# Weyl Curvature

1. Aug 10, 2015

### Markus Hanke

As an amateur, I am just testing my understanding on the following, since there is nothing worse than harbouring misconceptions.

Suppose we have a space-time ( e.g. of FRW type ) which is conformally flat :
$$C{^{\mu \nu }}_{\rho \sigma }=0$$
but not Ricci flat : $$R_{\mu \nu }\neq 0$$
Would that physically mean that, if I was to release a ball of test particles in such a space-time, that ball would retain its spherical shape along its geodesic, but either increase or decrease its volume ?

Just wanting to confirm my understanding on this. Thanks in advance !

2. Aug 10, 2015

### Markus Hanke

Also - and I'm not sure if I should open another thread on this - I am aware that the contraction of the Weyl tensor across two of its indices always vanishes ( i.e. it is trace free ). I can see why that would be the case purely algebraically, but I'm not sure about the physical significance of this - does it simply mean that Weyl curvature preserves volume, in a manner of speaking ?

3. Aug 10, 2015

### Staff: Mentor

Yes. [Edit: yes to the volume increase/decrease. The shape question is more complicated, see follow-up posts in this thread.]

Yes.

Last edited: Aug 11, 2015
4. Aug 10, 2015

### Markus Hanke

Great, thank you !

5. Aug 10, 2015

### WannabeNewton

No.

If the geodesic congruence has a 4-velocity $\xi^a$ then for vanishing Weyl tensor the shear evolves according to the following: $$\xi^c \nabla_c \sigma_{ab} = -\frac{2}{3}\theta \sigma_{ab} - \sigma_{ac}\sigma^{c}{}{}_b - \omega_{ac}\omega^c{}{}_b + \frac{1}{3}h_{ab}(\sigma^2 - \omega^2) + \frac{1}{2}\hat{R}_{ab}$$ where $\theta$ is the expansion, $\omega_{ab}$ is the twist, and $\hat{R}_{ab}$ is the spatial projection relative to $\xi^a$ of the trace-free part of the Ricci tensor. Clearly this is non-zero in general, even if we demand that the geodesic ball have vanishing twist. C.f. Wald section 9.2.

6. Aug 10, 2015

### Staff: Mentor

No to the part about retaining spherical shape, correct? If the Ricci tensor is nonzero, then the volume of a small ball of test particles will have to change. In other words, the expansion of a timelike geodesic congruence must be nonzero if the Ricci tensor is nonzero. (More precisely, $\dot{\theta}$ must be nonzero, which means that $\theta$ can't be zero for more than an instant.)

7. Aug 10, 2015

### WannabeNewton

Indeed.

8. Aug 11, 2015

### Markus Hanke

Ok, I see that my understanding must be off somewhere. Unfortunately I haven't studied Wald ( it's been sitting on my shelf for the past three years or so - it seems like a very "heavy" text for an amateur ), so I am finding it hard to make ( geometric ) sense of the expression above - what physical significance can I attribute to the non-vanishing shear ? In other words, in what way exactly would the shape of the ball change along its geodesic ? How can I visualise that change ? Also, what is $$h_{ab}$$ in the above expression ?

I appreciate all your help !

9. Aug 11, 2015

### Mentz114

$h_{ab}$ is the spatial projection tensor $g_{ab}+\xi_a \xi_b$ where $\xi^\mu$ is a timelike congruence. ( I'm not sure about the sign there).
The expression is the rate of change of the shear tensor $\sigma_{ab}$ projected in the direction $\xi^\mu$. I have to admit I would like some explanation.
If we have a rigid ($\theta=0$) shear and twist free ($\sigma=0, \omega=0)$ congruence then we are left with the projected $R_{ab}$ part.

So the rate of change is aways there but does this mean anything if the shear was zero in the first place ?

Yes, I am confused.

10. Aug 11, 2015

### Markus Hanke

Ok, thank you, this makes sense. But like yourself, I am confused about the physics here - I was always under the impression that vanishing Weyl curvature ( i.e. conformal flatness ) implies the conservation of angles, meaning the shape of the ball wouldn't change. There is obviously something going on that I don't know about yet.

11. Aug 11, 2015

### Mentz114

I don't think those (global ?) conformal properties affect the local kinematics of a congruence.

But I just worked out the expansion scalar for a basic FLRW ($k=0$) and I get $\theta=3\ddot{a}/a$ and the tidal tensor $T_{ab}=R_{ambn}\xi^m \xi^n$ ($\xi^\mu=\partial_t$) has three equal spatial components $\ddot{a}/a$. No shape change, even though $R_{\hat{a}\hat{b}}\ne 0$.

Last edited: Aug 11, 2015
12. Aug 11, 2015

### Markus Hanke

Ok, so in FRW space-time there won't be a shape change. My immediate question then would be - is this true in general whenever the Weyl tensor vanishes ? If not, what attributes does a space-time need to have in addition to a vanishing Weyl tensor to preserve shapes ?

13. Aug 11, 2015

### Mentz114

I'm not sure what 'preserving shapes' means globally. Probably it means that if we have a metric $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ the multipling $g_{ab}$ by a constant will preserved angles ( and so 'shapes').

But the behaviour of a ball of particles on a congruence is given by the locally projected quantities $\sigma_{ab}, \omega_{ab}, \theta$ amongst others.

Look for the definitions of those things and you will have the answer.

14. Aug 11, 2015

### Markus Hanke

Yes, that seems like a reasonable definition.

Ok, I will have to take a closer look at these, as I am not really sure what their meaning is, geometrically speaking. It seems like the more you learn about GR, the more you realise all the stuff you don't know yet !

15. Aug 11, 2015

### Mentz114

It never ends.

Section 9.2 of Wald is very good.

But I first learned this stuff from Stephani's little book and he explains that what we are doing is analysing ( decomposing) the part of the acceleration of $\xi^\mu$, $\nabla_b \xi_a$ that is orthogonal to $\xi^\mu$. We have brought ourselves to rest wrt the 'flow' and now look at what happens to nearby curves.

That is what those (locally spatially projected tensors) tell us.

16. Aug 11, 2015

### Markus Hanke

Thanks, Mentz114 I suppose sooner rather than later I should really put that copy of Wald, that has been sitting unread on my shelf for so long, to good use !