Does Energy Gravitate? A Check of Understanding

[russ_watters]

Fairly straightforward question/check of my understanding:

It is my understanding that energy gravitates (is that proper phrasing?), but not potential energy, since that is energy of a system, not internal energy of an object.

This would mean that as a star collapses and potential energy is converted to heat and pressure energy, its gravitational field measured from a distance would get stronger. Is that true? It seems to contradict what is typically said about black holes having exactly the same gravitational field strength at a distance after collapse as before - unless that's just a simplification due to the effect being small.

 

[PeterDonis]

It is my understanding that energy gravitates (is that proper phrasing?), but not potential energy, since that is energy of a system, not internal energy of an object.

No, in so far as energy gravitates, potential energy does too.

This would mean that as a star collapses and potential energy is converted to heat and pressure energy, its gravitational field measured from a distance would get stronger. Is that true?

No. If the collapse process doesn't radiate any energy away to infinity, then the gravitational field at a distance remains constant. If energy is radiated away to infinity, then the gravitational field at a distance weakens by the mass equivalent of the energy radiated away.

 

[PAllen]

You can see what Peter refers to for potential energy in the formula for Komar mass. The 'locally measured mass' gets discounted in its contribution to mass measured at a distance by the time dilation factor, which can be taken as a measure of its loss of potential energy.

 

[Naty1]

PeterDonis and PAllen have taught me the source of gravity, the stress energy tensor, includes all forms of energy...the energy of all types of fields.

As long as a star collapse is symmetric, the components of collapse offset as observed from an exterior distance.

The other 'trick' they have taught me is that the SET is properly in the frame of the center of mass...so for example, the kinetic energy of a laterally moving star does not reflect in it's gravitational nature. Spacetime in special relativity is usually described as flat spacetime...Minkowski spacetime as an example. Yet objects in relative motion 'distort' space and time in SR...We call that time dilation and length contraction. And objects in SR can move in curved worldlines...that is, curved paths in spacetime. So there IS a type of curvature in SR, but it is not gravitational in nature. In other words, bodies in relative motion may curve spacetime but not in a gravitational way.

It was explained to me elsewhere in these forums some years ago that you can picture world line [path] curves in SR as you would curves on a flat graph paper. When gravitational curvature is involved, as in GR, the graph paper itself on which the curved worldlines are drawn is itself curved.

 

[WannabeNewton]

PeterDonis and PAllen have taught me the source of gravity, the stress energy tensor, includes all forms of energy...the energy of all types of fields.

We have to be careful here. One of the physical reasons for why the always valid local conservation of energy $\nabla^{a} T_{ab} = 0$ fails to give, in general, a globally conserved energy is because there is no known physically meaningful notion of the local stress-energy of the gravitational field in general relativity i.e. $T_{ab}$ includes only non-gravitational field energy; however the total energy of space-time should surely take into account the self-energy density of the gravitational field. There have been objects constructed that try and represent the gravitational field energy for arbitrary space-time solutions to the Einstein equations but there have been issues tied with such constructions.

For example one can construct what is known as the Bel-Robinson tensor $T_{abcd} = C_{aecf}C_{b}{}{}^{e}{}{}_{d}{}{}^{f} -\frac{3}{2}g_{a[b}C_{jk]cf}C^{jk}{}{}_{d}{}{}^{f}$ where $C_{abcd}$ is the Weyl curvature tensor. One can show that $T_{abcd} = T_{(abcd)}$ and that $\nabla^{a}T_{abcd} = 0$. As you can see, $T_{abcd}$ has properties similar to that of $T_{ab}$ and is actually analogous in form to the stress-energy tensor of the electromagnetic field except instead of the electromagnetic field tensor we are using the space-time curvature (as we would expect if this quantity is to be related to the gravitational field). Regardless, $T_{abcd}$ does not even have the appropriate units.

Something else that is done is the introduction of the "Landau-Lifgarbagez pseudo tensor" $t_{ab}$ which is constructed out of the Einstein tensor but $t_{ab}$ fails to be gauge invariant, owing again to the comments made in the first paragraph.

 

[Naty1]

wannabe posts:

...a globally conserved energy is because there is no known physically meaningful notion of the local stress-energy of the gravitational field in general relativity i.e. Tab includes only non-gravitational field energy; however the total energy of space-time should surely take into account the self-energy density of the gravitational field.

