作者： Finn

分类
_NOTES_ General Relativity tensor

Newtonian Tidal Effect

This note can be use as an example of tensor analysis

This article shows the derivation of Newtonian tidal effect from Newton’s equation of motion and gravity, with the notation of tensor analysis and Einstein summation convention. This may serve as an example for beginners in tensor analysis to check their comprehension, meanwhile as a baby version or lead-in to geodesic deviation of general relativity.


Newton’s second law (componentwise)

$$
\frac{d^{2} x^{i}}{d t^{2}} \equiv a^{i} \equiv \frac{F^{i}}{m}
$$

Newtonian equation of gravitation

$$
\frac{F^{i}}{m}=-\eta^{i j} \partial_{j} \phi
$$

combine two equations give:

$$
\frac{d^{2} x^{i}}{d t^{2}}=-\eta i j\left[\partial_{j} \phi\right]_{\vec{x}}
$$

where the subscription $\vec{x}$ denotes the derivative is operated at position $\vec{x}$ similarly we have at $\vec{x}+\vec{n}$ :

$$
\frac{d^{2}\left(x^{i}+n^{i}\right)}{d t^{2}}=-\eta^{i j}\left[\partial_{j} \phi\right]_{\vec{x}+\vec{n}}
$$

make subtraction give:

$$
\frac{d^{2} n^{i}}{d t^{2}}=-\eta^{i j}\left(\left[\partial_{j} \phi\right]{\vec{x}+\vec{n}}-\left[\partial{j} \phi\right]_{\vec{x}}\right)
$$

given $\vec{n}$ is infinitesimal, we have RHS:

$$
\text { RHS }=-\eta^{i j} n^{k}\left[\partial_{k}\left(\partial_{j} \phi\right)\right]_{\vec{x}}
$$

set Cartesian coordinates originated from centre of earth orient the $z$-axis so that is consists with $\vec{x}$ (position vector) then $\vec{X}=\left(X^{1}, X^{2}, X^{3}\right)=(0,0, z)$.

gravitational potential is given by:

$$
\phi=-\frac{G M}{r}=-\frac{G M}{\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}
$$

and thus

$$
\begin{aligned}
\partial_{j} \phi & =\frac{\partial \phi}{\partial x^{j}} \\
& =-\frac{d \phi}{d r} \frac{\partial r}{\partial x^{j}} \\
& =\frac{G M}{r^{3}} x^{j}
\end{aligned}
$$

and

$$
\begin{aligned}
\partial_{k} \partial_{j} \phi & =\frac{\partial}{\partial x^{k}}\left(\frac{G M}{r^{3}} x^{j}\right) \\
& =\frac{\partial}{\partial x^{k}}\left(\frac{G M}{r^{3}}\right) x^{j}+\delta_{k}^{j} \frac{G M}{r^{3}} \\
& =-\frac{3 G M}{r^{4}} x^{j}+\delta_{k}^{j} \frac{G M}{r^{3}}
\end{aligned}
$$

thus.
$$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} n^{x}=-\frac{G M}{r^{3}} n^{x} \\
& \frac{d^{2}}{d t^{2}} n^{y}=-\frac{G M}{r^{3}} n^{y} \\
& \frac{d^{2}}{d t^{2}} n^{z}=\frac{2 G M}{r^{3}} n^{z}
\end{aligned}
$$

分类
_笔记_ 分析力学

由拉格朗日方程导出欧拉动力学方程

本文整理自俞千野同学的工作，预备役物理民工经授权搬运

千野哥哥太强了

Lagrangian of free rigid body in its comoving reference frame is given by:

$$
L=\sum_{i} \frac{1}{2} I_{i} \omega_{i}^{2}
$$

where angular velocities can be expressed in Eulerian angles: $\omega_{1}=\dot{\varphi} \sin \theta \sin \psi+$ $\dot{\theta} \cos \psi, \omega_{2}=\dot{\varphi} \sin \theta \cos \psi-\dot{\theta} \sin \psi, \omega_{3}=\dot{\varphi} \cos \theta+\dot{\psi}$.

