What is the source of curvature in an accelerating frame?

[Adel Makram]

I can understand that a free falling frame in a curved space with a non-zero Riemann tensor has a zero Ricci tensor but I have a doubt about the opposite; does an accelerating frame in a vacuum space have a non-zero Ricci or Riemann tensor? if so, where do components of energy momentum tensor come from then?
Also what is the law of general transformation between the curved space-time and the flat space-time in a free-falling inertial frame?

 

[PeterDonis]

does an accelerating frame in a vacuum space have a non-zero Ricci or Riemann tensor?

No. Both are zero.

 

[PeterDonis]

what is the law of general transformation between the curved space-time and the flat space-time in a free-falling inertial frame?

You're going to have to be more specific about what you mean here. There are an infinite number of possible coordinates in curved spacetime.

 

[Adel Makram]

No. Both are zero.

So if Ricci tensor is zero, the solution of the metric equation means a flat space. But the accelerating frame coordinates are not flat. Is it possible to have another solution to a zero Ricci?

 

[PeterDonis]

the accelerating frame coordinates are not flat.

"Flat" isn't a property of coordinates. It's a property of spacetime. The coordinates of an accelerating frame are not inertial, and don't have all the same properties as inertial coordinates; but that has nothing to do with the flatness or lack thereof of spacetime itself.

If you want to check this, compute the Ricci tensor from the coordinates of an accelerating frame in flat spacetime--for example, from Rindler coordinates. You will find that it is zero.

 

[Adel Makram]

"Flat" isn't a property of coordinates. It's a property of spacetime. The coordinates of an accelerating frame are not inertial, and don't have all the same properties as inertial coordinates; but that has nothing to do with the flatness or lack thereof of spacetime itself.

If you want to check this, compute the Ricci tensor from the coordinates of an accelerating frame in flat spacetime--for example, from Rindler coordinates. You will find that it is zero.

If it is zero, how does the observer in that accelerating frame explain the gravity which exists according to the equivalence principle.

Also if the flattness is a property of a space, does this mean the space of an accelerating frame is also flat?

 

[A.T.]

If it is zero, how does the observer in that accelerating frame explain the gravity which exists according to the equivalence principle.

Gravitational coordinate acceleration doesn't require intrinsic spacetime curvature (which is related to tidal effects). For example, the cone spacetime in the below video is intrinsically flat (negligible tidal effect on small scale):



See also this post:

This is my own non-animated way of looking at it:

• A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[cyrilus]

By operating any arbitrary coordinate transformation from a free falling inertial frame you get the geodesic equation - expressed in terms of the Christofflel symbol

If the frame is free falling in a gravitational field, you find the generated inertial force to be the gravitational force in the new fixed referential (laboratory referential)

 

[Nugatory]

If it is zero, how does the observer in that accelerating frame explain the gravity which exists according to the equivalence principle.

You are misunderstanding the equivalence principle. It doesn't say that acceleration is gravity, it says that the local effects of the two are equivalent. You can have acceleration without gravity.

Also if the flattness is a property of a space, does this mean the space of an accelerating frame is also flat?

Saying "the space of an accelerating frame" makes no sense. The frame is just a rule we using for assigning coordinates to points on space, so different frames are still working with the same space. In the absence of gravity, space is flat no matter which frames we use describe it.

 

[A.T.]

In the absence of gravity, space is flat no matter which frames we use describe it.

In the absence of gravity, spacetime is flat. As for space, it depends how you define it. For example, the spatial geometry measured with rulers at rest in a rotating frame is not flat.

 

[Nugatory]

In the absence of gravity, spacetime is flat. As for space, it depends how you define it. For example, the spatial geometry measured with rulers at rest in a rotating frame is not flat.

Thank you - yes - that was typing my reply too quickly. It's always spacetime curvature that we're talking about here, not space.

 

[Adel Makram]

Gravitational coordinate acceleration doesn't require intrinsic spacetime curvature (which is related to tidal effects). For example, the cone spacetime in the below video is intrinsically flat (negligible tidal effect on small scale)

So, my two concerns here in this thread can be demonstrated in those nice graphs;
1) Graph C shows how a curved space-time (gravity) with non-vanishing Riemann tensor may have a vanishing Ricci tensor.
2) Graph B shows how an accelerating frame with non-inertial coordinates can exit in a flat space-time (with no gravity).

Regarding the first point, will the vanishing Ricci tensor imply an inertial frame, ignoring tidal effect? Can this be shown mathematically? For example, will vanishing all second derivatives of the metric imply automatically that we are ignoring the tidal effect?

