Gravity

[S.Owais ismail]

If the gravity is not force but curvature in space fabric what about the things that are attracted by the massive body just beneath its contact with the depression or well formed by the massive body . How to explain that .

[StringJunky]

33 minutes ago, S.Owais ismail said:

If the gravity is not force but curvature in space fabric what about the things that are attracted by the massive body just beneath its contact with the depression or well formed by the massive body . How to explain that .

The trampoline concept is a 2 dimensional representation of a 3 dimensional phenomenon. The way to look at that image is that it says gravity gets stronger the deeper you go in the well,.and it's harder to escape. It's actually acting like that all around the massive object. Its called an "analogy", which is a way of describing things that cannot be directly visualised, like spacetime warping, so, scientists try to use familiar things that behave in approximately the same way. You shouldn't take them literally as being an exact description or representation.

Edited November 26, 2018 by StringJunky

[studiot]

Ismail do you understand the concept of vectors and adding vectors together?

Edited November 26, 2018 by studiot

[Strange]

It isn't really possible to represent the curvature of a 4D thing, especially not in 2 dimensions!

Here is an attempt to give a slightly more 3D view:

From: https://scienceatyourdoorstep.com/2017/09/28/einstein-space-time-curvature/

But we also need to say:

From: https://xkcd.com/895/

[Eise]

1 hour ago, S.Owais ismail said:

If the gravity is not force but curvature in space fabric what about the things that are attracted by the massive body just beneath its contact with the depression or well formed by the massive body . How to explain that .

I suggest this youtube video for a much better analogy. Still an analogy, but you get a better feeling for the fact that gravity is the curving of spacetime, not space. The rubbersheet analogy depicts only curving of space, and so is inherently faulty.

 

 

[quiet]

Notice: The intention of this post is to ask about the possibility or impossibility of demonstrating a specific theorem. There is no intention to present a speculative theory. To put the question in context I need more than one paragraph. That does not mean affirming or denying something outside current knowledge. Let's go to the context with the question included.

---------

The surface vector is defined at each point of a surface.

Think of a closed surface, for example a Gaussian surface. At each point, the surface vector belongs to a line that extends to the outside and to the inside of the surface. To simplify the vocabulary, let's call the surface segment the part of the line that extends inwards.

Is there a point where all the surface segments converge in the interior region? If it existed, a theorem could be deduced, which serves to distinguish between a flat surface and a curved surface. If the surface is flat, convergence at a point of two or more surface segments is impossible. And when there is convergence, then there is curvature.

Why does my question refer to the possibility or impossibility of proving this theorem? Because I suppose that the lines of the gravitational field can never be rigorously parallel. That is to say, I suppose that the gravitational field can not have divergence equal to zero. And something else. I suppose that the lines of the gravitational field always converge on a point. This idea, as it has been expressed, includes only the 3 dimensions of space. That is, it corresponds to a static gravitational field.

In 4 dimensions we do not have visual images. And the gravitational fields in 4D can be variable. There is also something that could facilitate understanding. If it were possible, for gravitation, to demonstrate mathematically a convergence in 4 dimensions, then we could say that convergence allows to treat spacetime as a case of curved geometry in 4 dimensions. It would not be a visual description, but the convergence theorem would allow formulating gravitation in a slightly less counterintuitive way.

Would such a theorem be possible or impossible?

Edited December 8, 2018 by quiet

[studiot]

9 minutes ago, quiet said:

Think of a closed surface, for example a Gaussian surface. At each point, the surface vector belongs to a line that extends to the outside and to the inside of the surface. To simplify the vocabulary, let's call the surface segment the part of the line that extends inwards.

Is there a point where all the surface segments converge in the interior region? If it existed, a theorem could be deduced, which serves to distinguish between a flat surface and a curved surface. If the surface is flat, convergence at a point of two or more surface segments is impossible. And when there is convergence, then there is curvature.

Hello quiet.

I suggest you have another go at the English of this to try to explain better what you mean.

How about a diagram?

 

Have a look at my question to ismail, who seems to have deserted us.

There is an important point behind it which shows why the trampoline picture is a bad one.

[quiet]

I can not connect the vectors and the vector addition with the ismail question. I will try to do some diagram and see what the Google translator can do to improve the expression.

[quiet]

Hi. Here I am again.

Let's see a figure referring to a spherical surface, with the surface vector indicated in some points .

The extensions of all the surface vectors converge on a point, which is the center of the sphere.

Let's see now the representation of a closed surface that is not spherical.

Could it happen that, regardless of the shape, the extensions of all the vectors of a closed surface converge at an interior point?

I do not know the answer. Is there any theorem that answers the question?

[Strange]

2 minutes ago, quiet said:

Could it happen that, regardless of the shape, the extensions of all the vectors of a closed surface converge at an interior point?

How are you defining these vectors? If they are surface normals, then the answer is fairly obviously, no. 

[quiet]

33 minutes ago, Strange said:

How are you defining these vectors? If they are surface normals, then the answer is fairly obviously, no. 

Help me please understand the obviousness.

[studiot]

18 minutes ago, quiet said:

Help me please understand the obviousness.

No the inward normals do not converge to a point except in the case of a sphere.

However the normals have two (or three) components.

One of these components will converge to a point whatever the surface shape.

The others will cancel when summed over the whole of the surface.

This is Gauss' Theorem.

https://en.wikipedia.org/wiki/Divergence_theorem

[Strange]

36 minutes ago, quiet said:

Help me please understand the obviousness.

Consider a cylinder: the surface normals converge on the axis of the cylinder, not a point.

[quiet]

1 hour ago, studiot said:

the normals have two (or three) components.

One of these components will converge to a point whatever the surface shape.

Very interesting property. Thank you.

1 hour ago, Strange said:

Consider a cylinder: the surface normals converge on the axis of the cylinder, not a point.

Thanks. This motivates another question.

Are a cylindrical surface (with a sharp contour) and a smooth surface (without sharp contour) topologically equivalent?

Edited December 9, 2018 by quiet

[studiot]

5 hours ago, quiet said:

Very interesting property. Thank you.

Thanks. This motivates another question.

Are a cylindrical surface (with a sharp contour) and a smooth surface (without sharp contour) topologically equivalent?

Read this from

Surface Topology

by Firby and Gardiner

It may help.

[Strange]

6 hours ago, quiet said:

Are a cylindrical surface (with a sharp contour) and a smooth surface (without sharp contour) topologically equivalent?

Yes. The degree of curvature is irrelevant. All that matters is the number of holes. Hence the joke about a topologist being someone who doesn’t know the difference between a doughnut and a coffee cup.