Why is $\mathbf F = m \, \mathbf a$ only valid in inertial reference frames?

Earth is not an inertial reference frame, even if we ignored its rotation.

On its surface everything is subjected to a uniform gravity. According to Einstein’s equivalence principle, being still in a uniform gravitational field is the same as being accelerated in the opposite direction, by means of a constant force.

So we are in an accelerated reference frame, and nevertheless Newton’s laws hold, provided we included the force of gravity, which according to us plays the same role as an apparent force would play in any other non-inertial reference frame.

So something is missing here. Either Newton’s laws hold in (uniformly) accelerated reference frames, or something else.

“Gravity is not a force” sounds a lot like a cheat.

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    $\begingroup$ Note that the definition of an inertial frame of reference in classical mechanics is not the same definition as an inertial frame of reference in general relativity. $\endgroup$ Nov 12, 2021 at 0:30

2 Answers 2


Newton’s laws hold, provided we included the force of gravity, which according to us plays the same role as an apparent force would play in any other non-inertial reference frame

That is precisely the matter. To add a fictitious force is to modify Newton's second law according to $$\mathbf{F} + \mathbf{F}_{\text{fic}} = m \mathbf{a}.$$ Hence, you can view Newtonian gravity as either a force or a fictitious force, and the equations will be the same nonetheless.

The problem is more difficult once we move to Einstein's gravity. This time, we also need to take into consideration that massless particles, for example, experiment gravity. Such particles can't be described within the framework of Classical Mechanics and forces, and a simple argument in this sense is that Newton's Law would become $\mathbf{F} = \mathbf{0}$. We see then that something is failing in the description.

Within Special Relativity, I'm quite certain it would be possible to formulate gravity as some sort of force between particles. Newton's Second Law would be written $$\frac{\textrm{d}^2x^{\mu}}{\textrm{d}\tau^2} + \sum_{\nu, \sigma} \Gamma^\mu{}_{\nu\sigma}\frac{\textrm{d}x^{\nu}}{\textrm{d}\tau}\frac{\textrm{d}x^{\sigma}}{\textrm{d}\tau} = 0,$$ which is known as the geodesic equation and describes the motions of particles in curved spacetime. However, this is not a convenient description. In particular because Special Relativity enforces that nothing can go faster than light and, as a consequence, one gets a deep relation between how light moves and causality. In other words, the way light moves determines which events in spacetime can influence which events. To understand these features of the theory in terms of forces would be quite complicated, but it becomes quite natural if we see gravity as a geometric effect rather than a force. Furthermore, I believe thinking in terms of forces would also make it very difficult to understand how the electromagnetic field interacts with gravity in contexts more general than just a light particle moving around.

It is important to notice that there are usually many ways one can describe a theory mathematically, but it does not mean all of them are simple, or convenient. For example, one could argue that the other planets orbit around the Earth, provided we include the effects of epicycles, and there would be nothing wrong with that description. In fact, this description was more accurate than Copernicus' description of circular orbits around the Sun. Nevertheless, understanding the movement of planets in terms of an inverse square law force eventually proved to be simpler than to consider the effects of many epicycles.

In Physics, we usually aim for the simpler possible description, in accordance with Occam's razor. For Newtonian Physics, it is easier to work with forces, and that is why every Physics 101 class will discuss gravity in these terms. Nevertheless, once problems get more complex, differential geometry becomes more convenient than descriptions in terms of forces.


Rotation is easy to ignore. The acceleration is very small.

$$a = \frac{\omega^2}{r}$$

$$\omega = 1 \frac{rotation}{day} = \frac{2\pi \space radians}{86400 \space sec} = 7.27 \cdot 10^{-5} \frac{radians}{sec}$$

$$r = 6.38 \cdot 10^6 \space meters$$

So $$a = 8.28 \cdot 10^{-16} m^2/s$$

When speaking of the surface of the Earth, often people are working in $2$-D in a horizontal plane.

An object sitting on ice just sits there motionless. It is not accelerated. Or rather the force of gravity is cancelled by the reaction force from the ice that prevents the object from penetrating the ice' surface. If you slide it, it moves at a constant velocity, if you ignore friction.

As you can see, this is a bit idealized. We are ignoring big forces that cancel and little forces that don't have a big effect.

Another case where we truly do have an inertial frame is free fall. Again idealizing. We can ignore wind resistance for a few seconds before we move too fast.

This is true precisely because gravity is not a force. For more about this, see Why can't I do this to get infinite energy?.

In a small region around a freely falling object, the laws of physics are precisely the same as for an object floating in space far from gravity.

In particular, if you let got of a rock, it stays motionless or moves in a straight line relative to you.


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