underset{0}{\underline{\left(\frac{\delta L}{\delta q}\delta q - \frac{\text{d}}{\text{d}t}\frac{\delta L}{\delta \dot{q}}\right)}}-\frac{\text{d}}{\text{d}t}\left(\frac{\delta L}{\delta \dot{q}}\delta q\right)\right]+\delta S\_2\\\\
&= \left.\frac{\delta L}{\delta\dot{q}}\delta q\right|^{t\_2}*{t\_1}+\delta S\_2\\\\
&= \left\[-\frac{\delta L}{\delta \dot{q}}\dot{q}+L(q(t),\dot{q}(t))\right]^{t\_2}*{t\_1} = 0
\end{aligned}
$$
:::