Positive curvature of a 2D surface

[Apashanka]

If considering a 2D surface (plane) having polar coordinate r,θ (where r is the distance from the origin and θ is the anticlockwise angle from the base line as usual)
The metric is now actually ds²=dr²+r²dθ²
If now this 2D surface is given a positive curvature of +1 (equivalent to the surface of a sphere) in which the the azimuthal angle is θ and the radial coordinate r is measured from the north pole of the sphere to the point on the surface along a great circle and R is the radius of the sphere.
In that case the metric of a 2D surface having a positive curvature becomes
ds²=dr²+R²sin²(r/R)dθ²
and for large radius of curvature R>>r the metric matches with the metric of a plane 2D surface(e.g without curvature).
Hence for a plane 2D surface being given a positive curvature of +1 it's actually equivalent to the curved surface of a 3D sphere.
My question is if a 3D surface (say for exm sphere) is given a positive or negative curvature then whether it's metric will be in some 4D minkowski space??
Or equivalent to some 3d metric in 4D minkowski space as in the case of a 2D plane.
Thanks

 

[PeterDonis]

for a plane 2D surface being given a positive curvature of +1 it's actually equivalent to the curved surface of a 3D sphere.

More precisely, the boundary of a 3-ball in 3D space will be a 2-sphere, the "2-D surface with positive curvature +1". However, that does not mean that every 2-sphere is the boundary of a 3-ball in some 3D space. Nothing requires that.

if a 3D surface (say for exm sphere) is given a positive or negative curvature then whether it's metric will be in some 4D minkowski space??

No. What will be true is that the boundary of a 4-ball in 4D space (not spacetime, space--i.e., with a Riemannian metric, not a Minkowski metric) will be a 3-sphere, a "3-D surface with positive curvature +1". But, just as above, that does not mean that every 3-sphere is the boundary of a 4-ball in some 4D space. A spacelike surface in a 4D spacetime can be a 3-sphere without being the boundary of anything in a 4D space.

 

[Apashanka]

A spacelike surface in a 4D spacetime can be a 3-sphere without being the boundary of anything in a 4D space.

Will you please explain why space like
surface ,for which the two events can be connected earlier than compared to light (since v>c) why not time like or light like surface??

 

[PeterDonis]

Will you please explain why space like
surface

Because that's the only kind of 3-surface in spacetime that is a "space" in the usual sense of that term (having a positive definite metric).

 

[PeterDonis]

will you please explain what actually positive definite metric

A positive definite metric means that the squared interval $ds^2$ between two distinct points is always positive. For example, the metric of a 2D surface that you gave in post #1 is positive definite. So is the metric of ordinary 3-space that you get if you add one more dimension. But the metric of 4D spacetime is not positive definite; squared intervals between distinct points can be positive, negative, or zero.

 

[Apashanka]

Because that's the only kind of 3-surface in spacetime that is a "space" in the usual sense of that term (having a positive definite metric).

For space like surface (hypersurface) e.g at any definite tim

A positive definite metric means that the squared interval $ds^2$ between two distinct points is always positive. For example, the metric of a 2D surface that you gave in post #1 is positive definite. So is the metric of ordinary 3-space that you get if you add one more dimension. But the metric of 4D spacetime is not positive definite; squared intervals between distinct points can be positive, negative, or zero.

For space like surface(hypersurface) e.g at any definite time t,square of the interval between two points on that space like surface is negative according to r^μ={ct,vec(r)} ,and that surface is actually a 3-dim hypersurface in 4-D space (since three coordinates are required say x,y,z) to define a point on that hypersurface of definite time t.
And according to your post 3 the space like hypersurface is actually a 3 sphere(3-D space with curvature +1)
If that's the case then why not time like hypersurface ??

Am I right??

 

[PeterDonis]

For space like surface(hypersurface) e.g at any definite time t,square of the interval between two points on that space like surface is negative

That depends on which sign convention you use. The convention you describe is called the "timelike convention", where timelike squared intervals are positive and spacelike ones are negative. It is more common in introductory textbooks and sources that are mainly concerned with special relativity (i.e., flat spacetime). The other convention, which I was implicitly using, is called the "spacelike convention", where spacelike squared intervals are positive and timelike ones are negative. This convention is more common in more advanced textbooks and sources that are mainly concerned with general relativity (i.e., curved spacetime).

