How fictitious are fictitious forces?

More specifically, in a rotating reference frame i.e. on the surface of the earth does an object that is 'stationary' and in contract with the ground feel centrifugal and Coriolis forces? Or are these forces purely fictional and used to account for differences in observed behaviour relative to an inertial frame?

To give a practical example a turreted armoured vehicle is sitting stationary and horizontally somewhere in the UK. The turret is continually rotating in an anti-clockwise direction. Do the motors that drive the turret's rotation require more power as the turret rotates from east to west and less power as the turret rotates from west to east? i.e. are the turret motors cyclically assisted and hindered by the earths rotation?

  • $\begingroup$ Here's one that always confuses me: Alice is falling freely under gravity. For Bob, on observer on earth, Alice experiences a force mg, and thus accelerates towards earth with an acceleration g. In Alice's frame, she experiences a force mg downwards, but because we are in a non-inertial frame, there is a 'pseudo' force mg upwards, so two forces cancel out and in her frame she is not accelerating. Everything was fine until here. But according to Einstein's principle of equivalence, an inertial frame is equivalent to a frame falling freely under gravity. So, does this make the 'pseudo' force tha $\endgroup$
    – Spot
    Apr 24, 2012 at 16:16
  • $\begingroup$ It's not necessary to make this pseudo force construction here - the last line explains why the third line is no problem. In the beginning claim "she experiences a force mg downwards" you'd have to explain what "experiences" means because if you consider her to be a point particle, then as you said, she doesn't effectively feel any acceleration. The principle says exactly that, namely that if you're in free call, you locally don't know that there is a gravitational field around. Notice there is no global inertial frame in that example. Also, don't post questions as an answer (this is no forum) $\endgroup$
    – Nikolaj-K
    Apr 24, 2012 at 16:28
  • $\begingroup$ You could probably post this as a separate question, since it's not really an answer to the question posted here. $\endgroup$
    – tmac
    Apr 24, 2012 at 16:28

4 Answers 4


No, they are not real forces.

Quoting from my answer here

Whenever we view a system from an accelerated frame, there is a "psuedoforce" or "false force" which appears to act on the bodies. Note that this force is not actually a force, more of something which appears to be acting. A mathematical trick, if you will.

Let's take a simple case. You are accelerating with $\vec{a}$ in space, and you see a little ball floating around. This is in a perfect vacuum, with no electric/magnetic/gravitational/etc fields. So, the ball does not accelerate.

But, from your point of view, the ball accelerates with an acceleration $-\vec{a}$, backwards relative to you. Now you know that the space is free of any fields, yet you see the particle accelerating. You can either deduce from this that you are accelerating, or you can decide that there is some unknown force, $-m\vec{a}$, acting on the ball. This force is the psuedoforce. It mathematically enables us to look at the world from the point of view of an accelerated frame, and derive equations of motion with all values relative to that frame. Many times, solving things from the ground frame get icky, so we use this. But let me stress once again, it is not a real force.

And here:

The centrifugal force is basically the psuedoforce acting in a rotating frame. Basically, a frame undergoing UCM has an acceleration $\frac{mv^2}{r}$ towards the center. Thus, an observer in that rotating frame will feel a psuedoforce $\frac{mv^2}{r}$ outwards. This psuedoforce is known as the centrifugal force.

Unlike the centripetal force, the centrifugal force is not real. Imagine a ball being whirled around. It has a CPF $=\frac{mv^2}{r}$, and this force is the tension in the string. But, if you shift to the balls frame (become tiny and stand on it), it will appear to you that the ball is stationary (as you are standing on it. The rest of the world will appear to rotate). But, you will notice something a bit off: The ball still has a tension force acting on it, so how is is steady? This balancing of forces you attribute to a mysterious "centrifugal force". If you have mass, you feel the CFF, too (from the ground, it is obvious that what you feel as the CFF is due to your inertia)

What really happens when you "feel" psuedoforces is the following. I'll take the example of spinning on a playground wheel.

From the ground frame, your body has inertia and would not like to accelerate(circular motion is acceleration as the direction of velocity changes).

But, you are holding on to the spinning thingy so you're forced to accelerate. Thus, there is a net inward force--centripetal force--a true force since it's from "holding on". In that frame, though, you don't move forward. So your body feels as if there is a balancing backwards force. And you feel that force acting upon you. It really is your body's "inertia" that's acting.

Yes, the turret's wheels are affected. Again, this is due to inertia from the correct perspective, psuedofoces are just a way to easily explain inertia.

Remember, Newton's definition of a force is only valid in an inertial frame in the first place. Psuedoforces make Newton's laws valid in non inertial frames.

  • $\begingroup$ I believe I understand the use of psuedo forces now. They are required to account for the effects of accelerations on the frame we are observing from in order to allow Newton's laws to be used effectively. Does the magnitude of the acceleration affect their use though? On Earth we are not aware of the fact that we are in a non-inertial frame as the accelerations we are experiencing are so small. What if the earth was spinning much faster and we could physically feel this centrifugal force? What if the earth is spinning so fast that friction can no longer maintain our 'stationary' position? $\endgroup$ Apr 26, 2012 at 10:36
  • $\begingroup$ @Ben yep. Psuedoforces are equal to the mass of the body in question times the acceleration of the frame, in the opposite direction. And yes, Earth would be a strange place. $\endgroup$ Apr 26, 2012 at 11:33
  • $\begingroup$ OK then let's get practical, back to the turreted vehicle on earth. The designer of the turret's traverse motors has a requirement to rotate the turret mass at a certain rate under all conditions. This requirement is stringent enough that the turret designer has to account for the effect of the coriolis force during design. If this is the case isn't that enough for us earth-bound folk to consider the force real in earth's frame? $\endgroup$ Apr 26, 2012 at 11:55
  • $\begingroup$ @Ben its still not a force, so not a real force. But, it has the same effects of a force, so you consider it, and treat it like a force. It's more of a technicality that its's fictitious. $\endgroup$ Apr 26, 2012 at 11:59
  • $\begingroup$ I think the statement "Psuedoforces are equal to the mass of the body in question times the acceleration of the frame, in the opposite direction" is the most enlightening comment so far. So on Earth, the centrifugal 'force' always acts directly 'upwards' and is equal to mv^2/r. The coriolis force however, Earth's tangential vecolicity is constant and therefore there is no tangential pseudoforce. Therefore where does Corioils originate? I assume it's something to do with the fact that the earth's radius about its axis of spin is not constant? $\endgroup$ Apr 27, 2012 at 13:39

