Recent content by TerryW

BTW thanks for the link to Wolfram - I tried it with numbers of four significant figures and it worked.

I looked at my workings for verifying the figures of 1.75 and 0.0036 and found that I was still using an early estimate for ##F(\frac{\pi}{2},k)## on in the workings for 1.75. Using Google AI to generate the value resulted in much closer result - 1.86 vs 1.75. As my result depends on my choice...

Yep! Still have my slide rule. I agree that Darwin's expressions and mine are the same, I'm only questioning why it was necessary to introduce Q in the first place when everything can be derived directly from the roots of the cubic. But I also thought my post may be useful if anyone else ever...

Hi TSny, I've had a look at the paper you recommended and yes it shows the relationship between the roots of the cubic and Q and R, but it still leaves me with the question of why bother to do it when you can just use the roots directly as I have demonstrated. I can't take all the credit for...

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 =...

You're right. I just wanted to indicate the integral I needed to deal with so I wasn't really paying attention to the precise format. But thanks for making it clear. I'm going to see if I can get hold of a copy of the 1959 paper by Ward and Wheeler which might be of use in finding the equation...

Hi JimWhoKnew! Yes. I'm currently working on Fig 25.7 in MTW. and I think this integral might be relevant. TerryW

That's great! Thanks for all your help with this. TerryW

Which brings me back to what was my original problem, which is to work out how to get my cubic expression (##\int\frac{du}{\sqrt{((c+u^2(-1+2u))}}##) into one of the forms of elliptic integral which has a solution in the form of ##F(sin^{-1}\theta|\alpha)##. I'll see if Abromowitz and Stegun can...

I've found a pdf of Abromowitz and Stegun. I'll see if it clarifies things for me. Looking at the result from Mathematica, I wonder if it is actually correct? For example, the term ##(r2-r3)^2## on the denominator of the second square root cancels out the term ##(r2-r3)## at the beginning of...

Many thanks for this. I think I can console myself that I didn't miss an easy route to a solution! :smile: It's interesting that the elliptic integral of the first kind appears in the solution, but where the rest of it comes from may remain a mystery to me. I'm still puzzled by the fact that...

I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2...

Hi Tony, You really are a star! Thanks. TerryW

Hi TSny I've now figured out the reasoning behind equations 28 & 29 in Box 25.4 - I just have one issue left which I haven't been able to resolve. It comes from the picture you posted in #6 of the orbit of the particle, which is a line that is normal to the wavefronts it crosses. I thought it...

Sorry, that was a bit sloppy on my part! I feel that there isn't much else to be extracted from this problem - but I have just one question I'm going to look at before I move on. That is: Why does ##\frac {\partial{ \tilde S}}{\partial{ \tilde L}}## produce 'the shape of the orbit' and why is...