I try to do basics computations of SR with the heavier formalism of GR to see if I understand it well.

Change of coordinates is spacetime: changes of coordinates in space time are change of coordinate maps in the $(\mathbb{R}^4,\eta)$ Lorentzian manifold. For the cartesian coordinates we have one global map and it's the identity. If we want to go to other coordinate we take another atlas of $(\mathbb{R}^4,\eta)$ then we perform the changes of coordinate as seen in differential geometry courses.

Change of coordinates in tangent spaces: tangent spaces have a natural basis given by the coordinate on the manifold. To change coordinates in tangent spaces, it's the same thing as for general vector spaces: we do a linear combination of the basis vectors, then deduced how the components change, etc.

Problem: when we are talking about Lorentz boosts is the $x$ direction in RR, we usually write \begin{equation} \begin{bmatrix} \gamma&-\beta\gamma&0&0\\ -\beta\gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} t\\x\\y\\z \end{bmatrix}= \begin{bmatrix} t'\\x'\\y'\\z' \end{bmatrix} \end{equation} with usual notations. We usually say that we "change of coordinate" from $(t,x,y,z)$ to $(t',x',y',z')$.

  1. Since it's a linear transformation between four-vectors, is it a change of coordinates in a tangent space?

  2. Aren't the $(x,y,z,t)$ suppose to be the coordinates in spacetime? I always saw $x^\mu$ as being the $\mu$-th component of a coordinate map $x:U\subset\mathcal{M}\to\mathbb{R}^4$, $\mathcal{M}$ a manifold.

  3. More generally, what is really the role of Lorentz transformation in curve spacetime? What does "reference frame" really means is this context?

I would love to read about that but I didn't see anything in the classical GR references.

  • $\begingroup$ Lorentz transformations are to be interpreted as a special set of chart transition maps: the ones that are linear. You may ask why vectors (like velocity, acceleration etc) transform this way as well in SR? Given a chart transition map $x'(x)$, vectors (which live in the tangent space) transform under the action $\frac{\partial x'}{\partial x}$. In the case of linear transformations, this matrix happens to be identical to the matrix that implements the change of chart. $\endgroup$
    – Anonjohn
    Commented Nov 13, 2020 at 17:09
  • $\begingroup$ Related: physics.stackexchange.com/q/544193/2451 $\endgroup$
    – Qmechanic
    Commented Nov 13, 2020 at 18:20
  • $\begingroup$ Regarding your last question, the most precise definition of a reference frame I have seem is that by Sachs & Wu in the book General Relativity for Mathematicians. A reference frame is a timelike future-directed unit vector field. It gives you a splitting between space and time at each tangent space of the open set in which it is defined. An alternative definition is to define a reference frame to be a section of the orthonormal frame bundle, i.e., a choice of orthonormal basis of vector fields. $\endgroup$
    – Gold
    Commented Nov 20, 2022 at 20:46

2 Answers 2


I think your question is resolved by the following observation. In a curved space $(t,x,y,z)$ do not in general make a 4-vector, but $(dt,dx,dy,dz)$ do make a 4-vector.

In GR you can use a Lorentz transformation to switch between local inertial frames in the vicinity of any given event. You only apply such a transformation locally.


Lorentz transformation is a change of basis of tangent space. Tangent space is defined to mean a vector space associated with a point in spacetime. One should not confuse this with coordinates. Coordinates do not define vectors in general curved spacetimes. Thus we can apply Lorentz transformation to vectors, like momentum, but not to coordinates.

A reference frame consists of the reference matter, the apparatus, and the procedures, required to determine a spacetime coordinate system.

A coordinate system is a mapping from physical events to coordinates with the form $(x^0, x^1, x^2, x^3)$

Usually $(x^0, x^1, x^2, x^3) = (t, x, y, z)$where $t$ is the time of the event and $(x, y, z)$ describes the position of the event, or we may be using polar coordinates $(r,\theta, \phi)$, and we may also use more general coordinates.

  • $\begingroup$ Note that, because the minkowski metric is flat, the tangent space and the original space have the same structure, and any change of basis in the tangent space is has a corresponding global coordinate transformation. So any lorentz boost or 3-rotation in the tangent space has an equivalent coordinate transformation, that is just equivalent to applying the boost/rotation to every point in the spacetime. $\endgroup$ Commented Nov 13, 2020 at 19:33
  • $\begingroup$ @JerrySchirmer, there is perhaps ambiguity in the original question. If we are talking of gr, we should not assume that the original spacetime is flat. If we are assuming that spacetime is flat, there is indeed no difference between sr and gr, and there is not even a question. I do not think that is what the OP intended, and it is certainly not assumed in my answer. $\endgroup$ Commented Nov 13, 2020 at 20:43
  • $\begingroup$ Oh, I'm just adding flavor explaining why this can get confusing to people, because for flat spacetimes, the space and the tangent space have the same structure. $\endgroup$ Commented Nov 14, 2020 at 0:15
  • $\begingroup$ What about @Anonjohn 's comment? Is Lorentz transformation also a particular combination of charts to change coordinate (in spacetime, not in the tangent space)? $\endgroup$
    – xpsf
    Commented Nov 14, 2020 at 17:49
  • 1
    $\begingroup$ @xpsf, Lorentz transformation applies only to inertial frames in flat spacetime. In general relativity it refers to transformations in a tangent space. $\endgroup$ Commented Nov 14, 2020 at 17:58

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