I have a hard time understanding the subject of fictitious forces. Let's discuss a few examples:

1) I'm sitting inside a vehicle which is accelerating in a straight line. I feel like someone is pushing me to the seat. So, on the one hand, I'm told that this happens according to the third Newton's law: this pressure is the result of me pushing the seat as a reaction to the seat pushing me (because it is accelerating with the same acceleration as the car). On the other hand, these forces are acting on different objects and I'm told that there is another fictitious force acting on me in an opposite direction to the acceleration. So what is right? And if there is a fictitious force, then why some call it a "math trick" when they are real and I can feel them?

2) I do not understand why some call centrifugal force a fictitious force. The earth is pulling the earth with its "invisible" string called gravitation. That's why the moon is still there. And this is the centripetal force. However the moon is also pulling the earth according to the Newton's third law, and that's why we have tides. This is the centrifugal force. So why it is fictitious? What it has to do with frame of reference? When we observe this in non-inertial frame of reference (such as the moon), does it simply mean that we can't call it anymore a reaction force according to the Newton's third law? But why if it is virtually the reaction force?

3) Accelerating elevator - similar to the first example - let's say the elevator is accelerating upwards. So we get that $\vec{N}=m(g+a)$, and that's the same as me pushing the floor. That is why I feel heavier. Then why some add to here the fictitious force?

I will appreciate any answer.


2 Answers 2


1) You surely feel the pressure when you accelerate. Whether you attribute it to fictitious forces or other forces depends on your choice of the "reference frame" (vantage point). From the viewpoint of your body's reference frame, which is not an inertial frame, there exist fictitious forces (inertia and/or centrifugal and/or Coriolis' force) that are pushing your body towards the seat. In an inertial reference frame, such as the vantage point of people who stand on the sidewalk and watch you, the pressure is exerted by the seat because it's accelerating i.e. pushing (you are pushing on the seat as well, by the third Newton's law) and there are no additional fictitious forces. Both of these descriptions are OK but the description from the inertial systems (e.g. the sidewalk system) is described by simpler, more universal equations. Without a loss of generality, we may describe all of physics from these frames and these frames never force us (and never allow us) to add any fictitious forces. The frame of your (accelerating) body may be considered "unnatural" and therefore all the forces that appear in that frame are artifacts of the frame's being unnatural, and therefore they are called "fictitious". They may be avoided.

2) Centrifugal forces are the textbook examples of fictitious forces; they have to be added if you describe the reality from the viewpoint of rotating systems. They are avoided if you use non-rotating frames. However, the tides have nothing to do with centrifugal forces. The tidal forces appear because the the side of the Earth that is further from the Moon is less strongly attracted to the Moon than the side that is closer to the Moon. In other words, the tidal forces totally depend on the non-uniformity of the gravitational field around the Moon – the force decreases with the distance. You could create the same attractive force as the Moon exerts by using a heavier body that is further than the Moon. The attractive i.e. "centripetal" force would be the same but the tidal forces would be weaker!

3) In an inertial system – connected with the Earth's surface, for example – the force acting on you is $mg$ downwards from the Earth's attraction plus $ma$ from the extra upwards accelerating elevator. The part $ma$ has a clear new source, object that causes it, namely the elevator. However, in a freely falling frame, for example, the gravitational downward $mg$ force cancels against the fictitious inertial force $mg$ upwards. However, the material of the elevator is now accelerating by the acceleration $g+a$ upwards so the total force is $m(g+a)$ again.

As you can see, whether there are fictitious forces depends on the reference frame. What I feel is your trouble is that you're not used to describe processes from the viewpoint of inertial reference frames. Take a spinning carousel. There is a centripetal force acting on the children and this force, $F=mr\omega^2$, is the reason why the children aren't moving along straight paths with the uniform velocity (as Newton's first law would suggest). Instead, they're deviating from the uniform straight motion and move along circles. The centripetal, inwards directed force $mr\omega^2$ from the pressure from the seats is the reason. (For planets, the centripetal force is the gravitational one.) There are no fictitious forces, in particular no centrifugal force, in the description using the inertial system (from the viewpoint of the sidewalk). However, from your rotating viewpoint, there is a centrifugal force $mr\omega^2$ acting outwards that's always there because the frame is rotating. This force is cancelled against the pressure from the seat, a centripetal force $mr\omega^2$, and the result is zero which implies that in the rotating frame, the coordinates stay constant in this case, especially the distance $r$ from the axis of the spinning carousel. Both frames are possible: one of them forces you to add fictitious forces, the other one (inertial frame) doesn't contain any such forces.