Substitute into EL equations:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}\right)=\frac{\partial \mathcal{L}}{\partial q_{i}}
$$

where $q_{i}$ take $q_{1}=\theta, q_{2}=\varphi, q_{3}=\psi$.

Apply chain rule to both sides of (2) :

$$
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial q_{i}} & =\sum_{j=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{j}} \frac{\partial \omega_{j}}{\partial q_{i}} \\
\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}} & =\sum_{j=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{j}} \frac{\partial \omega_{j}}{\partial \dot{q}_{i}}
\end{aligned}
$$

Substitute $q_{3}$ by $\psi$ in (2), RHS:

$$
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial \psi} & =\sum_{i=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{i}} \frac{\partial \omega_{i}}{\partial \psi} \\
& =\sum_{i=1}^{3} I_{i} \omega_{i} \frac{\partial \omega_{i}}{\partial \psi} \\
& =I_{1} \omega_{1}(\dot{\varphi} \sin \theta \cos \psi-\dot{\theta} \sin \psi)+I_{2} \omega_{2}(-\dot{\varphi} \sin \theta \sin \psi-\dot{\theta} \cos \psi)+0 \\
& =I_{1} \omega_{1} \omega_{2}+I_{2} \omega_{2}\left(-\omega_{1}\right) \\
& =\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}
\end{aligned}
$$

*this note is rearranged based on Qianye YU’s work LHS:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\psi}}\right)=I_{3} \dot{\omega}_{3}
$$

Combing (2)(4)(5) gives the first Euler dynamic equation:

$$
\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}-I_{3} \dot{\omega}_{3}=0
$$

Similarly substitute $q_{2}$ by $\varphi$ into equation(2). Since $\frac{\partial \mathcal{L}}{\partial \varphi}=0$, we have:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\varphi}}\right)=0
$$

Apply chain rule to LHS of (7) gives:

$$
\begin{aligned}
0= & I_{1} \dot{\omega}_{1} \sin \theta \sin \psi+I_{2} \dot{\omega}_{2} \sin \theta \cos \psi+I_{3} \dot{\omega}_{3} \cos \theta+I_{1} \omega_{1}(\cos \theta \sin \psi \dot{\theta} \\
& +\sin \theta \cos \psi \dot{\psi})+I_{2} \omega_{2}(\cos \theta \cos \psi \dot{\theta}-\sin \theta \sin \psi \dot{\psi})-I_{3} \omega_{3} \sin \theta \dot{\theta}
\end{aligned}
$$

We introduce $A=I_{1} \omega_{1}(\cos \theta \sin \psi \dot{\theta}+\sin \theta \cos \psi \dot{\psi})$ and $B=I_{2} \omega_{2}(\cos \theta \cos \psi \dot{\theta}-$ $\sin \theta \sin \psi \dot{\psi})$ to reduce the calculation.

Notice $I_{3} \dot{\omega}_{3}=\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}$. So (8) is equivalent to:

$$
\begin{aligned}
\sin \theta\left[I_{1} \dot{\omega}_{1} \sin \psi+I_{2} \dot{\omega}_{2} \cos \psi\right] & =\left(I_{2}-I_{1}\right) \omega_{1} \omega_{2} \cos \theta+I_{3} \omega_{3} \sin \theta \dot{\theta}-A-B \\
& =I_{1}\left(-\omega_{1} \omega_{2} \cos \theta-\omega_{1} \cos \theta \sin \psi \dot{\theta}-\omega_{1} \sin \theta \cos \psi \dot{\psi}\right) \\
& +I_{2}\left(\omega_{1} \omega_{2} \cos \theta-\omega_{2} \cos \theta \cos \psi \dot{\theta}+\omega_{2} \sin \theta \sin \psi \dot{\psi}\right) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta} \\
& =I_{1} \omega_{1}(-\dot{\varphi} \sin \theta \cos \psi \cos \theta+\dot{\theta} \sin \psi \cos \theta-\dot{\theta} \sin \psi \cos \theta\\&-\sin \theta \cos \psi \dot{\psi})  +I_{2} \omega_{2}(\dot{\varphi} \sin \theta \sin \psi \cos \theta+\dot{\theta} \cos \psi \cos \theta\\&-\dot{\theta} \cos \psi \cos \theta+\sin \theta \sin \psi \dot{\psi})  +I_{3} \omega_{3} \sin \theta \dot{\theta} \\
& =I_{1} \omega_{1}(-\dot{\varphi} \sin \theta \cos \psi \cos \theta-\sin \theta \cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \theta \sin \psi \cos \theta+\sin \theta \sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta}
\end{aligned}
$$