I roughly get the gist of that. But I had thought the non linearity of T_ab reflected the self energy density...I guess not...

Can you suggest a better way to phrase this:

Quote by Naty1
... the source of gravity, the stress energy tensor, includes all forms of energy...the energy of all types of fields.

 

[Naty1]

I just checked wikipedia...and I find this statement conflicted..

The stress–energy tensor (sometimes stress–energy–momentum tensor) is a tensor quantity in physics that describes the density and flux of energy and momentum in spacetime, generalizing the stress tensor of Newtonian physics. It is an attribute of matter, radiation, and non-gravitational force fields. The stress–energy tensor is the source of the gravitational field in the Einstein field equations of general relativity

http://en.wikipedia.org/wiki/Stress-energy_tensor

Sounds like they are saying the SET is the source of the gravitational field, but not its energy??

 

[WannabeNewton]

The stress energy tensor is the source of the gravitational field in general relativity but that does not mean it has the ability to codify the energy density of the gravitational field itself for arbitrary space-times. There is no known general prescription for extending the Newtonian energy density for the gravitational field to arbitrary space-times within the framework of general relativity, especially if the general covariance of general relativity is to be preserved. See here: http://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor

 

[PeterDonis]

Sounds like they are saying the SET is the source of the gravitational field, but not its energy??

You should know by now that Wikipedia is not a good source of info about questions with any degree of subtlety to them.

The SET appears on the RHS of the Einstein Field Equation; that's the sense in which it is the "source" of gravity. There is no "energy of the gravitational field" in the SET because, from the standpoint of the EFE, the gravitational field is the "effect", not the "source"; it's on the LHS of the EFE, not the RHS.

(Of course you could try to reshuffle terms from one side of the EFE to the other; but the standard way of writing the EFE has the unique property that the covariant divergence of the LHS, the Einstein tensor, is identically zero, because of the Bianchi identities. That means the covariant divergence of the RHS, the standard SET, is also zero, and *that* means the SET is locally conserved; the "stuff" the SET describes is not created or destroyed at any point in spacetime. That's a highly desirable property.)

Whether or not you consider the SET to be the only "energy" present depends on how you define energy; as WannabeNewton pointed out, there are ways to define "energy in the gravitational field", even though that isn't included in the SET, but all of them have issues. (One key issue is that all of these definitions amount to reshuffling terms from one side of the EFE to the other, which, as noted above, destroys the highly desirable property of local conservation of "energy".)

Btw, I blogged about this on PF some time back, in response to another thread where this came up:

https://www.physicsforums.com/blog.php?b=4287

That series of posts goes into more detail about the stuff I discussed above, as well as other relevant issues.

 

[49ers2013Champ]

Naty:

I'm not going to dispute anything Donis has taught you--I know better--but are you saying that there is spacetime curvature that has nothing to do with gravity? I'm a little confused here. I thought bodies in relative motion, at least for the most, were following curvatures in spacetime (gravity) being created by the mass of other bodies.

 

[nitsuj]

You should know by now that Wikipedia is not a good source of info about questions with any degree of subtlety to them.

As a user/abuser of wikipedia, I like your diplomatic wording

The SET appears on the RHS of the Einstein Field Equation; that's the sense in which it is the "source" of gravity. There is no "energy of the gravitational field" in the SET because, from the standpoint of the EFE, the gravitational field is the "effect", not the "source"; it's on the LHS of the EFE, not the RHS.

lol so we can't "harness" the effect of a gravitational field and extract energy from it

 

[PAllen]

lol so we can't "harness" the effect of a gravitational field and extract energy from it

Well, per the so called Penrose process, you can extract energy from a rotating BH until it eventually vanishes.

 

[nitsuj]

Well, per the so called Penrose process, you can extract energy from a rotating BH until it eventually vanishes.

WHAT! Is that for real? Rotating specifically; so is that energy spitting out as gravitational waves?

Can't wait to read about that...heading to wikipedia, sorry PeterDonis

 

[PAllen]

WHAT! Is that for real? Rotating specifically; so is that energy spitting out as gravitational waves?

Can't wait to read about that...heading to wikipedia, sorry PeterDonis

Well, I misspoke about one thing. The Penrose process can proceed not until the BH vanishes, but only until it has lost all angular momentum. Effectively, the angular momentum is extracted from the BH.

