Because of my mathematical background, I've been finding it hard to relate the physics-talk I've been reading, with mathematical objects.

In (say special) relativity, we have a Lorentzian manifold, $M$. This manifold has an atlas with local coordinates.

In differential geometry, when people talk about a `change of coordinates' they mean switching from one local coordinate system in this atlas, to another. For example, one coordinate system in this atlas is a map $\phi_1: U_1 \rightarrow V$ where $V$ is an open set of $M$, and $U$ is an open set of $\mathbb{R}^4$; and if another is $\phi_2: U_2 \rightarrow W$ is another ($U_2$ and open in $\mathbb{R}^4$, and $W$ an open in $M$), then $\phi_1^{-1} |_{V\cap W}\circ \phi_2|_{\phi_2^{-1}(V\cap W)}$ is a coordinate change.

However, in physics it seems that the meaning is different. Indeed if $p \in M$ then you can have a reference frame at $p$, but you can also have a reference frame that is accelerated at $p$. I'm not sure how to interpret this mathematically! What is the mathematical analogue of having an accelerated frame of reference at a point, as opposed to having an inertial frame of reference at a point?

  • 1
    $\begingroup$ For what it's worth, I think the "mathematical-physics" tag might be more appropriate than the "mathematics" tag for the sorts of questions you've been asking. To be honest, I'm not exactly sure whether the "mathematics" tag is really useful for this site... but anyway, your questions are very good, don't stop ;-) $\endgroup$
    – David Z
    Jul 12, 2011 at 23:23

5 Answers 5



  1. In the following explanation 4-dimensional space-times $M$ equipped with a metric of signature (3,1) are considered.

  2. There are several Wikipedia pages treating frames (sometimes called tetrads or Vielbeins) in GR. See for example, here, here and here

  3. There is a very good introductory chapter on the subject in chapter 5 of these notes by: R. Aldrovandi and J. G. Pereira.

A frame in GR means a set of four vector fields $\mathbf{e}_a: M \rightarrow TM$, $a=0, 1,2,3$ satisfying the constraint equation:

$\mathbf{g} = \eta^{ab} \mathbf{e}_a \mathbf{e}_b$,

where $\mathbf{g}$ is the inverse metric tensor and $\eta^{ab}$ is the flat Lorentzian metric.

These vector fields can be thought of as the mapping of the coordinate vectors of some given Mikowski space through the local coordinate system to the tangent space. In physical terms, we associate each such a frame with a local observer.

Now, basically, we can work with the components of the frame vector fields instead of the metric, but one observes that the frame fields have 16 components, while the metric has (due to its symmetry) only 10 components. This redundancy is due to the fact that the frame fields are not unique and a new set of frame fields $\mathbf{e}^{\prime}_a$ satisfying

$\mathbf{e}^{\prime}_a = M_a^b(x) \mathbf{e}_b$

satisfies the same constraint, where $M_a^b(x)$ is a Lorentz transformation matrix (i.e. satisfying $M_a^b(x) M_c^d(x) \eta_{bd} = \eta_{ab}$)

Please observe that we can choose a non-constant Lorentz transformation depending on the location on the manifold, for this reason, these transformations are called local Lorentz transformations.

Now the dimension count checks: 16 frame components = 10 Metric components + 6 Lorentz transformations at every point.

This formalism may seem merely a change of variables, but this is not the whole story.

First, the local Lorentz transformations can be viewed as sections of a principal $SO(3,1)$ bundle over $M$ (This bundle is called $SO(M, \mathbf{g})$. Thus this formulation is a formulation of GR as a gauge theory.Now, since we can allow the local Lorentz transformations to depend on the coordinates, this formalism allows to define accelerating frames, simply by taking thelocal Lorentz transformations to depend on time.

