here I come with a question about inertial frames as defined in General Relativity, and how to prove that the general definition is consistent with the particular case of Special Relativity.

So to contextualize, I have found that one can define inertial frames in General Relativity, as follows: given a 4-dimensional lorentzian manifold $M$ with metric tensor $g$ and signature (-1,1,1,1), a frame field is defined to be a set of four vector fields $\{e_{0},e_{1},e_{2},e_{3}\}$ such that

$g(e_{i},e_{j}) = \eta_{ij}$ ($\eta$ being the Minkowski matrix with the signature given above)

Further, one says that the frame field in question is inertial and nonrotating if the following condition is satisfied:

$\nabla_{e_{0}}e_{i} = 0$ for $i=0,1,2,3$

where $\nabla$ is the Levi-Civita connection on the manifold (let me call this "Definition A"). All this straight out of Wikipedia, and quite elegant and nice.

Now, my question is how does this correspond to the definition of inertial frames in Special Relativity as usually found in textbooks. So now assume that g is the Minkowski metric and that we are in Minkowski spacetime, where there is a given global coordinate chart $(t,x,y,z)$ (I am setting $c=1$ for simplicity) such that:

$ g = -dt\otimes dt + dx\otimes dx + dy\otimes dy + dz\otimes dz$

Then it is straightforward to see that the vector fields $\partial/\partial t, \partial/\partial x, \partial/\partial y, \partial/\partial z$ are indeed an inertial nonrotating frame field in the abstract sense of Definition A. But in Minkowski spacetime one has Definition B (the one usually found in textbooks), where we are given an initial "precursor" inertial frame (the one used by the scientist writting the textbook, or whatever), which in this context is a coordinate chart with certain physical properties, namely that there are no inertial forces (i.e. the Christoffel symbols vanish identically), and can of course be identified with the coordinate chart $(t,x,y,z)$ I mentioned earlier. And then -crucially- we are told that "inertial frames" are all those coordinate charts related to this precursor coordinate chart by a Lorentz transformation, which is a linear coordinate transformation $\Lambda$ such that $\eta_{ij} = \Lambda^{k}_{i}\Lambda^{\ell}_{j}\eta_{k\ell}$ (this guarantees that the spacetime interval is preserved).

So, here comes my question: in Minkowski spacetime, take an arbitrary nonrotating inertial frame in the sense of Definition A; that is, a set of four vector fields $e_{0},e_{1},e_{2},e_{3}$ satistying the requisites of Definition A. Is it the case, then, that there exists a coordinate chart $x^{0},x^{1},x^{2},x^{3}$ and a suitable Lorentz transformation $\Lambda$ from $(t,x,y,z)$ (the "precursor coordinate chart" inherent to the construction of Minkowski spacetime) to $(x^{0},x^{1},x^{2},x^{3})$ such that:

$e_{i} = \partial/\partial x^{i}$ for $i=0,1,2,3$ ?

  • $\begingroup$ In definition A, aren't the vector fields $e_i$ only defined on the worldline of a single observer, not on the whole spacetime? So I don't quite understand what you mean when you write $e_i = \partial / \partial x^i$. $\endgroup$
    – Kenny Wong
    Commented Mar 1, 2017 at 15:05
  • $\begingroup$ If you only want $e_i = \partial / \partial x^i$ to hold on your observer's worldline, you could try Fermi normal coordinates, en.wikipedia.org/wiki/Fermi_coordinates $\endgroup$
    – Kenny Wong
    Commented Mar 1, 2017 at 15:09
  • $\begingroup$ No, the $e_{i}$ are vector fields, so they are not attached to any worldline... think of them as a family of inertial observers rather than a single inertial observer (en.wikipedia.org/wiki/Frame_fields_in_general_relativity) I want to prove that $e_{i} = \partial / \partial_{x^{i}}$ for some inertial coordinate system $x^{i}$ on the entire Minkowski spacetime (not an individual worldline) since that would make Definition A (inertial frames in GR) equivalent to the standard definition of inertial frames in the context of Special Relativity, if we are in Minkowski spacetime. $\endgroup$ Commented Mar 2, 2017 at 15:48
  • $\begingroup$ Then you want the worldlines of the observers to be related in some way, I guess? Observers moving at "equal velocity", perhaps? $\endgroup$
    – Kenny Wong
    Commented Mar 2, 2017 at 15:49
  • $\begingroup$ $\nabla_{e_{0}}e_{i}=0$ means that from your point of view you have a triad of orthonormal vectors at each point of your spacelike slice of the spacetime at any given time (measured by you), and that all these triads are paralell transported along the timelike direction. $\endgroup$ Commented Mar 2, 2017 at 15:55