Dividing (9) through $\sin \theta$ gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} \sin \psi+I_{2} \dot{\omega}_{2} \cos \psi & =I_{1} \omega_{1}(-\dot{\varphi} \cos \psi \cos \theta-\cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \psi \cos \theta+\sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \dot{\theta}
\end{aligned}
$$

We keep equation (10) for later use

Substituting $q_{3}$ by $\theta$ into equation $(2)$ :

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right)=\frac{\partial \mathcal{L}}{\partial \theta}
$$

So,

$$
\begin{aligned}
\frac{d}{d t}\left(I_{1} \omega_{1} \cos \psi-I_{2} \omega_{2} \sin \psi\right)= & I_{1} \omega_{1} \dot{\varphi} \cos \theta \sin \psi+I_{2} \omega_{2} \dot{\varphi} \cos \theta \cos \psi \\
& +I_{3} \omega_{3} \dot{\varphi}(-\sin \theta)
\end{aligned}
$$

which gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} \cos \psi-I_{2} \dot{\omega}_{2} \sin \psi= & I_{1} \omega_{1}(\dot{\varphi} \cos \theta \sin \psi+\sin \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \cos \theta \cos \psi+\cos \psi \dot{\psi}) \\
& -I_{3} \omega_{3} \dot{\varphi} \sin \theta
\end{aligned}
$$

Since (10) and (13) are obtained from EL equation of generalized coordinates $\varphi$ and $\theta$ respectively, along with (6), we have used all three independent EL equations, the other two Euler equations must be obtained directly from combinations of (10) and (13):

Firstly, (10) $\cdot \sin \psi+(13) \cdot \cos \psi$ gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} & =I_{2} \omega_{2}(\dot{\varphi} \cos \theta+\dot{\psi})+I_{3} \omega_{3}(\dot{\theta} \sin \psi-\dot{\varphi} \sin \theta \cos \psi) \\
& =I_{2} \omega_{2} \omega_{3}+I_{3} \omega_{3}\left(-\omega_{2}\right) \\
& =\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}
\end{aligned}
$$

And $(10) \cdot \cos \psi-(13) \cdot \sin \psi$ gives:

$$
\begin{aligned}
I_{2} \dot{\omega}_{2} & =I_{1} \omega_{1}(-\dot{\varphi} \cos \theta-\dot{\psi})+I_{3} \omega_{3}(\dot{\theta} \cos \psi+\dot{\varphi} \sin \theta \sin \psi) \\
& =I_{1} \omega_{1}\left(-\omega_{3}\right)+I_{3} \omega_{3} \omega_{1} \\
& =\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1}
\end{aligned}
$$

Rearranging (6),(14),(15) gives the Eulerian dynamic equations:

$$
\left\{\begin{array}{l}
\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}-I_{3} \dot{\omega}_{3}=0 \\
\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}-I_{1} \dot{\omega}_{1}=0 \\
\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1}-I_{2} \dot{\omega}_{2}=0
\end{array}\right.
$$

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分类
_NOTES_ Analytical Mechanics

Derivation of Euler Dynamic Equation for Free Rigid Body from EL Equation

This article is rearranged based on Qianye YU’s previous work, Physics Reserved Labour is fully authorized to make the reprint.

qyggtql

Lagrangian of free rigid body in its comoving reference frame is given by:

$$
L=\sum_{i} \frac{1}{2} I_{i} \omega_{i}^{2}
$$

where angular velocities can be expressed in Eulerian angles: $\omega_{1}=\dot{\varphi} \sin \theta \sin \psi+$ $\dot{\theta} \cos \psi, \omega_{2}=\dot{\varphi} \sin \theta \cos \psi-\dot{\theta} \sin \psi, \omega_{3}=\dot{\varphi} \cos \theta+\dot{\psi}$.

