Why a sphere doesn't necessarily have a center in curved space?

I guess it depends on what you mean by a sphere. You can take a point and draw straight lines out from it all of the same proper distance, but the surface formed by the ends of those lines won't in general look spherical. If you take a surface that looks spherical then in general there is no point where straight lines from that point to the surface all have the same proper length.

This is the context in which it was written in Wald.

@JohnRennie Is this what he was meaning?

"r does not bear any relation to the distance to the center"---This I know via the example of Schwarzschild solution.

But the object still have a "center".

@JohnRennie So maybe he is meaning this...Is that it?

@ManasDogra Read it again, he even gives an explicit example: If your spacetime is like $\mathbb{R}\times S^2$ (this is just $\mathbb{R}^3$ with the origin missing), then there is no center to the $S^2$ that would be part of the spacetime - the notion of its center only makes sense when you embed a sphere $S^2$ in an $\mathbb{R}^3$, and Wald's point is that in a general manifold there's no reason to assume all spheres are embedded like that

@ManasDogra And actually, in exactly the same sense the Schwarzschild solution has no center - the singularity is not part of manifold in most treatments

If you want an example pick the 2D case

The cylinder has circles without a center

This generalized easily for $\mathbb{R} \times S^n$

@Slereah What is the meaning of that? I was thinking like---The surface of the cylinder doesn't have the circle's center. But then the circle alone also does not contain the circle's center?

Well yes, it is also an example

The Schwarzschild metric is an example of such a case too

Maximally extended anyway

The sphere of the event horizon doesn't have a center, it just continues all the way into the event horizon on the other side

Hm, is it a good example actually

It does have that topology but I don't know if the SO(3) of the black hole doesn't have a center

Wormholes are better examples

Wald says that in a curved space a sphere need not have a center...In flat space also the sphere itself does not have the center. The center simply does not lie on the sphere...So what's special about the curved case?

In Euclidian space for instance, a sphere has a center that is part of the space itself

But this isn't true for general spaces

The inside of a sphere $S^n$ simply does not exist

There isn't a space inside the sphere that you could point to

@Slereah Oh I get this part now maybe,Wald is not saying that the center is on the sphere...rather it is in the embedding space $\mathbb{R}^3$

Yes

The center is never on the sphere, unless it has radius zero

Just by the definition of a sphere

Now I get the para and ACuriousMind's answer completely except one part

How is $\mathbb{R}\times S^1$ isomorphic to $\mathbb{R^2}$ without the origin...I am visualizing it to be taking the Euclidean plane and identifying two points along say x-axis, then I am getting a cylinder...How is the origin of the plane getting removed?

Or is some other kind of construction involved?

That's just polar coordinates

The polar coordinates are a product of R and S, for r > 0

Also on a more intuitive level, you can "stretch" a plane with a hole into a cylinder

@Slereah How to stretch?

I mean it might be a bit difficult to explain in words but still :)

@ManasDogra $S^2$ is a sphere. If we remove the origin, then every point in $\mathbb{R}^3 - \{0\}$ lies on such a sphere of some radius, and for every radius $r\in(0,\infty)$ there are points, so $\mathbb{R}^3 - \{0\}\cong (0,\infty) \times S^2$

If you want an actual function, take the cylinder (0,1) x S, and make a function that sends the bottom circle to a point, while the top circle is sent to infinity

now $(0,\infty)\cong \mathbb{R}$ and we're done

@ACuriousMind Oh that was so simple!

