What is an inertial frame in terms of acceleration?

Question

After learning about the difference between coordinate and proper acceleration, I am now confused on the precise definition of an inertial reference frame in terms of proper and coordinate acceleration, in the context of special and general relativity.

In special relativity, an inertial frame is a frame with zero acceleration (zero proper acceleration?) and where Newton's laws hold and there are no fictitious forces.

In general relativity, it seems like gravity is now viewed as another kind of fictitious force and an inertial frame is now defined as a frame with zero proper acceleration and fictitious forces including gravity is absent (?) Under this definition, any frame where you are in weightless state (e.g. stranded in outer space or undergoing free fall in some gravitational field) is considered as inertial frames, right? So a frame where you feel gravity (without other fictitious forces) is considered as inertial frame in special relativity but not in general relativity?

Answer 1

In terms of acceleration an inertial frame is a reference frame where all objects with zero coordinate acceleration also have zero proper acceleration.

This condition implements Newton’s first law. Force-free objects are identified by zero proper acceleration and traveling at a constant speed in a straight line is identified by zero coordinate acceleration. Hence such frames implement Newton’s concept of inertia and are therefore called inertial frames.

The difference between SR and GR is only that in SR inertial frames are global while in GR inertial frames are local. Gravity is locally indistinguishable from a fictitious force in GR, and in SR it is globally indistinguishable from a fictitious force (and limited to fields with no tidal effects).

Under this definition, any frame where you are in weightless state (e.g. stranded in outer space or undergoing free fall in some gravitational field) is considered as inertial frames, right?

Yes. Again, locally in GR and globally in SR, and with the clarification that we are talking about the weightless object's rest frame.

So a frame where you feel gravity (without other fictitious forces) is considered as inertial frame in special relativity but not in general relativity?

Such a frame is not inertial in either GR or SR. Also, there isn't a way to identify frames where you feel gravity but not other fictitious forces. So that requires a bit of handwaving, but I think we can get away with it if we don't look too hard at edge cases.

Answer 2

In special relativity, an inertial frame is a frame with zero acceleration (zero proper acceleration?) and where Newton's laws hold and there are no fictitious forces.

EDIT: In light of the comment by @ProfRob I have rewritten this first section:

If by the Newtonian laws of physics, we restrict ourselves to just Newton's 3 laws of motion, then the statement is essentially correct with a small caveat for the 2nd law. Newton's 2nd law can be stated as $F = ma$ or $F = \dfrac{\mathrm{d}p}{\mathrm{d}t}$. When stated as $F = ma$, this law does not hold, but when stated in the form of change in momentum per unit time ($F = \mathrm{d}p/\mathrm{d}t$), the 2nd law does hold in an inertial frame.

It is not clear (to me) if the 2nd and 3rd laws fail in an accelerating reference frame. I think it is reasonable to conclude that if the observer feels no proper acceleration and the 1st law holds, then that is sufficient to define a inertial reference frame.

So a frame where you feel gravity (without other fictitious forces) is considered as inertial frame in special relativity but not in general relativity?

No. A frame where you feel gravity is not considered an inertial frame in either SR or GR. On the surface of the Earth, you can measure your own proper acceleration by standing on some bathroom scales, so you are not in an inertial reference frame when standing on the surface of the Earth in either SR or GR.

What is an inertial frame in terms of acceleration?

If an observer measures their own proper acceleration to be zero then they are locally in an inertial reference frame. If they set out an extended grid of clocks that are all at rest with respect to themselves and all the clocks appear to be running at the same rate and the speed of light is measured to be constant in all directions over that extended region, then they can consider themselves to be in an extended inertial reference frame.

There is a notion of a momentary inertial reference frame for an observer that is accelerating relative to an inertial observer. If the instantaneous velocity of the accelerating observer is $u$, then the time dilation of accelerating observer is equal to that of a momentarily co-moving inertial observer ($t = t'\sqrt{1-u^2/c^2}$) and over a very short time and space interval we can treat the accelerating observer as if they were moving inertially.

Answer 3

„In classical physics and special relativity, an inertial frame of reference (also called inertial space, or Galilean reference frame) is a stationary or uniformly moving frame of reference.“ Wiki.

The problem with this definition is the use of the words uniform and moving. The former implicitly assumes that time passes at the same rate in all frames, and the latter describes a relation "to move with respect to", which is not an equivalence relation. The remedy for this problem is to replace the word move with the word rest. The relation "to rest with respect to" is an equivalence relation that divides the open set of all reference frames into equivalence classes called frames of rest. Remarkably, the constancy of time plays no role here, and the relation "to move with respect to" is just the negation of the rest relation. Obviously, the open set of all frames of rest can only be associated with tardyons. Its closure, however, includes the luxons and tachyons, i.e. particles that do not rest in any frame of reference.

So what is an inertial frame in terms of acceleration?

The inertial frame is the frame of rest with zero acceleration, i.e. a frame that remains in its equivalence class.

Answer 4

Ok, let me summarize what I learned so far:

Theorem. The following statements are equivalent:

1. You are standing on (locally) inertial frame.

2. Your accelerometer reading is zero.

3. You feel zero proper acceleration.

4. There are no fictitious or pseudo forces (including gravitation) in your frame.

5. Christoffel symbols vanish (locally).

6. All laws of physics are satisfied.

Regarding (5), in general relativity, 4-acceleration $A=(A^0, A^1, A^2, A^3)$, in general (pun intended), can be written as

$$A^\alpha=\frac{dU^\alpha}{d\tau}+\Gamma_{\mu\nu}^\alpha U^\mu U^\nu$$

where $\Gamma_{\mu\nu}^\alpha U^\mu U^\nu$ represents acceleration contribution from spacetime geometry (this is source of all the fictitious forces like coriolis force for example).

Suppose you are at rest in some inertial frame and are grasping an apple in your hand. When you release the apple, it either will remain stationary or move in a straight line (yes, the apple feels zero proper acceleration and the frame of the apple is inertial). Why? The equation of motion for the apple of mass $m$ can be written as

$$f^\alpha=mA^\alpha=m\frac{dU^\alpha}{d\tau}+m\Gamma_{\mu\nu}^\alpha U^\mu U^\nu=0$$

where $U$, $A$, and $f$ are apple's 4-velocity, 4-acceleration and 4-force respectively. Since we are in an inertial frame, Christoffel symbols vanish, i.e. $\Gamma_{\mu\nu}^\alpha=0$ for all $\mu=0,1,2,3$ and $\nu=0,1,2,3$. So the fictitious force contribution is zero: $m\Gamma_{\mu\nu}^\alpha U^\mu U^\nu=0$. The equation of motion is now $m\frac{dU^\alpha}{d\tau}=-m\Gamma_{\mu\nu}^\alpha U^\mu U^\nu=0$. So the apple will remain in rest or move in a straight line.

If we are standing on the surface of Earth (assume the Earth is an infinite plane with uniform gravitational field $g$), which is non-inertial frame, then the apple's equation of motion (in z-direction) will be $m\frac{dU^3}{d\tau}=-m\Gamma_{\mu\nu}^3 U^\mu U^\nu\approx-mg$ (for other components, $m\frac{dU^0}{d\tau}=m\frac{dU^1}{d\tau}=m\frac{dU^2}{d\tau}=0$) instead. The apple falls downwards with acceleration of $g$, in agreement with Newtonian mechanics.

Let me know if I made any misleading statements!