Secondly, in the standard formulation of GR allows to can define classical fields as sections of bundles whose local transformations are functions of the coordinate transformation (diffeomorphisms) of the base manifold. These bundles are called natural bundles, for example the coordinate transformation of the tangent bundle is the Jacobian matrix of the coordinate transformations of the base manifold. (Similarly, the inverse Jacobian matrix for the cotangent bundle). Thus the standard formulation of GR allows the definition of vector fields, tensor fields etc. but not spinor fields, which are very important in physics. Spinor bundles are not natural, but there is no natural way to define a general coordinate transformation of a spinor field given a diffeomorphism of the base manifold.

However, if the base manifold $M$ has a spin structure, then the frame formalism allows to define spinor fields as follows: Since $M$ is spin, $SO(M, \mathbf{g})$ can be lifted to a spin bundle $Spin(M, \mathbf{g})$ , then a spinor bundle is the associated bundle corresponding to a fundamental spinor representation, and spinor fields are sections of the spinor bundle.

This construction can be performed in local coordinates follows:

First, we can form the dual frame $\mathbf{e}^a: M \rightarrow T^{*}M$ by requiring:

$\langle \mathbf{e}^a, \mathbf{e}_b \rangle = \delta^a_b$

The dual frame can be used to define the frame components of any vector field $\mathbf{V}$:

$V^a = \langle \mathbf{e}^a, \mathbf{V} \rangle$

Conversly, one can form the "curved" components of a vectors using the original frame. For example, consider the Dirac matrices $\{\gamma^a\}$ generating the Clifford algebra $Cl(3,1)$. Then their curved components are given by:

$\gamma^{\mu} = \gamma^a e_a^{\mu}$

More generally, one uses the metric $\mathbf{g}$ to lower "curved indices", the inverse metric to raise "curved indices". and similarly, the Lorentz metric $\mathbf{\eta}$ for the flat indices. One uses the frame vectors and their dual to replace curved indices with flat indices and vice-versa.

Next ,the spin connection

is defined as:

$\omega_{\mu}^{ab} = e^a_{\nu}(\partial_{\mu} e^{\nu b}+ e^{\sigma b}\Gamma^{\nu}_{\sigma \nu})$

where, $\Gamma^{\nu}_{\sigma \nu}$ is the Levi-Civita connection.

It is not difficult to verify (by looking at the local Lorentz transformations) that $\omega_{\mu}^{ab}\sigma_{ab}$ is a connection on $Spin(M, \mathbf{g})$, where $\sigma_{ab}$ are the generators of the fundamental spinor representation.

  1. Using the above data, the fully covariant Dirac equation on $M$ takes the form:

$-i \gamma^{\mu} D_{\mu} \psi + m \psi = 0$,

where $D_{\mu}$ is the covariant derivative associated with the spin connection

$ D_{\mu} = \partial_{\mu}-i\omega_{\mu}^{ab}\sigma_{ab}$

Thus the fully covariant Dirac equation looks just like the Dirac equation coupled to a gauge field given by the spin connection.

Classical fields where this construction is possible are sections of bundles called "gauge natural bundles".

It is important to mention that the solution of the fully covariant Dirac equation depends on the frame fields, but observable quantities such as the number of bound states for example, depend only on the metric.


Since local observers are identified with points on the fibers of the frame bundle, then all frames are inertial because they can be obtained from the action of a Lorentz transformation on a single frame (i.e. point on the fiber). The parameters of the lorentz transformation are the velocity vector and the orientation of the frame. It is explicit in the equations that we can allow variable Lorentz transformations, i.e., Lorentz transformations dependent on the local coordinates of the base manifold, in particular on the time coordinate. Now I'll divide my answer into two parts:

Particles: Suppose the four velocity vector of a particle moving on a geodesic is given by $V^{\mu} = \frac{dx^{\mu}}{d\tau}$, ($\tau$ is any parameter along the path) then the frame coordinates of this vector are: $V^a = e^a_{\mu} V^{\mu}$ and the components of the velociy measured by an observer moving with a velocity defined by the Lorenntz matrix $M(x)$ are $V^{\prime b} = M^b_a(x) V^a$. Again, a variable $M$ indicates an accelerating frame.

