# How the heck do I integrate this?

1. Nov 9, 2008

### yuiop

Hi,

I am trying to find a solution for this equation in terms of r or t:

$$\frac{d^2r}{dt^2} = \frac{m}{r(r-2m)}\left( \left( \frac{dr}{dt}\right)^2-\left(1-\frac{2m}{R}\right)\right)$$

It seems I should integrate it twice with respect t, but I have no idea how to do that with a dr/dt term in the middle of the right hand side. I need to get rid of all dr and dt variables and obtain an answer that contains just r,t and R variables, if that is possible.

One possible (given) solution that I am not convinced is correct is:

$$t=\frac{R}{2}\sqrt{\frac{R}{2m}}\left(cos^{-1}\left(\frac{2r}{R}-1\right)+2\sqrt{\frac{r(R-r)}{R^2}}\right)$$

but I have no idea how the solution was obtained.

2. Nov 9, 2008

### George Jones

Staff Emeritus
Since the independent variable $t$ doesn't appear, try the substitutions $dr/dt = p$ and

$$\frac{d^2 r}{dt^2} = \frac{dp}{dt} = \frac{dr}{dt} \frac{dp}{dr} = p \frac{dp}{dr}.$$

3. Nov 12, 2008

### yuiop

Hi George,

I am still getting nowhere with this derivation and maybe you could provide some more hints?

The equation that I am looking for the derivation of is actually given as a parametric equation

$$r = \frac{R}{2}(1+cos(\alpha)) , \tau = \frac{R}{2}\sqrt{\frac{R}{2m}}(\alpha+sin(\alpha))$$

on this mathpages page http://www.mathpages.com/rr/s6-04/6-04.htm

and the equation it is derived from appears to be equation (5) on that page.

Any ideas?

Last edited: Nov 12, 2008
4. Nov 12, 2008

### HallsofIvy

Staff Emeritus
Putting aside the complicated coefficients, letting p= dx/dt as George Jones suggested,
the equation becomes pdp/dx= A(p2- B) which is equivalent to
$$\frac{p dp}{Ap^2- B}= dx$$ which is easy to integrate.
Once you have found p= dx/dt, integrate again.

Last edited: Nov 13, 2008
5. Nov 12, 2008

### yuiop

Is this on the right track?

$$p \frac{dp}{dx} = A(p^2-B)$$

$$p^2 = \int{A(p^2-B)}dx = C[x,p]$$

where C[x,p] is some function of x and p.

Solve for p

$$p = D[x] = \frac{dx}{dt}$$ (assume t is shorthand for proper time tau)

$$x = \int{ D[x] } dt = E[x,t]$$

Solve for x or t

$$x = F[t]$$ or $$t = G[x]$$

I also need to know if a indefinite integration is OK for the first integration or if a definite integration is more rigorous. When I try a definite integral

$$\int_{0}^{R} {A(p^2-B)} dx$$ or $$\int_{2m}^{R} {A(p^2-B)} dx$$

the software complains about division by zero errors. Presumably that is singularity problems at x=0 and x=2m?

Should I allow for the constant of integration at the first integration or just add one in at the final integration?

I hope that makes some sort of sense. I am a bit out of my depth here and need all the help I can get from you guys ;)

P.S. I am aware some of these problems can be made a lot simpler if a parametric form can be found before integration. Is the parameter $\alpha$ in the mathpages parametric equation the same as the p = dr/dt introduced by George?

Last edited: Nov 12, 2008