Substitute into EL equations:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}\right)=\frac{\partial \mathcal{L}}{\partial q_{i}}
$$

where $q_{i}$ take $q_{1}=\theta, q_{2}=\varphi, q_{3}=\psi$.

Apply chain rule to both sides of (2) :

$$
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial q_{i}} & =\sum_{j=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{j}} \frac{\partial \omega_{j}}{\partial q_{i}} \\
\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}} & =\sum_{j=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{j}} \frac{\partial \omega_{j}}{\partial \dot{q}_{i}}
\end{aligned}
$$

Substitute $q_{3}$ by $\psi$ in (2), RHS:

$$
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial \psi} & =\sum_{i=1}^{3} \frac{\partial \mathcal{L}}{\partial \omega_{i}} \frac{\partial \omega_{i}}{\partial \psi} \\
& =\sum_{i=1}^{3} I_{i} \omega_{i} \frac{\partial \omega_{i}}{\partial \psi} \\
& =I_{1} \omega_{1}(\dot{\varphi} \sin \theta \cos \psi-\dot{\theta} \sin \psi)+I_{2} \omega_{2}(-\dot{\varphi} \sin \theta \sin \psi-\dot{\theta} \cos \psi)+0 \\
& =I_{1} \omega_{1} \omega_{2}+I_{2} \omega_{2}\left(-\omega_{1}\right) \\
& =\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}
\end{aligned}
$$

*this note is rearranged based on Qianye YU’s work LHS:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\psi}}\right)=I_{3} \dot{\omega}_{3}
$$

Combing (2)(4)(5) gives the first Euler dynamic equation:

$$
\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}-I_{3} \dot{\omega}_{3}=0
$$

Similarly substitute $q_{2}$ by $\varphi$ into equation(2). Since $\frac{\partial \mathcal{L}}{\partial \varphi}=0$, we have:

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\varphi}}\right)=0
$$

Apply chain rule to LHS of (7) gives:

$$
\begin{aligned}
0= & I_{1} \dot{\omega}_{1} \sin \theta \sin \psi+I_{2} \dot{\omega}_{2} \sin \theta \cos \psi+I_{3} \dot{\omega}_{3} \cos \theta+I_{1} \omega_{1}(\cos \theta \sin \psi \dot{\theta} \\
& +\sin \theta \cos \psi \dot{\psi})+I_{2} \omega_{2}(\cos \theta \cos \psi \dot{\theta}-\sin \theta \sin \psi \dot{\psi})-I_{3} \omega_{3} \sin \theta \dot{\theta}
\end{aligned}
$$

We introduce $A=I_{1} \omega_{1}(\cos \theta \sin \psi \dot{\theta}+\sin \theta \cos \psi \dot{\psi})$ and $B=I_{2} \omega_{2}(\cos \theta \cos \psi \dot{\theta}-$ $\sin \theta \sin \psi \dot{\psi})$ to reduce the calculation.

Notice $I_{3} \dot{\omega}_{3}=\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}$. So (8) is equivalent to:

$$
\begin{aligned}
\sin \theta\left[I_{1} \dot{\omega}_{1} \sin \psi+I_{2} \dot{\omega}_{2} \cos \psi\right] & =\left(I_{2}-I_{1}\right) \omega_{1} \omega_{2} \cos \theta+I_{3} \omega_{3} \sin \theta \dot{\theta}-A-B \\
& =I_{1}\left(-\omega_{1} \omega_{2} \cos \theta-\omega_{1} \cos \theta \sin \psi \dot{\theta}-\omega_{1} \sin \theta \cos \psi \dot{\psi}\right) \\
& +I_{2}\left(\omega_{1} \omega_{2} \cos \theta-\omega_{2} \cos \theta \cos \psi \dot{\theta}+\omega_{2} \sin \theta \sin \psi \dot{\psi}\right) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta} \\
& =I_{1} \omega_{1}(-\dot{\varphi} \sin \theta \cos \psi \cos \theta+\dot{\theta} \sin \psi \cos \theta-\dot{\theta} \sin \psi \cos \theta-\sin \theta \cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \theta \sin \psi \cos \theta+\dot{\theta} \cos \psi \cos \theta-\dot{\theta} \cos \psi \cos \theta+\sin \theta \sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta} \\
& =I_{1} \omega_{1}(-\dot{\varphi} \sin \theta \cos \psi \cos \theta-\sin \theta \cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \theta \sin \psi \cos \theta+\sin \theta \sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \sin \theta \dot{\theta}
\end{aligned}
$$

Dividing (9) through $\sin \theta$ gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} \sin \psi+I_{2} \dot{\omega}_{2} \cos \psi & =I_{1} \omega_{1}(-\dot{\varphi} \cos \psi \cos \theta-\cos \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \sin \psi \cos \theta+\sin \psi \dot{\psi}) \\
& +I_{3} \omega_{3} \dot{\theta}
\end{aligned}
$$

We keep equation (10) for later use

Substituting $q_{3}$ by $\theta$ into equation $(2)$ :

$$
\frac{d}{d t}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right)=\frac{\partial \mathcal{L}}{\partial \theta}
$$

So,

$$
\begin{aligned}
\frac{d}{d t}\left(I_{1} \omega_{1} \cos \psi-I_{2} \omega_{2} \sin \psi\right)= & I_{1} \omega_{1} \dot{\varphi} \cos \theta \sin \psi+I_{2} \omega_{2} \dot{\varphi} \cos \theta \cos \psi \\
& +I_{3} \omega_{3} \dot{\varphi}(-\sin \theta)
\end{aligned}
$$

which gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} \cos \psi-I_{2} \dot{\omega}_{2} \sin \psi= & I_{1} \omega_{1}(\dot{\varphi} \cos \theta \sin \psi+\sin \psi \dot{\psi}) \\
& +I_{2} \omega_{2}(\dot{\varphi} \cos \theta \cos \psi+\cos \psi \dot{\psi}) \\
& -I_{3} \omega_{3} \dot{\varphi} \sin \theta
\end{aligned}
$$

Since (10) and (13) are obtained from EL equation of generalized coordinates $\varphi$ and $\theta$ respectively, along with (6), we have used all three independent EL equations, the other two Euler equations must be obtained directly from combinations of (10) and (13):

Firstly, (10) $\cdot \sin \psi+(13) \cdot \cos \psi$ gives:

$$
\begin{aligned}
I_{1} \dot{\omega}_{1} & =I_{2} \omega_{2}(\dot{\varphi} \cos \theta+\dot{\psi})+I_{3} \omega_{3}(\dot{\theta} \sin \psi-\dot{\varphi} \sin \theta \cos \psi) \\
& =I_{2} \omega_{2} \omega_{3}+I_{3} \omega_{3}\left(-\omega_{2}\right) \\
& =\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}
\end{aligned}
$$

And $(10) \cdot \cos \psi-(13) \cdot \sin \psi$ gives:

$$
\begin{aligned}
I_{2} \dot{\omega}_{2} & =I_{1} \omega_{1}(-\dot{\varphi} \cos \theta-\dot{\psi})+I_{3} \omega_{3}(\dot{\theta} \cos \psi+\dot{\varphi} \sin \theta \sin \psi) \\
& =I_{1} \omega_{1}\left(-\omega_{3}\right)+I_{3} \omega_{3} \omega_{1} \\
& =\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1}
\end{aligned}
$$

Rearranging (6),(14),(15) gives the Eulerian dynamic equations:

$$
\left\{\begin{array}{l}
\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2}-I_{3} \dot{\omega}_{3}=0 \\
\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}-I_{1} \dot{\omega}_{1}=0 \\
\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1}-I_{2} \dot{\omega}_{2}=0
\end{array}\right.
$$

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