Regarding the second point, will be any corresponding vanishing Ricci tensor in 2D? For example, with plane polar coordinates in a flat space, the line elements;
$$ds^2=dr^2+r^2 d\theta^2$$
here the metric $g_{22}$ is not zero, and the second derivatives of it with respect to $r$ is not zero, will Ricci tensor be still zero?

 

[PeterDonis]

\

will the vanishing Ricci tensor imply an inertial frame

You can always construct a local inertial frame in any spacetime, whether the Ricci tensor vanishes or not.

will vanishing all second derivatives of the metric imply automatically that we are ignoring the tidal effect?

No; ignoring all second derivatives of the metric implies that we are ignoring all tidal effects. This is what we do in a local inertial frame: we choose a particular event of interest, and we construct coordinates such that the metric $g_{\mu \nu}$ at the chosen event is equal to the Minkowski metric $\eta_{\mu \nu}$, and all first derivatives of the metric at the chosen event are zero. This can always be done. But we cannot make all of the second derivatives of the metric vanish, unless the spacetime as a whole is flat. The best we can do in a curved spacetime is to restrict attention to a patch of spacetime around the chosen event that is small enough that the second derivatives of the metric, within that patch, can be ignored, to whatever level of accuracy of measurement we are using. This amounts to choosing the patch of spacetime such that within it, tidal effects are small enough that we can ignore them.

here the metric $g_{22}$ is not zero, and the second derivatives of it with respect to $r$ is not zero

But the Riemann tensor is zero; the flat plane is still a flat plane. (Try computing the Riemann tensor explicitly; it's not as simple as just computing the second derivatives of the metric.) So the Ricci tensor is obviously zero; but that's because the space is flat.

 

[Adel Makram]

No; ignoring all second derivatives of the metric implies that we are ignoring all tidal effects. This is what we do in a local inertial frame: we choose a particular event of interest, and we construct coordinates such that the metric $g_{\mu \nu}$ at the chosen event is equal to the Minkowski metric $\eta_{\mu \nu}$, and all first derivatives of the metric at the chosen event are zero. This can always be done. But we cannot make all of the second derivatives of the metric vanish, unless the spacetime as a whole is flat. The best we can do in a curved spacetime is to restrict attention to a patch of spacetime around the chosen event that is small enough that the second derivatives of the metric, within that patch, can be ignored, to whatever level of accuracy of measurement we are using. This amounts to choosing the patch of spacetime such that within it, tidal effects are small enough that we can ignore them.

I like the bold very much.

 

[vanhees71]

But the Riemann tensor is zero; the flat plane is still a flat plane. (Try computing the Riemann tensor explicitly; it's not as simple as just computing the second derivatives of the metric.) So the Ricci tensor is obviously zero; but that's because the space is flat.

In this case, it's very easy, because we know how we can find new coordinates, where the metric components become constant, namly Cartesian coordinates
$$x=r \cos \vartheta, \quad y=r \sin \vartheta.$$
We also know that in the new coordinates the metric components are just given by $\delta_{jk}$. This is easy to prove, because
$$g_{jk}' \mathrm{d} {q'}^j \mathrm{d} {q'}^k=\delta_{jk} \mathrm{d} {q'}^j \mathrm{d} {q'}^k =\mathrm{d} x^2 + \mathrm{d} y^2.$$
Indeed, it's easy to prove. We have
$$\mathrm{d}x = \mathrm{d}r \cos \vartheta - r \sin \vartheta \mathrm{d} \vartheta, \quad \mathrm{d}y = \mathrm{d}r \sin \vartheta + r \cos \vartheta \mathrm{d} \vartheta,$$
and thus indeed
$$\mathrm{d} x^2 + \mathrm{d} y^2 = \mathrm{d} r^2 + r^2 \mathrm{d} \vartheta^2.$$
Since in the new coordinates the metric components are constant the curvature tensor vanishes and does so in any coordinate system.

 

[Adel Makram]

In this case, it's very easy, because we know how we can find new coordinates, where the metric components become constant, namly Cartesian coordinates

Nice trick, so the lesson is; wherever the space is flat, Ricci tensor will be always zero no matter how the metric is a function of coordinates because we can always find a new coordinates-system in the same space where the metric is constant.
But this is an indirect proof right? the complete proof is to calculate the components of Ricci tensor in the former coordinates.

 

[Nugatory]

But this is an indirect proof right? the complete proof is to calculate the components of Ricci tensor in the former coordinates.