Am I right??

No. See above.

 

[Apashanka]

That depends on which sign convention you use. The convention you describe is called the "timelike convention", where timelike squared intervals are positive and spacelike ones are negative. It is more common in introductory textbooks and sources that are mainly concerned with special relativity (i.e., flat spacetime). The other convention, which I was implicitly using, is called the "spacelike convention", where spacelike squared intervals are positive and timelike ones are negative. This convention is more common in more advanced textbooks and sources that are mainly concerned with general relativity (i.e., curved spacetime).
No. See above.

A spacelike surface in a 4D spacetime can be a 3-sphere without being the boundary of anything in a 4D space

One thing I want to clarify ,whether the hypersurface in 4-D minkowski space corresponding to space like intervals is the 3-sphere.
And why is time like surface not taken since it's interval squared is -ve and doesn't make sense??

 

[PeterDonis]

whether the hypersurface in 4-D minkowski space corresponding to space like intervals is the 3-sphere.

No, it's ordinary Euclidean 3-space.

why is time like surface not taken since it's interval squared is -ve and doesn't make sense??

I don't understand what you're asking.

 

[Apashanka]

No, it's ordinary Euclidean 3-space.

So sir what's the difference in 3-dim sphere (without curvature) and 3-sphere(3-D surface with curvature +1) ,if both are in ordinary euclidean 3-space.

 

[PeterDonis]

what's the difference in 3-dim sphere (without curvature) and 3-sphere(3-D surface with curvature +1) ,if both are in ordinary euclidean 3-space.

A 3-dimensional sphere is just a subset of ordinary Euclidean 3-space. The correct term for it is actually a 3-ball.

A 3-sphere cannot exist in ordinary Euclidean 3-space; ordinary Euclidean 3-space is a different 3-manifold from a 3-sphere.

 

[Apashanka]

Metric of plane 2-D surface is
ds²=dr²+r²dθ².
For positively curved 2-D space
Metric:ds²=dr²+R²sin²(r/R)dθ²
Where R(radius of curvature).
Hence for the term for positive curvature is S_k(r)=Rsin(r/R) where in 2-D case the curvature term is added to the θ coordinate not r coordinate.
Similarly if for a 3-D euclidean flat space the metric is dr²+r²dθ²+r²sin²θdΦ².
If this space is positively curved then what will be the metric and where would the curvature term gets added and what does the term R signify here.??
Thank you

 

[PeterDonis]

@Apashanka I moved your post in a new thread into this thread, since it's really part of the same topic.

If this space is positively curved then what will be the metric

If the curvature is constant (which you implicitly assumed it was for the 2-D case), it will be the metric of a 3-sphere, which is easy to find online.

 

[Apashanka]

@Apashanka I moved your post in a new thread into this thread, since it's really part of the same topic.
If the curvature is constant (which you implicitly assumed it was for the 2-D case), it will be the metric of a 3-sphere, which is easy to find online.

I have made a schematic diagram and calculated the metric of a positively curved 2-D surface.
Pardon for image quality.

Using this I have calculated the metric (distance between A and B )
ds²=dr²+S_k²(r)dθ²
Where S_k(r)=Rsin(r/R)
The metric for 3-D positively curved space is given as dr²+S_k(r)²(dθ²+sin²θdΦ²)
Sir will you please help me in make me understand how it is coming (if possible diagrammatically) and its's connection with the diagram I have given above.
Thank you.

 

[PeterDonis]

will you please help me in make me understand how it is coming

How what is coming? I don't understand. You've already made the diagram; what else is there to understand?

 

[Apashanka]

How what is coming? I don't understand. You've already made the diagram; what else is there to understand?

For the 3-D positively curved space how is the metric coming ,that's what I am asking about??
e.g ds²=dr²+S_k²(r)(dθ²+sin²θdΦ²)
for the 2-D case it is well visualised and that's what I tried to understand diagrammatically.
Will you please help me in understanding this if possible diagrammatically.
Thanks

 

[PeterDonis]

For the 3-D positively curved space how is the metric coming

The same way as for the 2-D surface, just with 2-spheres at each radius $r$ instead of circles. For the 2-D surface, the term $S_k(r)^2 d\theta^2$ represents distance around the circle at radius $r$. For the 3-sphere, the term $S_k(r)^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$ represents distance around the 2-sphere at radius $r$.

if possible diagrammatically.

