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In the context of special & general relativity there exists the notion of "rest frame" w.r.t. an object or an observer.

The latter are represented in spacetime by a timelike worldline (or rather by a timelike worldtube when taking the viewpoint that they have a spatial extension).

Now the notion of object's "rest frame" requires its worldline being "at rest" in that frame/chart. That means it has to be represented in this frame/chart by fixed spacelike coordinate values and varying timelike coordinate $t$.

The point I'd make is that, therefore, the notion of object's rest frame isn't uniquely defined though. Any frame/chart fulfilling the requirement above can be declared to be a "rest frame" of a given object.

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In 3+1 dimensional spacetime, any rotation of 3-space converts your rest frame into another, equally good rest frame. So if $SO(3)$ is the group of 3-space rotations, you have an entire $SO(3)$'s worth of rest frames.

The probable reason for your confusion is that most people first learn relativitity in 1+1 dimensional spacetime, where you have an $SO(1)$'s worth of rest frames. But $SO(1)$ has only one element --- the identity --- so you have only one rest frame. (Or two, if we also allow reflection, which, for pedagogical reasons, we usually don't.) Thus you have a unique rest frame in 1+1 dimensions.

When you move up to more than one spatial dimension, there is no natural convention that lets us pick out one of your rest frames as somehow "special", so yes, you have many rest frames and there's no natural way to choose among them.

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  • $\begingroup$ I do not think this is true. Special relativity does not have to happen in one global chart. That is only true for Minkowski spacetime. Special relativity spacetime on a cylinder or a torus cannot (with the usual definitions of charts) be covered by a single global chart. $\endgroup$ Commented Mar 18 at 19:14
  • $\begingroup$ @MarkusKlyver : The tangent space to a 4-manifold at a point looks exactly like the tangent space to any other 4-manifold at any other point. $\endgroup$ Commented Mar 18 at 19:16
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    $\begingroup$ @MarkusKlyver : They are isomorphic vector spaces, with, a fortiori, isomorphic Stiefel manifolds. Anything one can say about choosing a point in one of those Stiefel manifolds is exactly mirrored by something one can say about choosing a point in the other. As far as its internal structure goes, $T_pM$ knows nothing specific about $M$. It cannot tell the difference between Minkowski space and your cylindrical spacetime. $\endgroup$ Commented Mar 18 at 19:36
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    $\begingroup$ @CarloC The "curvature" is entirely extrinsic, one can cut a cylinder up and "unfold" it to a planar surface. So there is no meaningful (intrinsic) "curvature" to speak of. Compare that to a sphere, that cannot be "un-curved". $\endgroup$ Commented Mar 19 at 21:41
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    $\begingroup$ @CarloC The notation $S^n$ refers to the $n$-sphere (in $\mathbb R^n$). It is the set defined by $x_1^2 + \cdots x_{n+1}^2 = 1$. It is equipped with the subset topology, the submanifold smooth structure and the submanifold metric (all from $\mathbb R^n$). In this context, we mean that $M$ is diffeomorphic to $R \times S^1$ with $\mathbb R_t$ signifying that $M$ is time-orientated with the "obvious" time orientation ($t$ increases). $\endgroup$ Commented Mar 19 at 22:09
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A rest frame is simply one in which the spatial coordinates of the object or observer do not change, so you are free to choose any coordinate system that's stationary with respect to the object or observer. That flexibility can help simplify calculations- for example, if you have two observers moving inertially relative to each other, you might take the line of their relative motion to be the common direction of their x axes- in that case their velocity components in the y and z directions are zero and can drop out of the calculations.