If the tensor is zero in one coordinate system, then it will be zero in all coordinate systems. Thus, if you can show that it's zero in any coordinate system, your proof is complete.

 

[Adel Makram]

\

You can always construct a local inertial frame in any spacetime, whether the Ricci tensor vanishes or not.

I am still confused about one thing, having said that Ricci tensor is nill does not help in calculating the equation of motion or geodescics, right? I mean to calculate the trajectory of particle in graph C which represents the gravity, Ricci must not be zero because the motion of the particle is a function of $T_{ij}$. So having that a local inertial frame with a vanishing Ricci tensor is just to show the correctness of the equivalence principle in a small scale but it is not related to the equation of motion.

 

[PeterDonis]

having said that Ricci tensor is nill does not help in calculating the equation of motion or geodescics, right?

I'm not sure what you mean, but as far as I can tell, this is incorrect. See below.

to calculate the trajectory of particle in graph C which represents the gravity, Ricci must not be zero because the motion of the particle is a function of $T_{ij}$.

No, it isn't. We're talking here about the motion of a test particle, not the motion of the source. The motion of a test particle, assuming it is geodesic, is determined entirely by the spacetime geometry, which in the case under discussion is a vacuum solution, with zero Ricci tensor. "Vacuum" means the stress-energy tensor is zero.

having that a local inertial frame with a vanishing Ricci tensor is just to show the correctness of the equivalence principle in a small scale but it is not related to the equation of motion.

If the Ricci tensor vanishes in one frame, it vanishes in all frames. You can't use a zero Ricci tensor in one frame and a nonzero one in another; that's inconsistent.

 

[Adel Makram]

I will try to be more clear. Einstein field equation equates the left hand side which is represented by Einstein tensor, or Ricci tensor, with the right hand side which is represented by energy momentum tensor.
When Einstein in his famous paper, spoke about vanishing Ricci tensor in a matter-free field, in what context did he mention that? Was that in the context of explaining an accelerating frame in a flat space with no matter? If so, what does this have to do with derivation of geodesics in a field with matter that requiring non-vanishing energy momentum tensor and consequently non-vanishing Ricci tensor?

 

[PeterDonis]

Einstein field equation equates the left hand side which is represented by Einstein tensor, or Ricci tensor, with the right hand side which is represented by energy momentum tensor.

Yes. And if the stress-energy tensor vanishes, then the Einstein tensor must also vanish; and if the Einstein tensor vanishes, it's simple to show that the Ricci tensor must also vanish.

Note, though, that the EFE is local; in other words, it's perfectly possible to have a spacetime in which the stress-energy tensor and Einstein tensor vanish in one region, but do not vanish in another. For example, in a spacetime describing an isolated massive body surrounded by vacuum, the spacetime region inside the body will have a non-vanishing SET and Einstein tensor, while the region outside the body will have a vanishing SET and Einstein tensor.

When Einstein in his famous paper, spoke about vanishing Ricci tensor in a matter-free field, in what context did he mention that?

In the context of a vanishing stress-energy tensor. That's what "matter-free" means.

Was that in the context of explaining an accelerating frame in a flat space with no matter?

No. A vanishing stress-energy tensor does not mean spacetime must be flat. There are non-flat vacuum (vanishing stress-energy tensor) solutions to the Einstein Field Equation--i.e., there are non-flat metrics which still have a vanishing Einstein tensor. The Schwarzschild metric is the most obvious example.

what does this have to do with derivation of geodesics in a field with matter that requiring non-vanishing energy momentum tensor and consequently non-vanishing Ricci tensor?

Obviously if you have a non-vanishing stress-energy tensor, you have a non-vanishing Einstein tensor, and so you must have a metric that leads to a non-vanishing Einstein tensor, which must be a different metric from any metric that leads to a vanishing Einstein tensor. But the methods for deriving geodesics from the metric are the same; you write down the geodesic equation for that metric and solve it.

 

[Adel Makram]

Note, though, that the EFE is local; in other words, it's perfectly possible to have a spacetime in which the stress-energy tensor and Einstein tensor vanish in one region, but do not vanish in another. For example, in a spacetime describing an isolated massive body surrounded by vacuum, the spacetime region inside the body will have a non-vanishing SET and Einstein tensor, while the region outside the body will have a vanishing SET and Einstein tensor.

I start to understand the point but still more to go!
Consider a test particle in the spacetime region outside a massive body, such as a planet around the Sun, should the space at a distance $r$ from the center of the Sun but outside the surface of the Sun have a vanishing SET and Ricci tensor? I thought only a free falling test particle has a vanishing Ricci tensor in its local inertial frame!

