# Thread: Relation of Rubber Sheet Analogy to GR

1. I'm planning a simple experiment, which will test, if General Relativity explains the force of gravity correctly. I want to use the rubber surface model of gravity, to see, if changing the size of an object really has no effect on the force of attraction towards the center of a gravitational field...

According to our current knowledge, force of gravity depends only on the mass of an object. Science tells, that if Sun would shrink enough, to become a black hole, it wouldn't have no effect on the orbital motion of planets in Solar System.

Sadly I'm a scientific sceptic and I don't believe in anything, what seems to be incorrect. Since science is about facts and not beliefs, this is why I want to observe and measure, if changing the density of a central body in a gravitational field won't affect the acceleration and forces, which are induced on other bodies in this field. To make it happen, I will use the model of gravity, which is proposed by the theory of General Relativity:

And this is, what we can expect to see, if we will change the size of an object, without changing it's mass:

According to science - force of gravity, induced on the attracted object won't change even a bit...
According to me - depending on the distance between the source of gravity, object can experience 2 (possibly 3) kinds of effects:
- if the distance is small, force of attraction towards the center of field, will grow significantly
- if the distance is big, force of attraction will become weaker and eventually completely fade
- it's possible, that at a certain distance from the center, force of attraction won't change

What I need to do, to prove my idea, is to measure the acceleration of a body, which is attracted towards the center for different densities of central bodies - but better will be to measure the force of attraction itself, as acceleration can be affected by friction between the surface and the body.

Another thing, which I can check, is to see, what will happen, if I change the density of attracted body - combined, it gives me 4 possible measurements to compare.

You need to know, that I've made already some attempts, only to check, if it's possible, that Einstein could be possibly wrong - but before I will show you some initial results, I would love to hear, what do you think about my experiment and what do you expect to happen...

2.

3. Those images explain, why I expect to observe the effects, which I described in previous post:

4. Originally Posted by AstralTraveler
I'm planning a simple experiment, which will test, if General Relativity explains the force of gravity correctly. I want to use the rubber surface model of gravity, to see, if changing the size of an object really has no effect on the force of attraction towards the center of a gravitational field.
By testing the rubber surface model of gravity, you are not testing General Relativity. The rubber surface model of gravity is a mere analogy of spacetime curvature and a poor one at that. Therefore, any result you get from an experiment based on the rubber surface model will have no bearing on the validity of General Relativity.

5. Originally Posted by AstralTraveler
Those images explain, why I expect to observe the effects, which I described in previous post:]
I can't work out if you are a troll or a total moron. Can you give me a clue?

(I should point out that, although this tool has only just joined here, he has already wasted hundreds of posts on another forum. He is unwilling or unable to learn and just posts the same ignorant drivel repeatedly.)

6. Well... how bouncy is spacetime? Is your rubber sheet too firm or too flimsy?

7. Originally Posted by Daecon
Well... how bouncy is spacetime? Is your rubber sheet too firm or too flimsy?
Mine is non-stick.

8. Moved to Pseudoscience.

9. I've been studying this excellent vid....with a bit more analysis I'll be able to show Einstien was wrong for sure. What you think?

10.

11. MOD NOTE: Since I am aware of your posting this very same thing on other forums, along with your responses in the course of the ensuing discussions, I feel the need to reiterate once again what has already been said to you many times: the „rubber sheet analogy“ is just precisely that - an analogy. It is intended only to convey the most general of ideas about the notion of „curvature“, for the beginning student without any prior knowledge of GR or differential geometry. As such, the analogy has critical flaws, omissions, and drawbacks, and is hence not a model of gravity, nor is it an accurate and meaningful representation of what GR as a model is about. To put it succinctly - an analogy is not the model itself; the two are completely different things.

Therefore, you cannot test GR by testing the rubber sheet analogy, because the two are not the same things. Not even close, in fact. To give just an isolated example, gravity in GR is not a force, and hence bodies in free fall under the influence of gravity do not experience any proper acceleration. You can test this yourself by dropping an accelerometer off a high tower etc, and then study the recorded values. You will find the accelerometer reading exactly zero at all times during free fall. This is precisely what GR models.

In order not to waste anyone‘s time here, I need you to answer me this very simple question before I allow the thread to continue: do you fully understand that GR as a model is not the same thing as the rubber sheet as an analogy?

It should also be mentioned that the analogy itself refers to only one particular solution to the field equations, which is not even the one that describes gravity on earth in the most accurate manner. So - not only is the rubber sheet just an analogy, it is also an incomplete one.

Please answer the above question in your next post; if no response to this is forthcoming, the thread will be locked and moved to the Trash Can immediately. There is no sense in repeating a discussion here that was already had on other forums.

12. "do you fully understand that GR as a model is not the same thing as the rubber sheet as an analogy? "

Of course - this is why I use it, as analogy. I don't want to make any precise measurements of the
rubber sheet geometry. I want to see, what happens, when mass is concentrated in a point and when it is distributed over some area - that's all... Because it is rather obvious, that the effects will be different in both cases

13.

14. Originally Posted by AstralTraveler
"do you fully understand that GR as a model is not the same thing as the rubber sheet as an analogy? "

Of course - this is why I use it, as analogy.
Then there's no point to continuing this thread, is there? You have admitted that the rubber sheet model is only an analogy, so you have admitted that testing the rubber sheet model in no way tests GR.

Are you merely using this forum to drive traffic to your youtube vids?

15. It should be noted that the Shell Theorem in Newton's gravity and Birkhoff's Theorem in General Relativity specify a spherical distribution of matter.

16. https://www.quora.com/Does-space-tim...ies-more-space
It seems, that everything, what I said in the beginning is absolutely correct. Check this out:
"The distortion or curvature of spacetime depends on the distribution of the mass of a body. As Leo C. Stein accurately explains in his answer, the Einstein's field equations explain how the curvature of spacetime, which is governed by the Einstein tensor increases, if the energy density, or the amount of mass per unit area increases.The basic idea here is that mass and energy are two sides of the same coin, and that basically energy causes distortion of space-time. The distribution of mass around a body is the mass density, or inevitably the energy density of the body.
If you squeeze the mass of a body into a small area, the mass/energy density will increase, and thus the curvature of spacetime will be more around a body in which thee energy density is high.
Intuitively, you can realise, why a massive star, say of 5 solar mass, does not behave as a black hole, while a similar black hole of 5 solar masses is. Now the difference between a 5 solar mass black hole, and a star of same mass is that, a black hole of 5 solar mass has the entire mass collapsed onto a single, infinitesimally small- 'singularity', because of which the energy density is so high that the curvature of spacetime is almost infinity (we don't know for sure, but our equations predict)."

17.

18. Effects of General Relativity on the Earth
"When entering the surface of the planet, the horizonal curvature keeps continuous values but the vertical ones change : in the case of a planet with constant density, they become equal to the horizontal curvature 2T/c2; in any case the sum of 3 spatial curvatures is proportional to the density. The coefficient of proportionality is thus 6T/ρc2 = 6×4πG/3c2= 8πG/c2.
Inside a theoretical planet with constant density, thus, we have the uniform space curvature 2T/c2=8πGρ/3c2, with coefficient the distance (per density in g/cm3)
8πG/3c2 = 6.221×10−24 m-2(g/m3)-1.
For a density of 1 g/m3, this curvature has radius
(6.221×10−24 )-1/2=400,930,000 km = 4.0093×1011 m = (c/√2)× 1891.3 s = 1337 light-seconds = 22.29 light-minutes.
For the Earth's density, this radius of geometrical curvature is 569.5 light-seconds = 9.49 light-minutes = 170,720,000 km
to be compared to the Earth's mean distance to the Sun, 149,598,000 km = 8.317 light-minutes.
In practice, it means that a "planet" with the same density as the Earth but as big as the Earth's orbit, would immediately become a black hole with that size.
"

19. Chp 4.6
"The amount of distortion or curvature will depend on the amount of matter - more accurately the density of matter - in the universe. It is fairly easy to relate the shape of space to the density of matter in the universe."

20. https://newsroom.unsw.edu.au/news/sc...rstand-gravity
"It was not until 1915 that Einstein’s unparalleled genius was demonstrated when he published his general theory of relativity. This theory made the claim that space–time curvature is proportional to, and is caused by, the “energy-momentum density”, that is, the energy and the momentum associated with all and any kind of matter in a unit volume of space."

21. Originally Posted by AstralTraveler
https://www.quora.com/Does-space-tim...ies-more-space
It seems, that everything, what I said in the beginning is absolutely correct. Check this out:
"The distortion or curvature of spacetime depends on the distribution of the mass of a body. As Leo C. Stein accurately explains in his answer, the Einstein's field equations explain how the curvature of spacetime, which is governed by the Einstein tensor increases, if the energy density, or the amount of mass per unit area increases.The basic idea here is that mass and energy are two sides of the same coin, and that basically energy causes distortion of space-time. The distribution of mass around a body is the mass density, or inevitably the energy density of the body.
If you squeeze the mass of a body into a small area, the mass/energy density will increase, and thus the curvature of spacetime will be more around a body in which thee energy density is high.
Intuitively, you can realise, why a massive star, say of 5 solar mass, does not behave as a black hole, while a similar black hole of 5 solar masses is. Now the difference between a 5 solar mass black hole, and a star of same mass is that, a black hole of 5 solar mass has the entire mass collapsed onto a single, infinitesimally small- 'singularity', because of which the energy density is so high that the curvature of spacetime is almost infinity (we don't know for sure, but our equations predict)."
This isn't saying that what you said in the opening post is correct. Nor is it in conflict with Birkhoff's Theorem. Yes, curvature does increase with density. Nobody said it doesn't. But if the total mass (of a spherically symmetric distribution of matter) is unchanged, and all that has changed is the density, then at a given distance from the centre (outside of the matter in both cases), the curvature will be the same.

