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： L = 1 2 m v → ⋅ v → − m φ ( x → ) {\displaystyle L={\frac {1}{2}}m{\vec {v}}\cdot {\vec {v}}-m\varphi ({\vec {x}})} 其中 v → {\displaystyle {\vec {v}}} —粒子速度， φ ( x → ) {\displaystyle \varphi ({\vec {x}})} —牛顿引力势 粒子运动方程由最小作用量原理 δ S = ∫ t 1 t 2 δ L d t = 0 {\displaystyle \delta S=\int \limits _{t1}^{t2}{\delta L}dt=0} 决定： 0 = δ S = ∫ t 1 t 2 δ L d t {\displaystyle 0=\delta S=\int \limits _{t1}^{t2}{\delta L}dt} = ∫ t 1 t 2 δ ( 1 2 m v → ⋅ v → − m φ ( x → ) ) d t {\displaystyle =\int \limits _{t1}^{t2}{\delta \left({\frac {1}{2}}m{\vec {v}}\cdot {\vec {v}}-m\varphi ({\vec {x}})\right)}dt} = ∫ t 1 t 2 ( m v → ⋅ δ v → − m δ φ ( x → ) ) d t {\displaystyle =\int \limits _{t1}^{t2}{\left(m{\vec {v}}\cdot \delta {\vec {v}}-m\delta \varphi ({\vec {x}})\right)}dt} = ∫ t 1 t 2 ( m v → ⋅ d δ x → d t − m ∇ φ ( x → ) ⋅ δ x → ) d t {\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} = m v → ⋅ δ x → | t 1 t 2 − ∫ t 1 t 2 ( m d v → d t ⋅ δ x → + m ∇ φ ( x → ) ⋅ δ x → ) d t {\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} = − ∫ t 1 t 2 ( m d v → d t + m ∇ φ ( x → ) ) ⋅ δ x → d t {\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： m d v → d t + m ∇ φ ( x → ) = 0 {\displaystyle m{\frac {d{\vec {v}}}{dt}}+m\nabla \varphi ({\vec {x}})=0} 即： a → = − ∇ φ ( x → ) {\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： L = ∫ ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) ) d V {\displaystyle L=\int {\left({\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}-\rho \varphi ({\vec {x}})\right)dV}} 其中： ρ {\displaystyle \rho } —流体质量密度， d V {\displaystyle dV} —体积元。 牛顿引力场本身的拉格朗日量为： L g = ∫ ( − 1 8 π G ∇ φ ⋅ ∇ φ ) d V {\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： L = ∫ ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ ) d V {\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 φ ( x → ) {\displaystyle \varphi ({\vec {x}})} 取变分我们有： 0 = δ S = ∫ t 1 t 2 δ L d t {\displaystyle 0=\delta S=\int \limits _{t1}^{t2}{\delta Ldt}} = ∫ t 1 t 2 δ ∫ ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ ) d V d t {\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}} = ∫ t 1 t 2 ∫ ( − ρ δ φ ( x → ) − 1 4 π G ∇ φ ⋅ δ ( ∇ φ ) ) d V d t {\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}} = ∫ t 1 t 2 ∫ ( − ρ δ φ ( x → ) − 1 4 π G ∇ φ ⋅ ∇ ( δ φ ) ) d V d t {\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}} = ∫ t 1 t 2 ∫ ( − ρ δ φ ( x → ) − 1 4 π G ( ∇ ⋅ ( δ φ ∇ φ ) − δ φ ∇ 2 φ ) ) d V d t {\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}} = ∫ t 1 t 2 ∫ Σ ( − 1 4 π G δ φ ∇ φ ⋅ d S → ) + ∫ ( − ρ δ φ ( x → ) + 1 4 π G δ φ ∇ 2 φ ) d V d t {\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 Σ {\displaystyle \Sigma } -包围体积V的边界 = ∫ ( − ρ + 1 4 π G ∇ 2 φ ) δ φ ( x → ) d V d t {\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 ∇ 2 φ = 4 π G ρ {\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： ⧸ L = 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ {\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： φ ˙ = ∂ φ ∂ t {\displaystyle {\dot {\varphi }}={\frac {\partial \varphi }{\partial t}}} ）为： H = ∫ ( ∑ i = 1 3 v i ∂ ⧸ L ∂ v i + φ ˙ ∂ ⧸ L ∂ φ ˙ − ⧸ L ) d V = ∫ ∑ i = 1 3 v i ∂ ∂ v i ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ ) d V + ∫ φ ˙ ∂ ∂ φ ˙ ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ ) d V − ∫ ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ ) d V = ∫ ( ρ v → ⋅ v → ) d V − ∫ ( 1 2 ρ v → ⋅ v → − ρ φ ( x → ) − 1 8 π G ∇ φ ⋅ ∇ φ ) d V = ∫ ( 1 2 ρ v → ⋅ v → + ρ φ ( x → ) + 1 8 π G ∇ φ ⋅ ∇ φ ) d V {\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 ρ φ ( x → ) {\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： ρ = 1 4 π G ∇ 2 φ {\displaystyle \rho ={\frac {1}{4\pi G}}{{\nabla }^{2}}\varphi } ，代入上式得： H = ∫ ( 1 2 ρ v → ⋅ v → + 1 4 π G φ ∇ 2 φ + 1 8 π G ∇ φ ⋅ ∇ φ ) d V = ∫ ( 1 2 ρ v → ⋅ v → + 1 4 π G ∇ ( φ ∇ φ ) − 1 4 π G ∇ φ ⋅ ∇ φ + 1 8 π G ∇ φ ⋅ ∇ φ ) d V {\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}}} = ∫ ( 1 2 ρ v → ⋅ v → − 1 8 π G ∇ φ ⋅ ∇ φ ) d V + 1 4 π G ∫ Σ φ ∇ φ ⋅ d S → {\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： Σ {\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 φ ∇ φ → O ( 1 r 3 ) {\displaystyle \varphi \nabla \varphi \to O({\frac {1}{{r}^{3}}})} ， d S → → O ( r 2 ) {\displaystyle d{\vec {S}}\to O({{r}^{2}})} ，其积 φ ∇ φ ⋅ d S → → O ( 1 r ) {\displaystyle \varphi \nabla \varphi \cdot d{\vec {S}}\to O({\frac {1}{r}})} ，因此 ∫ Σ φ ∇ φ ⋅ d S → = 0 {\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： H = ∫ ( 1 2 ρ v → ⋅ v → − 1 8 π G ∇ φ ⋅ ∇ φ ) d V {\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： E = ∫ ( 1 2 ρ v → ⋅ v → − 1 8 π G ∇ φ ⋅ ∇ φ ) d V {\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 d H d t = 0 {\displaystyle {\frac {dH}{dt}}=0} 也即 d E d t = 0 {\displaystyle {\frac {dE}{dt}}=0} 。 从上面的结果我们看到： 1 2 ρ v → ⋅ v → {\displaystyle {\frac {1}{2}}\rho {\vec {v}}\cdot {\vec {v}}} 代表理想流体的动能密度 T m {\displaystyle {{T}_{m}}} ， 1 8 π G ∇ φ ⋅ ∇ φ {\displaystyle {\frac {1}{8\pi G}}\nabla \varphi \cdot \nabla \varphi } 代表引力能密度 T g {\displaystyle {{T}_{g}}} ，这时我们看到总能量密度是 ε = T m − T g {\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 T g {\displaystyle {{T}_{g}}} 。 返回