Space-time curvature

[awaterpon]

How space-time curvature works ?

If space-time exists everywhere including the mass itself, in this case a mass can't curve space-time because all space-time to be curved is inside it.Also if space time exists inside mass  then  existence or non-existence of mass are the same.

If space-time doesn't exists inside mass or mass displace space-time, then when a mass move it will leave a broken space-time.

[swansont]

If you had a spherical shell, the would be no gravity inside the shell. If we are equating gravity with curvature, it only exists on the outside.

Spacetime exists everywhere. It’s your coordinate system. 

[joigus]

6 hours ago, awaterpon said:

in this case a mass can't curve space-time because all space-time to be curved is inside it

Why do you say all space-time to be curved is inside the mass? (My emphasis.)

I'm interested in your question, but I need clarification of the previous point.

[awaterpon]

2 hours ago, joigus said:

Why do you say all space-time to be curved is inside the mass? (My emphasis.)

If I put a piece of sponge  in water the sponge  can't displace water , instead water will fill all the sponge .This is an example of space-time inside mass, the mass can't curve space-time because it is filled with it. No displace in water by the sponge no curvature.by mass.

on the other hand , If I put an empty bucket in water , this empty bucket represent mass with no space-time inside it, this bucket will displace water "curvature" , but in this case there will be no space-time points in the universe.
 

Edited October 12, 2020 by awaterpon

[swansont]

7 minutes ago, awaterpon said:

If I put a piece of sponge  in water the sponge  can't displace water , instead water will fill all the sponge .This is an example of space-time inside mass, the mass can't curve space-time because it is filled with it. No displace in water by the sponge no curvature.by mass.

on the other hand , If I put an empty bucket in water , this empty bucket represent mass with no space-time inside it, this bucket will displace water "curvature" , but in this case there will be no space-time points in the universe, This will result in many problems in physics.
 

1. if you do an experiment, you will find that a sponge does displace water

2. Spacetime is a coordinate system, not a substance

[awaterpon]

1 hour ago, swansont said:

2. Spacetime is a coordinate system, not a substance

Mass curve space-time .What if  a piece "m" is inside M,  what  does it  do  to space-time? does it curve the space-time inside M? or the piece m doesn't affect space-time and the curvature is done by the whole M?

 

Edited October 12, 2020 by awaterpon

[MigL]

The model that we have, General Relativity, has a coordinate system whose 'curvature' is what we commonly call gravity.
This agrees to a very high degree with observations and experiments made of the actual world around us.

Do you really think that the concept we call space-time has an actual 'fabric' which can be curved ???

[swansont]

16 hours ago, awaterpon said:

Mass curve space-time .What if  a piece "m" is inside M,  what  does it  do  to space-time? does it curve the space-time inside M? or the piece m doesn't affect space-time and the curvature is done by the whole M?

 

Any mass element curves spacetime outside of that element. The effects of multiple mass elements combine (i.e reinforce each other)

[Markus Hanke]

19 hours ago, awaterpon said:

Mass curve space-time

Any form of energy-momentum will be a source of curvature, not just mass. For example, if you have an electromagnetic field in an otherwise empty region of space, then this will have a gravitational effect too. 

Also, curvature has nothing to do with displacement, it’s purely a geometric phenomenon - it’s about geodesic deviation, i.e. the failure of initially parallel geodesics to remain parallel.

[Xerxes]

2 hours ago, Markus Hanke said:

Also, curvature has nothing to do with displacement, it’s purely a geometric phenomenon - it’s about geodesic deviation, i.e. the failure of initially parallel geodesics to remain parallel.

Actually, curvature is defined as the second derivative of the metric field. So if the metric field is constant the curvature field is zero - elementary calculus. Non-zero otherwise

[MigL]

The metric field is a mathematical construct.
So are derivatives of it, and so is space-time.
They are components of a mathematical ( geometric ) model.

There is nothing to actually curve, although we do observe test particles following geodesic paths as IF there was.

[Xerxes]

2 minutes ago, MigL said:

The metric field is a mathematical construct.

You have a problem with mathematics? How else would you describe the Theory of General Relativity?

[MigL]

With a MATHEMATICAL model.
Does that change the fact that it is, in fact, the model, and not the reality ?
 

[Xerxes]

And what, pray, is this this called "reality"? Surely not snooker balls on trampolines?

Edited October 13, 2020 by Xerxes

[MigL]

Maybe you don't realize how many models of gravity there actually are...

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

Quite a few reduce to, or are equivalent to the well established GR.
What are they describing ?
 

GR alsoceases to have valid predictive power at certain limits, such as singularities.
Does that mean reality ( like snooker balls and trampolines ) also ceases to exist at those limits ?
Or has the MODEL exceeded its applicable range.

[Lan Todak]

14 hours ago, Markus Hanke said:

Any form of energy-momentum will be a source of curvature, not just mass. For example, if you have an electromagnetic field in an otherwise empty region of space, then this will have a gravitational effect too. 

Also, curvature has nothing to do with displacement, it’s purely a geometric phenomenon - it’s about geodesic deviation, i.e. the failure of initially parallel geodesics to remain parallel.

Do you mean that we can generate gravitational effect by using EM instead of mass?

Edited October 14, 2020 by Lan Todak

[ALine]

I actually just watched a video describing an understanding of relativity, not sure if it was general though. 

It goes about stating that every accelerating body, no matter where you are is equivalent. So if you are falling from a building somewhere or in the middle of space they are equivalent instances. In regards to being on a spacecraft surrounding a massive body you as not actually being "attracted" to the massive body, whereas spacetime is actually curving i thiiink outside of your inertial referance frame, do not quote me on that one. It is like not moving while at the same time things are moving "upwards", however I beliieeeve one of einsteins equations states that this is allowed due to a certain part of his equation. 

If you wish to learn more about this, or get the full understanding, Veritesium on youtube gives a much better rendition of it than I. Check him out!

10 hours ago, MigL said:

With a MATHEMATICAL model.
Does that change the fact that it is, in fact, the model, and not the reality ?

Yep, from what I understand everything is represented by a given model of the framework of reality, not reality itself. I do not believe that you can fully and accurately represent reality with 100 percent accuracy and precision because once ya do it becomes 100 percent wrong the next pico second.

Thats just the way it works, no existing explanations of reality is ever or will ever be "correct," they just match up with observations more and more.

[Markus Hanke]

11 hours ago, Xerxes said:

Actually, curvature is defined as the second derivative of the metric field.

Riemann curvature is a rank-4 tensor, so it's little bit more complicated than this - but yes, its components in a particular coordinate basis can be calculated via derivatives of the metric. The formal definition of this object is usually given as the commutator of the covariant derivative, which is much more intuitive.

11 hours ago, Xerxes said:

So if the metric field is constant the curvature field is zero

Sure. But what does this have to do with my original comment you quoted me on? I fail to see the connection (pun intended).

42 minutes ago, Lan Todak said:

Do you mean that we can generate gravitational effect by using EM instead of mass?

Yes of course. But you'd need extremely strong fields to get any kind of appreciable gravitational effect.

 

[Markus Hanke]

On 10/14/2020 at 6:34 AM, ALine said:

So if you are falling from a building somewhere or in the middle of space they are equivalent instances

Note that this is true only so long as the falling frame is small enough - the equivalence principle is a purely local statement. Once the frame becomes large enough, it will be possible to detect tidal effects, which are inherent in a gravitational field that is due to sources of energy-momentum.

[ALine]

thank you for the correction, learned something new today