Proof of Existence of Locally Inertial Frames

This is my own version of the proof. There is a version in Bernard Schutz's book, "A First Course in General Relativity" (Cambridge University Press) but the proof there seems overly complicated and also has some gaps relating to consistency of the the systems of linear equations he sets up.

First, we need a fact from linear algebra: if -,- is an inner product on the vector space L, then there exists a basis {V(1), V(2), . . . , V(n)} for L such that

(To prove this, use the fact that any symmetric matrix can be diagonalized using a P-PT type operation.)

To start the proof, fix any chart xi near m, with xi(m) = 0 for all i, and choose a basis {V(i)} of the tangent space at m such that they satisfy the above condition. With our bare hands, we are now going to specify a new coordinate system be i = i(xj) such that

The functions i = i(xj) will be specified by constructing their inverse xi = xi(j) using a quadratic expression of the form:

where A(i,j) and B(i,j,k) are constants. It will follow from Taylor's theorem (and the fact that xi (m) = 0 ) that

so that

where all the partial derivatives are evaluated at m.

Note These partial derivatives are just (yet to be determined) numbers which, if we differentiate the above quadratic expression, turn out to be its actual partial derivatives evaluated at m.

In order to specify this inverse, all we need to do is specify the terms A(i,j) and B(i,j,k) above.. In order to make the map invertible, we must also guarantee that the Jacobean (xi/j)m is invertible, and this we shall do.

We also have the transformation equations

and we want these to be specified and equal to V(i), V(j) when evaluated at m. This is easy enough to do: Just set

For then, no matter how we choose the B(i,j,k) we have

as desired. Notice also that, since the {V(i)} are a basis for the tangent space, the change-of-coordinates Jacobean, whose columns are the V(i), is automatically invertible. Also, the V(i) are the coordinate axes of the new system.

Next, we want to kill the partial derivatives ij/a by choosing appropriate values for the B(i, j, k) (that is, the second order partial derivatives 2xi/jk).By the lemma, it suffices to arrange that

hpk(m) = 0.
But
hpk(m) =
rti(m)
p

xt
xr

h
xi

k
+
p

xt
2xt

hk
(m)
=
p

xt
rti(m)
xr

h
xi

k
+
2xt

hk
(m)
so it suffices to arrange that
2xt

hk
(m)= -rti
xr

h
xi

k
(m).
That is, all we need to do is to define
B(t, h, k)= -rti
xr

h
xi

k
(m).
and we are done.
 

 

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