1 Answer 1


Consider this smooth vector field: $$ e_0 = \partial_t, \ \ \ e_1 = \cos z \partial_x - \sin z \partial_y, \ \ e_2=\sin z \partial_x + \cos z \partial_y, \ \ e_3 = \partial_z. $$ This obeys $\nabla_{e_0} e_0 = \nabla_{e_0} e_i = 0$.

However, there exists no coordinate system $(x^0, x^1, x^2, x^3)$ such that $e_i= \partial_{x^i}$ for $i = 0,1,2,3$.

Indeed, suppose for contradiction that such a coordinate system exists. Then the commutator $[e_1,e_3] = \partial_{x^1}\partial_{x^3} - \partial_{x^3}\partial_{x^1}$ is identically zero, since partial derivatives commute. However, by explicit calculation, $$ [e_1, e_3] = \sin z \partial_x + \cos z \partial_y \neq 0. $$

  • $\begingroup$ Interesting example! And it illustrates something: the $e_{i}$'s you give are linear combinations of the partials, but the coefficients are non-constant smooth functions... So lets suppose in general that you are in Minkowski spacetime $(M,g)$ and you define: $$e_{i} := \Lambda^{j}_{i}\partial_{j} $$ and impose that $\nabla_{e_{0}}e_{i} = 0$ for $i=0,1,2,3$ and that $g(e_{i},e_{j}) = \eta_{ij}$ for $i,j=0,1,2,3$, and that $[e_{i},e_{j}] = 0$. I would expect that the only way to make this so is to demand that the functions $\Lambda^{j}_{i}$ be constants... is this an "iff" ? $\endgroup$ Commented Mar 2, 2017 at 18:57
  • $\begingroup$ Interesting question. I'm not sure what the answer is. Maybe try experimenting? $\endgroup$
    – Kenny Wong
    Commented Mar 2, 2017 at 20:20
  • $\begingroup$ The answer is yes! If $g(e_{i},e_{j}) = \eta_{ij}$ and that $[e_{i},e_{j}] = 0$, then $g(\nabla_{e_{i}}e_{j},e_{k}) = 0$ by Koszul's formula and hence $\nabla_{e_{i}}e_{j} = 0$ for all $i,j$. Then: $$ 0 = \nabla_{e_{i}}e_{j} = \nabla_{\Lambda^{k}_{i}\partial_{k}}\Lambda^{\ell}_{j}\partial_{\ell} = \Lambda^{k}_{i}\partial_{k}(\Lambda^{\ell}_{j})\partial_{\ell} $$ which implies $\Lambda^{k}_{i}\partial_{k}\Lambda^{\ell}_{j} = 0$. Multiply this by the inverse $\tilde{\Lambda}^{i}_{m}$ and you get: $$\partial_{m}\Lambda^{\ell}_{j} = 0$$ as desired. The converse should be trivial :) $\endgroup$ Commented Mar 3, 2017 at 14:24
  • $\begingroup$ Interestingly enough, requiring that $[e_{i},e_{j}]=0$ and that $g(e_{i},e_{j})=\eta_{ij}$ automatically guarantees that the frame is inertial. $\endgroup$ Commented Mar 3, 2017 at 14:29
  • $\begingroup$ Also interestingly enough, the result doesn't depend on the fact that we are in Minkowski spacetime with the canonical basis of the $\partial_{i}$'s, since we could replace them with $e^{\prime}_{i}$'s such that $g(e^{\prime}_{i},e^{\prime}_{j})=\eta_{ij}$ and $[e^{\prime}_{i},e^{\prime}_{j}]=0$, and the same procedure gives $e^{\prime}_{m}[\Lambda^{\ell}_{j}]=0$, thus the coefficients are constants. So it seems that in any spacetime, constant Lorentz transformations connect frames with vanishing commutator... and I'm beginning to suspect that such frames only exist in Minkowski spacetime. $\endgroup$ Commented Mar 3, 2017 at 14:41

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