There's no way to draw a diagram of a 3-sphere, since you can't embed one in 3-dimensional space and that's all we can draw diagrams in.

 

[Apashanka]

The same way as for the 2-D surface, just with 2-spheres at each radius rrr instead of circles

Sir will you please explain the term 2-sphere

 

[Apashanka]

. For the 2-D surface, the term Sk(r)2dθ2Sk(r)2dθ2S_k(r)^2 d\theta^2 represents distance around the circle at radius r

Sir in the diagram I have taken r to be arc length from the north pole and the radius of curvature I have taken to be R for which S_k(r)=Rsin(r/R).

 

[PeterDonis]

will you please explain the term 2-sphere

You already know what a 2-sphere is. It's the positively curved 2-surface with constant curvature.

Based on your posts, I am re-labeling this thread as "B". An "A" level thread requires graduate level knowledge of the subject matter.

 

[PeterDonis]

in the diagram I have taken r to be arc length from the north pole and the radius of curvature I have taken to be R for which Sk(r)=Rsin(r/R).

Yes, I understand all that. And with all that, the term $S_k(r)^2 d\theta^2$ represents (squared) distance around a circle at $r$, i.e., holding $r$ constant and varying $\theta$.

 

[Ibix]

ds²=dr²+S_k²(r)dθ²
Where S_k(r)=Rsin(r/R)

You can imagine a Euclidean plane as an infinite set of concentric circles - that's what circular polar coordinates do. There's a very simple relationship between the circumference of the circles and the distance to the origin. However, if we make the relationship more complicated, the resulting surface doesn't lie in a Euclidean plane - it can't because the circles won't fit.

Similarly, you can split a Euclidean volume up as nested concentric spheres - this is what spherical polar coordinates do. Again, there's a simple relationship between the area of the spheres and the distance to the origin from that sphere. If you mess around with that relationship, the nested spheres won't fit inside each other in a Euclidean volume. This is a curved space.

Relating this to your diagram, you have shown a small part of a circle on the surface of a sphere. The analogous diagram is a small part of a sphere on the surface of a hypersphere - but that's not possible to draw clearly.

 

[Apashanka]

For the 3-sphere, the term Sk(r)2(dθ2+sin2θdϕ2)Sk(r)2(dθ2+sin2⁡θdϕ2)S_k(r)^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right) represents distance around the 2-sphere at radius rrr.

Sir if any proof available please suggest that would be more helpful
Thank you

 

[Ibix]

$r^2(d\theta^2+\sin^2\theta d\phi^2)$ should be familiar as the metric for the surface of a sphere of radius $r$. Changing it to $S_k(r)(d\theta^2+\sin^2\theta d\phi^2)$ just means that the sphere at a distance $r$ from the origin doesn't have radius $r$, instead it has radius (strictly, areal radius) $\sqrt{S_k(r)}$.

This is just like your circles. The radius of your circles on the surface of a sphere is not the distance, $r$, along the surface to the pole. It's $R\sin(r/R)$.

 

[Apashanka]

$r^2(d\theta^2+\sin^2\theta d\phi^2)$ should be familiar as the metric for the surface of a sphere of radius $r$. Changing it to $S_k(r)(d\theta^2+\sin^2\theta d\phi^2)$ just means that the sphere at a distance $r$ from the origin doesn't have radius $r$, instead it has radius (strictly, areal radius) $\sqrt{S_k(r)}$.

This is just like your circles. The radius of your circles on the surface of a sphere is not the distance, $r$, along the surface to the pole. It's $R\sin(r/R)$.

Am I right in telling that for 2-D positively curved space the elementary length of the circle changes as the radius becomes S_k(r) .
And hence also for 3-D positively curved space the radius of the sphere becomes S_k(r) ,since 2-D plane consists of infinite concentric circle and for positive curvature their radius gets affected.
And 3-D Euclidean space consists of infinite concentric spheres and for positive curvature their radius gets affected
Will the R in S_k(r) =Rsin(r/R) will be the same for both 2-D and 3-D.