Line element under coordinate transformation to get polar form

[cedricyu803]

Homework Statement

Hello Guys, I am reading Hobson's General Relativity and I have come across an exercise problem, part of which frustrates me:
3.20 (P. 91)
In the 2-space with line element
ds^2=\frac{dr^{2}+r^{2}d\theta^{2}}{r^{2}-a^{2}}-\frac{r^{2}dr^{2}}{{(r^{2}-a^{2})}^{2}}<br />

and given r{\frac{d\theta}{dr}}=tan\phi
show that the space is mapped to a Euclidean plane in which (r, phi) are taken as polar coordinates.

Homework Equations

The Attempt at a Solution

So I attempted to express d\theta as a l.c. of dr and dphi, but I don't know how to handle the \frac{d\theta}{dr} the given relation r{\frac{d\theta}{dr}}=tan\phi

to express the d\theta in given line element in terms of dphi and dr

Thanks for any help =]

 

[dpopchev]

I can't finished the problem too.

We can write the matrix representation of the metric:
g_{\mu\nu} = -\dfrac{a^2}{(r^2-a^2)^2} dr^2 + \dfrac{r^2}{r^2-a^2} d\theta^2

From which we can write the matrix representation and the inverse and use

\Gamma^a_{bc} = \dfrac{1}{2}g^{af}...

to obtain the following connections:

\Gamma^r_{rr}=-\dfrac{2r}{r^2-a^2} \; \Gamma^r_{\theta\theta}=-r
\Gamma^\theta_{r\theta}=\Gamma^\theta_{\theta r}=-\dfrac{a^2}{r(r^2-a^2)}

When I go to geodesic eq: \dfrac{d^2 x^a}{d\lambda^2} + \Gamma^a_{\alpha\beta} \dfrac{dx^\alpha}{d\lambda} \dfrac{dx^\beta}{d\lambda}=0

and substitute I can't find a way to obtain the equation wanted in the problem:
a^2 \left(\dfrac{dr}{d\theta}\right)^2 + a^2r^2 = Kr^4
where K is a constant such if K = 1 then geodesic is null...

Thanks in advance.

 

[dpopchev]

Sorry, forgot to post the second part of the answer.

We can see that the equation a^2 \left(\dfrac{d^2r}{d\theta^2}\right)^2 + \ldots is a line in polar coordinates by substituting directly with \dfrac{dr}{d\theta}=cotg\phi.

We obtain something like cotg\phi + 1 = K\left(\dfrac{r}{a}\right)^2

If we write the equation for line in Cartesian y = ax + b with a,b parameters and make the coordinate change to polar coordinates we get \dfrac{b}{r} = 1 - acotg\phi

We can see the two equations have the same form to error of free parameters a,b