Fields: The equations of motion will be covariant with respect to these transformations, because for sections of natural bundles, the frame vectors do not appear in the equations of motion, while in the case of gauge natural sections such as spinors these (variable Lorentz) transformations will appear as gauge transformations and the equations of motion are constructed to be gauge invariant. Thus, the equations of motion are not affected by local Lorentz transformations, or in other words, Physics looks the asme to all observers even if they are accelerating.

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    $\begingroup$ I guess, it is discussed an approach originated by Ashtekar and widely used in loop quantum gravity, but in general we may talk about tangent bundle without need to talk about spin structure ... yet without that matter could not exist. $\endgroup$ Jul 13, 2011 at 15:03
  • $\begingroup$ I'm not sure I understand. You're saying that a frame of reference is a section of $(TM)^4$ (satisfying some conditions)? But then what is meant by saying that it is a frame of reference at a point? It seems that the definition that you gave is global in nature. Is this an abstraction of the phrase `frame of reference' to the point of it not being dependent on the location of the observer? I think I'm missing something basic. $\endgroup$
    – Wesley
    Jul 13, 2011 at 15:17
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    $\begingroup$ Yes, a frame is a section of a bundle over $M$, which is actually called the frame bundle, (becuase of the extra constraints it is not just $(TM)^4$). The frame bundle is an associated bundle to $SO(M, \mathbf{g})$. A local frame at a manifold point is just a point in the fiber of this bundle over the given point. Local observers corresponding to different inertial frames constitute of different points on the same fiber $\endgroup$ Jul 13, 2011 at 15:38
  • $\begingroup$ Alright, let me make sure I understand. A frame of reference at a point is a choice of a fiber in the $SO(M,g)$-bundle on $M$. This, via the connection, is equivalent locally (near our point in $M$) to a section of the $SO(M,g)$-bundle. Now, an inertial reference frame, is such a choice so that locally near our point, the frame (section) doesn't change along the time component. An accelerated reference frame is a non-inertial reference frame. Did I get that right? $\endgroup$
    – Wesley
    Jul 13, 2011 at 16:09
  • $\begingroup$ Or is the point, perhaps, that "change along the time component" doesn't really make sense (change compared to what?); so that saying "inertial" or "accelerated" makes sense only compared to another reference frame? $\endgroup$
    – Wesley
    Jul 13, 2011 at 16:48

David Bar Moshe has given a very complete answer at a high level of sophistication in both math and physics. If that exactly meets the needs of the OP and others who read this page, that's great. I would just like to take a shot at addressing the OP's question in simpler language.

GR does not have global frames of reference the way SR does. (When David Bar Moshe says, "A frame in GR means ...," what he's defining isn't a global frame of reference, it's a collection of local frames of reference, each defined at a different point.) As a concrete example, in SR we expect to be able to say that in object A's frame of reference, distant object B has some well-defined velocity. In GR, we can't do this. For example if A is our galaxy and B is some cosmologically distant galaxy, then there is no uniquely well-defined way of defining B's velocity in A's frame. We can say that B is moving relative to A, or we can say that both A and B are at rest and the space between them is expanding. GR doesn't say that one of these verbal descriptions is to be preferred over the other.

The OP is asking about charts and atlases and how they relate to changes of frames of reference. They don't. All of the interesting issues can be discussed without ever having to deal with a manifold that requires more than one chart. For example, in FRW cosmological models, you can cover the entire spacetime with a single chart using coordinates $(t,x,y,z)$, and you would typically do this in such a way that a galaxy at rest relative to the Hubble flow had constant x, y, and z. An example of a change of coordinates would be $t\rightarrow 2t$. In this spacetime, there is no notion of a global frame of reference, and there is nothing that plays the role of a Lorentz boost as there would be in SR. If you try to apply the usual Lorentz transformation equations to $(t,x,y,z)$, you get a set of coordinates $(t',x',y',z')$ that are perfectly valid but utterly uninteresting physically, and you cannot interpret the new ones as a frame that is moving at certain speed relative to some frame defined by the old ones.

