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
$ds^2=\frac{dr^{2}+r^{2}d\theta^{2}}{r^{2}-a^{2}}-\frac{r^{2}dr^{2}}{{(r^{2}-a^{2})}^{2}}$

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 $a,b$