Can Metrics Reveal Flat Spaces Without Computing the Riemann Tensor?

[Morgoth]

If you are given a metric g_αβ and you are asked to find if it describes a flat space, is there any way to answer it without calculating the Riemman Tensor R^λ_μνσ?
and how can I find for that given metric the coordinate transformation which brings it in conformal form?

For example I'll give you the way I'm thinking on my problem.
The metric I'm given is in :
ds²= exp(ax+by) (-dx²+dy²)

If I want to show that the space is flat, I would compute the Riemman tensor and so it vanishes... (is there a faster way to do it?).
Then in order to find the transformation that would bring it in obvious conformal form, I would say that I want to write:
ds²=exp(w(x',y')) (dx'²+dy'²)

(translated to g_αβ=λ² n_αβ)
Is that a correct approach?
In the way that will it give me the x'(x,y) and y'(x,y) form?

 

[Ben Niehoff]

You can't, in general, bring a metric to conformal form unless it is a 2-dimensional metric.

Generally speaking, a metric is (locally) conformally flat if and only if the Weyl tensor vanishes vanishes. In 2d and 3d, however, the Weyl tensor vanishes identically, so they are exceptions. In 2d, every metric is (locally) conformally flat. In 3d, a metric is (locally) conformally flat if and only if the Cotton tensor vanishes.

Also, in general, a metric is flat if and only if the Riemann tensor vanishes. So your options are either A) Get lucky and notice that there is a coordinate transformation that puts a metric in an obviously-flat form, or B) Calculate the Riemann tensor.

But if you're working in 2 dimensions, it is really easy to calculate the Riemann tensor.

 

[bcrowell]

Did you mean to switch the signature from -+ to ++? A change of coordinates can't change the signature.

 

[Morgoth]

oops...yes it can't...that was my mistake...
still =_= it's obviously in the simplest form

 

[kevinferreira]

In addition to what Ben Niehoff wisely said, I have to say that the Riemann tensor does not always tell you if the space is flat: there are coordinate systems in flat space where the Riemann tensor does not vanish. This is because it is not an invariant, as its Lorentz indices tell. On the other hand the curvature scalar (trace of the Ricci) is a scalar, so it really tells you about the curvature.

There is another theorem that works in any dimension, called Bach's theorem. If the Bach tensor vanishes, then the space is conformaly flat. The Bach tensor is roughly the curl of the Ricci, you should check it.

 

[Nugatory]

I have to say that the Riemann tensor does not always tell you if the space is flat: there are coordinate systems in flat space where the Riemann tensor does not vanish.

Could I ask you for an example?

 

[TrickyDicky]

I have to say that the Riemann tensor does not always tell you if the space is flat: there are coordinate systems in flat space where the Riemann tensor does not vanish. This is because it is not an invariant, as its Lorentz indices tell.

If you really meant this as stated you have an important confusion here. Curvature is coordinate-independent and the Riemann tensor always tells you if there is Riemann curvature.

 

[Morgoth]

The Riemann tensor is a tensor...Which means that if it vanishes in a Coordinate System, it will vanish in any others after a coord.transformation...

But still what is the most obvious conformal form for a metric?
I've seen that it needs to be written in the form:
ds²= Φ(x,y) {dx²+dy²}
of course now I cannot change the sign, I had forgotten of that theorem...
So the needed form should just be:
ds²= Φ(x,y) {-dx²+dy²}

-_- however, it's already in that form...with Φ(x,y)=exp(ax+by)
if I say the transformation:
u= ax
v= by

du²=a²dx²
dv²=b²dy²

so
ds'²= exp{u+v} (- (du/a)² + (dv/b)² )

is that of any better?

 

[Dale]

I have to say that the Riemann tensor does not always tell you if the space is flat: there are coordinate systems in flat space where the Riemann tensor does not vanish.

I also think this is wrong. Please provide an example, or preferably a reference.

 

[WannabeNewton]

Please provide an example, or preferably a reference.

There is no example unfortunately. If the Riemann tensor vanishes identically in one coordinate system it will vanish in all coordinate systems by virtue of it being a tensor.