Re the OP's question about accelerating versus nonaccelerating, GR is relatively agnostic about this. We do need to distinguish frames from coordinates, however. Any smooth change of coordinates (diffeomorphism) is allowed, and has no effect on the form of the laws of physics (Einstein field equations). For example, you can describe flat spacetime in accelerating coordinates, where it appears to have a gravitational field: http://en.wikipedia.org/wiki/Rindler_coordinates In terms of frames, we do have preferred frames in GR, which are the frames of observers who are free-falling. In GR, the frame of a rock sitting on the ground is considered noninertial, and the frame of a falling rock is considered inertial; it's exactly the opposite of what we'd say in Newtonian physics.


I want to add my personal understanding of the concept of reference frame.

In the articles:

  • Marmo, G., Preziosi, B. (2006). The Structure Of Space-Time: Relativity Groups. International Journal of Geometric Methods in Modern Physics, 03(03), 591-603.
  • Marmo, G., Preziosi, B. (2005) Objective existence and relativity groups. Symmetries in Science XI, 445-458.

it is presented an approach toward reference frames I found physically useful.

Essentially a reference frame $\mathcal{R}$ is defined as a (1,1) tensor on the spacetime manifold $\mathcal{M}$ such that:

  • it is of rank one, i.e. it is decomposable as $\mathcal{R}=\theta\otimes T$, where $\theta\in\Lambda^{1}(\mathcal{M})$ and $T\in\mathfrak{X}(\mathcal{M})$;

  • it is such that $\theta(T)=1$.

In a general relativistic setting we can choose $T$ to be timelike and $\theta$ to be its dual form. This is close enough to the approach by means of triad.

The integral curves of $T$ defines the worldlines of different observers in $\mathcal{R}$.

The important thing to note is that the tangent bundle $T\mathcal{M}$ of $\mathcal{M}$ is splitted as:

$$ T\mathcal{M}=\mathcal{R}^{t}\oplus\mathcal{R}^{s} $$

where the time distribution is $\mathcal{R}^{t}=span(T)$ and the space distribution is $\mathcal{R}^{s}=Ker(\theta)$.

This is, in general, only a local splitting, reflecting the fact that in general there are no global reference frames.

If we have $\theta\wedge d\theta=0$ then $\theta$ gives rise to an integrable foliation (by means of the Frobenius theorem), and the leaves of this foliations are the local rest spaces of the reference frame.

If moreover $\theta=df$ then the reference frame is synchronizable. This means that there is a one-to-one correspondence between the leaves of the spatial foliation (local rest spaces) and the evolution parameter of the integral curves of $T$, thus different observers in this reference frame possess a common notion of Time.

One can speak of accelerated or inertial reference frame by choosing different vector fields $T$.

For example in Minkowski spacetime the (global) reference frame:

$$ \mathcal{R}=dx^{0}\otimes\partial_{x^{0}} $$ is inertial in the sense that, being $T=\partial_{x^{0}}$ a geodesic vector field:

$$ \nabla_{T}T=0 $$ with respect to the flat metric connection on Minkowski spacetime, its integral curves $\gamma(\tau)$ satisfy the equations:

$$ \frac{d^{2}\gamma^{\mu}(\tau)}{d\tau^{2}}=0 $$ The integral curves are straight lines.

Geodesics reference frames have no acceleration in the sense that the vector field $T$, being geodesic, is such that $a_{T}=\nabla_{T}T=0$, where $a_{T}$ is the acceleration vector field. This does not mean that in a non-flat spacetime a geodesic reference frame is inertial, in fact if the metric connection is not flat then the integral curves of the geodesic vector field $T$ are not $\frac{d^{2}\gamma^{\mu}(\tau)}{d\tau^{2}}=0$, and thus are not straight lines.

I know that this answer is not as clear and profound as the others, nevertheless I think that this approach toward reference frames can be useful because it can be used in contexts different from the general relativistic one, while the triad formalism is naturally tied to general relativity.

At this purpouse there is another approach toward reference frames as almost product structures, i.e. starting from the splitting $T\mathcal{M}=\mathcal{R}^{t}\oplus\mathcal{R}^{s}$ without using the tensor field $\mathcal{R}$, however I do not remeber the references for this (maybe it was something by D. H. Delpenich).

