Two degrees of freedom in GR after boundary conditions specified?

[Fractal matter]

TL;DR

2 degrees of freedom of GR after specification of boundary conditions

I read there are 2 degrees of freedom in GR after boundary conditions specified. Does that mean 2 equations are enough for EFE equivalent? Those two seem like the amplitude and a phase.

 

[Ibix]

Where did you read that? It doesn't sound right, at least as a general statement. There are usually four degrees of freedom, expressing your freedom to choose coordinate systems in 4d. With a better reference we may be able to comment further

 

[PeterDonis]

I read

As @Ibix said, you need to give a specific reference.

there are 2 degrees of freedom in GR after boundary conditions specified.

Whether this is a correct statement depends on what the degrees of freedom are degrees of freedom of.

There are 2 physical degrees of freedom in a gravitational wave. But that is not the same as "in GR" in general.

The metric tensor in GR has ten independent components, since it is a symmetric 2nd rank tensor. A general metric has no symmetries that would reduce this.

The Riemann tensor has twenty independent components; a fourth rank tensor in 4D spacetime has a total of 256 components, but various symmetries of the Riemann tensor reduce this to 20 independent ones.

Does that mean 2 equations are enough for EFE equivalent?

No.

 

[PeterDonis]

There are usually four degrees of freedom, expressing your freedom to choose coordinate systems in 4d.

Those aren't degrees of freedom. And the freedom to choose coordinates doesn't have four independent components; you have to look at how coordinate transformations work and what they specify. A useful way to do that is to look at a coordinate transformation from a general coordinate chart to Riemann normal coordinates, in which the metric at a given point is Minkowski and all of the first derivatives of the metric are zero.

A coordinate transformation can specify 16 independent coefficients (4 times 4) for the relationships between the coordinates themselves. But 6 of these correspond to the 6 parameter group of Lorentz transformations at a point (boosts plus spatial rotations), all of which are equivalent in terms of making the metric Minkowski at a point. That leaves 10 coefficients, which is exactly the number needed to fix the ten independent components of the metric tensor to their Minkowski values at a point.

The transformation can specify 40 coefficients for the relationships between the first derivatives of the metric coefficients. This is also exactly the number of first derivatives of the ten independent metric coefficients; so the coordinate transformation can fix all of those to be zero at a point.

The transformation can specify 80 coefficients for the relationships between the second derivatives of the metric coefficients. But there are 100 second derivatives of the ten independent metric coefficients. So a coordinate transformation must leave 20 second derivatives of the metric at a point nonzero; those correspond to the 20 independent components of the Riemann tensor.

 

[chirag1]

But 6 of these correspond to the 6 parameter group of Lorentz transformations at a point (boosts plus spatial rotations), all of which are equivalent in terms of making the metric Minkowski at a point. That leaves 10 coefficients, which is exactly the number needed to fix the ten independent components of the metric tensor to their Minkowski values at a point.

I don't understand how these 6 coefficients are equivalent in terms of making the metric Minkowski at a point.

The transformation can specify 40 coefficients for the relationships between the first derivatives of the metric coefficients. This is also exactly the number of first derivatives of the ten independent metric coefficients; so the coordinate transformation can fix all of those to be zero at a point.

Also, I couldn't understand how did you get the numbers? Can you please show the calculations for the number of coefficients and the number of first derivatives of the 10 components of the metric? Same for the second derivative.

I'm having a hard time following the number of coefficients and fixing the metric and its derivatives calculation.
Thank you very much in advance.

 

[PeterDonis]

I don't understand how these 6 coefficients are equivalent in terms of making the metric Minkowski at a point.

"Making the metric Minkowski at a point" does not specify a unique coordinate chart; it only specifies a 6 parameter group of coordinate charts. The 6 parameters are the parameters of the group of proper orthochronous Lorentz transformations (heuristically, 3 spatial rotation parameters and 3 boost parameters).

Can you please show the calculations for the number of coefficients

A coordinate transformation can be thought of as a series of matrices, or sets of coefficients, relating coordinates in the two charts. Mathematically, it looks like this:

$$
x'^\alpha = M^\alpha{}_\mu x^\mu + \frac{1}{2} N^\alpha{}_{\mu \nu} x^\mu x^\nu + \frac{1}{6} P^\alpha{}_{\mu \nu \rho} x^\mu x^\nu x^\rho + ...
$$

There are four possible values for the $\alpha$ index (four coordinates in the new chart), and four possible values for the $\mu$, $\nu$, $\rho$ indexes (four coordinates in the old chart), but you also have to consider that the terms with more than one coordinate in the old chart have to be symmetric under permutations of the indexes, which reduces the number of independent coefficients. So you end up with 16 $M^\alpha{}_\mu$ coefficients (no permutations, so just 4 x 4), 40 $N^\alpha{}_{\mu \nu}$ coefficients (4 $\alpha$ times 10 independent permutations of $\mu$ and $\nu$ because of symmetry) and 80 $P^\alpha{}_{\mu \nu \rho}$ coefficients (4 $\alpha$ times 20 independent permutations of $\mu$, $\nu$, and $\rho$ because of symmetry).

See Misner, Thorne & Wheeler, Exercise 13.3 for more information.

and the number of first derivatives of the 10 components of the metric?

You have 10 independent components of the metric. Each of them has four first derivatives (one with respect to each of the four coordinates). That's 10 times 4 = 40 first derivatives.

Same for the second derivative.

There are 10 possible second derivatives for each of the 10 independent metric components (partial derivatives commute so there are 10 independent possible pairs of coordinates to take the derivatives with respect to). That's 10 times 10 = 100 second derivatives.

 

[martinbn]

There are 10 indipendent components of the metric tensor. But the choice of 4 coordinates is a gauge, so that brings it down to 6. There are 4 identities coming from the Bianchii identities, which are coordinate indipendent, thus these are new. That brings it down to 2 indipendent parameters.

Another way to see it is to look at the initial value problem. Too lazy to type it on my phone, but there is the initial data and the constraint equations. When you count it gives you two independent parameter.