ECT理论-牛顿引力理论

返回 在牛顿引力场中，粒子运动的拉格朗日量为：

• ${\displaystyle L={\frac {1}{2}}m{\vec {v}}\cdot {\vec {v}}-m\varphi ({\vec {x}})}$

其中 ${\displaystyle {\vec {v}}}$—粒子速度， ${\displaystyle \varphi ({\vec {x}})}$—牛顿引力势 粒子运动方程由最小作用量原理${\displaystyle \delta S=\int \limits _{t1}^{t2}{\delta L}dt=0}$决定：

${\displaystyle 0=\delta S=\int \limits _{t1}^{t2}{\delta L}dt}$

${\displaystyle =\int \limits _{t1}^{t2}{\delta \left({\frac {1}{2}}m{\vec {v}}\cdot {\vec {v}}-m\varphi ({\vec {x}})\right)}dt}$

${\displaystyle =\int \limits _{t1}^{t2}{\left(m{\vec {v}}\cdot \delta {\vec {v}}-m\delta \varphi ({\vec {x}})\right)}dt}$

${\displaystyle =\int \limits _{t1}^{t2}{\left(m{\vec {v}}\cdot {\frac {d\delta {\vec {x}}}{dt}}-m\nabla \varphi ({\vec {x}})\cdot \delta {\vec {x}}\right)}dt}$

${\displaystyle =m{\vec {v}}\cdot \delta {\vec {x}}|_{t1}^{t2}-\int \limits _{t1}^{t2}{\left(m{\frac {d{\vec {v}}}{dt}}\cdot \delta {\vec {x}}+m\nabla \varphi ({\vec {x}})\cdot \delta {\vec {x}}\right)}dt}$

${\displaystyle =-\int \limits _{t1}^{t2}{\left(m{\frac {d{\vec {v}}}{dt}}+m\nabla \varphi ({\vec {x}})\right)}\cdot \delta {\vec {x}}dt}$

因此有：${\displaystyle m{\frac {d{\vec {v}}}{dt}}+m\nabla \varphi ({\vec {x}})=0}$即：${\displaystyle {\vec {a}}=-\nabla \varphi ({\vec {x}})}$，这是牛顿引力场中的粒子运动方程。 考虑在牛顿引力场中无压理想流体的运动，则拉格朗日量变为：

• ${\displaystyle L=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})\right)dV}}$

其中： ${\displaystyle \rho }$—流体质量密度， ${\displaystyle dV}$—体积元。 牛顿引力场本身的拉格朗日量为：

• ${\displaystyle {{L}_{g}}=\int {\left(-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}}$

同时考虑引力场和无压理想流体，其总拉格朗日量为：

• ${\displaystyle L=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}}$

为了得到引力场的运动方程，只对${\displaystyle \varphi ({\vec {x}})}$取变分我们有：

${\displaystyle 0=\delta S=\int \limits _{t1}^{t2}{\delta Ldt}}$

${\displaystyle =\int \limits _{t1}^{t2}{\delta \int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}dt}}$

${\displaystyle =\int \limits _{t1}^{t2}{\int {\left(-\rho \delta \varphi ({\vec {x}})-{\frac {1}{4\pi G}}\nabla \varphi \cdot \delta (\nabla \varphi )\right)dV}dt}}$

${\displaystyle =\int \limits _{t1}^{t2}{\int {\left(-\rho \delta \varphi ({\vec {x}})-{\frac {1}{4\pi G}}\nabla \varphi \cdot \nabla (\delta \varphi )\right)dV}dt}}$

${\displaystyle =\int \limits _{t1}^{t2}{\int {\left(-\rho \delta \varphi ({\vec {x}})-{\frac {1}{4\pi G}}(\nabla \cdot (\delta \varphi \nabla \varphi )-\delta \varphi {{\nabla }^{2}}\varphi )\right)dV}dt}}$

${\displaystyle =\int \limits _{t1}^{t2}{\int \limits _{\Sigma }{\left(-{\frac {1}{4\pi G}}\delta \varphi \nabla \varphi \cdot d{\vec {S}}\right)}+\int {\left(-\rho \delta \varphi ({\vec {x}})+{\frac {1}{4\pi G}}\delta \varphi {{\nabla }^{2}}\varphi \right)dV}dt}}$，其中${\displaystyle \Sigma }$-包围体积V的边界