Finally I want to add an easy application of the concept of reference frame in Electrodynamics I find very interesting.

In the Classical Theory of Electromagnetism on Minkowski Spacetime $\mathcal{M}$ the Electromagnetic Field $F$ is seen as the curvature $2$-form of the pullback of a $U(1)$-principal connection on $\mathcal{M}$, then it is easily seen that the choice of the reference frame $\mathcal{R}=dx^{0}\otimes\partial_{x^{0}}$ allows for the definition of a differential one-form (electric field):

$$ E:=i_{T}F\equiv - E_{1}dx^{1} - E_{2}dx^{2} - E_{3}dx^{3} $$ and a differential two-form (magnetic field):

$$ B:=F - \theta\wedge i_{T}F\equiv B_{3}dx^{1}\wedge dx^{2} - B_{2}dx^{1}\wedge dx^{3} + B_{1}dx^{2}\wedge dx^{3} $$ such that Maxwell's equations $dF=0$ and $d\star F=0$ can be written as:

$$ 0=\vec{\nabla}\times \vec{E} + \frac{d\vec{B}}{dt}=\left\{\begin{array}{c} \partial_{2}E_{3} - \partial_{3}E_{2} + \partial_{0}B_{1} \\ \partial_{3}E_{1} - \partial_{1}E_{3} + \partial_{0}B_{2} \\ \partial_{1}E_{2} - \partial_{2}E_{1} + \partial_{0}B_{3}\end{array}\right. $$

$$ 0=\vec{\nabla}\times \vec{B} - \frac{d\vec{E}}{dt} =\left\{\begin{array}{c} \partial_{2}B_{3} - \partial_{3}B_{2} - \partial_{0}E_{1} \\ \partial_{3}B_{1} - \partial_{1}B_{3} - \partial_{0}E_{2} \\ \partial_{1}B_{2} - \partial_{2}B_{1} - \partial_{0}E_{3}\end{array}\right.\;\;\; $$

$$ 0=\vec{\nabla}\cdot \vec{B}=\partial_{1}B_{1} + \partial_{2}B_{2} + \partial_{3}B_{3} $$

$$ 0=\vec{\nabla}\cdot \vec{E}=\partial_{1}E_{1} + \partial_{2}E_{2} + \partial_{3}E_{3} $$ which is nothing but the formulation of Maxwell equations originally proposed by Maxwell before Special Relativity.

  • $\begingroup$ Thanks for the explanation of what synchronizable means! $\endgroup$
    – Stan
    Jul 21, 2014 at 17:19

I believe the equivalent mathematical terminology to an inertial frame is "normal coordinates" at the point $p$. Meaning, in your coordinates the metric at $p$ is just the flat Minkowski metric and all the first derivatives of the metric vanish at $p$. Conversely an accelerated frame would be any coordinates that are not normal.

  • $\begingroup$ This answer (v1) is correct. However, it should be stressed that normal coordinates does not guarantee that higher partial derivatives of the metric vanish at $p$. Only the single partial derivatives of the metric (and the Christoffel symbols) vanish at $p$. This implies that the geodesic equation at $p$ implements Newton's first law at $p$. $\endgroup$
    – Qmechanic
    Jul 13, 2011 at 18:07
  • $\begingroup$ @Qmechanic I think normal coordinates means local frame (at a point of spacetime), and an inertial frame should be the Fermi coordinates of a geodesic. (which makes sense since an inertial frame is a frame of freely falling observer) Or I am wrong? $\endgroup$ Jan 25, 2015 at 10:44

It is likely mean a consideration of a map to reference frame of accelerated observer discussed, e.g., in ch. 13.6 of Misner, Thorne, Wheeler, “Gravitation”. Maybe it is simple to start with consideration of coordinate system corresponding to motion with constant acceleration in ch. 26 of Pauli “Theory of relativity” about hyperbolic motion.

  • $\begingroup$ This isn't really an answer. $\endgroup$
    – user4552
    Jul 30, 2017 at 17:13

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