# Weyl curvature and tidal forces

#### TrickyDicky

I'm a bit confused about this and would like for someone to help me get this straight.
I read in wikipedia that a manifold with more than three dimensions, like spacetime, is conformally flat if its Weyl tensor vanishes. I think all FRW metrics are conformally flat, so I guess our universe is conformally flat and that means a Weyl curvature=0 in our spacetime manifold, doesn't it?

But this seems at odds with the wikipedia definition of the Weyl tensor that says it expresses the tidal force a body feels in a geodesic, specifically the part about how the shape of the body distorts, i.e., the classic example of the sphere of balls initially at rest that changes its shape to an elipsoid in the presence of gravity. So there seems to be a Weyl curvature there different from 0.
Also in Penrose's Road to reality book, he explains that the outward deflection of light rays outside the sun's rim is an example of the Weyl curvature.

What am I not getting right here?

#### bcrowell

Staff Emeritus
Gold Member
I'm a bit confused about this and would like for someone to help me get this straight.
I read in wikipedia that a manifold with more than three dimensions, like spacetime, is conformally flat if its Weyl tensor vanishes. I think all FRW metrics are conformally flat, so I guess our universe is conformally flat and that means a Weyl curvature=0 in our spacetime manifold, doesn't it?

But this seems at odds with the wikipedia definition of the Weyl tensor that says it expresses the tidal force a body feels in a geodesic, specifically the part about how the shape of the body distorts, i.e., the classic example of the sphere of balls initially at rest that changes its shape to an elipsoid in the presence of gravity. So there seems to be a Weyl curvature there different from 0.
Also in Penrose's Road to reality book, he explains that the outward deflection of light rays outside the sun's rim is an example of the Weyl curvature.

What am I not getting right here?
When I read the first paragraph, I wasn't sure if it was right that FRW solutions were conformally flat, and the way I then convinced myself that they *were* was exactly by considering the WP explanation given in your second paragraph. If you put a ball in an FRW spacetime, there is no way that the ball can experience any tidal distortion into an ellipsoid. This follows from symmetry. If it distorted into an ellipsoid, then there would be preferred directions in space, e.g., the line containing the longest principal axis of the ellipse. But the FRW solutions are isotropic, so there can't be any such preferred direction.

#### TrickyDicky

When I read the first paragraph, I wasn't sure if it was right that FRW solutions were conformally flat, and the way I then convinced myself that they *were* was exactly by considering the WP explanation given in your second paragraph. If you put a ball in an FRW spacetime, there is no way that the ball can experience any tidal distortion into an ellipsoid. This follows from symmetry. If it distorted into an ellipsoid, then there would be preferred directions in space, e.g., the line containing the longest principal axis of the ellipse. But the FRW solutions are isotropic, so there can't be any such preferred direction.
But this example is in many texts on GR, normally they put the sphere of balls near the earth and then release them, they become ellipsoid, and that explains the shapes of the sea tides too.
Maybe the problem is that these are not examples of Weyl curvature and the Ricci curvature also produces this shape effect that don't affect volume? I say it because I've just read that the Weyl curvature only happens in in empty spacetime, without any gravitational source nearby.
If this is the case I think that the wikipedia article and the paragraph from Penrose book are quite misleading.

#### pervect

Staff Emeritus
There are definitely tidal forces in most FRW space-times, but they are spherically symmetric - and they're not due to the Weyl, they're due to the matter density, i.e. the Ricci.

If you consider the FRW basis the tidal forces are just a'' / a, where a(t) is the scale factor.

Technical details: the variabiles are t,chi,theta, and phi,

the basis one forms are:
[1,0,0,0], [0,a(t),0,0], [0,a(t)f(chi),0], [0,0,0,a(t)chi f(chi)sin(theta)]

a(t) is the time dependent scale factor for the expansion
f(chi) = chi, sin(chi), or sinh(chi) depending on if it's closed, flat, or open

and
$$R_{\hat{t}\hat{\chi}\hat{t}\hat{\chi}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} = \frac{d^2 a}{dt^2} / a$$

The ideal FRW space-time is homogeneous, so the matter distribution is uniform, and it's not a vacuum solution. Without a cosmological constant, the acceleration is always negative, gravity serves to brake the expansion, and you get tidal compressive forces on a small rod, I think of them as being due to the matter and pressure (rho+3P) contained in a sphere co-located with the rod. With a cosmological constant, you might have an accelerating expansion, in which case the tidal forces tend to pull your rod apart.

