# The vacuum Einstein equations and flat spacetime

1. Aug 22, 2009

### xiaomaclever

Rencently, I found myself confused by some fundamental concepts in GR. I hope someone can help me with that.

We all know the vacuum Einstein equation (VEE) without the cosmological constant \Lambda is Rab=0. Since I learn GR the words " matters bend the spacetime " been told again and again. I know the zero Rici tensor does not mean flat spacetime while the zero Riemann tensor do. So the VEE means spacetime may be still curved without matter. Who curves it?
Until now we have many solutions for the VEE, which one is the real metric if there are really this vacuum state? Is this a paradox? I feel a flat spacetime is better in vacuum so that it is the only one .

Another question. If we are in flat spacetime, e.g. Minkowski spacetime,the Ricci tensor is zero. According to the general Einstein equation the energy momentum tensor should also be zero. Whether that means the total energy must be zero in flat spacetime. What about curved spacetime?

Thanks !!!

2. Aug 22, 2009

### Mentz114

Hi xiaomaclever,

The matter nearby curves it. Just think of classical fields. The matter creates a field $g_{\mu\nu}$ which extends into the space around it. This field isn't the same as an electric fileld, but it shares that property.

I'm not sure what you mean by 'total energy'. Are you using a quantum field analogy ?

3. Aug 22, 2009

### xiaomaclever

Do you mean the fields curve the spacetime? I know in QFT the vacuum is not nothing but with the fields which all reside in the ground states, namely no excitations. So this case it will still have vacuum energy which will curve the spacetime. But you should know at the time of GR's generation the QFT was not found. Certainly I don't know what is the vacuum in Einstein's mind. I want to know if there is noting in vacuum,no matter,no fields,Is spacetime curved?

4. Aug 23, 2009

### atyy

Yes, spacetime can be curved in the local absence of matter. Take an isolated spherical star. Inside the star there is matter, and the solution is not the vacuum Schwarzschild solution. Outside the star, there is no matter, and by Birkhoff's theorem, the solution is the vacuum Schwarzschild solution, which is curved.

5. Aug 23, 2009

### xiaomaclever

you do not explain who curves the spacetime in vacuum, that's my question.

6. Aug 23, 2009

### robphy

7. Aug 23, 2009

### Staff: Mentor

As an analogy, an electric field can exist at a certain point in space even though there is no charge at that point. The field at that point is determined by the field at nearby points according to Laplace's equation.

8. Aug 23, 2009

### Jonathan Scott

It might help to note that "curved" has two distinct meanings which can be illustrated in 3D by the curvature of part of a cone (which can be locally matched by a flat piece of paper) and the curvature of part of a sphere (which cannot).

Loosely speaking the space-time curvature of the gravitational field outside a mass-energy distribution is like the curvature of part of a cone, and the space-time curvature inside a mass-energy distribution is like the curvature of part of a sphere.

9. Aug 23, 2009

### Amanheis

Curvature can be an intrinsic property of spacetime, it is not necessary to have mass. Think of the Friedmann-Robertson-Walker-metric with rho=0 and k=+-1. Also, there is no a priori reason why the background metric of spacetime shouldn't be the Schwarzschild metric. M would then be an arbitrary parameter instead of some Newtonian mass. Of course this does not agree with our observations of an homogeneous universe, but theoretically it's thinkable.
You can also think of it as living on a sphere (of any dimension). It is curved, even if there is no mass on it. Or (anti) de Sitter-spaces.

So what I'm trying to say: Mass does bend spacetime, but that doesn't mean spacetime can't be bent without mass or energy.

10. Aug 24, 2009

### Mentz114

amanheis raises an interesting case. For the FLRW the first two Ricci components are

$$R_{00}=-\frac{3\,\partial t\left( \partial t R\left( t\right) \right) }{R\left( t\right) }$$

$$R_{11}=-\frac{R\left( t\right) \,\partial t\left( \partial t\left( R\left( t\right) \right) \right) +2\,{\partial t\left( R\left( t\right) \right) }^{2}+2\,k}{k\,{r}^{2}-1}$$

and the scalar curvature is

$$R=\frac{6\,\left( R\left( t\right) \,\partial t\left( \partial t\left( R\left( t\right) \right) \right) +{\partial t\left( R\left( t\right) \right) }^{2}+k\right) }{{R\left( t\right) }^{2}}$$

where k is -1,0 or 1.

Draw your own conclusions. If all the derivatives are zero ( static universe ) then there is still spatial curvature and a non-zero scalar curvature if k is not 0.

The greatest effect comes from the expansion itself.