22. Of course - this is why I use it, as analogy.
Good, so you understand then that the table top experiment you allude to in your OP is not in fact a test of whether GR explains gravity correctly; it’s just a demonstration of standard Newtonian mechanics, and has no bearing on the validity of GR at all. The analogy is not the model, so you can’t test the model by testing the analogy.

I will, for now, let this thread continue, but I must ask everyone to please keep it civil. I will also change the thread title to more accurately reflect what this is about, and reassign it to Alternative Hypotheses.

23. According to our current knowledge, force of gravity depends only on the mass of an object.
This is incorrect. The geometry of spacetime (not: force) depends on the distribution of energy-momentum, which is a rank-2 tensor field, and not just a single scalar value such as mass. In vacuum regions then, the geometry will depend on mass, angular momentum, and electric charge of the distant gravitational source, as well as on the value of the cosmological constant. If there is any other background curvature (e.g. gravitational radiation), then this needs to be taken into account as well. In addition to all this, the Einstein equations are highly non-linear, so in some sense the “gravitational field” is itself also a source of gravity (this is very sloppily put, but you get the idea) - unlike in Newtonian gravity, which is linear.

Be careful not to conflate Newtonian gravity with GR - the former uses only masses, distances, and densities, but in GR the situation is far more complex, and many other factors are taken into account as well.

24. Originally Posted by Markus Hanke
According to our current knowledge, force of gravity depends only on the mass of an object.
This is incorrect. The geometry of spacetime (not: force) depends on the distribution of energy-momentum, which is a rank-2 tensor field, and not just a single scalar value such as mass. In vacuum regions then, the geometry will depend on mass, angular momentum, and electric charge of the distant gravitational source, as well as on the value of the cosmological constant. If there is any other background curvature (e.g. gravitational radiation), then this needs to be taken into account as well. In addition to all this, the Einstein equations are highly non-linear, so in some sense the “gravitational field” is itself also a source of gravity (this is very sloppily put, but you get the idea) - unlike in Newtonian gravity, which is linear.

Be careful not to conflate Newtonian gravity with GR - the former uses only masses, distances, and densities, but in GR the situation is far more complex, and many other factors are taken into account as well.
With all that complexity is great use made of computers to run through the actual calculations? Somewhat in the manner of weather forecasting ? Would quantum computers bring anything new to the party?

I am trying to understand tensors at the moment. Is the rank-2 tensor field you have mentioned the highest level of field (or tensor) that is involved in GR? (I have heard that these tensors are the same no matter what ones frame of reference-is that just inertial?)

25. Originally Posted by geordief
Is the rank-2 tensor field you have mentioned the highest level of field (or tensor) that is involved in GR?
No, the Riemann curvature tensor is rank-4.

Originally Posted by geordief
I have heard that these tensors are the same no matter what ones frame of reference-is that just inertial?
No, that's not quite correct. The correct term is "covariant" (as distinct from "invariant"), which means that that tensors change in a particular way under a change of coordinate system. This particular way guarantees that a tensor that is zero in one coordinate system is zero in every coordinate system, and also guarantees that an equation that is valid in one coordinate system is valid in every coordinate system.

26. I am trying to understand tensors at the moment.
Just think of them as machines in a production line - you feed in raw material on one side, the machine processes this, and out comes the finished product on the other side.

Tensors are similar to this - you “feed” them one mathematical object, and out comes another. Take for example the metric tensor - you feed it two vectors, the tensor crunches them up, and spits out their (squared) inner product, which is a scalar (i.e. a real number), as the end product. What you can feed in, and what comes out as a result, is determined by the structure of the tensor’s indices (if written in index notation).

Input >> TENSOR >> Output

That’s the general idea. Now here’s the neat thing - the mechanism inside the machine looks different to each different observer, BUT the difference is such that the end product which the machine produces is always the same one. In the case of the metric tensor - the components of the pair of vectors which you feed in are different in different coordinate systems (the raw material); the components of the tensor itself are different in different coordinate systems (the mechanism); but the output which the tensor produces is always guaranteed to be the squared inner product of the two vectors, regardless of who performs the operation!

So far as physics is concerned: when you go from one observer to another, the individual components of a tensor will change, but they will change in precisely such a way that any physical law written in terms of that tensor retains its overall form. To put it more succinctly: if a tensor equation is valid for one observer, it is automatically guaranteed to be valid for all observers. That is what makes tensors so useful.

Hope that makes sense. It’s really not easy to find the right balance between intuitive understanding and mathematical rigour here.

27. Originally Posted by Markus Hanke
I am trying to understand tensors at the moment.
Just think of them as machines in a production line - you feed in raw material on one side, the machine processes this, and out comes the finished product on the other side.

Tensors are similar to this - you “feed” them one mathematical object, and out comes another. Take for example the metric tensor - you feed it two vectors, the tensor crunches them up, and spits out their (squared) inner product, which is a scalar (i.e. a real number), as the end product. What you can feed in, and what comes out as a result, is determined by the structure of the tensor’s indices (if written in index notation).

Input >> TENSOR >> Output

That’s the general idea. Now here’s the neat thing - the mechanism inside the machine looks different to each different observer, BUT the difference is such that the end product which the machine produces is always the same one. In the case of the metric tensor - the components of the pair of vectors which you feed in are different in different coordinate systems (the raw material); the components of the tensor itself are different in different coordinate systems (the mechanism); but the output which the tensor produces is always guaranteed to be the squared inner product of the two vectors, regardless of who performs the operation!

So far as physics is concerned: when you go from one observer to another, the individual components of a tensor will change, but they will change in precisely such a way that any physical law written in terms of that tensor retains its overall form. To put it more succinctly: if a tensor equation is valid for one observer, it is automatically guaranteed to be valid for all observers. That is what makes tensors so useful.

Hope that makes sense. It’s really not easy to find the right balance between intuitive understanding and mathematical rigour here.
Well (trying to get the lie of the land here )if we are looking at a minute volume of spacetime how many vectors would "available" to be usefully fed into this "tensor machine"?

Are any of these vectors perpendicular to the surface?

28. Duplicate

29. Originally Posted by KJW

Originally Posted by geordief
I have heard that these tensors are the same no matter what ones frame of reference-is that just inertial?
No, that's not quite correct. The correct term is "covariant" (as distinct from "invariant"), which means that that tensors change in a particular way under a change of coordinate system. This particular way guarantees that a tensor that is zero in one coordinate system is zero in every coordinate system, and also guarantees that an equation that is valid in one coordinate system is valid in every coordinate system.
Including non-inertial?

30. Originally Posted by geordief
Well (trying to get the lie of the land here )if we are looking at a minute volume of spacetime how many vectors would "available" to be usefully fed into this "tensor machine"?

Are any of these vectors perpendicular to the surface?
The vectors are tangent to the space, and there are four of them at any particular point in any particular coordinate system (the space is four-dimensional), but there are an infinite number of coordinate systems (each with their own tetrad at each point).

31. Originally Posted by geordief
Including non-inertial?
A coordinate system is more general than a frame of reference, so the question of "non-inertial" doesn't apply (though the answer would be "yes" if we are talking about observers' frames of reference).

32. Originally Posted by KJW
Originally Posted by geordief
Including non-inertial?
A coordinate system is more general than a frame of reference, so the question of "non-inertial" doesn't apply (though the answer would be "yes" if we are talking about observers' frames of reference).
It's worth remarking that it is in curvilinear coordinate systems that tensors really come into their own.

33. <edit> I've posted the same answer twice... sorry...

34. This isn't saying that what you said in the opening post is correct. Nor is it in conflict with Birkhoff's Theorem. Yes, curvature does increase with density. Nobody said it doesn't. But if the total mass (of a spherically symmetric distribution of matter) is unchanged, and all that has changed is the density, then at a given distance from the centre (outside of the matter in both cases), the curvature will be the same.
https://www.quora.com/Does-space-tim...ies-more-space
"Yes, the curvature depends on how much space your mass takes up." and "If you squeeze the same amount of mass into a smaller space, the energy density increases, so the Einstein tensor will have to increase, which quantifies part of the curvature of spacetime."?

I've noticed as well, that there's a big difference between Newton's gravity and Einstein's field equations. In the most known equation for gravity, only 4 values are being used: constant G, mass of 2 objects and distance between them. But in GR, gravitational field is based on the metric tensor, which describes the distribution of mass/energy in space and includes the density, as one of the main factors. However science still uses the first one - which doesn't include the concentration of mass in space and seems to forget about GR in this case...