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#### TrickyDicky

There are definitely tidal forces in most FRW space-times, but they are spherically symmetric - and they're not due to the Weyl, they're due to the matter density, i.e. the Ricci.

If you consider the FRW basis the tidal forces are just a'' / a, where a(t) is the scale factor.

Technical details: the variabiles are t,chi,theta, and phi,

the basis one forms are:
[1,0,0,0], [0,a(t),0,0], [0,a(t)f(chi),0], [0,0,0,a(t)chi f(chi)sin(theta)]

a(t) is the time dependent scale factor for the expansion
f(chi) = chi, sin(chi), or sinh(chi) depending on if it's closed, flat, or open

and
$$R_{\hat{t}\hat{\chi}\hat{t}\hat{\chi}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} = \frac{d^2 a}{dt^2} / a$$

The ideal FRW space-time is homogeneous, so the matter distribution is uniform, and it's not a vacuum solution. Without a cosmological constant, the acceleration is always negative, gravity serves to brake the expansion, and you get tidal compressive forces on a small rod, I think of them as being due to the matter and pressure (rho+3P) contained in a sphere co-located with the rod. With a cosmological constant, you might have an accelerating expansion, in which case the tidal forces tend to pull your rod apart.
Thanks, Pervect, I think I get it now. Probably what I took as meaning that tidal forces that don't affect volume are due to Weyl curvature, actually meant that Weyl curvature when present acts like tidal forces on shape.

What I need to understand better now is this, the Weyl tensor is explained as the curvature that equals the Riemannian curvature when there is no gravitational source (no Ricci curvature, vacccum solution-like).
Now, even in the vaccum solution the Riemannian curvature doesn't vanish (othervise we would be in flat Minkowski spacetime), and that would seem to lead to the Weyl curvature not vanishing, but as we have seen this seems not to be the case in conformally flat manifolds like ours. What am I not getting right?

#### bcrowell

Staff Emeritus
Gold Member
I guess there's something I'm not understanding here, because I would have considered a vanishing Weyl tensor to be the definition of "no tidal forces," and yet I'm sure pervect is right, especially because another knowledgeable PF'er sent me a message telling me that my #2 was wrong.

#### Passionflower

Now, even in the vaccum solution the Riemannian curvature doesn't vanish (othervise we would be in flat Minkowski spacetime), and that would seem to lead to the Weyl curvature not vanishing, but as we have seen this seems not to be the case in conformally flat manifolds like ours. What am I not getting right?
The FLRW solution is not a vacuum solution, there is no empty space in this model, thus there is no Weyl curvature. Ricci curvature is of course also tidal (tidal effect is really nothing else than spacetime curvature), in that it usually is volume reducing. So a sphere of volume x can become < x due to Ricci curvature.

#### pervect

Staff Emeritus
The way I think of it is that if you have the sphere of coffee grounds, matter outside the sphere will change its shape, but not the volume. This shape distortion can be ascribed to the Weyl curvature, though I don't recall the reference offhand. Matter inside the sphere will tend to cause the sphere to contract - Baez talks about this some in his tutorial paper "The meaning of Einstein's equation" http://math.ucr.edu/home/baez/einstein/ - and that's due to the Ricci.

#### TrickyDicky

The way I think of it is that if you have the sphere of coffee grounds, matter outside the sphere will change its shape, but not the volume. This shape distortion can be ascribed to the Weyl curvature, though I don't recall the reference offhand.
But you said before that the tidal forces in FRW spacetimes are not due to the Weyl tensor but to the Ricci, since Weyl tensor vanishes in these conformally flat spacetimes.
So I'm still confused about the Riemann curvature in empty space that is supposed to be equal to the Weyl curvature in this setting. Is this right or not?
Passionflower said:
The FLRW solution is not a vacuum solution, there is no empty space in this model, thus there is no Weyl curvature.
Hmm, so you would say there is Weyl curvature in Schwartzschild metric? The problem is that according to the definition of conformally flat manifolds, the spacetimes of constant sectional curvature like Schwartzschild are also conformally flat and therefore have a vanishing Weyl curvature.
Let's see if someone can clear up these confusions.

