Null curve coordinate system

In summary, the conversation was about difficulties in solving a problem related to 2-dimensional Riemannian manifolds and conformal flatness. The problem involved using null curves as coordinate curves and making a coordinate transformation to prove the reduction of the line element. The person also had a conceptual question about measuring positions in a coordinate system composed of null curves. The expert summarizer explains that in most commonly used coordinate systems, coordinates are measured by distances from the origin along coordinate curves, but in a flat, null coordinate system, the coordinates indicate a multiple of the basis vector that points along the axis, and lengths are not additive in non-Riemannian manifolds.
  • #1
Antarres
200
99
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.

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 = edλdν
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 e (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'abab.
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:
,cν ,d gcd01

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 :)
 
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  • #2
Antarres said:
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
In most coordinate systems that are commonly used, yes, but not in all. Consider polar coordinates ##(r,\theta)##. The second coordinate does not measure distance from the origin, as the origin has no ##\theta## coordinate.

In a flat, null coordinate system the coordinates in each axis direction will indicate a multiple of the basis vector that points along the axis. Consider ##\mathbb R^2## with metric ##\pmatrix{1&0\\0&1}## and let coordinate system C' be produced from basis vectors ##\vec e_1=\pmatrix{1\\1},\vec e_2=\pmatrix{1\\-1}##. Then the point ##\vec v## with coords (1,0) in the usual system has coords ##(1/\sqrt2,1/\sqrt2)## in system C'. We can write ##\vec v=\vec e_1 + \vec e_2## and note that, even though the vecs ##\vec e_1, \vec e_2## have zero length, their sum does not. That is because lengths are not additive and, more generally, the triangle inequality does not hold, in a non-Riemannian manifold.
 
  • #3
I guess you meant -1 as the first coordinate of the second base vector of system C'. And yeah, thanks for the reply, I figure now that by changing to other coordinate system, I change the metric, so null curves won't stay null curves if I take them as axes, I suppose. I'm still confused about solving the exercise though. That might be more of a technical problem, but some pointers would be appreciated. Thanks :)

Edit: Actually -1 there wouldn't work as those vectors wouldn't be independent but I get what you mean, so that second question is probably solved for me, I'll look into it a bit more(transforming into light cone coordinates which I haven't tried), but yeah, the main problem still stands.
 
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  • #4
Antarres said:
Actually -1 there wouldn't work as those vectors wouldn't be independent
That's right. The basis vectors I gave are a clockwise 45 degree rotation of the usual basis vectors.

Now, looking back at the OP, I can't make sense of the statement:
Antarres said:
every 2-dimensional Riemannian manifold is conformally flat in the case of a metric of signature 0
If the manifold has a metric of signature zero it is not Riemannian, as a condition for a manifold being Riemannian is that its metric be positive definite. It would have to be pseudo-Riemannian.
 
  • #5
Oh, that is true sorry. In the GR book I'm reading, from which this exercise is, Riemannian manifold is any manifold with a metric, the author doesn't distinct positive metrics from others. But yeah, it would be pseudo-Riemannian actually since the metric can be both positive, negative or indefinite. Just a mix of terminology, sorry.

Edit: Oh and a bit of clarification. Null curves exist only for indefinite metrics. So in a standard euclidean space with standard metric we don't have null curves. However if we take a 2d Minkowski space with diag(+1 -1) metric, null curves would be defined as light rays, so those vectors you pointed out would be on these curves. However, they would have norm zero in this system which was the point of my second question. Null curves are defined as curves on which tangent vector has squared norm zero in each point along the curve.
 
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  • #6
Alright, after some time I managed to solve this problem, so I will post my solution for everyone to check, and for some people in the future, if they need help with this exercise, since I couldn't find anything relevant to help me with it when I was doing it.

We will first get the form of our transformed metric, that is, the one in null coordinates. Using the equations for null curves from the hint to the exercise, we get:

$$g'^{00} = \frac{\partial\lambda^0}{\partial x^a}\frac{\partial\lambda^0}{\partial x^b}g^{ab} = 0$$
$$g'^{11} = \frac{\partial\lambda^1}{\partial x^a}\frac{\partial\lambda^1}{\partial x^b}g^{ab} = 0$$

So now, we can find the form of covariant metric by taking inverse of contravariant metric(just 2x2 matrix inverse), and we find that it also must have zeros on the main diagonal, while on off-diagonal it has two same elements(metric is a symmetric tensor) ##g'_{01}##. From this we can find the transformed line element:

$$(ds)^2 = g'_{ab}d\lambda^a\lambda^b = 2g'_{01}d\lambda^0d\lambda^1$$

Now we would like this element ##g'_{01}## to be positive, so that we can give it exponential form that was hinted in the exercise hint. But, it doesn't seem like it would always have to be positive. Therefore, we do a little sort of geometric analysis.
When we draw the null curves in our coordinate system, we can deduce that they cut the space in four areas(we will observe the problem locally, so our null curves don't intersect more than once and so on), and those four areas are precisely defined as spacelike or timelike, that is, two of them are spacelike and two of them are timelike. Now, our line element is invariant under coordinate transformations, so areas which are timelike/spacelike, will remain timelike/spacelike in the new, null curve coordinate system. Now, since our new metric is off-diagonal, we can see that the sign of our line element depends on the off-diagonal metric element, as well as the directions of our null curve axes. But since sign of line element is fixed under coordinate transformations, and directions along the axes can be chosen arbitrarily, we can choose our axes such that their positive directions enclose a ##s>0## area in this coordinate system. That way we directly observe that our off-diagonal metric element is positive, and can therefore be transformed into an exponential form. Now we have:
$$(ds)^2=e^{2\mu}d\lambda^0d\lambda^1$$
where : ##\mu = \frac{1}{2}\ln{2g'_{01}}##.

Then the last transformation: ##\lambda^0 \rightarrow \lambda'^0 = (\lambda^0 + \lambda^1)/2##, ##\lambda^1 \rightarrow \lambda'^1 = (\lambda^0 - \lambda^1)/2## transforms our line element to:
$$(ds)^2 = e^{2\mu}((d\lambda'^0)^2 - (d\lambda'^1)^2)$$
And we clearly see from this form that our metric is conformally flat, which was what was required of us to prove.
 

What is a Null Curve Coordinate System?

A Null Curve Coordinate System is a mathematical coordinate system used to describe the geometry of a curved space. It is based on a set of null curves, which are curves that have a tangent vector with zero length at every point.

How is a Null Curve Coordinate System different from other coordinate systems?

A Null Curve Coordinate System differs from other coordinate systems, such as Cartesian or polar coordinates, in that it is specifically designed to describe the geometry of a curved space. It takes into account the curvature of the space and uses null curves, rather than straight lines or circles, as its basis.

What are some applications of Null Curve Coordinate Systems?

Null Curve Coordinate Systems have various applications in the field of general relativity, where they are used to describe the geometry of spacetime. They are also used in other areas of theoretical physics, such as string theory and quantum gravity.

How are Null Curve Coordinate Systems related to the theory of relativity?

Null Curve Coordinate Systems are closely related to the theory of relativity, as they are used to describe the geometry of spacetime, which is a central concept in relativity. In fact, null curves are used to define the light cones that play a crucial role in the theory of relativity.

Are there limitations to using Null Curve Coordinate Systems?

While Null Curve Coordinate Systems are useful in describing the geometry of curved spaces, they are not always the best choice for practical applications, such as calculating distances or angles. In these cases, other coordinate systems may be more suitable and easier to work with.

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