Besides Poisson's equation also includes the density of objects - and can be used to calculate both: the curvature of space-time or the geometry of rubber sheet...

35. Originally Posted by KJW

The vectors are tangent to the space,
I forgot to ask "What is that space?" Is it ,as I suspected a 2D hypersurface with perhaps one dimension being time and the other one or other of the spacial dimensions?

36. again it posted my response twice :/

37. Originally Posted by Markus Hanke
According to our current knowledge, force of gravity depends only on the mass of an object.
This is incorrect. The geometry of spacetime (not: force) depends on the distribution of energy-momentum, which is a rank-2 tensor field, and not just a single scalar value such as mass. In vacuum regions then, the geometry will depend on mass, angular momentum, and electric charge of the distant gravitational source, as well as on the value of the cosmological constant. If there is any other background curvature (e.g. gravitational radiation), then this needs to be taken into account as well. In addition to all this, the Einstein equations are highly non-linear, so in some sense the “gravitational field” is itself also a source of gravity (this is very sloppily put, but you get the idea) - unlike in Newtonian gravity, which is linear.

Be careful not to conflate Newtonian gravity with GR - the former uses only masses, distances, and densities, but in GR the situation is far more complex, and many other factors are taken into account as well.
Exactly! However Newton's equation doesn't include densities - but GR does. And it makes me wonder, why science continues to use the Newton's equation, if it contradicts in some way the GR field equations. In GR gravity is directly connected with the distribution of mass/energy in space - objects are attracted towards a high concentration of mass. But according to Newton, force of attraction is directed towards the center of mass and not towards the space with high density of matter. It's a huge difference, as a force works differently, when it's concentrated in a point and when it's distributed over some area. Einstein's and Poisson's equations include the density of interacting objects and they are most likely much more adequate, than rather old Newton's concept of gravity. So why in 90% of sources, gravity is still explained with the old formula?

Thanks for not closing the thread. I always remain civil

38. Originally Posted by geordief
Originally Posted by KJW

The vectors are tangent to the space,
I forgot to ask "What is that space?" Is it ,as I suspected a 2D hypersurface with perhaps one dimension being time and the other one or other of the spacial dimensions?
The "space" of relativity is four-dimensional spacetime. Note that the term "space" has a very general meaning although I concede that confusion may have arisen because it also means the 3-dimensional part of spacetime.

39. Be careful not to conflate Newtonian gravity with GR - the former uses only masses, distances, and densities, but in GR the situation is far more complex, and many other factors are taken into account as well.
I think, that it is the source of entire disagreement. My claims seem to be fully consistent with GR an Poisson's equations, but are contradicted by Newton's law of universal gravitation.

40. Originally Posted by AstralTraveler
This isn't saying that what you said in the opening post is correct. Nor is it in conflict with Birkhoff's Theorem. Yes, curvature does increase with density. Nobody said it doesn't. But if the total mass (of a spherically symmetric distribution of matter) is unchanged, and all that has changed is the density, then at a given distance from the centre (outside of the matter in both cases), the curvature will be the same.
https://www.quora.com/Does-space-tim...ies-more-space
"Yes, the curvature depends on how much space your mass takes up." and "If you squeeze the same amount of mass into a smaller space, the energy density increases, so the Einstein tensor will have to increase, which quantifies part of the curvature of spacetime."?

I've noticed as well, that there's a big difference between Newton's gravity and Einstein's field equations. In the most known equation for gravity, only 4 values are being used: constant G, mass of 2 objects and distance between them. But in GR, gravitational field is based on the metric tensor, which describes the distribution of mass/energy in space and includes the density, as one of the main factors. However science still uses the first one - which doesn't include the concentration of mass in space and seems to forget about GR in this case...

Besides Poisson's equation also includes the density of objects - and can be used to calculate both: the curvature of space-time or the geometry of rubber sheet...
Looking at your reply, I feel as if my statement has been summarily dismissed. But I strongly urge you to read the article on Birkhoff's theorem (relativity). This theorem shows that the shell theorem from Newtonian gravity also applies to non-rotating spherically symmetric solutions in General Relativity.

41. Looking at your reply, I feel as if my statement has been summarily dismissed. But I strongly urge you to read the article on Birkhoff's theorem (relativity). This theorem shows that the shell theorem from Newtonian gravity also applies to non-rotating spherically symmetric solutions in General Relativity.
Ok, but it's still just a theorem - and still needs some empirical confirmation. But I have some problems with accepting things, which don't seem logical for me. For example here:
https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity

"We can conclude (by using a "Gaussian pillbox") that for an infinite, flat plate (Bouguer plate) of any finite thickness, the gravitational field outside the plate is perpendicular to the plate, towards it, with magnitude 2πG times the mass per unit area, independent of the distance to the plate[2] (see also gravity anomalies).More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2πG times (the mass per unit area above z, minus the mass per unit area below z)."

So, in the case of such plate, we include the distribution of mass in the object - and as a result, force of attraction is directed perpendiculary towards the surface(s) and not towards the center of it's mass. So why the same rules can't be used in the case of an uniform sphere? While this still makes sense to me:

"
In the case of a spherically symmetric mass distribution we can conclude (by using a spherical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of G/r2 times only the total mass within a smaller distance than r. All the mass at a greater distance than r from the center can be ignored."I just can't understand, why the change of r won't affect the gravitational field outside the sphere. Change of the volume will have direct influence on the stress-energy tensor and result in different curvature of space-time. Why the uniform sphere is treated in a completely different way? All other shapes, create attraction directed towards the object (or a fragment of space-time, filled with mass), so why only the uniform sphere causes attraction directed towards a single point?

42. Looking at your reply, I feel as if my statement has been summarily dismissed. But I strongly urge you to read the article on Birkhoff's theorem (relativity). This theorem shows that the shell theorem from Newtonian gravity also applies to non-rotating spherically symmetric solutions in General Relativity.
Ok, but it's still just a theorem - which still needs some empirical confirmation. However I have some problems with accepting things, which don't seem logical for me. For example here:
https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity

"We can conclude (by using a "Gaussian pillbox") that for an infinite, flat plate (Bouguer plate) of any finite thickness, the gravitational field outside the plate is perpendicular to the plate, towards it, with magnitude 2πG times the mass per unit area, independent of the distance to the plate[2] (see also gravity anomalies).More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2πG times (the mass per unit area above z, minus the mass per unit area below z)."

So, in the case of such plate, we include the distribution of mass in the object - and as a result, force of attraction is directed perpendiculary towards the surface(s) and not towards the center of it's mass. So why the same rules can't be used in the case of an uniform sphere? While this still makes sense to me:

"
In the case of a spherically symmetric mass distribution we can conclude (by using a spherical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of G/r2 times only the total mass within a smaller distance than r. All the mass at a greater distance than r from the center can be ignored."

I just can't understand, why the change of r won't affect the gravitational field outside the sphere. Change of the volume will have direct influence on the stress-energy tensor and result in different curvature of space-time. Why the uniform sphere is treated in a completely different way? All other shapes, create attraction directed towards the object (or a fragment of space-time, filled with mass), so why only the uniform sphere causes attraction directed towards a single point?

43. Exactly! However Newton's equation doesn't include densities - but GR does. And it makes me wonder, why science continues to use the Newton's equation, if it contradicts in some way the GR field equations.
A very valid question. You see, even though GR is the more comprehensive and accurate model of gravity, many of its relativistic effects and corrections become relevant only in situations where we deal with strong gravitational fields, high velocities, or forms of energy other than mass. On the other hand, in cases where only weak gravitation and no other forms of energy are involved, relativistic effects are so small as to be negligible for most purposes.

If you look at it in terms of computational effort, you will see that Newtonian gravity is a very simple model, and many problems can be solved by the pen-and-paper method, i.e. algebraic manipulations done manually on a sheet of paper, or via standard CAM software. This is why it is still taught in high school physics. GR on the other hand is a very different story - it is the more precise and accurate model, but in all but the most simplistic of cases the Einstein equations are extremely difficult to solve; even numerical methods (which do exist) require substantial computing power.

You can probably now see the answer to your question - Newtonian gravity continues to be used in cases where relativistic effects can be neglected, simply because it is much easier to handle and solve, mathematically speaking. For example, if you look at the Earth - which has weak gravity in the bigger scheme of things -, you will see that Newtonian gravity gives very accurate predictions under most circumstances. On the other hand, if you look at the gravity of our sun (or even stronger sources), you will notice that Newton gives predictions that are ever so slightly off, so GR is needed. You get the idea.

Formally, Newtonian gravity is the asymptotic limit of GR - for example, in the Schwarzschild metric, if you make r very large, you will recover Newtonian gravity. Hence, GR and Newtonian gravity aren’t really different and separate - Newton is a subset of GR, that is still valid under certain restricted circumstances, and hence continues to be used due to its simplicity and ease of application.