#### atyy

The way I think of it is that if you have the sphere of coffee grounds, matter outside the sphere will change its shape, but not the volume. This shape distortion can be ascribed to the Weyl curvature, though I don't recall the reference offhand. Matter inside the sphere will tend to cause the sphere to contract - Baez talks about this some in his tutorial paper "The meaning of Einstein's equation" http://math.ucr.edu/home/baez/einstein/ - and that's due to the Ricci.
A supplement from the same site Pervect recommends: http://math.ucr.edu/home/baez/gr/ricci.weyl.html

(Maybe it will help to carefully track which statements are amdew about FRW versus Schwarzschild spacetimes.)

#### atyy

But you said before that the tidal forces in FRW spacetimes are not due to the Weyl tensor but to the Ricci, since Weyl tensor vanishes in these conformally flat spacetimes.
So I'm still confused about the Riemann curvature in empty space that is supposed to be equal to the Weyl curvature in this setting. Is this right or not?

Hmm, so you would say there is Weyl curvature in Schwartzschild metric? The problem is that according to the definition of conformally flat manifolds, the spacetimes of constant sectional curvature like Schwartzschild are also conformally flat and therefore have a vanishing Weyl curvature.
Let's see if someone can clear up these confusions.
p413 of http://books.google.com/books?id=SiWXP8FjTFEC&dq=interior+schwarzschild+conformally+flat&source=gbs_navlinks_s says the interior Schwarzschild solution (ie. one which contains matter) is conformally flat. Do you think that might be what you're thinking of, instead of the vacuum Schwarzschild solution?

#### atyy

When I read the first paragraph, I wasn't sure if it was right that FRW solutions were conformally flat, and the way I then convinced myself that they *were* was exactly by considering the WP explanation given in your second paragraph. If you put a ball in an FRW spacetime, there is no way that the ball can experience any tidal distortion into an ellipsoid. This follows from symmetry. If it distorted into an ellipsoid, then there would be preferred directions in space, e.g., the line containing the longest principal axis of the ellipse. But the FRW solutions are isotropic, so there can't be any such preferred direction.
I guess there's something I'm not understanding here, because I would have considered a vanishing Weyl tensor to be the definition of "no tidal forces," and yet I'm sure pervect is right, especially because another knowledgeable PF'er sent me a message telling me that my #2 was wrong.
http://math.ucr.edu/home/baez/gr/ricci.weyl.html

Perhaps in FRW it's no tidal distortion of shape (since Weyl is zero), but there is tidal distortion of volume (since Ricci is not zero)?

In contrast in Schwarzschild it's tidal distortion of shape (since Weyl is not zero), but no tidal distortion of volume (since Ricci is zero)?

#### TrickyDicky

p413 of http://books.google.com/books?id=SiWXP8FjTFEC&dq=interior+schwarzschild+conformally+flat&source=gbs_navlinks_s says the interior Schwarzschild solution (ie. one which contains matter) is conformally flat. Do you think that might be what you're thinking of, instead of the vacuum Schwarzschild solution?
All constant curvature and also all FRW metric spacetimes seem to be conformally flat which according to what I read implies a vanishing Weyl curvature. What I'm looking for is a way to reconcile this with tidal forces that only affect shape that are described by some as Weyl curvature and others as Ricci curvature, and I'm also trying to reconcile a not vanishing Riemann curvature in empty space and a vanishing Weyl curvature( for our type of spacetime) that in textbooks are equated.

#### atyy

I think all FRW metrics are conformally flat, so I guess our universe is conformally flat and that means a Weyl curvature=0 in our spacetime manifold, doesn't it?

Also in Penrose's Road to reality book, he explains that the outward deflection of light rays outside the sun's rim is an example of the Weyl curvature.
All constant curvature and also all FRW metric spacetimes seem to be conformally flat which according to what I read implies a vanishing Weyl curvature. What I'm looking for is a way to reconcile this with tidal forces that only affect shape that are described by some as Weyl curvature and others as Ricci curvature, and I'm also trying to reconcile a not vanishing Riemann curvature in empty space and a vanishing Weyl curvature( for our type of spacetime) that in textbooks are equated.
Penrose is talking about the vacuum Schwarzschild solution, which as far as I understand has non-zero Weyl curvature.

