Coordinates adapted to two vector fields

[smallphi]

I am considering two vector fields in the spacetime of General Relativity. One is spacelike, the other is timelike, they are normalized and orthogonal:

U.U = -1
V.V = +1
U.V = 0

where dot denotes scalar product.

In addition, it is known the integral curves of U and V always remain in some two dimensional submanifold of the spacetime.

I know for one vector field it's always possible to construct the so called comoving coordinate system in which it has components (1, 0, 0, 0). Is it always possible to find a coordinate system in which the two fields have upper index (contravariant) components:

U -> (1,0,0,0)
V -> (0,1,0,0)

If possible give references where such topics are discussed.

 

[Peeter]

I was curious to see the replies on this thread, but there haven't been any, so add my 1c answer (I don't really know much of the differential geometry that you are using ... it's in my TODO list of things to learn;)

This sounds kind of like you want a moving reciprocal frame vector calculation at the point of interest on your surface.

The book Geometric Algebra for Physicists shows one way of calculating this, but requires learning their algebra techniques first. I thought I also had some notes typed up on how to do such a calculation with projection related matrixes and a metric tensor too, but they must be on paper at home somewhere.

 

[Doodle Bob]

This is actually a differential topology problem and can be solved -- to some degree -- without regard to the given spacetime geometry. The crucial theorem is Frobenius Theorem, one version of which says that a set of pointwise linearly independent vector fields spans a submanifold if and only if the Lie derivatives of pairs of these vector fields are in the span of the original set of vector fields.

In your case, you only need to write U and V as vector fields with respect to the coordinates on R^3 and check to see if [U,V] is in span{U,V}, i.e., if we write U=(U1,U2,U3) and V=(V1,V2,V3) where the Uj's and Vj's are functions, then [U,V]=(U1(d/dx(V1))+U2(d/dy(V1))+U3(d/dz(V1)), U1(d/dx(V2))+U2(d/dy(V2))+U3(d/dz(V2)),U1(d/dx(V3))+U2(d/dy(V3))+U3(d/dz(V3))-{same expression except switch the U's and V's}. If you can find functions a and b such that [U,V]=aU+bV, then U and V will span a submanifold of the spacetime.

 

[Doodle Bob]

"In addition, it is known the integral curves of U and V always remain in some two dimensional submanifold of the spacetime."

I just read this. So, you are good: As long as the above statement is true, then the spacetime is foliated by U-V leaves, i.e., each point of the spacetime has a unique surface passing through it that is tangent to both U and V. In particular, each point has a local coordinate system for which this surface is given as the zero set of two of the coordinate variables. Similarly, within that surface, since both U and V as non-zero vector fields of the surface, there is a local coordinate system for which the U-curves and V-curves are each given as the zero set of one of the two coordinate variables on the surface.

By continuity and reducing the domains of the coordinate systems if necessary, we can assume that these two coordinate systems (the local one on the spacetime and the local one on the UV-leaves) "match up". Which gives you the result that you wanted.

Now what exactly those coordinate systems are... I have no idea.

 

[OrderOfThings]

Assume $u$ and $v$ are coordinates with basis vectors

$$\mathbf{e}_u = U,\qquad\mathbf{e}_v = V$$

The covariant way to say the same thing is to state that

$$\begin{array}{ll}du(U)=1 & dv(U)=0\\ du(V)=0 & dv(V)=1\end{array}$$

Then the line element becomes

$$ds^2 = -du^2+dv^2$$