# Homework Help: Algebra/small approx quick Q

1. Apr 5, 2015

### binbagsss

Question :

$u=\frac{E}{J}sin\phi + V$
Sustituting into $(\frac{du}{d\phi})^{2}=\frac{E^{2}}{J^{2}}-u^{2}+2Mu^{3}$, and keeping only linear terms in $V$ attain $cos \phi \frac{dV}{d\phi}=-sin \phi V+\frac{ME^{2}}{J^{2}}sin^{3}\phi$

My working: equalities given up to linear terms in V

$\frac{du}{d\phi}=\frac{E}{J}cos\phi+\frac{dV}{d\phi}$
$(\frac{du}{d\phi})^{2}=\frac{E^{2}}{J^{2}}cos^{2}\phi+\frac{2E}{J}cos\phi\frac{dV}{d\phi}$
$u^{2}=\frac{E^{2}}{J^{2}}sin^{2}\phi+2\frac{E}{J}sin\phi V$
$u^{3}=\frac{E^{3}}{J^{3}}sin^{3}\phi+3\frac{E^{2}}{J^{2}}sin^{2}\phi V$

Subbing in gives $\frac{E^{2}}{J^{2}}cos^{2}\phi+\frac{2E}{J}cos\phi\frac{dV}{d\phi}=\frac{E^{2}}{J^{2}}-\frac{E^{2}}{J^{2}}sin^{2}\phi-2\frac{E}{J}sin\phi+2M\frac{E^{3}}{J^{3}}sin^{3}\phi+6M\frac{E^{2}}{J^{2}}sin^{2}\phi V$

Upon taking the $cos^{2}\phi$ to the other side and cancelling the $E^{2}{J^{2}}$ I have the correct answer but with the addition of the $sin^{2} \phi$ term.

But I can't see a way to justify getting rid of it.
Can anyone see where I've gone wrong?

Thanks very much !

Last edited by a moderator: Apr 6, 2015
2. Apr 6, 2015

### MarcusAgrippa

Check your algebra. You have dropped a factor.

3. Apr 6, 2015

### binbagsss

Really can't spot it :/ when substituting in or the expressions previous to that?

4. Apr 6, 2015

### binbagsss

or should I be keeping $d^{2}V/dphi^{2}$ terms?

5. Apr 6, 2015

### MarcusAgrippa

You have evaluated u^2 correctly, but you have not substituted it correctly. Try it again.

6. Apr 6, 2015

### binbagsss

Oh, typo, I missed a $V$. This doesnt address my question in the OP.

7. Apr 6, 2015

### MarcusAgrippa

Ok. I see. I stopped working through the equation when I encountered your typo and assumed that was the problem. I too get the additional term. From where did you copy this problem? It looks like the equation for orbits in a Schwarzschild geometry.

8. Apr 6, 2015

### binbagsss

Indeed. Introduction to General Relativity L.P Hughston and K.P Tod.

9. Apr 6, 2015

### MarcusAgrippa

The only thing that comes to mind, perhaps, is that M is small (obvious from the fact that the u^3 term is known to be a small correction term to the Newtonian equation, obtained by ignoring it). That means MV is a second order term and is therefore negligible.

Almost every textbook has a very different method for solving this equation. Most are content to convert the equation to second order and then to use an iterative method (usually Picard's method) to find the solution. The only other text I know that solves this equation by a method similar to your book is by Foster and Nightingale, A short course in GR, p 148-150. Give or take a few changes of notation, their solution appears to be identical to yours. However, I have not worked through it in sufficient detail to be able to say with certainty that the methods coincide.

Last edited: Apr 6, 2015
10. Apr 7, 2015

### binbagsss

I see, we neglect $2mu^{3}$ to attain the Newtonian equation. But, this is relative to the magnitude of $u^{3}$ to $M$ . $M$ is the mass of the Sun in the solar system which is of the order $10^{30}kg$.

Figures on the previous page for the orbit of Earth around the sun, considering time-like geodesics are :

$2Mu$ ~ $10^{-8}$
$J^{2}u^{2}$~$10^{-8}$
$2MJ^{2}u^{3}$~$10^{-16}$

But there's no figures stated for $E$ terms.

An arguement that would work equally well is that $\frac{E^{2}}{J^{2}}$ is small compared to $E^{3}/J^{3}$, i.e. $E>J$?

Last edited: Apr 7, 2015
11. Apr 7, 2015

### MarcusAgrippa

I think 2M may the Schwarzschild radius, not the solar mass. It appears in the metric in the factor $\left( 1 -\frac{2M}{r} \right)$. It must therefore have the dimensions of a distance. I think it abbreviates the quantity $\frac{2GM_\odot}{ c^2}$, which is of the order of 3 km for the Sun.

Also, $u = 1/r$. For grazing incidence with the Sun, $r \approx 700\ 000$km, so $u \approx 10^{-6} km^{-1}$. These really are small numbers.

Approximate solution (18.16) shows that u is of the same order of magnitude as E/J, and solution (18.18) shows that V is of the order of Mu^2. So my conjecture was right: MV is of the order of V^2.

To see this in advance, rather than retrospectively, note that the approximate solution (18.16) shows that u is of the order of E/J. Further, we expect the correction term V to this solution to be very small since, in equation (18.15), the term 2Mu^3 = (2Mu) u^2, which is a term much smaller than the u^2 term because 2Mu is of the order of 10^{-8}. So, when constructing the second approximation to the solution u by putting u = (E/J) sin \phi + V, we expect V to be a small correction of size about 2Mu which is about 2M J/E. So, when substituting (18.17) into (18.15), we can neglect the term containing MV since it is of the order V^2. That then gives you the answer that you sought.

Last edited: Apr 7, 2015