The FRW solution with uniform matter only approximates our universe when stuff is averaged out over very large scales. Clearly, matter is not uniform in our solar system, it is bunched up into planets and sun, so on this smaller scale, the curved empty space outside our sun is approximated by the vacuum Schwarzschild solution (with mass parameter being the sun's mass). On an even smaller scale, the curved empty space outside the earth is approximated by the vacuum Schwarzschild solution (with mass parameter being earth's mass).

The "exact" solution describing our non-uniform universe must somehow contain corrections to all these approximations which must be joined up to each other somehow.

#### TrickyDicky

Penrose is talking about the vacuum Schwarzschild solution, which as far as I understand has non-zero Weyl curvature.

The FRW solution with uniform matter only approximates our universe when stuff is averaged out over very large scales. Clearly, matter is not uniform in our solar system, it is bunched up into planets and sun, so on this smaller scale, the curved empty space outside our sun is approximated by the vacuum Schwarzschild solution (with mass parameter being the sun's mass). On an even smaller scale, the curved empty space outside the earth is approximated by the vacuum Schwarzschild solution (with mass parameter being earth's mass).

The "exact" solution describing our non-uniform universe must somehow contain corrections to all these approximations which must be joined up to each other somehow.
So how does this answer my questions? is deflection of light outward the sun's rim produced by Weyl curvature or Ricci's? if vaccum solutions have constant sectional curvature (they are static) so they have also vanishing Weyl curvature, how can they not have Riemann curvature vanishing if Weyl's and Riemann's are equal in vacuum?

#### sheaf

So how does this answer my questions? is deflection of light outward the sun's rim produced by Weyl curvature or Ricci's
The deflection of light around a star must be produced by Weyl curvature, since the light rays are travelling through the vacuum where there's no Ricci curvature.

if vaccum solutions have constant sectional curvature (they are static) so they have also vanishing Weyl curvature,
Take the Schwarschild solution for example. It's a vacuum solution, but does not have vanishing Weyl curvature. Why do you think a static solution necessarily has vanishing Weyl curvature ?

#### TrickyDicky

Take the Schwarschild solution for example. It's a vacuum solution, but does not have vanishing Weyl curvature. Why do you think a static solution necessarily has vanishing Weyl curvature ?
Schwarschild solution has constant sectional curvature (that of a parabole) so from wikipedia:
http://en.wikipedia.org/wiki/Conformally_flat_manifold

-Every manifold with constant sectional curvature is conformally flat.

-An n-dimensional pseudo-Riemannian manifold for n ≥ 4 is conformally flat if and only if the Weyl tensor vanishes.

#### Passionflower

Schwarschild solution has constant sectional curvature (that of a parabole) so from wikipedia:
http://en.wikipedia.org/wiki/Conformally_flat_manifold

-Every manifold with constant sectional curvature is conformally flat.

-An n-dimensional pseudo-Riemannian manifold for n ≥ 4 is conformally flat if and only if the Weyl tensor vanishes.
All conformally flat vacuum solutions must be flat, so yes if they are flat then obviously the Weyl curvature tensor vanishes as well.

Two examples of a conformally flat spacetime are the FLRW and Schwarzschild interior solution. The Schwarzschild vacuum solution is not conformally flat. If wikipedia says differently then it is stated wrong.

#### sheaf

Schwarschild solution has constant sectional curvature (that of a parabole)
I'm a bit rusty on relativity, and I'm not really familiar with the term "sectional curvature". Does it mean the curvature like in spaces of constant curvature (S3 etc) ? i.e. the manifold is characterised by the curvature scalar ?

If so, then Schwarzschild isn't like that.

Maybe you can find some spacelike 3-slices of the Schwarzschild solution that have constant intrinsic curvature, but that doesn't compeletely characterise the Riemann tensor of the spacetime. There is also the question of how the slices are embedded in the spacetime, which brings in the extrinsic curvature.

If you want to be sure though, sit down and compute the components of the Weyl tensor (approx 3 hours work, allowing for mistake time !)

#### sheaf

Ah I see Passionflower has just responded. I should point out that my remarks pertain to the Schwarzchild vacuum solution (the one everybody knows and loves), not the interior solution

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