Chapter 6.5, Problem

Interpretation Introduction

Interpretation:

The relativistic equation for the orbit of a planet around the sun is $\frac{{\text{d}}^{\text{2}}\text{u}}{{\text{dθ}}^{\text{2}}}{\text{ + u = α + εu}}^{\text{2}}$

Rewrite this equation as a system in the $\left(\text{u, v}\right)$ phase plane; find all the equilibrium points of the system. Also, show that one of the equilibria is a center in the $\left(\text{u, v}\right)$ phase plane. Show that the equilibrium point found above corresponds to a circular planetary orbit.

Concept Introduction:

Equation of orbit $\frac{{\text{d}}^{\text{2}}\text{u}}{{\text{dθ}}^{\text{2}}}{\text{ + u = α + εu}}^{\text{2}}$ is the second-order differential equation.

Here, $\text{u}$ is $\frac{\text{1}}{\text{r}}$, $\text{α}$ is any parameter having a positive value, ${\text{εu}}^{\text{2}}$ is Einstein’s correction, $\text{ε}$ is any parameter having a very small positive value and $\text{r , θ}$ are the polar coordinates of the planet.

A conserved quantity is a real-valued continuous function that is constant on trajectories.

Answer

Solution:

The equation as a system in phase plane $\left(\text{u, v}\right)$ is $\frac{\text{dv}}{\text{dθ}}={\text{α + εu}}^{\text{2}}\text{- u}$.

The equilibrium points come out to be $\left(\frac{\text{1 ± }\sqrt{{\text{b}}^{\text{2}}\text{- 4αε}}}{\text{2ε}},0\right)$

It is proved that the center is a non-linear center.

The equilibrium point $\left(\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$ corresponds to a planetary orbit is shown below.

Explanation of Solution

For a given solution $\text{x (t)}$, the total energy E can be expressed as,

$\text{E = }\frac{\text{1}}{\text{2}}{\text{m}\dot{\text{x}}}^{\text{2}}\text{ + V(x)}$

Here, $\text{V(x)}$ is the potential energy and $\frac{\text{1}}{\text{2}}{\text{m}\dot{\text{x}}}^{\text{2}}$ is the kinetic energy.

This kinetic energy is also called a conserved quantity because it is constant as a function of time.

The relativistic equation for the orbit of a planet around the sun is given as,

$\frac{{\text{d}}^{\text{2}}\text{u}}{{\text{dθ}}^{\text{2}}}{\text{ + u = α + εu}}^{\text{2}}$

Here, $\text{u}$ is $\frac{\text{1}}{\text{r}}$, $\text{α}$ is any parameter having a positive value, ${\text{εu}}^{\text{2}}$ is Einstein’s correction, $\text{ε}$ is any parameter having a very small positive value and $\text{r , θ}$ are the polar coordinates of the planet.

The objective is to find the equation of the given system is to be rewritten in the $\text{(u, v)}$ phase plane. It is given that,

$\text{v = }\frac{\text{du}}{\text{dθ}}$

Differentiate it with respect to $\text{θ}$,

$\frac{\text{dv}}{\text{dθ}}=\frac{{\text{d}}^{2}u}{{\text{dθ}}^2}$

Substitute, ${\text{α + εu}}^{\text{2}}\text{- u}$ for $\frac{{\text{d}}^{2}u}{{\text{dθ}}^2}$ in the above equation and solve,

$\frac{\text{dv}}{\text{dθ}}={\text{α + εu}}^{\text{2}}\text{- u}$.

The objective is to find the system can be rewritten as,

$\dot{\text{u}}\text{ = v}$

And,

${\dot{\text{v}}\text{ = (α + εu}}^{\text{2}}\text{- u)}$

Fixed points occur for the following conditions,

$\dot{\text{u}}\text{ = 0}$

And,

$\dot{\text{v}}\text{ = 0}$

Substitute, $\text{0}$ for $\dot{\text{v}}$ in ${\text{(α + εu}}^{\text{2}}\text{- u)}=0$ and solve,

${\text{(α + εu}}^{\text{2}}\text{- u) = 0}$

The roots of the equation can be calculated as,

$\text{u = }\frac{\text{- b ± }\sqrt{{\text{b}}^{\text{2}}\text{- 4ac}}}{\text{2a}}$

Substitute, $\text{-1}$ for $\text{ε}$ for $\text{a}$ and $\text{α}$ for c and solve,

$\text{u = }\frac{\text{1 ± }\sqrt{{\text{b}}^{\text{2}}\text{- 4αε}}}{\text{2ε}}$

Thus, the equilibrium points come out to be $\left(\frac{\text{1 ± }\sqrt{{\text{b}}^{\text{2}}\text{- 4αε}}}{\text{2ε}},0\right)$

The objective is to be shown that one of the equilibrium is a center in the $\text{(u, v)}$ phase plane, according to the linearization.