44. Originally Posted by AstralTraveler
Be careful not to conflate Newtonian gravity with GR - the former uses only masses, distances, and densities, but in GR the situation is far more complex, and many other factors are taken into account as well.
I think, that it is the source of entire disagreement. My claims seem to be fully consistent with GR an Poisson's equations, but are contradicted by Newton's law of universal gravitation.
GR, Poisson, and Newton seem to be different things or the surface, but really they are just descriptions of gravity in different regimes. GR is the most general model, which, according to current consensus, accurately describes gravity in the entire classical regime (i.e. in all cases when quantum effects can be neglected). The Poisson equation can be used to derive a potential field from a given mass density distribution, and from that potential you can then obtain the gravitational field via standard methods; it is valid only in non-relativistic situations though. Newton’s law of universal gravitation also applies only to the non-relativistic case, and in addition it is limited to situations were the masses can be considered as point masses, i.e. there are no density distributions involved, only total masses.

The latter is the most simplistic case, but of course it gives accurate descriptions only if you can actually treat the gravitational sources as point-like.

So - each of the above three approaches to gravity are valid, but they are valid only in specific domains of applicability. Which one to choose depends both on the specific situation, and on the level of accuracy you require for your solution.

However, our current understanding of gravity is of course based on the most general of these models, being GR - meaning, gravity is understood to be geodesic deviation, and hence a geometric property of spacetime. That is what gravity “is”. Other options like Newton and Poisson are simplifications that are available to us in specific scenarios, when the full description is not necessarily needed.

So I hope this clears up things a bit. If the claim you are referring to was that variables like density, spatial distribution etc are relevant for gravity, then of course you are correct in the general case. However, sometimes we have scenarios when things are particularly simple, and we need to worry only about total masses and their distances; in such cases, we can still safely use Newton, and will get a result that fits observation quite well. Of course you could still use the full machinery of GR even for such simple cases, if you wanted to - however, this would be computational overkill, because the difference between GR and Newton under such circumstances is negligibly small. I suppose it boils down to a trade-off between accuracy and ease of computation.

45. again posted a double...

46. Originally Posted by Markus Hanke
Exactly! However Newton's equation doesn't include densities - but GR does. And it makes me wonder, why science continues to use the Newton's equation, if it contradicts in some way the GR field equations.
A very valid question. You see, even though GR is the more comprehensive and accurate model of gravity, many of its relativistic effects and corrections become relevant only in situations where we deal with strong gravitational fields, high velocities, or forms of energy other than mass. On the other hand, in cases where only weak gravitation and no other forms of energy are involved, relativistic effects are so small as to be negligible for most purposes.

If you look at it in terms of computational effort, you will see that Newtonian gravity is a very simple model, and many problems can be solved by the pen-and-paper method, i.e. algebraic manipulations done manually on a sheet of paper, or via standard CAM software. This is why it is still taught in high school physics. GR on the other hand is a very different story - it is the more precise and accurate model, but in all but the most simplistic of cases the Einstein equations are extremely difficult to solve; even numerical methods (which do exist) require substantial computing power.

You can probably now see the answer to your question - Newtonian gravity continues to be used in cases where relativistic effects can be neglected, simply because it is much easier to handle and solve, mathematically speaking. For example, if you look at the Earth - which has weak gravity in the bigger scheme of things -, you will see that Newtonian gravity gives very accurate predictions under most circumstances. On the other hand, if you look at the gravity of our sun (or even stronger sources), you will notice that Newton gives predictions that are ever so slightly off, so GR is needed. You get the idea.

Formally, Newtonian gravity is the asymptotic limit of GR - for example, in the Schwarzschild metric, if you make r very large, you will recover Newtonian gravity. Hence, GR and Newtonian gravity aren’t really different and separate - Newton is a subset of GR, that is still valid under certain restricted circumstances, and hence continues to be used due to its simplicity and ease of application.
Thanks! At last something concrete Another thing is, that it's physically impossible to compress the matter in an uniform material or a liquid - so we can forget about the density in most of cases. But what would happen to orbits of planets in outer Solar System, if Sun would become a nebula (even if we assume no loss of total mass)? According to most of sources, it shouldn't have no effect on the gravitational field - just like Sun compressed into a black hole. But for me it seems incorrect. We don't see stars or planets moving around gas clouds (on the outside).

Also, if I can use the rubber sheet as an analogy and compare black holes to holes in the material of sheet - it's much easier, to make such hole, when you concentrate the mass/energy in a small space - so the distribution plays a key role in this case... I don't say, that Newton's equation is invalid - it just makes things easier. But it doesn't mean, that it can be used in the cases, which include difference of mass distribution - as there's no such value in the equation. But this is exactly, what is happening - and that's why in 90% of sources, we can read, that density has no effect on g. field, while GR tells something exactly opposite...

47. and that's why in 90% of sources, we can read, that density has no effect on g. field, while GR tells something exactly opposite
One needs to distinguish here between the region inside the mass distribution, and the vacuum region outside the mass distribution. The interior gravitation field will of course always depend on the density distribution - this is true both in GR and in Newtonian gravity.

If the source body has an irregular shape and/or irregular distribution of density, then even the exterior field will in some way depend on the density distribution, at least in a region very close to the central body.

However, if the gravitational source is perfectly and uniformly spherical, uncharged, and does not rotate, then its actual size, composition and density has no bearing on the external field; it is completely determined just by the total mass alone. Again, this is true both in GR and in Newton.

So it all depends on the specific scenario you are looking at. In many real-world situations, spherical symmetry is a very good approximation to what is happening, and it is also very easy to treat mathematically - that is why it is used so often in the literature.

But what would happen to orbits of planets in outer Solar System, if Sun would become a nebula (even if we assume no loss of total mass)?
The problem with this is that, if you retain the total mass of the sun, then the nebula would have to be spread out very widely (much further than the size of the solar system itself) in order not to collapse back into a star under its very own gravity. Again, remember that gravity acts not only external to the nebula, but also in its interior. This is why you don’t see planets orbiting nebulas - firstly, the conditions for planets to form in the first place aren’t there, and secondly, the nebula - even if it has the same total mass - cannot be considered a point source, so standard orbital mechanics don’t apply here.

48. Originally Posted by AstralTraveler
Ok, but it's still just a theorem - which still needs some empirical confirmation.
We don't have empirical confirmation, so how can this discussion continue? Clearly, you need to accept some theoretical statements. After all, you are using theoretical statements to explain why you think Birkhoff's theorem is wrong. Are you trying to argue that General Relativity is inconsistent?

Originally Posted by AstralTraveler
So, in the case of such plate, we include the distribution of mass in the object - and as a result, force of attraction is directed perpendiculary towards the surface(s) and not towards the center of it's mass. So why the same rules can't be used in the case of an uniform sphere?
In the case of an infinite plate, there is no centre. Symmetry demands that the "force of attraction" is normal to the surface. Similarly, for a spherically symmetric distribution, symmetry demands that the "force of attraction" be directed towards the centre. So, it isn't that the infinite plate and the sphere have different rules, but that a different principle is used to show that the rules are indeed the same in both cases.

Originally Posted by AstralTraveler
I just can't understand, why the change of r won't affect the gravitational field outside the sphere.
You do realise that the gravity inside a non-rotating hollow spherical shell is zero (the spacetime is flat)? Thus, the matter outside of radius r has no contribution to the field at or inside of radius r.

Originally Posted by AstralTraveler
why only the uniform sphere causes attraction directed towards a single point?
It's not just a uniform sphere. The spherical distribution of matter can have any radial distribution. It is this independence of the radial distribution that allows one to say that it is only the total mass that determines the field at a distant location.

49. Originally Posted by KJW

The "space" of relativity is four-dimensional spacetime. Note that the term "space" has a very general meaning although I concede that confusion may have arisen because it also means the 3-dimensional part of spacetime.
Are the tangent vectors on that space arrived at using parallel transport along the 4 axes and making the vectors smaller and smaller?

50. However, if the gravitational source is perfectly and uniformly spherical, uncharged, and does not rotate, then its actual size, composition and density has no bearing on the external field; it is completely determined just by the total mass alone. Again, this is true both in GR and in Newton.
But if I would use Poisson's equation, to calculate the external gravitational field for 2 uniform spheres with the same mass, but made of different materials - like metal and styrofoam - shouldn't I get different results, considering the difference of density?

51. In the case of an infinite plate, there is no centre. Symmetry demands that the "force of attraction" is normal to the surface. Similarly, for a spherically symmetric distribution, symmetry demands that the "force of attraction" be directed towards the centre. So, it isn't that the infinite plate and the sphere have different rules, but that a different principle is used to show that the rules are indeed the same in both cases.
And what about a finite uniform disc - even, if we include some thickness? All finite geometric shapes have their centers, so is the attraction directed towards the center of a disc, or towards it's surfaces?

