# Space-time curvature without mass-energy ?

1. Mar 30, 2014

### kderakhshani

Hi everybody,

As you know, the Einstein field equation

$R_{μ\nu} - 1/2Rg_{μ\nu} =κT_{μ\nu}$

implies that at any point with vanishing energy-momentum tensor the Ricci curvature also vanishes:

$T_{μ\nu} = 0 \Rightarrow R_{μ\nu} = 0$

hence a Ricci-flat space-time (the vacuum solutions of the field equation).

However, $R_{μ\nu}$ includes just 10 elements of the Riemann-Christofel curvature tensor $R_{μ\nuλρ}$. The remaining 10 elements are coded in the Weyl tensor (the conformal tensor) $C_{μ\nuλρ}$, which has all symmetries of $R_{μ\nuλρ}$ but is totally traceless, and does not vanish only for $n≥4$.

http://en.wikipedia.org/wiki/Weyl_tensor

The Weyl tensor is often associated with tidal forces or gravitational waves.

Thus, in an absolute (classical) vacuum (with no mass, energy, momentum densities), $R_{μ\nuλρ}$ can still be nonzero.

Does it mean a curved space-time (and a gravitation) without mass-energy?

2. Mar 30, 2014

### ofirg

Yes, A region of space-time can definitley be curved with a vanishing energy-momentum tensor in that region. Otherwise their would be no gravitational effects (i.e. , tidal forces) outiside the mass of a gravitating body, such as tidal forces caused by the sun and moon here on earth.

However, when the energy-momentum tensor vansihes everywhere, then space-time is flat everywhere. (I mean without a cosmological constant)

3. Mar 30, 2014

### Bill_K

No, the pp wave is an obvious counterexample to this.

4. Mar 31, 2014

### kderakhshani

@ofrig

I feel I should clarify my point. Imagine a point very far away from any source of gravitation (mass, energy, radiation, etc.). According to the Einstein's field eq. the space-time around this point is Ricci flat. But not totally flat. Because the Riemann curvature tensor can still be nonzero. The field eq. says nothing about the traceless part of the Riemann tensor (the Weyl conformal tensor).
Is this an example of source-free curvature of space-time?

If yes, there could be some "inherent" curvature of space-time, as well as the mass-generated curvature.

5. Mar 31, 2014

### Bill_K

The energy-momentum tensor produces both kinds of curvature. Locally it produces a nonzero Ricci tensor. Locally meaning at the same point. Nonlocally (i.e. at other points) it produces a nonzero Weyl tensor. The Ricci and Weyl tensors are algebraically independent, but they are connected by differential equations, namely the Bianchi identities. For example, a source such as a neutron star has a nonzero Ricci tensor everywhere within its body, but there is a nonzero Weyl tensor extending to infinity in all directions around it.

Think of an electromagnetic analogy. A charge density produces a nonzero E field at all points surrounding it.

You can certainly have source-free gravitational fields. They will be gravitational waves, such as the pp wave I mentioned.

Last edited: Mar 31, 2014
6. Mar 31, 2014

### kderakhshani

Dear Bill_K,

Thank you for reminding the Bianchi identities to me.

Regarding your last paragraph, the gravitational or pp waves have energy densities and momentum fluxes. Aren't they considered as "sources" ?

7. Mar 31, 2014

### kderakhshani

Dear Bill_K,

Thank you for reminding the Bianchi identities to me.

Regarding your last paragraph, the gravitational or pp waves have energy densities and momentum fluxes. Aren't they considered as "sources" ?

8. Mar 31, 2014

### Bill_K

The Wikipedia page tries to be as general as possible, so in addition to purely gravitational waves they have included waves with various sources: electromagnetic Tμν, null dust, etc. The pure gravitational case is further mentioned on https://en.wikipedia.org/wiki/Plane_gravitational_waves

9. Mar 31, 2014

### bcrowell

Staff Emeritus
In addition to Bill_K's #8, note that even in the case of a purely vacuum spacetime, the energy in gravitational fields is not incorporated into the stress-energy tensor T, and therefore it is not a source of gravitational fields in the sense that T is the source term in the Einstein field equations. Conceptually, one can say that the energy in the gravitational field is a source of gravitational fields, but this is built into the field equations through the nonlinearity of the equations, not as part of T.

10. Apr 1, 2014

### kderakhshani

Dear Friends,

Thank you for your useful replies.

OK, let me ask the question that is boggling my mind:

When $T_{μ\nu}=0$ then $R_{μ\nu}=0$ but $R_{μ\nuλρ}$ is not necessarily equal to zero. In fact $R_{μ\nuλρ} = C_{μ\nuλρ}$ , the Weyl tensor.

What is the "source" or cause of this curvature?
If the gravitational/pp/etc. waves, is there any math relation between the Weyl curvature and e.g. the energy density of those waves?

Thank you.

11. Apr 1, 2014

### bcrowell

Staff Emeritus
Waves don't need to have sources. For example, a plane wave in a vacuum is a solution of Maxwell's equations. It's a vacuum, so there are no charges or currents.