Find 2D Geometry of Line Element in Coordinates

[steve1763]

TL;DR

Geometry of given line element

i'm trying to find what sort of 2-d geometry this system is in, I've been given the line element

𝑑𝑠2=−sin𝜃cos𝜃sin𝜙cos𝜙[𝑑𝜃2+𝑑𝜙2]+(sin2𝜃sin2𝜙+cos2𝜃cos2𝜙)𝑑𝜃𝑑𝜙
where
0≤𝜙<2𝜋
and
0≤𝜃<𝜋/2

Im just not sure where to start. I've tried converting the coordinates to cartesian to see if it yields a recognisable line element (which didn't work), integrating over theta and phi (but I am not sure how one would recognise the geometry from this), any help would be much appreciated.

Thanks.

 

[jedishrfu]

Look at 3D coordinate systems that use two angles and a radius:
- spherical maybe
- cylindrical no it uses one angle, a radius and a z
- for other orthogonal coordinate systems available see MathWorld link below

Is it non-orthogonal?

https://mathworld.wolfram.com/OrthogonalCoordinateSystem.html

 

[steve1763]

Look at 3D coordinate systems that use two angles and a radius:
- spherical maybe
- cylindrical no it uses one angle, a radius and a z
- for other orthogonal coordinate systems available see MathWorld link below

Is it non-orthogonal?

https://mathworld.wolfram.com/OrthogonalCoordinateSystem.html

I've tried spherical, but (I don't think) the line element given can be converted successfully. Also the its meant to be a 2-D geometry (unless it means a 2-D geometry embedded in 3 dimensions). It also doesn't specify whether or not the coordinate system is orthogonal. I'm at a bit of a loss with it really.

 

[jedishrfu]

So if its 2D with two angles, is it like latitude and longitude on a sphere or something else?

$ds^2=−sin \theta cos \theta sin \phi cos \phi [d\theta^2+d\phi^2]+(sin^2\theta sin^2\phi +cos^2\theta cos^2\phi) d\theta d\phi$

Using Latex, is this the formula?

 

[pervect]

Summary:: Geometry of given line element

i'm trying to find what sort of 2-d geometry this system is in, I've been given the line element

𝑑𝑠2=−sin𝜃cos𝜃sin𝜙cos𝜙[𝑑𝜃2+𝑑𝜙2]+(sin2𝜃sin2𝜙+cos2𝜃cos2𝜙)𝑑𝜃𝑑𝜙
where
0≤𝜙<2𝜋
and
0≤𝜃<𝜋/2

Im just not sure where to start. I've tried converting the coordinates to cartesian to see if it yields a recognisable line element (which didn't work), integrating over theta and phi (but I am not sure how one would recognise the geometry from this), any help would be much appreciated.

Thanks.

The notation is confusing - I am assuming there is a missing exponent symbol ^ before every "2". Latex would be best, but you could also just add the "^" symbol where it's needed.

I did a quick plot with some computer tools, and assuming I've untangled the notation, it looks like the determinant of the metric tensor g is less than or equal to zero in the specified range, so it appears to be Lorentzian.

det(g) = 0 when $\tan\theta \, \tan\phi$ is +1 or -1, for instance when $\phi = -\pi /2 + \theta$ or $\phi = \pi / 2 - \theta$, that only covers the region where $0 < \phi < \pi$.

 

[steve1763]

The notation is confusing - I am assuming there is a missing exponent symbol ^ before every "2". Latex would be best, but you could also just add the "^" symbol where it's needed.

I did a quick plot with some computer tools, and assuming I've untangled the notation, it looks like the determinant of the metric tensor g is less than or equal to zero in the specified range, so it appears to be Lorentzian.

det(g) = 0 when $\tan\theta \, \tan\phi$ is +1 or -1, for instance when $\phi = -\pi /2 + \theta$ or $\phi = \pi / 2 - \theta$, that only covers the region where $0 < \phi < \pi$.

Apologies for the notation. Yes you are right in what I meant. I don't think I recall being taught that the determinant being less than or equal to zero means the metric is lorentzian. Would 'Lorentzian' actually count as the geometry then?

 

[jedishrfu]

What is the context behind your question? Is this the geometry of a black hole or more general than that?

 

[steve1763]

What is the context behind your question? Is this the geometry of a black hole or more general than that?

There isn't much unfortunately. All we were given was the line element and range, the fact that its a two dimensional geometry and told to calculate what geometry that is.