You do realise that the gravity inside a non-rotating hollow spherical shell is zero (the spacetime is flat)? Thus, the matter outside of radius r has no contribution to the field at or inside of radius r.
It makes sense in the case of a hollow sphere - changing it's volume won't affect the concentration of energy/mass in space

It's not just a uniform sphere. The spherical distribution of matter can have any radial distribution. It is this independence of the radial distribution that allows one to say that it is only the total mass that determines the field at a distant location.
But is there an attraction towards the center of mass, inside an uniform material? If there would be a spherical body made of a liquid or gas, would the concentration of matter grow, the closer we would get to the center? If we keep in mind, that every medium is made of particles, what would cause such attraction? Particles with the same mass would rather attract all other particles around them - so they should float freely in an uniform medium. And if there won't be no attraction towards the center of a medium, other bodies won't be attracted towards a point, but towards the medium, as a whole.

Similar, like in the case of a flat disc - objects are rather attracted towards the surfaces and not towards the center. If this would be the case, a ball dropped on an uniform flat disc, would place itself always right in the middle...

52. We don't have empirical confirmation, so how can this discussion continue? Clearly, you need to accept some theoretical statements. After all, you are using theoretical statements to explain why you think Birkhoff's theorem is wrong. Are you trying to argue that General Relativity is inconsistent?
Theories are theories - you can disagree with them. I can fully accept a theory, only when it becomes a fact, or when it seems to me as logical. There's no proof, if a pulsating star has or has not any effects on bodies, which are orbiting it - so why should I accept it, if I think differently?

53. Originally Posted by AstralTraveler
However, if the gravitational source is perfectly and uniformly spherical, uncharged, and does not rotate, then its actual size, composition and density has no bearing on the external field; it is completely determined just by the total mass alone. Again, this is true both in GR and in Newton.
But if I would use Poisson's equation, to calculate the external gravitational field for 2 uniform spheres with the same mass, but made of different materials - like metal and styrofoam - shouldn't I get different results, considering the difference of density?
Suppose we have two uniform balls of the same mass but different radius. Let the radius of one ball be R1 and the radius of the other ball be R2. Let R2 be greater than R1 (and therefore R2 is of the less dense ball). Then for both balls, the spacetime curvature at all locations outside the R2 radius will be identical. At locations inside the R1 radius, the spacetime curvature will be different due to the difference in density of the two balls. At locations between R1 and R2, the spacetime curvature will also be different because in one case, the curvature is due to the matter, whereas in the other case, the curvature is of the vacuum and due to pure gravitation.

Let me repeat: The spacetime curvature at all locations outside the R2 radius will be identical. The spacetime curvature at all locations inside the R2 radius will be different.

54. Originally Posted by AstralTraveler
But if I would use Poisson's equation, to calculate the external gravitational field for 2 uniform spheres with the same mass, but made of different materials - like metal and styrofoam - shouldn't I get different results, considering the difference of density?
You get different fields for the interior of those bodies, and the bodies themselves will have a different radius. However, in the exterior vacuum you will not be able to observe any difference at all.

Theories are theories - you can disagree with them.
Birkhoff’s theorem (in the context of physics) is a mathematical deduction based on the laws of gravity. Each step of this derivation can be shown to be mathematically rigorous - this means that, the only logical possibility for Birkhoff’s theorem to be wrong, would be for the laws of gravity to be wrong. And this means both GR and Newton, because spacetime being flat / the absence of a gravitation force in the interior of a hollow shell is also a consequence of Birkhoff’s theorem. In fact, the thin shell case logically implies that the exterior spacetime of a pulsating star must be stationary. This may not be immediately obvious, but can be shown in a mathematically rigorous way. But I do not blame you at all for questioning it, because this outcome seems at first glance both surprising and counterintuitive. But when you look at the logical basis behind the theorem, it becomes a very straightforward conclusion, much like the sides of a triangle being related via Pythagoras’ theorem. You can disagree with these things, but since there are rigorous proofs for these theorems, the disagreement is rather meaningless.

Since to date both Newton and GR are in excellent agreement with experiment and observation within their respective domains of applicability, the current consensus is that Birkhoff’s theorem is quite valid.

Obviously, experimentally testing a pulsating star scenario is difficult, but you could - at least in principle - test the thin shell case, which is the same phenomenon. Not sure if that experiment has been done. If you are inclined towards experimental physics, then this would be an interesting project for you to sink you teeth into

So essentially, what I am trying to say is that, in order to experimentally test Birkhoff’s theorem, you can test the laws of gravity it is based on, because its derivation from them is mathematically rigorous and hence not in dispute. There is a very large body of experimental tests for both Newton and GR.

P.S. Birkhoff’s theorem is a direct and logical consequence of the symmetries of Schwarzschild spacetime, which itself enjoys a large number of experimental confirmations over the past 100 years.

55. I can fully accept a theory, only when it becomes a fact, or when it seems to me as logical.
You need to be careful here, because “when it seems to me as logical” is not an objective criteria for something being true or false. It’s just a subjective opinion based on someone’s current level of knowledge and understanding, which may well be incomplete or flawed. What is logical is only a rigorous mathematical deduction.

For example, when I first started to learn relativity, I was completely convinced that spacetime in an accelerated frame must be curved, because at the time that was my understanding of the model. Of course, that understanding was flawed, and even though it “seemed logical” at the time, it is nonetheless wrong. It took me making a fool of myself on various forums, before I sat down and actually did the maths myself - and discovered that I was wrong, much to my amazement. I then had to go back and eat humble pie. That was a crucial and very important learning experience. Since that time, I have never trusted my personal version of “seems logical” ever again; instead, I look up the maths (if I can), and see for myself.

56. Originally Posted by AstralTraveler
In the case of an infinite plate, there is no centre. Symmetry demands that the "force of attraction" is normal to the surface. Similarly, for a spherically symmetric distribution, symmetry demands that the "force of attraction" be directed towards the centre. So, it isn't that the infinite plate and the sphere have different rules, but that a different principle is used to show that the rules are indeed the same in both cases.
And what about a finite uniform disc - even, if we include some thickness? All finite geometric shapes have their centers, so is the attraction directed towards the center of a disc, or towards it's surfaces?
A uniform disc does not have spherical symmetry (nor the symmetry of an infinite plate). All finite geometric shapes may have their centres, but they do not necessarily have the symmetries that constrain the direction of the "force of attraction". Note that the gravitational field external to a distribution of matter must have the same symmetries as the distribution of the matter itself. Thus, for a uniform disc, the direction of the "force of attraction" at any given point must lie in the plane that is perpendicular to the disc and containing both the given point and the symmetry axis.

57. And if there won't be no attraction towards the center of a medium, other bodies won't be attracted towards a point, but towards the medium, as a whole.
But this is actually not true. Particles in an uniform medium might float freely and don't "feel" any attraction towards a specified point, but things will change, if there will be a second object, located within the g. field - let's say that it's mass is smaller, so it will be attracted towards the source of gravity, but is made of different material.

For example a small metal ball with the mass of 10g is attracted towards a big sphere of water, with the mass of 100kg. What will happen, when it will reach the surface of water? It will pass through the surface, place itself in the center of mass and start to attract particles of water (what will increase the density of liquid around the ball). So, for the metal ball, there will be attraction towards the central point and for the particles, there will be attraction towards the central object. But what would happen, if instead of metal, second object would be made of oil? It would reach the surface and remain there, as a flattened sphere. Now let's consider a third option - 10g bubble of hydrogen, which is placed inside the 100kg sphere of water. Effect - hydrogen will be "pushed out" from the water towards the surface. However "bubble" of hydrogen, placed outside of water will "wrap itself" around the entire sphere. And finally, a scenario, where both objects are made of water. What happens? Smaller object becomes part of the medium and the attraction disappears.

In all scenarios, smaller object had the same mass, but different densities/concentrations of matter and experienced 4 different effects of the same force. If gravity can be explained as curvature of space-time which turns potential energy of mass into a kinetic force, then it's the difference of energy/mass distribution, what defines the direction and magnitude of that force. Question is, if all 4 objects will be attracted towards the water with the same force? Observation shows, that yes - all of them will experience the same gravitational pull, as they all will be "falling to the bottom" of the same gravitational well. However I guess, that there will be difference between the distances, at which those objects will start to experience the pull.

But will there be a difference, if instead of water, we would use 100kg of hydrogen? I say: absolutely! Different distribution of energy/mass creates different shape of gravitational well. In the case of a gas, this well should be much more "shallow", but will affect bigger volume of time-space.

I will leave for now the interaction between objects with the same mass and different densities, although I think, that they should be attracted towards eachother with the same force...

58. A uniform disc does not have spherical symmetry (nor the symmetry of an infinite plate). All finite geometric shapes may have their centres, but they do not necessarily have the symmetries that constrain the direction of the "force of attraction". Note that the gravitational field external to a distribution of matter must have the same symmetries as the distribution of the matter itself. Thus, for a uniform disc, the direction of the "force of attraction" at any given point must lie in the plane that is perpendicular to the disc and containing both the given point and the symmetry axis.
Yes. But what makes a sphere, a special case, is the fact, that any force, which is oriented perpendicullary to it's surface, is at the same time pointing towards it's center - and this is most likely allowing us to use a more simple equation.

Note that the gravitational field external to a distribution of matter must have the same symmetries as the distribution of the matter itself.
Ok, so a solid and uniform cube will create a field, distributed in a cube of time-space. This means, that the symmetry of gravitational well is directly connected with the shape of source. But isn't the size/volume one of the properties, which describe the shape of an object? Won't a bigger sphere create a g. well with bigger radius?