Centrifugal and Coriolis forces are indeed so called pseudo forces that account for differences in observed behaviour relative to an inertial frame.

So if you see an object standing on the surface of the Earth, you can be sure that static friction is holding it at rest relative to Earth's surface.

Great example of the effect of pseudo forces is so called Foucalt's pendulum. Since there is no static friction for pendulum, pendulum's plane of oscillation rotates. Foucalt's pendulum is also a proof that Earth is not an inertial frame of reference.

The problem of observing pseudo forces is in the fact they are very small compared to gravity. Centripetal acceleration due to rotation of the Earth around its axis is of the order $10^{-2}$ m/s$^2$ (depending on the position), while centripetal acceleration due to rotation of the Earth around Sun is $6 \times 10^{-3}$ m/s$^2$. So you have an effect when rotating a turret, but I doubt you would be able to measure it.

So what makes forces pseudo? Well, you might have heard that Newton's laws are valid only in inertial frame of reference. If you watch the movement of the turret from outside the Earth (inertial frame of reference), you can observe that turret is making complex movements and constantly accelerating. Gravitational and frictional forces acting on turret are responsible for these movements.

However if you are standing on the Earth it seems to you that turret is at rest. But gravitational and frictional forces are still acting on it, so this does not add up. The sum of forces different than zero, and turret at rest, breaks 2nd Newton's law! 2nd Newton's law is no longer valid because you are no longer in inertial frame of reference.

In order to "patch" 2nd Newton law in non-inertial frames of references, you introduce pseudo forces. After introduction of pseudo forces, 2nd Newton law is valid even if you are no longer in inertial frame of reference. You can feel those forces only because your intuition requires additional forces in order to explain your observations.

  • $\begingroup$ Then these forces are in fact very real? We are all constantly experiencing them yet they are so small they are virtually impossible for us to detect without precise measuring equipment? Is 'ficticious force' therefore a misleading term or does it have some other implication? $\endgroup$ Apr 24, 2012 at 10:52
  • $\begingroup$ I will add some text into my answer to attend your additional question. $\endgroup$
    – Pygmalion
    Apr 24, 2012 at 10:54
  • $\begingroup$ +1 for explaining the friction/etc aspect of it clearer than I did :) $\endgroup$ Apr 24, 2012 at 11:28
  • $\begingroup$ @NickKidman: Could you clarify that? (for one, you haven't logically defined $f$). And $\vec F\neq\frac{\mathrm d\vec p}{\mathrm dt}$ in a non-inertial frame, so Newton's laws are obviously invalid there. $\endgroup$ Apr 24, 2012 at 11:33
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    $\begingroup$ (edited) I just want to point out that "Newton's laws are valid only in inertial frame of reference" is common abuse of language (it always bugs me when I read it, sorry). The Second law says "In an inertial frame: F=ma", an axiom whose validaty doesn't depend on a reference frame f you're working with. To put it in logical terms, if $f$ means "We are now working in an interial frame" and the law is $(f→"F=ma")$ then $((f→"F=ma")∧(¬f)→¬"F=ma")$ is not false but you're saying $(¬f→¬(f→"F=ma"))$ which is not sound (it could only be true if never $f$). Its because $"F=ma"$ is not the law itself. $\endgroup$
    – Nikolaj-K
    Apr 24, 2012 at 11:50

In classical mechanics it makes sense to distinguish between fictitious forces caused by accelerating coordinate systems and "real" forces in inertial frames, but this is no longer the case in general relativity.

In general relativity except in simple cases there are generally no preferred global reference frames, and gravity in some sense becomes indistinguishable from the Newtonian concept of a pseudoforce.

You can take your pick as to whether this means that gravity is less real or pseudoforces are more real, but it's not a physics question to worry about the answer.


Place a stationary object on a piece of graph paper and accelerate the graph paper anyway you want over time, while recording the position of the object on the graph paper and keeping the object stationary relative to you:

Q: Did you see the object accelerate while you were moving the graph paper?

A: Nope, so there isn't a physical force on it.

Q: What is the trajectory of the object on the graph paper and your conclusion?

A: The trajectory is a curve and so it was accelerating in the coordinate system of the graph paper. We can model this as an unphysical force acting upon the object in this coordinate system. This fictitious force will depend upon how this coordinate system is accelerating wrt one moving at a constant velocity.

  • $\begingroup$ Why is the trajectory a curve? I may have only accelerated the graph paper in one direction for a brief moment. $\endgroup$ Apr 26, 2012 at 10:29
  • $\begingroup$ @ben well a curve is a generalization and a line is a special case of a curve. I'm sure you get the general idea ;) $\endgroup$ Apr 26, 2012 at 11:04
  • $\begingroup$ This example doesn't seem analogous to the example in my question. In my example, static friction is keeping the vehicle stationary in it's frame on the Earth, while in yours you are suggesting that that static friction is overcome and the object slips? Could you rephrase the example please? $\endgroup$ Apr 27, 2012 at 12:42

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