Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In PADMANABHAN, Gravitation (Foundations and Frontiers), Cambridge, p $304$, exercice $7.6$, an example of the Schwarzschild metric in a different coordinate system is given :

$$\mbox{d}s^2= -c^2\mbox{d}T^2 + \dfrac{4}{9}\left(\frac{9GM}{2(R-cT)}\right)^ {\large \frac{2}{3}} \mbox{d} R^2+\left(\frac{9GM}{2}(R-cT)^2\right)^ {\large \frac{2}{3}} \mbox{d}\Omega^2 \tag{7.52}$$

The second part of the exercise is :

"Consider observers located at fixed spatial coordinates. Show that they are on a free fall trajectory starting with zero velocity at infinite distances and that $T$ represents their proper time. This provides a simple interpretation of this coordinate system"

I am afraid I don't really know how precisely to see this...

share|improve this question
2  
Try looking for a conversion from this coordinate system back into Schwarzschild coordinates. –  Jerry Schirmer Sep 18 '13 at 21:15
1  
If any of our full time relativists could answer this I'd be very interested to see the answer. I'm guessing that T and R are the rain coordinates (en.wikipedia.org/wiki/…) but my attempts to find the appropriate transformations have covered several sheets of paper with much scribble but little information. –  John Rennie Sep 19 '13 at 16:15
    
@JohnRennie: I'll give it a try at lunch. This looked like a textbook problem, so I thought I'd just give Trimok a push along, but if it's actually looking nontrivial, I'm certainly interested in something fun to do. –  Jerry Schirmer Sep 19 '13 at 17:02
1  
@Trimok: you've solved for $t(T,R)$ and $r(T,R)$. the question presupposes that you're at constant $R$. Find the timelike curve that corresponds to this, and then find $v = \frac{\dot r}{\dot t}$ –  Jerry Schirmer Sep 20 '13 at 13:53
1  
@JerrySchirmer : OK. I finally found, when $r \to \infty$ , a dependence of $|v|$ in $r^{-\frac{1}{2}}$, so all seems OK. Thanks. –  Trimok Sep 20 '13 at 16:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.