 

[PeterDonis]

Consider a test particle in the spacetime region outside a massive body, such as a planet around the Sun, should the space at a distance rr from the center of the Sun but outside the surface of the Sun have a vanishing SET and Ricci tensor?

Yes, of course; by hypothesis the region is vacuum outside the massive body.

I thought only a free falling test particle has a vanishing Ricci tensor in its local inertial frame!

I don't know where you got this from, but it's obviously false. As I've said several times now, if the Ricci tensor vanishes in one frame, it vanishes in every frame. That's true of any tensor.

 

[PeterDonis]

I don't know where you got this from

It's possible that you are still confusing "vanishing Ricci tensor" with "spacetime is flat". As I said before, they are not the same; you can have a non-flat spacetime (or region of spacetime) with a vanishing Ricci tensor.

 

[Ibix]

I don't know if this needs to be pointed out, but a vanishing Riemann tensor corresponds to flat spacetime. The Ricci tensor is a contraction of the Riemann tensor and may be zero while allowing some curvature that isn't described by the Ricci tensor (this is called Weyl curvature, I believe).

 

[PeterDonis]

The Ricci tensor is a contraction of the Riemann tensor and may be zero while allowing some curvature that isn't described by the Ricci tensor (this is called Weyl curvature, I believe).

Yes. For example, all of the curvature in any vacuum solution of the EFE, such as the Schwarzschild solution, is Weyl curvature.

 

[Adel Makram]

It's possible that you are still confusing "vanishing Ricci tensor" with "spacetime is flat". As I said before, they are not the same; you can have a non-flat spacetime (or region of spacetime) with a vanishing Ricci tensor.

If the field is measured at distance $r$ from the center of the mass, then it should satisfy or be approximated to Poisson equation.
$$\bigtriangledown ^2 \Phi=4 \pi G \rho$$
Now $$\Phi=\frac{1}{2}g_{00}$$
So, given that $T_{ij}$ is a representative of $\rho$, $g_{00}$ not Ricci tensor should be the representative of $\Phi$.
But Ricci tensor is a function of second derivatives of $g_{00}$, which makes me wondering why the field dynamical variable is treated differently in both classical and relativity theories. In classical theory the dynamical variable is the field potential for which its second derivative is not vanishing at $r$ from the source, but in GR, the dynamic variable is the metric of the space time for which its second derivatives in the form of Ricci Tensor vanishes at the same distance from the source!

 

[PeterDonis]

$$
\Phi = \frac{1}{2} g_{00}
$$

Actually, in the weak field approximation, $g_{00} = 1 + 2 \Phi$.

given that $T_{ij}$ is a representative of $\rho$, $g_{00}$ not Ricci tensor should be the representative of $\Phi$.

And it is; $g_{00}$ corresponds to $\Phi$ (with the correction I gave above), and the Einstein tensor corresponds to $\nabla^2 \Phi$.

In classical theory the dynamical variable is the field potential for which its second derivative is not vanishing at $r$ from the source

Yes, it does. The equation you wrote down is $\nabla^2 \Phi = 4 \pi \rho$. But outside the source, $\rho = 0$, so $\nabla^2 \Phi$ is also zero.

 

[Adel Makram]

Yes, it does. The equation you wrote down is $\nabla^2 \Phi = 4 \pi \rho$. But outside the source, $\rho = 0$, so $\nabla^2 \Phi$ is also zero.

Just a bit of calculus, the equation of classical potential at a distance $r$ from the source is given by,
$$\Phi=\frac{-1}{r}$$
Taking the second derivatives with the respect of $r$ will give $\frac{-2}{r^3}$, why should it be zero?

 

[PeterDonis]

Taking the second derivatives with the respect of $r$ will give $\frac{-2}{r^3}$, why should it be zero?

I'll respond to this with another question: how can $\nabla^2 \Phi$ be nonzero when $\rho$ is zero, which it is everywhere outside the source? Doesn't that violate Poisson's equation?

(Hint: you might want to check the actual definition of the Laplacian operator $\nabla^2$ in spherical coordinates.)

 

[Adel Makram]

I'll respond to this with another question: how can $\nabla^2 \Phi$ be nonzero when $\rho$ is zero, which it is everywhere outside the source? Doesn't that violate Poisson's equation?

(Hint: you might want to check the actual definition of the Laplacian operator $\nabla^2$ in spherical coordinates.)

This is cool, yes it should violate it unless $\nabla^2\Phi$ is zero which it is according to the defintion of Lapalacisn operator. Thank you.