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    $\begingroup$ This is not what a "rest frame" is. You can perfectly define a chart $(U, \varphi)$ for an accelerated particle such that the local coordinates of the particle are $\varphi(p) = (t, x = 0)$. $\endgroup$ Commented Mar 18 at 23:44
  • $\begingroup$ @MarkusKlyver as accelerated particle you mean a (proper) accelerating one, i.e. a particle traveling along a non geodesic timelike curve. $\endgroup$ Commented Mar 20 at 6:44
  • $\begingroup$ @CarloC The answer didn't specify the nature of motion, but even if the curve is a geodesic (and you want an inertial chart) you still have to say that all geodesics ought be lines, it isn't enough to preserve only the time axis. Anyway, you do not work with inertial charts in general relativity. You work with inertial frames, and frames are sections of the frame bundle. $\endgroup$ Commented Mar 20 at 9:54
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    $\begingroup$ @CarloC If you want an inertial frame you have to use something like Fermi-Walker transport to prevent the frame from "rotating". An inertial frame would also require $\gamma$ (in my definition) to be a timelike geodesic as opposed to any timelike curve $\gamma$. The definition I gave you gives you a definition of an observer and not an inertial and non-rotating observer. You are correct that I did not include this in my definition. $\endgroup$ Commented Mar 21 at 18:42
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    $\begingroup$ @MarkusKlyver Ah ok, you mean to get an inertial frame (intended as a section of the frame bundle, not necessarily a chart) such a frame field is required to be Fermi-Walker transported along the timelike curves of the congruence (i.e. zero Fermi derivative of the frame's vector fields along the congruence). $\endgroup$ Commented Mar 21 at 20:56
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Inertial coordinates only exist in special relativity, these are per definition those charts $(U, \varphi)$ such that

$$g := \operatorname{diag}(1, -1, -1, -1)$$

for all events $p \in U$, expressed using local coordinates $\varphi$. These charts are "declared" by virtue that the pseudometric must be $\operatorname{diag}(1, -1, -1, -1)$ in local coordinates. To what extent you are "choosing your inertial charts" or whether you "choose g" (or both simultaneously) is a mere philosophical difference.

In special relativity, inertial charts are related by the group elements of the reduced Poincaré group.

In general relativity, you only have inertial frames. There need not to be any inertial chart on a general spacetime $M$.

In general relativity (as opposed to special relativity), inertial frames are related by the group elements of the orthochronous Lorentz group (and the proper orthochronous Lorentz group if onwe wants to forbid orientation flipping).

The reason for why you can pretend that your chart is inertial in special relativity is that, in special relativity, you can identify the tangent space $TpM$ with points on the manifold. This is absolutely not true in general relativity and it is a grave sin to ever think of tangent vectors as embedded in the manifold. We often do this, because we are so indoctrinated by how it works in (affine) Euclidean space. In (affine) Euclidean space, we do not make any real difference between points and vectors.

The Euclidean plane, for example, can be thought as either:

a) all elements of a set $\mathbb R^2$ equipped with a vector space structure, or

b) the set of all two-dimensional "vectors" or "directions" anchored at some point deemed "the origin".

But this viewpoint breaks down on manifolds.

We have to remember that the manifold and its extra structure is merely a mathematical model for a real physical system. So we of course can, by our whim, define anything we want to mathematically. But the physically significant part is when we identify parts of that mathematical structure to real physical objects.

When we define our inertial charts (or our inertial frames, as in GR), we do so with a physical interpretation of what it means. We interpret an inertial chart $(U, \varphi)$ as "lines in $(U, \varphi)$ are geodesics with no acceleration". In general relativity, we interpret an inertial frame operationally as "a set of physical geometrical vectors that do not accelerate".

EDIT:

Someone mentioned the exponential map, and I think it is worth to explain what it is and the philosophical differences that I think are confused here.

The key point is that the exponential map is not determined by the smooth structure alone. The exponential map requires you to know what a “straight line” is, because that is how the map is defined: fix a point $p \in M$, and given a way to talk about “geodesics”, every tangent vector $v \in T_pM$ determines a geodesic $\gamma_v$ solving

$$\gamma_v(0) = p,\qquad \dot\gamma_v(0) = v.$$

Then define

$$\exp_p(v) := \gamma_v(1),$$

provided that the geodesic exists up to time $t=1$

So $\exp_p$ maps a neighborhood of $0 \in T_pM$ into $M$, and it sends the zero vector to $p$.