The Jacobian matrix at a general point $\text{(u, v)}$ is given by,

$\text{A = }\left(\begin{array}{cc}\frac{\partial \dot{\text{u}}}{\partial \text{u}}& \frac{\partial \dot{\text{u}}}{\partial \text{v}}\\ \frac{\partial \dot{\text{v}}}{\partial \text{u}}& \frac{\partial \dot{\text{v}}}{\partial \text{v}}\end{array}\right)$ Substitute, $\text{v}$ for $\dot{\text{u}}$, ${\text{(α + εu}}^{\text{2}}\text{- u)}$ for $\dot{\text{v}}$ and solve,

$\text{A = }\left(\begin{array}{cc}\text{0}& \text{1}\\ \text{2uε}& \text{0}\end{array}\right)$

Evaluate the matrix at the fixed point $\left(\frac{\text{1 + }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$,

Substitute, $\frac{\text{1 + }\sqrt{\text{1- 4αε}}}{\text{2ε}}$ for $\text{x}$ and $\text{0}$ for $\text{y}$ in the Jacobian matrix,

${\text{A}}_{\left(\frac{\text{1 + }\sqrt{\text{1 - 4αε}}}{\text{2ε}},0\right)}\text{ = }\left(\begin{array}{cc}\text{0}& \text{1}\\ \sqrt{\text{1 - 4αε}}& \text{0}\end{array}\right)$

For the above Jacobian matrix determinant, $\Delta$ comes out to be $\text{- }\sqrt{\text{1 - 4αε}}$ and trace, $\tau$ comes out to be $0$. This implies that this fixed point is a saddle point.

Evaluate the matrix at the fixed point $\left(\frac{\text{1 ± }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$,

Substitute, $\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}}$ for $\text{x}$ and $\text{0}$ for $\text{y}$ in the Jacobian matrix,

${\text{A}}_{\left(\frac{\text{1 - }\sqrt{\text{1 - 4αε}}}{\text{2ε}},0\right)}\text{ = }\left(\begin{array}{cc}\text{0}& \text{1}\\ -\sqrt{\text{1 - 4αε}}& \text{0}\end{array}\right)$

For the above Jacobian matrix determinant, $\Delta$ comes out to be $\sqrt{\text{1 - 4αε}}$ and trace, $\tau$ comes out to be $0$. This implies that this fixed point is a linear center.

Thus, it is proved that one of the equilibrium is a center.

The theorem for non-linear centers states that,

A system is considered as $\dot{\text{x}}\text{ = f (x)}$. Here, $\text{x = (x, y)}\in {\text{R}}^2$ and f is continuously differentiable. Suppose a conserved quantity $\text{E (x)}$ exists and ${\text{x}}^{\text{*}}$ is an isolates fixed point.

If a local minimum of $\text{E}$ is ${\text{x}}^{\text{*}}$, then all trajectories close to ${\text{x}}^{\text{*}}$ are closed.

According to this above theorem, the center is a non-linear center provided that the fixed point is a local minimum of a conserved quantity.

The energy equation for the system can be expressed as,

$\text{E = }\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{ - }{\displaystyle \int \dot{\text{y}}\text{dx}}$

Substitute, ${\text{α + εu}}^{\text{2}}\text{- x}$ for $\dot{\text{y}}$ and solve,

$\text{E = }\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{ - }{\displaystyle \int {\text{(α + εu}}^{\text{2}}\text{ - x) dx}}$

The divergence of energy, $\nabla \text{E}$ for the above equation, will come out to be,

$\nabla \text{E = }\left({\text{α + εu}}^{\text{2}}\text{- x , }\dot{\text{x}}\right)$

For the calculated fixed point, the divergence is,

$\nabla E\left(\frac{\text{1 - }\sqrt{\text{1 - 4αε}}}{\text{2ε}},0\right)=\left(0,0\right)$

Determine the matrix for the above energy equation

$\left|\text{E}\right|\text{=}|\begin{array}{l}{\text{E}}_{\text{xx}}\\ {\text{E}}_{\text{yx}}\end{array}\begin{array}{l}{\text{E}}_{\text{xy}}\\ {\text{E}}_{\text{yy}}\end{array}|$

Here, the subscripts determine the partial derivatives.

Solve the above matrix by substituting the above value of E,

$\left|\text{E}\right|\text{=}|\begin{array}{l}\text{ -2ε}\\ {\dot{\text{x}}}^{\text{-α-εx}}\text{+x}\end{array}\begin{array}{l}{\dot{\text{x}}}^{\text{-α-εx}}\text{+x}\\ \text{ 1}\end{array}|$

For a fixed point $\left(\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$

${\left|\text{E}\right|}_{\left(\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}}\text{,0}\right)}\text{= - 2ε}$

Since, $(-2\epsilon <0)$, then $\left(\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$ is a local minimum.

Thus, the center is a non-linear center.

The objective is to be shown that the equilibrium point $\left(\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$ corresponds to a planetary orbit. It can be done by showing that the radius of this orbit is constant.

It is given that,

$\text{u = }\frac{\text{1}}{\text{r}}$

$\text{r = }\frac{\text{1}}{\text{u}}$

Substitute, $\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}}$ for $\text{u}$ and solve,

$\text{r = }\frac{\text{2ε}}{\text{1 - }\sqrt{\text{1 - 4αε}}}$

This generated value is a constant as both the parameters $\text{ε}$ and $\text{α}$ are constants.

Thus, the equilibrium point $\left(\frac{\text{1 - }\sqrt{\text{1- 4αε}}}{\text{2ε}},0\right)$ corresponds to a planetary orbit.