@Slereah Great...Thanks everyone

@Slereah I remember you mentioned GR can be formulated without using the metric tensor. What replaces it, though? Like in Einstein's FE

The typical formalism sans metric uses frame fields

Oh here it comes again. So there is an underlying metric but everything is formulated in terms of ff

The frame field is a map from the tangent bundle to $\mathbb{R}^n$, roughly

But $\mathbb{R}^n$ comes equipped with a canonical Minkowski metric

Hence you can do your GRin' there

physicists usually call the formulation in terms of frames the tetrad formalism

or the vielbein, if you're German

Oh yes, that was the thing I couldn't quite understand some days ago

The version with diads (or dyads, I'm not sure about the spelling)

In Rovelli's book there was this definition of Frame Field. The meaning of such object is quite clear to me but the definition of "local cartesian coordinates" used there seems handwavy

I'm afraid that is very much what it is

A frame is the germ of an orthonormal coordinate system, basically

If you have a frame $e$ at a point $p$, then there is a coordinate system $x$ centered on $p$ such that the metric is diagonal at $p$

such that the holonomic coordinates generate the frame

So you chose a random orthonormal basis on each tangent space and you orthogonally project (what does this mean without the metric)?

but because we're in curved space, the diagonality of the metric generally doesn't extend beyond that point

@Feynman_00 You do indeed!

The fact that it's a random orthonormal basis is actually very important

Because that is what the gauge freedom is

:O :O

This made my day :P

You can always rotate those frames (as long as you do it smoothly at every point), and this will still be a valid frame field

and in fact the same one, physically speaking

Oh, I see, now things are clear. The "projection" part was just a way to talk about going from this basis to the one induced by the map

The frame at a point gives you the components, basically

If you have a frame $e$ and a vector $v$, then you have that $e(v) = v^i$, a set of real numbers

Yes, the $e^i_j$ are basically transition functions on $TM$

the $e^a_\mu$ are a slightly different thing, since they map the general frames to the orthonormal frames

Frames aren't orthonormal in general, they could be any size and at any angle

this is what you generally get from random coordinates

I see what you're saying

Having a set of orthonormal frame gives you informations on 1) size (since every vector has length 1) 2) angle (since they are all orthogonal)

That's roughly why they are equivalent to the metric tensor

Isn't there a chart with the coordinates that induce such an orthonormal basis though?

Making them particular transition functions

you can always find coordinates which are exactly a frame at one point, but usually not beyond

If you can find a coordinate system that corresponds to the frame field at every point, then that's what is called an integrable frame

But frames are integrable specifically when the spacetime is flat

Integrable in the sense of Frobenius theorem, right?

Now I see where the word "flat chart" is coming from

this also relates to like normal coordinates

Anyhow, now that I know the basis is a random orthonormal basis on the tangent space the whole picture seems clear. The "orthogonal projection on the manifold" was a weird way to say project on the basis induced by $\varphi=(x^1...x^n)$. Thanks for helping, this was very useful @Slereah

You may have seen that normal coordinates look nice up til second order, after which they depend on the curvature

that's what prevents the integrability

@Feynman_00 Well, beware though

The problem of the frame field is that it's not holonomic frames

In general you can't have such coordinates

You just have germs of the coordinates

I might have studied this in DG when I learnt the condition under which a set of vector fields which are a basis at each point can be written as the canonical derivations fields

I.e. having all of them commuting (that should be the second order condition you mentioned above)

And this also makes me understand (and appreciate) why I had to study those things

Physicists do use frames a lot, they just don't do it explicitely :p

This is why some good Math makes things much easier

Studying differential geoemtry was basically having flashbacks of the past three years and saying "Aah, you were at my side, all along."

Only this time without a moonlight sword :P

orthonormal frames in a physical sense is pretty much what it means for GR to be "locally euclidian"

at any given point you can approximate Euclidian space arbitrarily close for a small enough neighbourhood

Up to some O(x²) terms

Wait, do you mean euclidean or lorentzian?

Lorentzian

But people usually say "Euclidian" as a catch-all

Nice tip

why am i not satisfied with the derivation of wave velocity from a mechanical wave on a string?

what if the tension force is non existent.. i've made waves on a string without a fixed end and weight hanging off the end before.

or does tension exist regardless of added tension?

Physics needs to be global ... I've noticed hbar is always silent around this time :(