${\displaystyle =\int {\left(-\rho +{\frac {1}{4\pi G}}{{\nabla }^{2}}\varphi \right)\delta \varphi ({\vec {x}})dV}dt}$

因此有引力场运动方程${\displaystyle {{\nabla }^{2}}\varphi =4\pi G\rho }$ 。 这样，我们有包含引力场和无压理想流体的总拉格朗日密度为：

• ${\displaystyle \not {L}={\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi }$

按照分析力学原理，我们有守恒量---哈密顿量（其中：${\displaystyle {\dot {\varphi }}={\frac {\partial \varphi }{\partial t}}}$）为：

${\displaystyle {\begin{aligned}&H=\int {\left(\sum \limits _{i=1}^{3}{{{v}_{i}}{\frac {\partial \not {L}}{\partial {{v}_{i}}}}}+{\dot {\varphi }}{\frac {\partial \not {L}}{\partial {\dot {\varphi }}}}-\not {L}\right)}dV\\&=\int {\sum \limits _{i=1}^{3}{{{v}_{i}}{\frac {\partial }{\partial {{v}_{i}}}}\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)}dV}\\&\mathop {} ^{}+\int {{\dot {\varphi }}{\frac {\partial }{\partial {\dot {\varphi }}}}\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}\\&\mathop {} ^{}-\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}\\&=\int {\left(\rho {\vec {v}}\cdot {\vec {v}}\right)dV}-\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}\\&=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}+\rho \varphi ({\vec {x}})+{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}\\\end{aligned}}}$

其中${\displaystyle \rho \varphi ({\vec {x}})}$代表理想流体与引力场的相互作用能，可以将它归为理想流体的能量，也可以把它归为引力场的能量，我们现在把它归为引力场的能量，这时需要从引力场运动方程解出：${\displaystyle \rho ={\frac {1}{4\pi G}}{{\nabla }^{2}}\varphi }$，代入上式得：

${\displaystyle {\begin{aligned}&H=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}+{\frac {1}{4\pi G}}\varphi {{\nabla }^{2}}\varphi +{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}\\&=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}+{\frac {1}{4\pi G}}\nabla (\varphi \nabla \varphi )-{\frac {1}{4\pi G}}\nabla \varphi \cdot \nabla \varphi +{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}\\\end{aligned}}}$

${\displaystyle =\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}+{\frac {1}{4\pi G}}\int \limits _{\Sigma }{\varphi \nabla \varphi \cdot d{\vec {S}}}}$

其中： ${\displaystyle \Sigma }$为包围体积V边界。体积V是全空间。 一般我们考虑有限区域的理想流体和引力场的情况，这时边界是无限远处，无限远处的边界条件是 ${\displaystyle \varphi \nabla \varphi \to O({\frac {1}{{r}^{3}}})}$，${\displaystyle d{\vec {S}}\to O({{r}^{2}})}$ ，其积${\displaystyle \varphi \nabla \varphi \cdot d{\vec {S}}\to O({\frac {1}{r}})}$ ，因此${\displaystyle \int \limits _{\Sigma }{\varphi \nabla \varphi \cdot d{\vec {S}}}=0}$ .考虑到有限区域的理想流体和引力场以及边界条件，我们有：

• ${\displaystyle H=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}}$

在分析力学中我们称哈密顿量为能量，因此又可写为：

• ${\displaystyle E=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-{\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi \right)dV}}$

哈密顿量是守恒量即${\displaystyle {\frac {dH}{dt}}=0}$ 也即${\displaystyle {\frac {dE}{dt}}=0}$ 。 从上面的结果我们看到： ${\displaystyle {\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}}$代表理想流体的动能密度${\displaystyle {{T}_{m}}}$ ， ${\displaystyle {\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi }$代表引力能密度${\displaystyle {{T}_{g}}}$ ，这时我们看到总能量密度是 ${\displaystyle \varepsilon ={{T}_{m}}-{{T}_{g}}}$，引力能贡献的是负能。当然，如果将相互作用能归为理想流体的能量，则引力能贡献的是正能，数值仍然是${\displaystyle {{T}_{g}}}$ 。 返回