59. Originally Posted by Markus Hanke
I can fully accept a theory, only when it becomes a fact, or when it seems to me as logical.
You need to be careful here, because “when it seems to me as logical” is not an objective criteria for something being true or false. It’s just a subjective opinion based on someone’s current level of knowledge and understanding, which may well be incomplete or flawed. What is logical is only a rigorous mathematical deduction.

For example, when I first started to learn relativity, I was completely convinced that spacetime in an accelerated frame must be curved, because at the time that was my understanding of the model. Of course, that understanding was flawed, and even though it “seemed logical” at the time, it is nonetheless wrong. It took me making a fool of myself on various forums, before I sat down and actually did the maths myself - and discovered that I was wrong, much to my amazement. I then had to go back and eat humble pie. That was a crucial and very important learning experience. Since that time, I have never trusted my personal version of “seems logical” ever again; instead, I look up the maths (if I can), and see for myself.
Thanks! This is the main problem with subjects, like time-space - which are more, like an abstract concept, lacking some physical substance to grasp. Magnetic field lines seem to be like granite monuments, compared to some parts of GR It sounds logical, that if because of acceleration (velocity) we are experiencing time dilation, time-space might be indeed curved in the accelerated frame of reference - but I wouldn't try to make such claims, as it all can be just well... relative... For one observer it can be curved, but for a secod it doesn't have to... Luckily gravity is connected with physical matter and we can observe it's effects - so there are some empirical ways to confirm/contradict a claim...

60. again a double...

61. Originally Posted by Markus Hanke

For example, when I first started to learn relativity, I was completely convinced that spacetime in an accelerated frame must be curved, because at the time that was my understanding of the model. Of course, that understanding was flawed, and even though it “seemed logical” at the time, it is nonetheless wrong. It took me making a fool of myself on various forums, before I sat down and actually did the maths myself - and discovered that I was wrong, much to my amazement. I then had to go back and eat humble pie. That was a crucial and very important learning experience. Since that time, I have never trusted my personal version of “seems logical” ever again; instead, I look up the maths (if I can), and see for myself.
I am stuck in that misunderstanding now( well have been since the thought occurred to me).

Have you a pointer or two since learning the maths for me is an excruciatingly long drawn out process and I can only imagine that happening fortuitously if at all ?

Gravity curves spacetime coordinates but acceleration doesn't?

62. I just came back from a walk with my dog, during which I was thinking about the mass within and outside a given radius, inside a solid and uniform sphere. Science tells, that entire mass beyond the distance of a point from the center of mass, doesn't affect that point gravitationally - and only the inner mass induces a force.

But what about a diver at the bottom of a sea? Doesn't he feel the weight of water, which is above him? If the mass, placed beyond the given radius would cancel itself out, we shouldn't feel it's influence. But it seems, that for the outer mass, attraction turns simply into pressure. Of course, a miner won't feel the pressure of the soil above him, until the mine won't collapse. But he will directly experience increased pressure of air...
https://physics.stackexchange.com/qu...elow-sea-level

63. Gravity curves spacetime coordinates but acceleration doesn't?
Well, you can slow down and cancel the effects of acceleration, but you can't simply get rid of the mass. Velocity is a relative value, while mass remain mass - no matter from which side you keep looking at it It actually makes sense...

64. Gravity curves spacetime coordinates but acceleration doesn't?
It actually makes a LOT of sense, when you think about it - when you are driving a car, space beyond the vehicle seems to be curved, while inside it remains "stright". Opposite effect will be noticed by someone, who's standing on the road - for him the car will seem to be distorted.
I have the feeling, that I think too much... :P

65. Originally Posted by AstralTraveler
A uniform disc does not have spherical symmetry (nor the symmetry of an infinite plate). All finite geometric shapes may have their centres, but they do not necessarily have the symmetries that constrain the direction of the "force of attraction". Note that the gravitational field external to a distribution of matter must have the same symmetries as the distribution of the matter itself. Thus, for a uniform disc, the direction of the "force of attraction" at any given point must lie in the plane that is perpendicular to the disc and containing both the given point and the symmetry axis.
Yes. But what makes a sphere, a special case, is the fact, that any force, which is oriented perpendicullary to it's surface, is at the same time pointing towards it's center - and this is most likely allowing us to use a more simple equation.
What makes a sphere a special case is all the symmetries it possess. Thus, the gravity must be directed along the radial line because any other direction breaks the spherical symmetry. Spherical symmetry implies that the magnitude of the gravity depends only on the distance from the centre.

Originally Posted by AstralTraveler
Note that the gravitational field external to a distribution of matter must have the same symmetries as the distribution of the matter itself.
Ok, so a solid and uniform cube will create a field, distributed in a cube of time-space. This means, that the symmetry of gravitational well is directly connected with the shape of source. But isn't the size/volume one of the properties, which describe the shape of an object? Won't a bigger sphere create a g. well with bigger radius?
The larger size is compensated by the lower density. Not only do we need to consider the geometric symmetry of the sphere, but also transformation properties of the gravitational field itself. For example, do you know how the gravitational field transforms under a change of scale? If one performs a scale transformation on the Schwarzschild metric¹, then a compensating coordinate transformation, the result is another Schwarzschild metric but with a different mass parameter. Thus, the single-parameter family of Schwarzschild metrics is symmetric to the combination of scale transformations and compensating coordinate transformations.

¹ Our entire discussion is about the properties of the Schwarzschild metric, which describes the geometry outside a non-rotating spherically symmetric distribution of matter (the vacuum part alone is described by the Schwarzschild metric, not the matter part).

66. Suppose we have two uniform balls of the same mass but different radius. Let the radius of one ball be R1 and the radius of the other ball be R2. Let R2be greater than R1 (and therefore R2 is of the less dense ball). Then for both balls, the spacetime curvature at all locations outside the R2 radius will be identical. At locations inside the R1 radius, the spacetime curvature will be different due to the difference in density of the two balls. At locations between R1 and R2, the spacetime curvature will also be different because in one case, the curvature is due to the matter, whereas in the other case, the curvature is of the vacuum and due to pure gravitation.

Let me repeat: The spacetime curvature at all locations outside the R2 radius will be identical. The spacetime curvature at all locations inside the R2radius will be different.
Here's why I have problems with accepting it, justy like this:

https://en.wikipedia.org/wiki/Gravity_well
"The rubber-sheet model

Consider an idealized rubber sheet suspended in a uniform gravitational field normal to the sheet. In equilibrium, the elastic tension in each part of the sheet must be equal and opposite to the gravitational pull on that part of the sheet; that is,

where k is the elastic constant of the rubber, h(x) is the upward displacement of the sheet (assumed to be small), g is the strength of the gravitational field, and ρ(x) is the mass density of the sheet. The mass density may be viewed as intrinsic to the sheet or as belonging to objects resting on top of the sheet. This equilibrium condition is identical in form to the gravitational Poisson equation

where Φ(x) is the gravitational potential and ρ(x) is the mass density. Thus, to a first approximation, a massive object placed on a rubber sheet will deform the sheet into a correctly shaped gravity well, and (as in the rigid case) a second test object placed near the first will gravitate toward it in an approximation of the correct force law."

And using the rubber sheet, it can be seen, that density of objects affects greatly the shape of gravity well...

67. Originally Posted by AstralTraveler
Poisson equation
The Poisson equation is based on Newtonian gravitation, so it cannot produce a result that contradicts Newtonian gravitation. It is fairly straightforward to prove the shell theorem from Newtonian gravitation. Therefore, the Poisson equation cannot produce a result that contradicts the shell theorem.

Why do you find it difficult to accept that the effect of greater density is cancelled by the effect of smaller radius?

68. again a double post

69. Originally Posted by AstralTraveler
But what about a diver at the bottom of a sea? Doesn't he feel the weight of water, which is above him?
That's an entirely different thing. The pressure exerted on the diver is caused by the attraction of the water towards the earth below, not the attraction of the diver towards the water above.

70. The larger size is compensated by the lower density. Not only do we need to consider the geometric symmetry of the sphere, but also transformation properties of the gravitational field itself. For example, do you know how the gravitational field transforms under a change of scale? If one performs a scale transformation on the Schwarzschild metric¹, then a compensating coordinate transformation, the result is another Schwarzschild metric but with a different mass parameter. Thus, the single-parameter family of Schwarzschild metrics is symmetric to the combination of scale transformations and compensating coordinate transformations.
Wait a second - I'm lost right now. Density of mass doesn't affect the Schwarzschild radius.
 rs = 2 G M / c2
It doesn't matter, if we deal with a nebula, or a neutron star - what matters, is to what size you need to compress the matter, to create a black hole - and initial density of an object doesn't have nothing to do with this. So, changing just the size or shape of a body, shouldn't affect it's Schwarzschild radius in any way.

Yes, it should change the properties of gravitational field itself, so the rest of variables in this equation may differ (I would expect it):
 ds2 = - ( 1 - rs / r ) dt2 + ( 1 - rs / r )-1 dr2 + r2 do2
However the mass of object remains the same and so should be the Schwarzschild radius. But from what you said, I can conclude, that by changing the size of a body, we change as well it's total mass... I don't get it...

