Let $(M,g)$ be a four-dimensional Lorentzian manifold representing spacetime. We define a reference frame on the open subset $U\subset M$ to be a unit timelike future-directed vector field $Z$.

We define then a naturaly adapted coordinate system to a reference frame $Z$ to be a chart $(V,x)$ with $V\subset U$, being $\dfrac{\partial}{\partial x^0}$timelike, $\dfrac{\partial}{\partial x^i}$ spacelike for each $i=1,2,3$ and such that on $V$ upon expanding $Z = Z^\mu \dfrac{\partial}{\partial x^\mu}$ the spacelike components of $Z$ vanish, that is, $Z = Z^0 \dfrac{\partial}{\partial x^0}$.

I want to show that such a naturaly adapted coordinate system always can be constructed from $Z$, so that it always exists, but I'm not being able to do this correctly.

My first guess was to use the well known fact (proven in Spivak's DG Vol 1, Page 148) that given a vector field $Z$ on $M$ and a point $p$ there is always a coordinate system around $p$ such that $Z$ is the first coordinate basis vector field.

But that is obviously not enough. The coordinate system so obtained might contain lightlike coordinates for example, or not be an orthonormal coordinate system more generaly.

The next idea would be to use the Fermi Walker coordinate constrction somehow. The point is that in that construction given one curve we can define a coordinate system adapted to it. This construction is outlined here.

Here we have a vector field, hence we have infinitely many curves. I think we just need to adapt this construction somehow. The problem is that each curve would give one coordinate system with one different domain, and there would be lots of overlaps, so we wouldn't be able to just apply this construction point by point.

So how can I understand intuitively the procedure of construction of the naturaly adapted coordinate system and how can I precisely construct it?


Say you want to construct coordinates around a point $p\in M$. You can start by taking any chart $(V,x)$ around $p$ with respect to which $\frac{\partial}{\partial x^0}=Z$. Now, choose three space like vectors $X_1,X_2,X_3\in T_pM$, such that $Z(p),X_1,X_2,X_3$ is a basis of $T_pM$. By composing the coordinate chart $x$ with a linear isomorphism of $\mathbb{R}^4$ that fixes one axis, you obtain a new chart, $(V,T\circ x)$, with respect to which you still have$$\frac{\partial}{\partial x^0}=Z,$$ and furthermore, $$\frac{\partial}{\partial x^i}(p)=X_i,\;i=1,2,3.$$ As being space like is an open condition, the vector fields $\frac{\partial}{\partial x^i},\;i=1,2,3,$ are space like in some neighborhood $p\in V'\subset V$, and you're done.

  • $\begingroup$ Nice approach! Just to be sure I got the idea: a vector field $X$ is spacelike at $p$ if $g(X,X)(p) > 0$. But $g$ is smooth, hence $g(X,X)\in C^\infty(M)$ and is continuous in particular. Sice it is greater than $0$ at $p$ it follows there's an open set containing $p$ where $g(X,X)$ keeps its sign. Hence if $X$ is spacelike at $p$ it is spacelike in this neighborhood. Then we just need to make sure we get spacelike vectors at $p$. It's this right? $\endgroup$ – user1620696 May 6 '17 at 3:45
  • $\begingroup$ @user1620696 Yes, exactly. $\endgroup$ – Amitai Yuval May 6 '17 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.