The exponential map essentially says:

  • pick a direction and speed at $p$,
  • follow the geodesic for unit time,
  • land at a point of $M$.

This gives a precise way to replace $M$ by its tangent space near $p$, but only up to the first place where geodesics start to overlap or fail to stay minimizing.

On a smooth manifold alone there is no canonical exponential map. The construction is canonical only after you have fixed extra geometric data, and even then the exponential map can fail to be defined because multiple geodesics connecting the same points, or geodesics just “ceasing to exist”.

The main point is that talking about “straight lines” is fundamentally different than talking about velocities, and both are needed in order to do the whole “tangent space to manifold” identification thing. The differential structure and the projective structure are different independent specifications of data.

Neither is really more fundamental than the other, and even if you have both you can only do an “identification” in limited cases geodesics may cease to exist, or overlap). Thus, it is rather the proper way to never confuse the two, i.e. the manifold and the tangent space.

A sort of identification works in special relativity (and in Minkowski space, in particular) because geodesics never overlap and are defined by affine lines in an inertial chart. In Minkowski space (with a global chart), the exponential map is defined on all of $T_pM$ and is globally bijective.

In fact, the geodesics in special relativity are sort of defined very clumpsy and out of thin air, the only way to know if a curve is (locally) a geodesic is to start from a chart you know is inertial and check if the curve is an affine line there. In general relativity, one adds the proper projective structure to talk about acceleration, in which charts does not matter again (as they “should not” do).

The cleanest chart-free version in special relativity is to use the fact that Minkowski spacetime is not just a manifold, but an an affine space with a translation-invariant Minkowski metric. That affine structure gives a canonical way to compare vectors at different points, so you can define acceleration without choosing coordinates. But the point is that one still has to "choose the inertial charts" (what charts counts as "inertial"). If one says "no inertial charts" and then immediately writes things $\nabla_u u$, one has only moved the machinery around. In special relativity, the connection is usually not "an extra physical structure" in the way it is in general relativity.

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  • $\begingroup$ "The reason for why you can pretend that your chart is inertial in special relativity is that, in special relativity, you can identify the tangent space TpM with points on the manifold. This is absolutely not true in general relativity..." First, this is just wrong. You can use the exponential map to identify points in the tangent space with points on the manifold in general relativity just as well as you can in special relativity. Second, even if it were right, I think you misspoke when you wrote "pretend your chart is inertial; it's frames, not charts that are inertial. $\endgroup$ Commented Mar 19 at 0:44
  • $\begingroup$ Third, identifying points in the tangent space with points on the manifold has absolutely nothing to do with frames, which are bases for the tangent space and have nothing to do with the rest of the manifold. If you had a disembodied vector space with no manifold in sight, you could still talk about frames, and if you had an inner product you could still talk about orthonormal (i.e. inertial) frames. Anything that mentions the rest of the manifold is irrelevant. (It becomes relevant if you want to talk about sections of the frame bundle, but that's a different matter.) $\endgroup$ Commented Mar 19 at 0:48
  • $\begingroup$ Charts can be inertial in special relativity. $\endgroup$ Commented Mar 19 at 0:50
  • $\begingroup$ Frames are per definition sections of the frame bundle, a basis for $T_pM$ at one $p\in M$ is not a frame. I also think you confuse and intertwine the differential structure and the projective structure of the manifold $M$, both are needed for any identification to happen. Only when the concepts of both velocity and geodesics exist, and the exponential map is built from them. Special relativity really do not have way to talk about acceleration, so the geodesics are kind of ad hoc defined. I expand on this in my edit above. $\endgroup$ Commented Mar 19 at 22:03
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A 'rest frame' is one at which the object is at rest, with no translation or rotation. For a bit more complication, its 4D Cartesian components are (x,y,z,t) with x, y, and z fixed for any t. How can you complicate that to have 'many rest frames'?

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  • $\begingroup$ As far as I understand it, from a rigorous perspective a frame of reference under SR also includes orientation, doesn't it? Thus WillO's answer. $\endgroup$ Commented Mar 19 at 19:38

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