I'm currently teaching a general-physics-for-engineers class, and we approached fictitious forces. As I was explaining them, students asked me how to discriminate "real" forces from fictitious ones, and I was giving them an explanation using reference frames (real forces are visible in every frame while fictitious ones are not), but they weren't convinced. A student then said:

Can we say that real forces cause accelerations, while fictitious forces are caused by acceleration?

I didn't know what to answer, because I never thought about it this way: to me, the "$=$" sign in "$F=ma$" does not have a cause-effect meaning, it's just saying that two quantities ($F$ and $a$) are proportional to each other, and the coefficient is $m$.

Does his reasoning make sense? Is there a rigorous way to include a cause-effect relationship in dynamics (and in fictitious forces in particular)?


10 Answers 10


I have found that assigning causality to forces and accelerations is tricky. It's intuitive to think that forces cause accelerations, and 99% of the time it won't get you into trouble. But its corner cases like these where its important to realize that $\Sigma F=ma$ is simply a statement of proportionality, not causality. Forces and accelerations go together.

The only reason we have to call these effects "fictitious forces" is because we are trained to think of $F=ma$ applying in all circumstances. It doesn't always apply. Specifically, it applies when the reference frame is "inertial." Or perhaps less circularly, we identify that in some frames, the relationship between forces and accelerations was simple, and those are the "inertial frames."1

In other frames (rotating frames), the relationship is less simple. Its no more right or wrong, just different. The accelerations involved include a few extra terms (such as coriolis effects). Calculating in those rotating frames proved to be tricky, and it was found that it was easier to move those effects from the acceleration side to the forces side. After doing that, one could compute the motion as if the frames were inertial (i.e. using $F=ma$).

The only reason they are "fictitious" is because we are using them to pretend that the motion is in an inertial frame, which we do because motion in an inertial frame is so well understood. We like the idea that if the sum of all forces is 0, an object is not accelerating (which is a false statement in a rotating frame unless you treat those extra effects as forces).

And, as anyone who has been on a Gravitron ride (or similar) knows, a fictitious force (centrifugal force), paired with its equal and opposite reactionary force (normal force from the wall), can crush you just as easily as any real force. It's only when you try to reason about the situation as-if you were inertial that they become fictional.

(1) My original answer credited Newton for these ideas. In light of the comments, I've reworded this slightly to give less credit to Newton for things that weren't formulated until decades or centuries after his time. He did indeed observe that sometimes there are centripetal "forces" and sometimes there are not, but the history of the connection between that and frames is a fascinating history lesson and worth the read for anyone interested in how we came to our modern understanding of physics.

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    $\begingroup$ "Newton identified" — was it he? The inertial forces were added by d'Alembert, in what we now call the d'Alembert's principle. $\endgroup$
    – Ruslan
    Nov 8, 2023 at 18:46
  • $\begingroup$ @Ruslan You raise a good history question. Was Newton aware of non-inertial frames? I thought he was, but your link has me questioning it. It does look like d'Alembert was the first to discover the transformation I described in the later paragraphs, but did Newton come across the rotating frame cases (and just not realize you could treat them as forces)? I'd always assumed he did, but now I have to do some research. $\endgroup$
    – Cort Ammon
    Nov 10, 2023 at 0:33
  • $\begingroup$ To check this, we can try finding his treatment of the two-body problem, which is typically solved by switching to polar coordinates, where one gets the term $\frac{L^2}{2mr^2}$ that directly corresponds to centrifugal force. I'd guess that Newton doesn't name this force, but I haven't read his works to confirm or deny this. $\endgroup$
    – Ruslan
    Nov 10, 2023 at 6:39
  • $\begingroup$ @Ruslan I've updated the answer to no longer claim Newton identified these things. I put a question about this on the History of Science and Math stack exchange, and got a wonderful history lesson out of the deal! Thank you for drawing attention to my incorrect assumptions about how physics was constructed over the centuries! $\endgroup$
    – Cort Ammon
    Nov 10, 2023 at 15:56
  • $\begingroup$ @Cleonis Thats interesting. Other than relativistic effects, I was unaware of any kinematic phenomena that cannot be conveyed in any arbitrary frame (and even the relativistic ones can be captured with GR). Mind you sometimes they may be terribly obscured, such as turning simple orbits into complex trajectories. $\endgroup$
    – Cort Ammon
    Nov 12, 2023 at 17:29

I find engineers to be fixated on coordinates, which makes sense: they calculate and build stuff. They make CAD drawings, too. Coordinates matter.

A key part of physics is: physics does not care about your coordinate system. This takes a while to ingest, as homework problems are full of "find the vertical component of force...what is the velocity in the $y$ direction"...the mysterious (at first) cross product, and so on.

So, in a Galilean/Newtonian worldview, objects on which no external (total) force is applied move in straight lines at constant velocities--so if your coordinate system is non-inertial, you introduce fictitious forces to handle that mathematically. Gravity also pulls them down....kinda like a fictitious force, but we say it's not, because Newton named and described it.

Now in General Relativity, physics really--it can not be overemphasized--REALLY doesn't care about your coordinate system. The Earth's gravity can be considered a fictitious force because one is accelerating through spacetime (i.e., not following an internal geodesic) just by stand still on the ground.

Well the same is true for the Coriolis Force and the centrifugal force. When you're playing catch on a merry-go-round, that ball curves laterally because of gravitation ($\vec v \times \vec G_M$, for weak fields).

Your spinning coordinates are just as valid as an Earth-Centered Internal (ECI) frame, it's just that the earth pulls down, and the universe spinning around you pulls outward, and there's a weird velocity dependent "force".

