Einstein's field equations

1. Jul 26, 2004

kurious

When people say EFEs are coupled does that mean that to solve
them you have to have a solution which conforms to them all
simulataneously.And if general relativity is such a great description of nature
why is there more than one solution to EFEs?

2. Jul 26, 2004

Staff Emeritus

The answer to the second is the partial differential equations often have multiple solutions. Th3e solutions are all mathematically valid, but the relationship tp physical validity needs to be checked for each one.

There are different physical situations in spacetime, and each one should be described by one of the solutions to the field equations. Spacetime around a static gravitating mass is described by the Schwartzschild solution. Around a rotating mass it would be described by the different Kerr-Newman solution.

3. Jul 26, 2004

kurious

So if we know how many physical situations there are in spacetime we should be able to say how many solutions Einstein's field equations have?

4. Jul 28, 2004

Russell E. Rierson

I will give my interpretation of the Einstein Field Equations and if anyone would like to correct any mistakes that I make, please do. I, and others, have much to learn

If my interpretation is correct, the fall of a solitary object to the
ground to one location on the Earth can be locally transformed away by
a correct choosing of a system of accelerating coordinates, but the
fall of objects all over the Earth cannot. Effectively, these objects
can be visualized as a spherical cloud of tiny particles, each
following its own geodesic path, and the multiplicity of paths are all
converging, such, that the cloud's volume is shrinking at an
accelerating rate, and, as the shell/cloud collapses toward the
Earth, the rate of acceleration, which is the second time derivative,
is proportional to the mass of the Earth, in accordance with the
Einstein field equations.

I apologize for the lack of tensor skills below ;)

Einstein says that for empty space, far from any gravitating object,
the spacetime will be Minkowskian [flat], requiring that the Riemann
curvature tensor R_abcd should vanish?

But in a spacetime near a gravitating object, there will be a
non-zero intrinsic curvature, because the total gravitational field of
an object, cannot be simply "transformed away" to the second order.
This necessary condition is given by the non-vanishing Riemann
curvature tensor. Although, at points where the full curvature tensor
R_abcd is non-zero, the contracted curvature tensor of the second rank
tensor, R_ab = g^ab R_abcd = R^c_acb also known as the "Ricci
tensor" , may ...vanish?

Therefore the equations for the gravitational field in the vacuum of
empty space are: Ricci tensor, R_ab = 0

The R_ab = 0 represents ten equations in the ten components of g_ab
at each point in empty spacetime, devoid of matter or electomagnetic
energy, while not eliminating the gravitational metric-field itself.
The equations are generally covariant, thus, given any single
solution, infinitely many others may also be constructed via applying
arbitrary, albeit continuous coordinate transformations. The
conditions to be satisfied by the field equations must be the
vanishing of the covariant derivatives - guaranteeing the conservation
of any energy-momentum source term, that may be placed on the right
side of the equation. Although it seems to make one think about the
energy of the energy of the....of the ... of the ...of the energy of
a gravitational field!

The divergence generalizes to the covariant derivative in tensor
calculus, such, that the covariant derivatives of the metrical field
equations must identically vanish. The Ricci tensor R_ab in and of
itself does not satisfy this necessary requirement, but a new tensor
can be created which does satisfy the requirement via a slight
modification of the Ricci tensor, and, without disturbing the
relation R_ab = 0 for the vacuum of empty space.

Subtracting half the metric tensor times the invariant Ricci scalar R
= g^ab R_ab gives the Einstein Tensor:

G_ab = R_ab - (1/2) g_ab R

Since R_ab = 0

G_ab = 0

The covariant energy-momentum tensor is T_ab , regarded as the cause,
or the "source" of the metric curvature. "Mass tells space how to
curve and space tells mass how to move." It gives the conservation of
energy-momentum, and it implies gravitational energy "gravitates"
just as all other forms of energy do.

G_ab = k T_ab

k = -8pi G where G is Newton's universal gravitational constant.

5. Jul 28, 2004

pmb_phy

At any point in a curved spacetime the gravitational field can be transformed away in all cases. What cannot be transformed away are tidal gradients. Tidal gradients is what you're refering to when you speak of second order.

Pmb

6. Jul 28, 2004

DW

No, the gravitational field which is the Riemann tensor a field of spacetime curvature in modern relativity, not the Newtonian acceleration field, can not be transformed away. And the tidal gradients can be transformed away, they just can not be transformed away simultaneously with the forces of affine connection. You can transform away either, just not both.

7. Jul 28, 2004

kurious

If electromagnetism dealt with charges that could only attract one another
would electromagnetism be a rank 2 field theory with ten coupled field equations?

8. Jul 29, 2004

Russell E. Rierson

Thanks for the help kurious, DW, and pmb_phy

If my interpretation is correct, the affine connections can always be transformed away locally, by transforming to any local free fall frame, which means basically that space-time is locally flat. The affine connections can be transformed away and the metric tensor will be reduced to that of special relativity.

Whenever matter is present, the Riemann curvature tensor is not zero. Even though the affine connections can always be locally transformed away to a "free fall" frame, the Riemann tensor is a tensor. One property of tensors is that if the tensor in question is not zero according to one frame, it is not zero according to any frame. Spacetime curvature, or a nonzero Riemann curvature tensor, is always present when, and where, there is any kind of matter, whether or not the affine connection has been locally transformed away. Matter intrinsically produces a field of curvature, not a field of Newtonian acceleration.

In the sense of spacetime curvature, gravity can never be transformed away, while the Newtonian acceleration field can always be transformed away. So from a general relativistic perspective, objects appear to accelerate because they follow geodesics on a curved spacetime. In general relativity, the field is actually a curvature field, i.e. the Riemann curvature tensor.

So when there is matter present, Enstein's "field" equations imply that the Riemann tensor is not zero in the presence of it. The Riemann tensor is then the "field" of spacetime curvature.

9. Jul 29, 2004

DW

Yes, that is exactly correct, and a very good account of the modern GR paradigm!

10. Jul 29, 2004

pmb_phy

You're most welcome
Absolutely 100% correct.
If there is matter at the point in spacetime X then the Riemann tensor will be non-zero at X. However that does not mean that if there is matter in a the region of space R that there will be curvature at points near R. That depends on the distribution of matter.

To see what I mean by this consider the Newtonian analogy (Or the weak field approximation in GR). Let there be a sphere which has uniform mass density and a spherical cavity cut out of its interior such that the center of the cavity does not coincided with the center of the sphere. See
http://www.geocities.com/physics_world/gr/grav_cavity.htm

The Newtonian tidal force tensor is the analogy to the Riemann tensor. See
http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm

That's why the Riemann tensor is sometimes refered to as the tidal force tensor. The tidal force tensor will be non-zero wherever there is matter. In this case at all points in the figure colored brown. Outside the sphere there are tidal forces. Inside the sphere, but not inside the cavity, (where there is matter) there are tidal forces. But inside the cavity there are no tidal forces, but there is a gravitational field.

That is correct. If you'll notice, the metric tensor is never zero so we don't use that criteria for determining the presence of a gravitational field or the presence of tidal forces.

Yup. That's the reason that many (if not most) GRists prefer to think of the non-vanishing of the Riemann tensor as the criteria for the presence of a gravitational field. However that was never Einstein's criteria.
No. Objects do not appear to accelerate due to spacetime curvature, that is what the affine connection does. The presence of curvature means that two nearby particles which are initially moving on geodesics which are parallel will eventually diverge, i.e. they will either move apart or move together.

Pete