kev said:
The following can be directly obtained from the Schwarzschild metric:
\frac{dr^2}{dt^2} = \alpha \left(\alpha - \frac{ds^2}{dt^2} - r^2\frac{d\phi^2}{dt^2} \right)
The second Euler-Langrange equation is
r^2 \frac{d\phi}{ds}=H
starthaus said:
You are learning. Notice that \frac{dH}{dr}=2r\frac{d\phi}{ds}
LOL. That is an elementary calculus blunder. Shame on you!
Using the power rule:
\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)
You need to prove
r^2 \frac{d}{dr}\left(\frac{d\phi}{ds} \right)= 0
That is only true if (d\phi/ds)=0 or r=0 which it generally not the case for orbital motion, or if (d\phi/ds)=0 is not a function of (r). We can easily show that (d\phi/ds) IS a function of r in the Newtonian limit when ds \approx t because it is well known that angular momentum is conserved in Newtonian physics and d\phi/dt changes with radius for a non circular orbit. You do realize GR has to agree with the Newtonian limit right? Try again.
kev said:
Hi Espen, I have double checked and triple checked all 3 lines of (57) now and can assure you it is perfect. Ignore Starthaus. It is another red herring.
starthaus said:
Err, wrong. I'll give you a hint, look at the coefficient for \frac{d\phi}{dt}. Not your day today.
The equations in section (57) of Espen's document using G=c=1 are:
------------------------------------
\frac{ds^2}{dt^2} \cdot \frac{d^2r}{ds^2} = \frac{ds^2}{dt^2} \left(-\frac{M}{r^2} + (r-3M)\frac{dt^2}{ds^2}\frac{d\phi^2}{dt^2}\right)
= -\frac{M}{r^2}\frac{ds^2}{dt^2} + (r-3M)\frac{d\phi^2}{dt^2}
= -\frac{M}{r^2}<br />
\alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} <br />
+ (r-2M)\frac{d\phi^2}{dt^2}
------------------------------------
What you are missing is that from the Schwarzschild metric:
\frac{ds^2}{dt^2} = \alpha - \frac{1}{\alpha}\frac{dr^2}{dt^2} -{\color{red} r^2\frac{d\phi^2}{dt^2}}
Using this information I can expand on Espen's work to make it clear that his work in section (57) contains no errors:
------------------------------------
\frac{ds^2}{dt^2} \cdot \frac{d^2r}{ds^2} = \frac{ds^2}{dt^2} \left(-\frac{M}{r^2} + (r-3M)\frac{dt^2}{ds^2}\frac{d\phi^2}{dt^2}\right)
= -\frac{M}{r^2}\frac{ds^2}{dt^2} + (r-3M)\frac{d\phi^2}{dt^2}
= -\frac{M}{r^2}\left(<br />
\alpha - \frac{1}{\alpha}\frac{dr^2}{dt^2} {\color{red} -r^2\frac{d\phi^2}{dt^2}} \right)<br />
+ (r-3M)\frac{d\phi^2}{dt^2}
= -\frac{M}{r^2}<br />
\alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} {\color{red} + M\frac{d\phi^2}{dt^2}}<br />
+ (r-3M)\frac{d\phi^2}{dt^2}
= -\frac{M}{r^2}<br />
\alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} <br />
+ (r-3M {\color{red} +M})\frac{d\phi^2}{dt^2}
= -\frac{M}{r^2}<br />
\alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} <br />
+ (r-2M)\frac{d\phi^2}{dt^2}
------------------------------------
Get it now? Will you now stop telling Espen that (57) is wrong?
So in one day you have posted a elementary calculus blunder, an elementary physics blunder and an elementary algebra blunder and you were trying to tell me I was having a bad day? LOL.
Sorry bout that 
