Testing Understanding: Conformally Flat Space-Time

[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 !

 

[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 ?

 

[PeterDonis]

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 ?

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

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 ?

Yes.

 

[Markus Hanke]

Great, thank you !

 

[WannabeNewton]

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 ?

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.

 

[PeterDonis]

No.

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.)

 

[WannabeNewton]

No to the part about retaining spherical shape, correct?

Indeed.

 

[Markus Hanke]

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.

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 !

 

[Mentz114]

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 !

$h_{ab}$ 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.

 

[Markus Hanke]

$h_{ab}$ 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 $\xi^\mu$.

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.

 

[Mentz114]

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.

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 $R_{\hat{a}\hat{b}}\ne 0$.

 

[Markus Hanke]

No shape change, even though $R_{\hat{a}\hat{b}}\ne 0$.

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 ?

 

[Mentz114]

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 ?

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.

 

[Markus Hanke]

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').

Yes, that seems like a reasonable definition.

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.

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 realize all the stuff you don't know yet !

 

[Mentz114]

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 realize all the stuff you don't know yet !

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.

 

[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 !