 

[WannabeNewton]

WHAT! Is that for real? Rotating specifically; so is that energy spitting out as gravitational waves?

Can't wait to read about that...heading to wikipedia, sorry PeterDonis

Well the penrose process deals with particles. There is a wave analogue called superradiance. This works for gravitational as well as electromagnetic and scalar waves for as long as the stationary black hole is rotating.

 

[PeterDonis]

are you saying that there is spacetime curvature that has nothing to do with gravity?

I don't think he is, but if he is, he's not correct. Spacetime curvature is always associated with gravity. (I would say spacetime curvature *is* gravity, but that's really only true for tidal gravity, and the term "gravity" itself is broader than that.)

I thought bodies in relative motion, at least for the most, were following curvatures in spacetime (gravity) being created by the mass of other bodies.

Gravity is not the only interaction: there's electromagnetism, and the weak and strong nuclear forces, as well. All of them can cause relative motion. But only gravity does so by spacetime curvature.

 

[PeterDonis]

is that energy spitting out as gravitational waves?

A single rotating BH can't emit gravitational waves. The energy extracted by the Penrose process comes out as additional kinetic energy of particles coming out of the ergosphere, or additional wave energy of (non-gravitational) waves coming out of the ergosphere.

 

[nitsuj]

Thanks for the explinations guys! I read a snip-it of the wiki too,

It described a piece of matter entering the "ergosphere" and then it splits into two, okay whatever, some magic happens.

And one piece of matter falls into the BH and the other is flung out into the universe with extra energy. But it clearly says this relates to the spinning of the black hole. Not an "extraction" of energy from the gravity field of the black hole.

My "WHAT!" has gone down to a "that makes sense". (only understanding where the energy comes from, and from what i read here and in the wiki it ain't the gravitational field of the BH)

 

[WannabeNewton]

The point is that the energy $E = -p_a \xi^a$ can be negative in the ergosphere. If we throw a particle at the ergosphere, we can arrange for it to blow up into two pieces: one which falls into the rotating black hole with negative energy and the other which escapes to infinity with a net positive energy greater than that of the original particle (by local conservation of energy). In doing this the negative energy of the infalling piece actually carries negative angular momentum into the black hole and we get an upper bound on this process after which it can no longer be done.

 

[pervect]

Fairly straightforward question/check of my understanding:

It is my understanding that energy gravitates (is that proper phrasing?), but not potential energy, since that is energy of a system, not internal energy of an object.

This would mean that as a star collapses and potential energy is converted to heat and pressure energy, its gravitational field measured from a distance would get stronger. Is that true?

No.

It seems to contradict what is typically said about black holes having exactly the same gravitational field strength at a distance after collapse as before - unless that's just a simplification due to the effect being small.

Issues like this are why we don't even have a definition for "energy" in GR, except for some important special cases.

To study the collapse of a star, you'd probably want to use one of those special cases, asymptotic flatness, in which case you'd have a couple of choices for energy, the ADM energy and the Bondi energy.

If the collapse is spherically symmetrical, it won't radiate any gravitational waves away, and the two masses will be identical, and constant.

If there is significant gravitational radiation, the Bondi mass will decrase, while the ADM mass won't.

There's an even simpler one, the Komar mass, that could also be useful, but strictly speaking it doesn't apply to collapse situations. I'm not sure how much trouble you can get into by trying to apply it to "slow" collapses, but since I haven't seen anyone try to propose such a thing formally in a paper, I have to suppose you can get into quite a bit of trouble. (It's possible it has been done successfully , and I've missed the paper, of course.)

Anyway, you'd have to get deep into the technicalities here to get a really good understanding - the ADM and Bondi masses in general don't come from solely the matter distribution, but come from the metric. This means that they can, in general, depend not only the matter, but the "gravitational fields" as expressed by the metric or any of the other tensor quantities that you can derive from it.

One example to illustrate - gravity waves carry "energy", in some sense, in that they have the ability to do work. (As per Fenyman's stickiy bead argument).

ADM mass includes the "energy" in gravity waves, Bondi mass doesn't, ,and you'd have a tough time making a static system that includes gravity waves, so I don't think it even applies.

 

[atyy]

Just a quick comment on the terminology.

In GR, we separate things into curved spacetime and matter. The gravitational field is spacetime.

When we say energy is the source of the gravitational field (spacetime curvature), we mean localizable energy of matter.

