Some work on MTW Figure 25.7

[TerryW]

Homework Statement

What is Q in Figure 25.7

Relevant Equations

Q^2 = (R-2)(R+6) etc

Figure 25.7 in MTW is unusual in that it introduces formulae with no real explanation as to how they have been derived. I refer to the use of $Q^2 = (R-2)(R+6)$ and $Sin^2 \theta = k^2 = (Q-R+6)/2Q$ and then $sin^2 \phi_{min} = (2+Q-R)/(6+Q-R)$ leading to $\Theta = 4*(R/Q)^{½}(F(\pi/2,\theta) - F(\phi_{min},\theta)) - \pi$

So I decided to to try find out how these formulae might be generated.

My first step was to solve: $\big(\frac{du}{d\phi}\big)^2 +(1-2u)u^2 = u_b^2$
ie $d\phi = \frac{du}{\sqrt2(u^3-1/2u^2+1/2U_b^2)^{½}}= \frac{du}{\sqrt2 (\sqrt(u-r_1)(r_2-u)(r_3-u))}$

A cubic of the form $u^3 +\alpha u^2+\beta u + \gamma$ can be factorised into $(u-r_1)(r_2-u)(r_3-u)$ using a standard process.

I built a spreadsheet to find $r_1$ ,$r_2$ and $r_3$ for the values of $\alpha = 0.5, \beta =0$ and a range of values for $\gamma = \frac{1}{2}u_b^2 =< \frac{1}{2}u^2_{b(min)}$ where $u_{b(min)} = \frac{1}{b(min)} = \frac{1}{3\sqrt3}$

The next stage is to get $\frac{du}{\sqrt2((u-r_1)(r_2-u)(r_3-u))^{½}}$ into a form which can be integrated.

Using the substitution $u = r_2-(r_2-r_1)cos^2\theta$ and $du = -2(r_2-r_1)Sin\theta cos\theta$
I arrive at $\frac{1}{\sqrt 2} \int _0^b\frac{du}{\sqrt ((u-r_1)(r_2-u)(r_3-u))} = \frac{1}{\sqrt 2} \int \frac{2d\theta}{\sqrt(r_3-r_1)\sqrt(1-k^2sin^2\theta)}$ where $k^2 = \frac{(r_2-r_1)}{(r_3-r_1)}$

I'm now ready to evaluate (1) the various formulae involving Q and R in Fig 25.7 with their equivalents from my analysis, taking the values for $b= \frac{1}{u_b} = 5.2065.$ and contrasting them with the equivalents derived from the roots of the cubic.

A (MTW) $b^2 = 5.2065^2 = \frac{R^3}{(R-2)}$ This is a cubic equation which has one root with the value of 3.11474.

A (Cubic) One of the roots of $u^3 - \frac{1}{2} u^2 + \frac{1}{2}u_b^2 = 0$ when u_b is set to 0.19207 (=$\frac{1}{5.2065}$) is 0.32105 =$\frac{1}{3.11474}$

So, solving the cubic $u^3 - \frac{1}{2} u^2 + \frac{1}{2}u_b^2 = 0$ provides a value for 1/R where R is the distance of closest approach.

B (MTW) $Q^2 = (R-2)(R+6) = (3.1147-2)(3.1147+6)= 10.16016, Q= 3.1874$
Q is then used to generate $k^2$
$k^2 = (Q-R+6)/2Q = 0.9526$

B (Cubic) $k^2 = \frac{(r_2-r_1)}{(r_3-r_1)} = \frac{(0.32105-(-0.16637))}{(0.34532-(-0.16637))} = 0.9526$

So the roots of $u^3 - \frac{1}{2} u^2 + \frac{1}{2}u_b^2 = 0$ are all that is needed to produce $k^2$.

C (MTW) The so-called amplitude of the elliptic function $sin^2\phi _{min} = (2+Q-R)(6+Q-R)$.
$(2+Q-R)(6+Q-R) = 2.0727/6.0727 = 0.34131.$
$Sin^{-1}(\sqrt(0.34131) = 35.748º$

C (Cubic) $\phi = cos^{-1}\big(\sqrt\frac{r_2}{r_2-r_1}\big) = 35.748º$


D (MTW) $4(R/Q)^\frac{1}{2} = 4 * \sqrt {(3.1147/3.1874)} = 3.9541$

D (Cubic) $\frac{4}{\sqrt 2 \sqrt (r_3-r_1)} = 3.9540$

I've confirmed that an impact parameter of $b/M = \sqrt 27 - 0.065$ gives a value of $\Theta = 2\pi$
The calculation for $b/M = \sqrt 27 - 0.000012$ gives $4\pi$

So why did MTW use the formulae involving Q and R instead of working through from the cubic equation?
I've tried to find some relationship between MTW's R&Q and the three cubic roots $r_1$, $r_2$ and$r_3$ but nothing is immediately obvious.


TerryW







.

 

[TSny]

The relations between the three root $(r_1, r_2, r_3)$ and the quantities $R$ and $Q$ are given in the paper by C. Darwin referenced in Fig. 25.7 of MTW. You can read this paper online here if you register for a free account. Section 8 of the paper deals with the orbits of light rays. There, he shows how to obtain $$r_1 =- \frac{Q-R+2}{4R}, \,\,\, r_2 = \frac 1 R, \,\,\,r_3 = \frac{Q+R-2}{4R}.$$ $Q$ is just an abbreviation for $\sqrt{(R-2)(R+6)}$. Darwin uses $P$ instead of $R$ for the distance of closest approach. And he uses $l$ for the impact parameter instead of $b$.

For a specified $R$, the roots are determined by these relations. Darwin also shows how to determine the deflection of the light rays.

The next stage is to get $\frac{du}{\sqrt2((u-r_1)(r_2-u)(r_3-u))^{½}}$ into a form which can be integrated.

Using the substitution $u = r_2-(r_2-r_1)cos^2\theta$ and $du = -2(r_2-r_1)Sin\theta cos\theta$
I arrive at $\frac{1}{\sqrt 2} \int _0^b\frac{du}{\sqrt ((u-r_1)(r_2-u)(r_3-u))} = \frac{1}{\sqrt 2} \int \frac{2d\theta}{\sqrt(r_3-r_1)\sqrt(1-k^2sin^2\theta)}$ where $k^2 = \frac{(r_2-r_1)}{(r_3-r_1)}$

This is a nice substitution. It reduces the integral to an elliptic integral. (I'm not sure of your upper limit of $b$ for the $u$ integral.)

These days, it is easy to use software to numerically evaluate the integral. Back when computers were not accessible, you could use numerical tables for elliptic integrals. Darwin's paper was published in 1957. Even at the time of the publication of MTW (1970) most students didn't have easy access to computers. This might partly explain MTW's use of Darwin's expression for the deflection angle in terms of the elliptic integral $F$: $$\Theta = 4(R/Q)^{1/2}\left[F(\pi/2, \theta) - F(\phi_{\rm min}, \theta)\right] - \pi$$