chirag1 said:
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).
chirag1 said:
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.
chirag1 said:
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.
chirag1 said:
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.