The gravitational field does not have localizable energy. "Energy" can be associated with the gravitational field under some circumstances, but this energy is not localizable to spacetime points, you have to associate it to a region of spacetime.

Although the gravitational field does not have localizable energy, which is usually considered the "source" of the gravitational field, GR allows spacetime curvature to be present even when no localizable energy due to matter is present, because the equations for the gravitational field are nonlinear.

The distant observer sits where no matter (localized energy) is present - he sits in a vacuum portion of the solution. This vacuum solution can be joined up to a solution containing matter, like the sun, or joined up to a complete vacuum solution all the way like a black hole. Because the solution where the external observer sits is the same regardless of which solution it is joined to, he cannot distinguish which solution has been used beyond the joint (unless he stops being faraway and falls and hits the surface if it is the sun, or start feeling massive tidal forces if it is a black hole).

 

[WannabeNewton]

Indeed the fact that the EP allows the gravitational field to vanish locally for freely falling observers demands that a meaningful notion of gravitational energy density must vanish locally as well.

Things like the Komar energy are more tied in with the global "gravitational energy". In fact the Komar energy comes right out of the force $F = -\int _{S}V^{-1}N^{b}\xi^{a}\nabla_{b}\xi_{a} dA$ that a stationary observer at infinity must exert in order to keep in place a unit surface mass density distributed over a 2-sphere $S$ (here $\xi^{a}$ is the time-like killing vector field of the stationary space-time, $V$ is the usual redshift factor, and $N^{b}$ is the unit outward normal to $S$ that is orthogonal to $\xi^{a}$). The general relativistic gravitational potential, for stationary space-times, is defined as $\varphi = \frac{1}{2}\ln (-\xi_{a}\xi^{a}) = \ln V$. Note that $\nabla_{b}\varphi = V^{-1}\nabla_{b}(-\xi_{a}\xi^{a})^{1/2} = -V^{-1}\xi^{a}\nabla_{b}\xi_{a}$ so the derivative of this gravitational potential does show up in the expression for $F$ and the Komar energy is proportional to $F$ so it is rather clear that the Komar integral for energy takes into account the global "gravitational energy" in that sense.

 

[WannabeNewton]

Because the solution where the external observer sits is the same regardless of which solution it is joined to, he cannot distinguish which solution has been used beyond the joint (unless he stops being faraway and falls and hits the surface if it is the sun, or start feeling massive tidal forces if it is a black hole).

What do you mean by this? An observer at spatial infinity can certainly distinguish between whether he is in the vacuum region of e.g. Schwarzschild spacetime as opposed to Kerr spacetime. He can, for example, take an extremely long massless string with a unit test mass attached to the other end and see the magnitude of the force he has to exert at infinity in order to keep that test mass stationary in the space-time.

The magnitude of the force he has to exert will be given by $F = V^{-1}[(\xi^{b}\nabla_{a}\xi_{b})(\xi_{c}\nabla^{a}\xi^{c})]^{1/2}$ and this will not be the same for the two space-times (particularly because in Schwarzschild space-time the stationary orbits are also locally non-rotating orbits whereas in Kerr space-time the stationary orbits are not locally non-rotating so the observer at spatial infinity will have to exert even more force on his end of the string in order to keep the unit test mass on the other end of the string stationary)

 

[russ_watters]

Thanks guys.

 

[PAllen]

What do you mean by this? An observer at spatial infinity can certainly distinguish between whether he is in the vacuum region of e.g. Schwarzschild spacetime as opposed to Kerr spacetime. He can, for example, take an extremely long massless string with a unit test mass attached to the other end and see the magnitude of the force he has to exert at infinity in order to keep that test mass stationary in the space-time.

I'm thinking he is just referring to Birkhoff's theorem. However, it can't be made as a general statement. For example, so far as I know, it s unknown, and believed not true that there is any solution in which a spinning fluid or mass solution has exactly the Kerr vacuum outside some surface. That is, there is no analog of Birkhoff's theorem for Kerr. Of course, as an approximation, Kerr is often used away from a spinning body. Yet, precise numeric solutions for specific spinning bodies show deviation from Kerr in the vacuum region.

 

[Naty1]

PeterDonis:

I blogged about this on PF some time back, in response to another thread where this came up:

https://www.physicsforums.com/blog.php?b=4287

Ah, I have only seen discussions of the 'yes' answer..gravity gravitates, non linearity...etc
Will have to read your blog several more times...