71. Originally Posted by AstralTraveler
However the mass of object remains the same and so should be the Schwarzschild radius. But from what you said, I can conclude, that by changing the size of a body, we change as well it's total mass... I don't get it...
I should remark that a scale transformation also changes the time scale.

72. That's an entirely different thing. The pressure exerted on the diver is caused by the attraction of the water towards the earth below, not the attraction of the diver towards the water above.
I know, but doesn't it mean, that the diver is attracted towards earth below, together with the entire mass above him? And wouldn't it mean, that we should add the mass, placed beyond a given radius (not whole, but the one placed between attracted object and the surface) to the mass of attracted body, to calculate the attraction? Jumping off a cliff with a backpack full of rocks will make you more flat, when you hit the surface...

73. I should remark that a scale transformation also changes the time scale.
I would expect it to. If time dilation is connected with the "slopes" of gravity well, then it should change, when the curvature becomes more "shallow"

74. Why do you find it difficult to accept that the effect of greater density is cancelled by the effect of smaller radius?
Because due to smaller radius, curvature of time-space gets "deeper". If I concentrate 100 tonnes in the size of a grain of sand, it will penetrate the crust of Earth and maybe (if it won't melt on the way) will even reach the core. But if the same mass will be distributed over a volume of 1000km3, it will fly up to the sky, until it won't enter a medium with the same (or similar) density. Even without the force of buoyancy, such size won't allow it to penetrate the crust.

Or because a soccer ball and a bullet can have the same mass and the same velocity, but the ball will hurt you slightly, while a bullet might kill you. Or because it's much easier to puncture solid matter using a nail, than with a metal plate. When an object induces a force on a second object, this force is distributed over the entire surface, which is in contact with another body...

Why should I think, that gravity is something completely different and those obvious rules can't be applied to it?

75. The Poisson equation is based on Newtonian gravitation, so it cannot produce a result that contradicts Newtonian gravitation. It is fairly straightforward to prove the shell theorem from Newtonian gravitation. Therefore, the Poisson equation cannot produce a result that contradicts the shell theorem.
Maybe not today, but in the near future I will have to try using the Poisson's equation to calculate the external g. field for some objects with different densities but the same masses. Luckily it doesn't look THAT complicated, compared to some other formulas, which I saw today If it is indeed consistent with the rubber sheet model, then I should get different results - just like I saw it on the rubber sheet...

I can guess it already, just by looking at the equation:

It seems, that the mass density is actually the only variable value. By changing the density I will most likely get a diffetent result.

76. Originally Posted by AstralTraveler
Why do you find it difficult to accept that the effect of greater density is cancelled by the effect of smaller radius?
Because due to smaller radius, curvature of time-space gets "deeper".
Does it? I'm not convinced.

Originally Posted by AstralTraveler
If I concentrate 100 tonnes in the size of a grain of sand, it will penetrate the crust of Earth and maybe (if it won't melt on the way) will even reach the core. But if the same mass will be distributed over a volume of 1000km3, it will fly up to the sky, until it won't enter a medium with the same (or similar) density. Even without the force of buoyancy, such size won't allow it to penetrate the crust.

Or because a soccer ball and a bullet can have the same mass and the same velocity, but the ball will hurt you slightly, while a bullet might kill you. Or because it's much easier to puncture solid matter using a nail, than with a metal plate. When an object induces a force on a second object, this force is distributed over the entire surface, which is in contact with another body...

Why should I think, that gravity is something completely different and those obvious rules can't be applied to it?
None of these examples are relevant to the issue at hand... the gravitation associated with a non-rotating spherical distributions of matter.

77. Originally Posted by AstralTraveler
Maybe not today, but in the near future I will have to try using the Poisson's equation to calculate the external g. field for some objects with different densities but the same masses.
Let me save you the time and tell you that the solution to Poisson equation for a spherical object (the only shape that matters for this discussion) will quite simply recover Newton's formula.

78. doubled post...

79. I know, but doesn't it mean, that the diver is attracted towards earth below, together with the entire mass above him? And wouldn't it mean, that we should add the mass, placed beyond a given radius (not whole, but the one placed between attracted object and the surface) to the mass of attracted body, to calculate the attraction? Jumping off a cliff with a backpack full of rocks will make you more flat, when you hit the surface...
Ok, I admit, that I made a mistake in this case. Pressure of the mass, which is placed between me and the outer surface will cause different effects, depending on it's density (again). If I would take a column of air, which induces the pressure on me, turn it into a piece of metal and drop it on my head, it would most likely kill me. But even, if entire atmosphere except this column, would disappear, dropping the gas from above wouldn't hurt me at all. Conclusion - it's not the mass, which should be added to the force of attraction, but just it's pressure, which depends mostly on the concentration of matter in space...

Ok, enough for today. I'm going sleep. C'ya!

80. Let me save you the time and tell you that the solution to Poisson equation for a spherical object (the only shape that matters for this discussion) will quite simply recover Newton's formula.
Maybe you're right - but I can't be sure, if I won't check it by myself. Earth still would be placed in the middle of Universe, if Copernicus would believe in everything, what was considered, as scientific facts at this time

81. :/ What's with all those doubles?

82. Does it? I'm not convinced.
https://en.wikipedia.org/wiki/Energy_density
Well, I won't force you to believe in anything, however from scientific point of view, there's not much to argue with...

None of these examples are relevant to the issue at hand... the gravitation associated with a non-rotating spherical distributions of matter.
I still remember the subject of this discussion. Who said, that the grain of sand, with concentrated 100 tonnes of mass can't be spherical - just as the object with volume of 1000km3... Let's leave the rest for now, as you are still not convinced about the concept of force/energy/mass distribution affecting the way of interaction with environment

83. Originally Posted by KJW
Originally Posted by AstralTraveler
Why do you find it difficult to accept that the effect of greater density is cancelled by the effect of smaller radius?
Because due to smaller radius, curvature of time-space gets "deeper".
Does it? I'm not convinced.
Ok, on further consideration, I realised that I dropped the ball on this one. Yes, the potential is "deeper" at the centre of the smaller more dense ball than at the centre of the larger less dense ball. But the "depth" of the potential at the centre is not the issue. It is the "depth" of the potential at a given distance far enough away from the ball that is issue, and these are the same for both balls (of equal mass). So the "deeper" potential at the centre of the smaller more dense ball does not suffice to answer my question: Why do you find it difficult to accept that the effect of greater density is cancelled by the effect of smaller radius? Bear in mind that you are fighting an uphill battle against both the shell theorem of Newtonian gravity and Birkhoff's theorem of general relativity.

84. Originally Posted by AstralTraveler
Let me save you the time and tell you that the solution to Poisson equation for a spherical object (the only shape that matters for this discussion) will quite simply recover Newton's formula.
Maybe you're right - but I can't be sure, if I won't check it by myself.
Fair enough. Do you know how to solve the Poisson equation? I'll give you a hint: Green's function. Are you familiar with the divergence theorem?

85. For one observer it can be curved, but for a secod it doesn't have to
The mathematical object that describes curvature - the Riemann tensor - is something that all observers agree on. It is not a relative quantity.

Gravity curves spacetime coordinates but acceleration doesn't?
Yes. In order for spacetime to be curved, you need a gravitational source, which mere acceleration in an otherwise empty region of space is not. This once again may be somewhat counterintuitive at first glance, but is actually rather straightforward to prove; you can write down the general form of the metric for an accelerated frame, and then use this to directly calculate the components of the curvature tensor. The result is that all components are identically zero, so spacetime in such a local frame is perfectly flat.

Velocity is a relative value
Yes, but proper acceleration is not; it’s an invariant.

86. Hate to butt in but always wanted to ask this question....

A layman's puzzlement....I once had something in a paper bag. I lifted the bag too quickly and the bottom gave out, spilling the contents. Made me think of the rubber sheet for some reason. If I were to place an object on the sheet and then lift it, wouldn't the well formed by the object deepen? The mass of the object wouldn't increase by lifting(accelerating?) the sheet upward would it? Does this deepening/stretching of the rubber sheet represent a fair example of what would occur if space time was in motion, something which I'm not sure can even be true? What's happening here?

87. Originally Posted by Markus Hanke

Gravity curves spacetime coordinates but acceleration doesn't?
Yes. In order for spacetime to be curved, you need a gravitational source, which mere acceleration in an otherwise empty region of space is not. This once again may be somewhat counterintuitive at first glance, but is actually rather straightforward to prove; you can write down the general form of the metric for an accelerated frame, and then use this to directly calculate the components of the curvature tensor. The result is that all components are identically zero, so spacetime in such a local frame is perfectly flat.
Not trying to be obtuse ,but what about artificial gravity? (that is just acceleration isn't it?)

Have you seen Janus' nice representations? They look "curved" to me

Artificial gravity... - Physics - Science Forums

Can acceleration curved spacetime on a local level?