 

[pervect]

It's been a while, but I think much of the relevant info is in wiki's discussion of metric signature, https://en.wikipedia.org/w/index.ph...an_manifold&oldid=973818631#Metric_signatures

Given a metric tensor g on an n-dimensional real manifold, the quadratic form q(x) = g(x, x) associated with the metric tensor applied to each vector of any orthogonal basis produces n real values. By Sylvester's law of inertia, the number of each positive, negative and zero values produced in this manner are invariants of the metric tensor, independent of the choice of orthogonal basis. The signature (p, q, r) of the metric tensor gives these numbers, shown in the same order. A non-degenerate metric tensor has r = 0 and the signature may be denoted (p, q), where p + q = n.

So in essence, what looking at the sign of det(g) does is gives us information on the signature.

For instance, we can observe that -dt^2 + dx^2 has one positive eigenvalue (+1) and one negative eigenvalue (-1), as the eigenvalues of a diagonal matrix are just the elements along the diagonal.

The determinant of a diagonal matrix is just the product of the diagonals.

So the value of the determinant gives us the product of the eigenvalues of the matrix when it's been diagonalized. By Sylvester's law of inertia, the signature is invariant under changes of basis which makes the metric non-diagonal.

You might look to see what your text/course notes say about metric signatures.

https://www.physicsforums.com/threads/how-to-diagonalize-a-warped-metric.405847/
may also be helpful.

It's definitely worth noting that this metric tensor is degenerate, i.e has zero eigenvalues in certain regions where the deteriminant vanishes. Perhaps this is all that was wanted, it's hard to tell from the (lack of) context. Quoting wiki again

In differential geometry, a pseudo-Riemannian manifold,[1][2] also called a semi-Riemannian manifold, is a differentiable manifold with a metric tensor that is everywhere nondegenerate

But this metric tensor is not everywhere nondegenerate.

 

[robphy]

Along the lines of @pervect ’s comment,
it might help to write the metric as a matrix,
then diagonalize it (find its eigenvectors and eigenvalues).

If needed, one can also compute the scalar curvature.

 

[PeroK]

If you try some trig identities, you'll see some sum and difference formulas appear. E.g. $$2\sin \theta \sin \phi = \cos(\theta - \phi) - \cos(\theta + \phi)$$ That gives you the idea to try: $$\theta = \alpha + \beta, \ \phi = \alpha - \beta$$
Assuming that diagonalises the metric, you could relatively easily compute the geodesics, Christoffel symbols and the curvature.

Or, perhaps you could then do a further coordinate transformation to further simplify the metric.

Or, do both and check you get the same answer.

 

[pervect]

Using Rob's technique, we can find an orthnonormal cobasis. Let

$$p = \frac{1}{2} \left( \sin^2 \theta \sin^2 \phi + \cos^2 \theta \cos^2 \phi \right) \quad q = -\sin\theta \sin\phi \cos\theta \cos\phi$$

p>=0 and p-q >= 0

Then an orthonormal cobasis da, db is

$$da = \sqrt{\frac{p+q}{2}}\left(d\theta+d\phi \right) \quad db = \sqrt{\frac{ p-q}{2}} \left( d\theta - d\phi \right)$$

the line element in the orthonormal cobasis is $da^2 - db^2 = q \left( d\theta^2 + d\phi^2 \right) + 2 p \, d\theta d\phi$

This comes from considering the eigenvectors [1,1] and [1,-1] of
$$\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$

 

[pervect]

Also, using a computer algebra GRTensor package, it appears the curvature tensor vanishes. So it's a reparameterization of flat 2d spacetime, ( 1 space + 1 time), though I don't know the mappings between $\theta, \phi$ in the given line element to $t, x$ with $ds^2 = -dt^2 + dx^2$.

Since the metric tensor does vanish for the given metric, there are removable coordinate singularities with the $\theta, \phi$ coordinates.

 

[PeroK]

Also, using a computer algebra GRTensor package, it appears the curvature tensor vanishes. So it's a reparameterization of flat 2d spacetime, ( 1 space + 1 time), though I don't know the mappings between $\theta, \phi$ in the given line element to $t, x$ with $ds^2 = -dt^2 + dx^2$.

Since the metric tensor does vanish for the given metric, there are removable coordinate singularities with the $\theta, \phi$ coordinates.

$$x = \frac 1 2 \sin(\theta + \phi), \ t = \frac 1 2 \sin(\theta - \phi)$$

 

[PeterDonis]

it's a reparameterization of flat 2d spacetime, ( 1 space + 1 time)

Yes, this is obvious from what @pervect showed.

I don't know the mappings between $\theta, \phi$ in the given line element to $t, x$ with $ds^2 = -dt^2 + dx^2$.

In what @pervect showed, his $a$ is $x$ and his $b$ is $t$ (assuming we are using the $- + + +$ metric signature convention, which you appear to be). I think your post #14 is consistent with that.