Timeline for Action Integral
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 18, 2015 at 5:52 | answer | added | Zurab Silagadze | timeline score: 1 | |
| May 16, 2015 at 15:35 | answer | added | Machinato | timeline score: 3 | |
| Jan 18, 2014 at 14:41 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced tag 'tag-removed' (if you make an edit to an old question, please make it count!)
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| S Jan 18, 2014 at 5:27 | history | suggested | Felix Marin | CC BY-SA 3.0 |
I put some \left and \right commands into the integral expressions and change it to a paragraph style.
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| Jan 18, 2014 at 4:11 | review | Suggested edits | |||
| S Jan 18, 2014 at 5:27 | |||||
| May 8, 2012 at 10:43 | comment | added | Yemon Choi | Well, my feeling on seeing the name Kepler and seeing the form of the final answer was that there must be an ellipse involved... | |
| May 8, 2012 at 4:10 | vote | accept | Raymond | ||
| May 8, 2012 at 4:09 | comment | added | Raymond | The transformation suggested by Yemon Choi is very useful. First of all, it gets rid of the square-root singularities at the endpoints, so the numerical integration is easier for arbitrary potentials. It is also closely related to the rational parameterization mentioned by the other commenters, since I think $\theta = \phi/2$ where $\phi$ is the azimuthal coordinate of the circle introduced by Igor. | |
| May 7, 2012 at 15:19 | answer | added | Igor Khavkine | timeline score: 10 | |
| May 7, 2012 at 12:20 | comment | added | Américo Tavares | To reduce an integral of the type $$\int R\left( x,\sqrt{ax^{2}+bx+c}\right) dx$$ to an integral of a rational function one may use the so called Euler substitutions. See Euler substitutions, L.D. Kudryavtsev (originator), Encyclopedia of Mathematics. URL: encyclopediaofmath.org/…. | |
| May 7, 2012 at 9:21 | comment | added | Yemon Choi | I've never tried this particular integral, but off the top of my head I would start by trying $r=a+(b-a)\sin^2 \theta$. Then double angle formulae. Did you already try this? | |
| May 7, 2012 at 4:55 | comment | added | David Roberts♦ | When you say 'derive this', what are referring to? The formula for $I$, or some general transformation technique? If the former, I'm afraid MO isn't the site for such questions, see the FAQ for a detailed explanation why. | |
| May 7, 2012 at 4:21 | history | asked | Raymond | CC BY-SA 3.0 |