88. Ok, on further consideration, I realised that I dropped the ball on this one. Yes, the potential is "deeper" at the centre of the smaller more dense ball than at the centre of the larger less dense ball. But the "depth" of the potential at the centre is not the issue. It is the "depth" of the potential at a given distance far enough away from the ball that is issue, and these are the same for both balls (of equal mass). So the "deeper" potential at the centre of the smaller more dense ball does not suffice to answer my question: Why do you find it difficult to accept that the effect of greater density is cancelled by the effect of smaller radius? Bear in mind that you are fighting an uphill battle against both the shell theorem of Newtonian gravity and Birkhoff's theorem of general relativity.
Ok, so at last we're going somewhere Most important is the fact, that gravity well leads to the surface of an object, not to it's center - until we're speaking about external gravitational field...

If we change just the size of Earth, it will affect the radius of gravity well - twice the size = twice the radius of well. But since we didn't change the mass/energy input, the well should get more "shallow" in order to maintain the ratio between mass and size of the gravity well. To get a gravity well of the same "depth", we would have to maintain the ratio size/mass.

89. Yesterday I spent the whole day on rethinking the subject and I got a nice idea. Problem is with the difference between the force of attraction and the pressure. According to science, gravity is caused by the pressure, which space-time curvature induces on mass/energy.

Mass and Gravity Explanation

But according to Newton's laws each action creates opposite reaction - so it should be absolutely valid to say, that the space-time curvature is created because of the pressure, which mass/energy induces on space-time. And it is a well known fact, that pressure depends on the density of energy/mass...

90. I found as well a nice publication:
https://www.ijser.org/researchpaper/...d-Modeling.pdf
Here's the conclusion:
"Building a library of identified shapes for direct gravimetryvoid detection technique for geophysical exploration in smallareas. It is clear the differentiation between the shape in caseof fixing the size, depth, and density. A complete geometricalsimulation was done and in the future a complete library ofshapes will be tested for direct techniques in archeologicalapplications with constrained environment for void detection"

And if this principle works in small scales, why it shouldn'w work with bigger objects?

91. Originally Posted by zinjanthropos
Hate to butt in but always wanted to ask this question....A layman's puzzlement....I once had something in a paper bag. I lifted the bag too quickly and the bottom gave out, spilling the contents. Made me think of the rubber sheet for some reason. If I were to place an object on the sheet and then lift it, wouldn't the well formed by the object deepen? The mass of the object wouldn't increase by lifting(accelerating?) the sheet upward would it? Does this deepening/stretching of the rubber sheet represent a fair example of what would occur if space time was in motion, something which I'm not sure can even be true? What's happening here?
Hmm, isn't this the exact reason why pilots experience stronger gravity while turning/accelerating the plane? BTW nice example/question...

92. The mathematical object that describes curvature - the Riemann tensor - is something that all observers agree on. It is not a relative quantity.
But what about time dilation as a result of the curvature? Someone, who is placed deep within a gravity well won't be able to notice, that he's experiencing the dilation - so for him the space-time looks flat. To see the difference of time "flow", he needs to compare his clock, to another one, which is placed beyond the gravity well. However from his perspective, clock placed inside the well shows correct time and it's the distant clock, which behaves incorrectly - so from his point of view, it's the time-space beond the well, which is curved...

So, if there are 2 observers, where one is located inside a g. well, both of them should agree, that there is a curvature of time-space - however both will tell, that it's the time-space around the second observer, which is curved... If the guy inside gravity well would be able to see the curvature in his local space-time, he should also notice, that his clock works too fast or too slow - but this is not, how time dilation works...

93. Originally Posted by AstralTraveler
Ok, on further consideration, I realised that I dropped the ball on this one.
This diagram makes it obvious how I dropped the ball. Suppose the radius of the earth is increased while maintaining the same mass. Then the red curves remain unchanged and the bigger earth would be higher up the potential well at a location where the surface of the earth touches the red lines. Because the red lines are unchanged, the gravity above the surface of the larger earth is unchanged and the radius of the geosynchronous orbit (the top blue line) is unchanged.

94. Ok, on further consideration, I realised that I dropped the ball on this one. Yes, the potential is "deeper" at the centre of the smaller more dense ball than at the centre of the larger less dense ball.
This is absolutely consistent with our current understanding of gravity. Science tells, that gravity on the surface of smaller and more dense object is stronger due to smaller distance from the center. As the effect, the gravity well will be indeed deeper - so, even if there are different reasons, we will get the same effects.

95. This diagram makes it obvious how I dropped the ball. Suppose the radius of the earth is increased while maintaining the same mass. Then the red curves remain unchanged and the bigger earth would be higher up the potential well at a location where the surface of the earth touches the red lines. Because the red lines are unchanged, the gravity above the surface of the larger earth is unchanged and the radius of the geosynchronous orbit (the top blue line) is unchanged.
Ok, good point. I have to think about it. Give me some time...

96. This diagram makes it obvious how I dropped the ball. Suppose the radius of the earth is increased while maintaining the same mass. Then the red curves remain unchanged and the bigger earth would be higher up the potential well at a location where the surface of the earth touches the red lines. Because the red lines are unchanged, the gravity above the surface of the larger earth is unchanged and the radius of the geosynchronous orbit (the top blue line) is unchanged.
Ok, I think, that I figured it out. There's a possible problem with the relation of size to the "depth" of gravity well. Looking at this diagram, it's visible that a small change of size would cause a significant difference of the "depth". But this ratio would get smaller, the bigger the planet would get. So - if the radius of Earth would change just by 5%, the "depth" of gravity well would change greatly. But the difference of "depth" would be much smaller, if the radius would change from 200% to 300%.

While "experimenting" with the rubber sheet, I didn't notice such significant differences of the "gravity well depths" for objects with the same mass and a small difference of size. This is because the difference of "depth" was equalized by the radius of "gravity well" - so, for objects with the same mass and a small difference of size, I observed small difference of "depth" and small difference of radius - instead of significant difference of depth and no difference of radius... On the second hand, changing the size of object from 200% to 300% didn't in fact change the "depth" too much, but it had a significant effect on the radius of "gravity well". So in both cases, all the proportions were maintained.

But still, I need to find some more scientific explanation...

97. Originally Posted by AstralTraveler
I need to find some more scientific explanation.
One difference between gravitation and the rubber sheet analogy that I don't think you are taking into consideration is that gravitation is a three-dimensional problem whereas the rubber sheet analogy is a two-dimensional problem.

98. But still, I need to find some more scientific explanation...
I have an idea... Let's take two spheres with the same mass, but with the difference of size at 5%. It should be possible to calculate, how much mass we would have to add to the bigger sphere, to keep the same "depth" of gravity well for both cases - with and without the changing gravity well radius.

If the radius of well wouldn't change, we would have to add much more mass to the bigger object, to "push" it to the same "depth", as the gravity well of smaller sphere. But with the changing radius of gravity well, it's depth should remain the same, if we keep the size/mass proportions for both objects... At least, that's what I think

99. One difference between gravitation and the rubber sheet analogy that I don't think you are taking into consideration is that gravitation is a three-dimensional problem whereas the rubber sheet analogy is a two-dimensional problem.
Yes, but it should be possible to make the same mathematical operations (like scaling) on 2D and 3D objects - you just need to add one dimension more...

100. In 3D you just get something, like this:

101. If the radius of well wouldn't change, we would have to add much more mass to the bigger object, to "push" it to the same "depth", as the gravity well of smaller sphere. But with the changing radius of gravity well, it's depth should remain the same, if we keep the size/mass proportions for both objects... At least, that's what I think

It should be: relation between "depth" and radius of gravity well would remain the same...

It doesn't matter that much in the case of planets and stars - but matters a LOT in the case of galaxies or clouds of gas. Gravitational fields of galaxies can't be calculated, with the same formulas, which apply to solid bodies. How can you think, that a galaxy and a solid body of equal mass create gravitational fieds with the same shape? It's something completely irrational for me...

Yes - planets and stars allow us to treat them as point-mass, but it doesn't mean, that we can apply this rule to everything else

I think, that "pressure" is the missing word in this concept. Time-space and energy/mass induce pressure on eachother - and this would solve the problem...

Pressure includes the distribution of energy/mass in space - and everything makes sense: objects with different distributions of energy/mass create gravitational fields with different shapes - gravity well gets bigger and more shallow with decreasing density of objects. And really, you don't need to be a genius, to figure it out...

One more thing - this applies to magnetic and electric fields as well. Distribution of energy/mass in space DOES matter...

I don't want to spam this thread, so I will keep editing this post

I want to explain, why the word pressure matters. With pressure comes the surface tension and buoyancy - and one should consider those properties of matter inside a gravitational field. Because the fun begins, when we move slightly out from the gravitational field of Earth

Only pressure, surface tension and buoyancy can explain, what's going on. On Earth a rock in water would go to the bottom due to it's density - but there's no bottom in space - so instead water would "wrap" itself around it. If we wouldn't disturb the sphere, rock would be placed in the middle of sphere. Those are things, which have to be considered, while speaking about gravity.

I have a test of your theory: explain me physically, what would happen, if we would drop a metal ball on a planet, which is made 100% of water? I can do it...

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