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So, I've been studying some tensor calculus for general theory of relativity, and I was reading d'Inverno's book, so out of all exercises in this area(which I all solved), this 6.30. exercise is causing quite some problems, so far. Moreover, I couldn't find anything relevant on the internet that could help me with it. I think there was even a question on these forums that was quite old, like 7 years, but OP was pretty clueless in his attempts(I could clearly see he wasn't on a right path), and there were no replies to him, so I hope I'll arrive at some concrete hints that would help me solve it, since I don't think the problem is too complicated, it's probably just me missing some details.
6.30. Prove the theorem that says that every 2-dimensional Riemannian manifold is conformally flat in the case of a metric of signature 0(diag. form is (+1 -1)).[Hint: use null curves as coordinate curves, that is, change to new coordinates:
λ=λ(x0, x1) ν=ν(x0, x1)
satisfying
gabλ ,aλ ,b = 0 = gabν ,aν ,b
and show that line element reduces to the form:
ds2 = e2μdλdν
and finally introduce new coordinates ½(λ+ν) and ½(λ-ν)].
Stardard tensor transformation laws.
λ ,a ≡ ∂λ/∂xa
[/B]
Alright, so, it is pretty obvious that if the reduction of the line element is proved, the last coordinate transformation will make our metric conformably flat with the factor e2μ (or so I think, since differentials would decouple to two parts which would indicate (+1 -1) diagonal metric tensor).
So I assumed that I have transformed into those new, null coordinates, and my line element looks like:
ds2 = g'abdλadλb.
Here I marked λ0=λ, λ1=ν.
So now I figure the idea is to use tensor transformation laws to the g' tensor and then use the null curve equations to eliminate dλ2 and dν2 terms in the expansion of the sum, leaving only mixed differential term, which could then maybe be proved to be positive(although I have no idea how), so that it would be equivalent to that exponential. In order to use the null curve equations from the hint, I lifted indexes on g' and lowered indexes on differentials, and then expanded the sum. Indeed, the non-mixed terms turn to be zero, but the mixed term is:
2λ ,cν ,d gcddλ0dλ1
Now I'm not even sure that this differential product is the same as dλdν since it has lowered indexes, but lifting indexes here would involved the primed metric again so I feel like there's no progress that way. And I'm not sure if I can prove the term next to the differential could be that exponential(that is, prove that it's a positive). So some hints and advices here would be very welcome if anyone has any idea.
Also, I have another question, more of a conceptual nature I guess, not directly related to the problem. And that is, if I have a coordinate system composed of null curves, then how do I measure positions in this coordinate system, if every line element along my coordinate curve is zero? It kinda looks hard to imagine to me. Cause in every coordinate system we have, we measure positions by looking at coordinates, which are measured by distances from the origin along coordinate curves(and that would be always zero in this case?). Thank you guys :)
Homework Statement
6.30. Prove the theorem that says that every 2-dimensional Riemannian manifold is conformally flat in the case of a metric of signature 0(diag. form is (+1 -1)).[Hint: use null curves as coordinate curves, that is, change to new coordinates:
λ=λ(x0, x1) ν=ν(x0, x1)
satisfying
gabλ ,aλ ,b = 0 = gabν ,aν ,b
and show that line element reduces to the form:
ds2 = e2μdλdν
and finally introduce new coordinates ½(λ+ν) and ½(λ-ν)].
Homework Equations
Stardard tensor transformation laws.
λ ,a ≡ ∂λ/∂xa
The Attempt at a Solution
[/B]
Alright, so, it is pretty obvious that if the reduction of the line element is proved, the last coordinate transformation will make our metric conformably flat with the factor e2μ (or so I think, since differentials would decouple to two parts which would indicate (+1 -1) diagonal metric tensor).
So I assumed that I have transformed into those new, null coordinates, and my line element looks like:
ds2 = g'abdλadλb.
Here I marked λ0=λ, λ1=ν.
So now I figure the idea is to use tensor transformation laws to the g' tensor and then use the null curve equations to eliminate dλ2 and dν2 terms in the expansion of the sum, leaving only mixed differential term, which could then maybe be proved to be positive(although I have no idea how), so that it would be equivalent to that exponential. In order to use the null curve equations from the hint, I lifted indexes on g' and lowered indexes on differentials, and then expanded the sum. Indeed, the non-mixed terms turn to be zero, but the mixed term is:
2λ ,cν ,d gcddλ0dλ1
Now I'm not even sure that this differential product is the same as dλdν since it has lowered indexes, but lifting indexes here would involved the primed metric again so I feel like there's no progress that way. And I'm not sure if I can prove the term next to the differential could be that exponential(that is, prove that it's a positive). So some hints and advices here would be very welcome if anyone has any idea.
Also, I have another question, more of a conceptual nature I guess, not directly related to the problem. And that is, if I have a coordinate system composed of null curves, then how do I measure positions in this coordinate system, if every line element along my coordinate curve is zero? It kinda looks hard to imagine to me. Cause in every coordinate system we have, we measure positions by looking at coordinates, which are measured by distances from the origin along coordinate curves(and that would be always zero in this case?). Thank you guys :)