 

[Naty1]

49ers2013Champ

Naty:

I'm not going to dispute anything Donis has taught you--I know better--but are you saying that there is spacetime curvature that has nothing to do with gravity? I'm a little confused here. I thought bodies in relative motion, at least for the most, were following curvatures in spacetime (gravity) being created by the mass of other bodies

yup. there is non gravitational spacetime curvature...
[edit: delete the prior sentence, replace with There is apparent curvature..see PeterDonis
correction in a later post].
A simple example. A particle with rectilinear motion passes a stationary [or inertial] observer in their common frame. A coincident accelerating observer sees the same particle following a curved worldline...it appears spacetime is curved...that is a coordinate effect related to the acceleration, not a gravitational characteristic of the particle.

Here is how DrGreg explained that to me in more general terms several years ago in these forums: [some restatements of the same points appear using different words]

Lets draw the flat spacetime of an inertial observer on on a flat piece of graph paper. [ She sees a particle with rectilinear motion as a straight line.] If we switch to a non-inertial frame, an accelerated observer but still in the absence of gravitation, we are now drawing a curved grid, but still on the same flat sheet of paper. [He is accelerating and sees a curved trajectory of the same particle.]

GRAVITATIONAL "spacetime curvature" refers to the curvature of graph paper itself, regardless of observer, whereas visible [apparent] curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer. In the absence of gravity, spacetime [graph paper] is always "flat" whether you are an inertial observer or not; non-inertial observers will draw a curved grid on flat graph paper.

DrGreg:

When we talk of curvature in spacetime (either curvature of a worldline, or curvature of spacetime itself) we don't mean the kind of curves that result from using a non-inertial coordinate system, i.e, non-square grid lines in my analogy.

Gravitational spacetime curvature is coordinate independent; the observer based curvature obviously is not.

 

[WannabeNewton]

Of course, as an approximation, Kerr is often used away from a spinning body. Yet, precise numeric solutions for specific spinning bodies show deviation from Kerr in the vacuum region.

Wow that's very interesting. Would you know of any articles etc. that talk about that in more detail? I'd be interested in the physical reasons for such deviations.

 

[atyy]

What do you mean by this? An observer at spatial infinity can certainly distinguish between whether he is in the vacuum region of e.g. Schwarzschild spacetime as opposed to Kerr spacetime. He can, for example, take an extremely long massless string with a unit test mass attached to the other end and see the magnitude of the force he has to exert at infinity in order to keep that test mass stationary in the space-time.

The magnitude of the force he has to exert will be given by $F = V^{-1}[(\xi^{b}\nabla_{a}\xi_{b})(\xi_{c}\nabla^{a}\xi^{c})]^{1/2}$ and this will not be the same for the two space-times (particularly because in Schwarzschild space-time the stationary orbits are also locally non-rotating orbits whereas in Kerr space-time the stationary orbits are not locally non-rotating so the observer at spatial infinity will have to exert even more force on his end of the string in order to keep the unit test mass on the other end of the string stationary)

I'm thinking he is just referring to Birkhoff's theorem. However, it can't be made as a general statement. For example, so far as I know, it s unknown, and believed not true that there is any solution in which a spinning fluid or mass solution has exactly the Kerr vacuum outside some surface. That is, there is no analog of Birkhoff's theorem for Kerr. Of course, as an approximation, Kerr is often used away from a spinning body. Yet, precise numeric solutions for specific spinning bodies show deviation from Kerr in the vacuum region.

Yes, that's what I was referring to. Just addressing Russ's question in the OP about why it is said that one can't distinguish a spherically symmetric star from a static black hole when one is far from it, without dropping a probe to the surface or through the event horizon. It's interesting that it does seem possible to distinguish rotating stars from rotating black holes by probing outside the event horizon.

 

[WannabeNewton]

Yes, that's what I was referring to.

Gotcha. It wasn't clear to me if you were referring only to static asymptotically flat solutions or asymptotically flat space-times in general, hence the juxtaposition I referred to. Cheers!

 

[PeterDonis]

yup. there is non gravitational spacetime curvature...
A simple example. A particle with rectilinear motion passes a stationary [or inertial] observer in their common frame. A coincident accelerating observer sees the same particle following a curved worldline...it appears spacetime is curved...that is a coordinate effect related to the acceleration, not a gravitational characteristic of the particle.

