# Is a local Lorentz frame a coordinate chart on a spacetime manifold?

I am just starting to learn GR. I'm alternating between studying physics books and studying math books. I keep seeing the term Lorentz frame and I'm not sure what it means mathematically.

Is a local Lorentz frame in mathematical terms a coordinate chart on a manifold?

• Is physics invariant to a displacement in your coordinate chart? – CuriousOne Dec 27 '14 at 21:46

I) Given a point $p\in M$ in a Lorentzian manifold $(M,g)$ a local Lorentz frame at $p$ usually refers to the tangent space $T_pM$ at $p$, not the manifold $M$ itself.

Note that the tangent space $T_pM$ at $p$ is naturally equipped with a constant metric $g_p: T_pM \times T_pM \to \mathbb{R}$, while the metric $g$ on the manifold $M$ in general could be curved.

II) However, in a neighborhood $\Omega \subseteq T_pM$ around the origin $0\in T_pM$ in the tangent space $T_pM$, there is an exponential map $\exp_p: \Omega \to M$ to the manifold $M$.

So in that sense, we can use the exponential map to view $\Omega$ as a coordinate chart for a neighborhood $U:=\exp_p(\Omega)\subseteq M$ on the manifold $M$.

The relevant curved metric $\exp_p^{\ast}(g|_U)$ in the coordinate chart $\Omega\subseteq T_pM$ is the pullback of the metric $g|_U$ in $U\subseteq M$.

• I don't know what an exponential map is but I can tell if I did this would answer my question. Thanks! – Stan Shunpike Dec 27 '14 at 23:04
• @StanShunpike: the exponential map just tells us that the tangent plane locally approximates the manifold. – Jerry Schirmer Dec 28 '14 at 15:39

In differential geometry, a frame for a manifold $M$ is a section of the frame bundle $FM$, the fiber bundle over $M$ whose fiber at $p$ is the set of all bases for $T_p M$, which is the same as $GL(\dim M)$. For any point $p$ in the domain of a coordinate chart $x^\mu$, since the set $\{\frac\partial{\partial x^\mu}\big|_p\}$ is a base, any coordinate chart gives you a frame.

When $M$ is equipped with a metric, we can consider only orthonormal bases. More generally, let the fiber over $p$ be the set of ordered tuples $v^i$ of tangent vectors at $p$, such that $$g(v^i, v^j) = a^{ij}$$ for some constant symmetric matrix $a^{ij}$ (with the same signature as the metric). This is the same as $O(g)$ -- the orthogonal group for a metric of signature $g$. In general relativity where $g$ has signature $(1,3)$, this is more commonly known as the Lorentz group, and so these are called Lorentz frames (or tetrads, or vierbein).

Note that we need take $a^{ij} = \operatorname{diag}(1,-1,-1,-1)$. In many applications it is convenient to take a null tetrad with $a^{01} = -1, a^{23} = 1$. Note also that a coordinate frame is a Lorentz frame if and only if the domain of the coordinate chart is flat. (Can you see why?)