  • $\begingroup$ By the way, you stated that "There are no fictitious forces, in particular no centrifugal force, in the description using the inertial system (from the viewpoint of the sidewalk).". However, there must be some force which will act against the centripetal force (otherwise the object will just move towards the center) - reactive centrifugal force which is described by third Newton's law - it is the reaction to the string or whatever wants to pull you to the center. Am I wrong? Thanks in advance. $\endgroup$
    – apdmn6072
    Feb 27, 2013 at 20:32
  • $\begingroup$ Hi, no, as I tried to emphasize, the centripetal force is exactly what's needed for the object to move along the circle in the inertial frame's language (not towards the center!). If no centripetal force were acting, the object would move along a straight line, away from the circle (outside to infinity), as declared by the first Newton's law (one that you seem unfamiliar with but it's the most important one)! $\endgroup$ Feb 28, 2013 at 6:21
  • $\begingroup$ So you basically say that there is no reaction force acting on the string? How it could possibly be? Then what is "reactive centrifugal force" and why it is described in Landau's books (as well as in other decent books)? And it is described from inertial frame of reference. And I am familiar with Newton's laws. You right when you said that my trouble was with reference frames. The whole issue of fictitious forces is just a way to describe things in non-inertial frames. $\endgroup$
    – apdmn6072
    Feb 28, 2013 at 11:05
  • $\begingroup$ Dear apdmn, there may also be something wrong about your understanding of the third law and the meaning of the reaction force. On a carousel, the seat acts on the baby with a centripetal force that keeps the baby on the circular path. Third Newton's law implies that there is also the opposite force by which the baby acts on the seat or the whole carousel. But this is a force acting on the seat or carousel, not the baby, so it doesn't contribute to the motion of the baby! If it did, all forces would just cancel which they surely don't. Reaction means that A acts on B and B acts on A. $\endgroup$ Mar 1, 2013 at 12:41
  • $\begingroup$ I didn't say that this force acts on the object. I said that there is a reaction force acting on the string/seat. As I said - "it is the reaction to the string or whatever wants to pull you". As I understood from books and from the words of a physicist, this force is called "reactive centrifugal force", so I am right. Of course, this has nothing to do with the centrifugal force in the rotating frame of reference - that's what could lead to yours misunderstanding of my question. Thanks anyway, I got the subject quite well. $\endgroup$
    – apdmn6072
    Mar 1, 2013 at 14:02

Luboš Motl already answered your subquestions very well, but I thought I'd work out an example in detail for your understanding.

Some people like to work with fictitious forces, most people don't. The thing is that Newton's laws don't work in reference frames that have a non-zero (finite) acceleration. In those reference frames we need to resort to mysterious fictitious forces to make things add up. In inertial frames (frames with zero acceleration), Newton's laws work fine and we need no hocus-pocus.

Now for my example, let's look at the situation of a satellite of mass $m$ orbiting the earth. In the reference frame $O'$ of the satellite, there's only one force exerted on it: the gravitational pull of the earth. In its own reference frame, the satellite is also stationary, so its acceleration with respect to $O'$ is zero. Now, we know this can't be the whole story, otherwise Newton's second law $m\mathbf{a} = \mathbf{F}$ would yield

$$\mathbf{a} = G\frac{M}{r^2}\mathbf{\hat{r}'}$$

where $G$ is the gravitational constant, $M$ is the mass of the earth and $\mathbf{r'}$ is the vector from the satellite to the earth's center. This would mean the acceleration is non-zero, which we know is not the case. So we have to add a fictitious centrifugal force to get $a$ to be zero:

$$m\mathbf{a} = G\frac{mM}{r'^2}\mathbf{\hat{r}'} + \mathbf{F}_{cf}$$

where the centrifugal force $\mathbf{F}_{cf}$ is equal to $-m\frac{v^2}{r'}\mathbf{\hat{r}'}$ where $v$ is the orbital velocity and to get $\mathbf{a}=\mathbf{0}$ we demand that this force cancels out with gravity (a condition from which we can derive the orbital speed!).

A different way of looking at this is by saying: well, we know the reference frame $O'$ isn't inertial, because the second law of Newton gives us nonsense (this doesn't necessarily mean you're not in an inertial frame but let's assume it does). So let's take the reference frame fixed to the center of the earth $O$, which is a fair approximation of an inertial frame for these purposes. From this reference frame we can clearly observe that the satellite's reference frame $O'$ moves with some finite acceleration $\mathbf{a}_{O'}$. Moreover, since it's moving (approximately) on a circle, we know an expression for this acceleration: $$\mathbf{a}_{O'} = -\frac{v^2}{r}\mathbf{\hat{r}}$$ where $v$ is the orbital velocity and $\mathbf{\hat{r}}$ is the vector from the center of the earth to the satellite. So let's insert this acceleration into the equation:

$$m\mathbf{a}_{O'} = -G\frac{mM}{r^2}\mathbf{\hat{r}}$$

(notice the sign change because $\mathbf{\hat{r}} = -\mathbf{\hat{r}'}$) We can rewrite this as follows:

$$\begin{align} \mathbf{0} &= -G\frac{mM}{r^2}\mathbf{\hat{r}} - m\mathbf{a}_{O'} \\ \mathbf{0} &= -G\frac{mM}{r^2}\mathbf{\hat{r}} + m\frac{v^2}{r}\mathbf{\hat{r}} \end{align}$$

This is the same equation as before. So we end up with the same physics, but the handling of the problem is much more natural in inertial frames of reference. We needed some artificial construct to get things right in the non-inertial frame, while we had no issues at all in the inertial frame.

Another way of looking at fictitious forces is with fictitious accelerations: the maths are all the same, but the mindset is different. Instead of adding a fictitious force, which you know isn't real, you add a "fictitious" acceleration, which is more true to the actual physics since it's the acceleration of the reference frame if you were to look at it from an inertial frame. If you add the fictitious acceleration $\mathbf{\hat{r}'}v^2/r'$ in the left hand side of the equation $m\mathbf{a} = \mathbf{F}$ for the reference frame $O'$ attached to the satellite (where $\mathbf{a}$ was zero), you get:

$$m\left(\mathbf{0}+\frac{v^2}{r'}\mathbf{\hat{r}'}\right) = G\frac{mM}{r^2}\mathbf{\hat{r}'}$$

This is again the same equation.

  • $\begingroup$ I appreciate your answer, sir. Thank you. It is indeed a great addition to the another answer. $\endgroup$
    – apdmn6072
    Feb 27, 2013 at 17:42

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