This isn't spacetime curvature. The Riemann curvature tensor is zero. Put another way, there is no tidal gravity present. Spacetime curvature is not a coordinate effect.

Lets draw the flat spacetime of an inertial observer on on a flat piece of graph paper. [ She sees a particle with rectilinear motion as a straight line.] If we switch to a non-inertial frame, an accelerated observer but still in the absence of gravitation, we are now drawing a curved grid, but still on the same flat sheet of paper. [He is accelerating and sees a curved trajectory of the same particle.]

If the sheet of paper is still flat, then there is no spacetime curvature. Spacetime is the sheet of paper, not the coordinate grid. I doubt if DrGreg intended any other interpretation.

GRAVITATIONAL "spacetime curvature" refers to the curvature of graph paper itself, regardless of observer

"Gravitational spacetime curvature" is the only kind of spacetime curvature there is. Whatever kind of curvature the curvature of the coordinate grid is, it isn't spacetime curvature. See above.

 

[Naty1]

Quote by Naty1

yup. there is non gravitational spacetime curvature...

PeterDonis

This isn't spacetime curvature.

we agree...and the rest of your post is of course correct...Thanks

my wording was misleading...I edited my earlier post to read 'apparent' curvature...

In that earlier discussion DrGreg also provided a related mathematical insight which further confirms your comments:

In Minkowski [flat] coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of the worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature".

In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

[I don't usually post mathematical descriptions because I know very well there are too many aspects, subtleties, that elude me.]

For those interested in a more detailed discussion, and a complete actual quote from DrGreg on the subject of 'apparent curvature' , try here:

space time curvature caused by fast electron [Nov 7, 2011]
https://www.physicsforums.com/showthread.php?t=548148

edit: One of the great things about these forums is that there are always experts ready to keep us non experts within proper bounds!

 

[WannabeNewton]

The extrinsic curvature of a smooth curve embedded in a Riemannian or semi-Riemannian manifold is in no way a coordinate dependent quantity. The only thing it depends on is how the embedding map is defined. A circle of radius $r$ naturally embedded in $\mathbb{R}^{2}$ has an extrinsic curvature of $1/r$; this is totally unrelated to whether we use polar, cartesian or w\e coordinates on $\mathbb{R}^{2}$.

The only "apparent" coordinate dependent behavior of curves I can possibly think of in relation to what is being talked about is the fact that the Christoffel symbols need not vanish identically for e.g. polar coordinates on $\mathbb{R}^{2}$ so when we write down the geodesic equation in polar coordinates we get some non trivial expression of the form $\frac{\mathrm{d} ^{2}x^{\mu} }{\mathrm{d} s^2} = f^{\mu}(x^{\nu})$ whereas in Cartesian coordinates we just get $\frac{\mathrm{d} ^{2}x^{\mu'} }{\mathrm{d} s^2} = 0$ but regardless the geodesics will be straight lines.

EDIT: Sort of unrelated but interesting nonetheless: http://www.phas.ubc.ca/~dwang/extrinsic.pdf

 

[A.T.]

Here is how DrGreg explained that to me in more general terms several years ago in these forums: [some restatements of the same points appear using different words]

Lets draw the flat spacetime of an inertial observer on on a flat piece of graph paper. [ She sees a particle with rectilinear motion as a straight line.] If we switch to a non-inertial frame, an accelerated observer but still in the absence of gravitation, we are now drawing a curved grid, but still on the same flat sheet of paper. [He is accelerating and sees a curved trajectory of the same particle.]

GRAVITATIONAL "spacetime curvature" refers to the curvature of graph paper itself, regardless of observer, whereas visible [apparent] curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer. In the absence of gravity, spacetime [graph paper] is always "flat" whether you are an inertial observer or not; non-inertial observers will draw a curved grid on flat graph paper.

DrGreg:

Gravitational spacetime curvature is coordinate independent; the observer based curvature obviously is not.

Here is the picture by DrGreg:

Here the full post with more explanations:
https://www.physicsforums.com/showthread.php?p=4281670&postcount=20

 

[Naty1]

AT...thanks for the post from DrGreg...My discussion with him is now likely three years ago or so w.o any diagrams; had not seen those VERY nice illustrations from this year...Thanks..How does one copy such diagrams from here to Microsoft WORD?? All I get with COPY is the text,
not the illustrations?