According to fictitious force wiki,

"Fictitious forces arise in classical mechanics and special relativity in all non-inertial frames. Inertial frames are privileged over non-inertial frames because they do not have physics whose causes are outside of the system, while non-inertial frames do. Fictitious forces, or physics whose cause is outside of the system, are no longer necessary in general relativity since these physics are explained with the geodesics of spacetime."

However, @VincentThacker disagreed with this in a comment to an answer. Upon being asked by me if fictitious forces are necessary in general relativity, his statement of two comments (collated) was as follows:

"Yes, because proper acceleration cannot be set to zero by a coordinate transformation. In GR, gravity is a result of curved spacetime and geodesics, so it is not a source of proper acceleration. However, if an observer has non-zero proper acceleration, objects nearby will appear to have a fictitious (coordinate) acceleration. The easiest example is the surface of the Earth. The surface of the Earth is accelerating radially outwards with proper acceleration g, so we see freely-falling objects "accelerate" at −g. See my answer here."

So, are fictitious forces necessary for correct calculations in general relativity or not?

Disclaimer: I don't understand general relativity at all. I only understand special relativity.


2 Answers 2


The generalization of Newton's 2nd law to general relativity is given by

$$ m \frac{d^2 x^\mu}{d\tau^2} + m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = f^\mu \qquad (\star)$$ where $\tau$ is the proper time along the particle's worldline, $f$ is the net 4-force acting on the particle, and $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols corresponding to your choice of coordinates.

In inertial Cartesian coordinates, all of the $\Gamma$s are equal to zero which means that $$m \frac{d^2x^\mu}{d\tau^2} = f^\mu$$ If you pick different coordinates in which the $\Gamma$s are not zero, then obviously you're going to need to include that extra term. This happens if you use polar coordinates, for example, but it also occurs when you use accelerated Cartesian coordinates. In the latter case, what you can do is simply move the extra terms to the other side of the equation and call them pseudoforces: $$m \frac{d^2 x^\mu}{d\tau^2} = f^\mu - \underbrace{m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau}}_{\text{pseudoforce}}$$

Example: The coordinates which correspond to a relativistic observer undergoing constant proper acceleration are the Rindler coordinates. In this coordinate system, assuming proper acceleration $a_0$ along the $x$-axis, the line element becomes $$ds^2 = -\left(1+ \frac{a_0 x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$$ The nonzero Christoffel symbols are $$\Gamma^0_{10} = \Gamma^0_{01} = \frac{a_0/c^2}{1+a_0 x/c^2}\qquad \Gamma^1_{00} = \left(1+\frac{a_0 x}{c^2}\right)a_0$$ which means that the relativistic generalization of Newton's 2nd law for $x$ is

$$ m \left(\frac{d^2x}{d\tau^2} + a_0 \frac{1+\frac{a_0 x}{c^2}}{1+ \frac{a_0 x}{c^2} - \frac{v^2}{c^2}}\right) = f_x$$

In the nonrelativistic limit, this becomes $$m \left(\frac{d^2 x}{dt^2} + a_0\right) = f_x \iff m \frac{dx^2}{dt^2} = f_x \underbrace{ -\ m a_0}_{\text{pseudoforce}}$$

which is exactly what we'd need to do in Newtonian mechanics if we wanted to switch to an non-inertial frame accelerating with $\mathbf a = a_0 \hat x$.

So in summary, you have the following options:

  1. Choose a coordinate system in which all the $\Gamma$s vanish - that is, a global inertial frame. Such frames generally do not exist in curved spacetime, so you can only do this in the context of SR.
  2. Recognize and accept that the left-hand side of $(\star)$ has additional terms due to the $\Gamma$s, which reflect the fact that your coordinate basis changes with position. If you've ever constructed Newton's laws in polar coordinates, this is what you do.
  3. Move the term $m\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau}$ to the right-hand side of the equation and call it a pseudoforce. If you've ever constructed Newton's laws in a non-inertial frame, this is what you probably did.

In general curved spacetime, you can't escape the $\Gamma$s so your options are limited to 2 and 3. When the Wiki article says that you don't need pseudoforces, it means that you can choose option 2, not that you can ignore them outright.

Finally, though in GR you cannot make the $\Gamma$s vanish everywhere, you can always make an instantaneous choice of coordinates to make them vanish at a point. As a result, at any given instant of time you can choose coordinates such that $f^\mu_{\mathrm{pseudo}} \equiv -m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = 0$. Because $f^\mu_{\mathrm{pseudo}}$ can be set to zero by a coordinate transformation, we know that it is not a 4-vector because if a 4-vector vanishes in one coordinate system, it must vanish in all coordinate systems. This is the mathematical distinction between real forces, which are 4-vectors and therefore coordinate-independent geometrical objects, and pseudoforces, which can be viewed as artifacts of your choice of coordinates.


You can define a co-ordinate transformation which makes proper acceleration locally zero, and so creates a local inertial reference frame. In SR, where spacetime is flat, this can be extended to a global inertial reference frame - a reference frame that is still inertial when you move away from the origin. But in GR you cannot in general extend a reference frame that is locally inertial to one that is globally inertial everywhere (although this may be possible with some specific spacetime metrics).

  • $\begingroup$ So, @VincentThacker is right and the Wikipedia article is wrong? $\endgroup$ Jul 8, 2021 at 15:43
  • 1
    $\begingroup$ @AbuSafwan Maybe Wikipedia is talking locally, whereas Vincent is talking globally. $\endgroup$
    – gandalf61
    Jul 8, 2021 at 15:57
  • $\begingroup$ @AbuSafwan (1) In flat (Minkowski) spacetime, there is nothing stopping you from choosing coordinates with fictitious forces. The catch is that you can always choose a frame where inertial laws hold everywhere and at all times. (2) However, in curved spacetime, you can only choose an inertial frame at one point in spacetime. This means not only one point in space, but also one point in time. The key point is that in curved spacetime, non-proper acceleration arise because of two reasons: spacetime curvature and choice of coordinates, which is explained in J. Murray's answer. $\endgroup$ Jul 9, 2021 at 0:45
  • $\begingroup$ @AbuSafwan According to my understanding, the first reason, spacetime curvature, is called gravity, while the second reason, choice of coordinates, is called "fictitious forces". However, both effects are encoded in the Christoffel symbols so it is hard to distinguish them from the symbols alone. $\endgroup$ Jul 9, 2021 at 0:51

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