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 $\Theta = 2pi$
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







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