That view works in weak fields. Near blackholes, it's so nonlinear it's hard to simplify to linear equations. See Vortex and Tendex lines for the ultimate fictitious forces.

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    $\begingroup$ What is $\vec{G}_M$? $\endgroup$
    – ions me
    Nov 8, 2023 at 1:33
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    $\begingroup$ Never mind, seems like it's the GravitoMagnetic Field. $\endgroup$
    – ions me
    Nov 8, 2023 at 1:43
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    $\begingroup$ @ionsme and that begs the question: why does the EM magnetic field look like a fictitious force? $\endgroup$
    – JEB
    Nov 8, 2023 at 3:29
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    $\begingroup$ @JEB After your comments on my answer (in which I was claiming that fictitious forces are not "real" because we cannot find their source as one of the 4 forces of nature) I have read the equivalence principle one more time and realized that I did not understand it correctly before. Thank you. This was a revelation. $\endgroup$ Nov 10, 2023 at 13:49

The idea of adding an extra "force" in some situations because it enables the use of Newton's laws.

Imaging a book on a frictionless horizontal table.
You are sitting at the table and observing that the book is stationary relative to the table.
You draw a free body diagram for the book showing the two forces acting on it, normal force on the book due to the table upwards and an equal magnitude gravitational force on the book due to the Earth acting downwards.

Suddenly you see the book accelerating horizontally, magnitude $A$, to the right with the same two forces acting on it.
How can that be?
Well the explanation is that you, the table and the train all have a horizontal acceleration, magnitude $A$, to the left relative to the ground.
All this is obvious to an observer standing outside the train on the ground.
You, at rest relative to a train which is accelerating relative to the ground, have observed the motion of the book.
You are in an accelerating frame of reference and Newton's laws do not work as the book is accelerating relative to the table with no force causing that acceleration.

To make Newton's laws work in the accelerating train one can add a horizontal force $MA$ to the right on the book acting at the centre of mass of the book, where $M$ is the mass of the book and then using $F=Ma \Rightarrow MA=ma$, the acceleration of the book relative to the train $a$ is $A$ exactly as you would measure relative to the train.

So whatever you call $MA$ it has been added so that Newton's laws can be used when taking measurement relative to the accelerating train.
Note also there is no Newton Third Law partner to $MA$.

You may say, well why bother as one can always go back to an inertial (non-accelerating) frame of reference.
The answer to that is that it often the case that doing the analysis relative to an accelerating (non-inertial) frame of reference is easier.

  • 1
    $\begingroup$ "...you have observed the motion of the book at rest relative to a train which is accelerating relative to the ground." Isn't it the other way around? Isn't the the book at rest relative to the ground and accelerating relative to the frictionless table? $\endgroup$
    – Bob D
    Nov 7, 2023 at 15:19
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    $\begingroup$ @BobD Thanks. I missed out a comma but think it better as follows, You, at rest relative to a train which is accelerating relative to the ground, have observed the motion of the book. $\endgroup$
    – Farcher
    Nov 7, 2023 at 22:36
  • 1
    $\begingroup$ I thought as much.đź‘Ť+1 $\endgroup$
    – Bob D
    Nov 7, 2023 at 23:16

In Newtonian mechanics, there are special reference frames where bodies, if sufficiently far from any other body in the universe, move with constant velocity (including the case of zero velocity).

These reference frames are said to be inertial.

There is actually a wide family of these reference frames: Given one of them, all the others are exactly the ones in translational motion with constant velocity with respect to it.

Performing dynamical experiments in an inertial reference frame, one sees that the motion of two bodies close to each other (but far from the other bodies) satisfy a certain pair of equations. Here the bodies are represented in terms of material points with mass $m_i>0$ and corresponding positions $\vec{x}_i$, velocity $\vec{v}_i$, and acceleration $\vec{a}_i$: $$m_1\vec{a}_1 = \vec{F}_{12}(\vec{x}_1, \vec{x}_2, \vec{v}_1, \vec{v}_2)$$ $$m_2\vec{a}_2 = \vec{F}_{21}(\vec{x}_2, \vec{x}_1, \vec{v}_2, \vec{v}_1)\:.$$ The functions $\vec{F}_{ij}$ are known and represent, in mathematical terms, the mechanical interaction between the two bodies. They are called forces exerted to each other.

There are many types of forces due to the specific nature of the interactions between the two considered points. From the Newtonian perspective, the main task of mechanical physics was to classify all these functions $\vec{F}_{13}(\vec{x}_1, \vec{x}_2, \vec{v}_1, \vec{v}_2)$ describing interactions. For many reasons, this very naive perspective foundered very quickly.

In case one deals with a number $N$ of similar material points, the forces computed pairs-by pairs superpose.

$$m_k\vec{a}_k = \sum_{j\neq k}\vec{F}_{kj}(\vec{x}_k, \vec{x}_j, \vec{v}_k, \vec{v}_j)\quad for \: k=1,2,\ldots, N \tag{1}$$

It is important to stress that, when viewing the resulting set of equations as a system of differential equations, an existence and uniqueness theorem for the motion of the points takes place as soon as you give initial conditions, i.e., the positions $\vec{x}_k(t_0)$ and the velocities $\vec{v}_k(t_0)$ at a given initial time $t_0$. (Some mild regularity assumption on the functions $\vec{F}_{kj}$ are also necessary.)

Forces as above satisfy a triple of fundamental facts.

1. They always come in pairs: if there is $\vec{F}_{12}$, there is also $\vec{F}_{21}$.

2. In each pair their sum is zero: $\vec{F}_{12}= - \vec{F}_{21}$. (Actually, it is mores trongly assumed that they are also directed along $\vec{x}_1- \vec{x}_2$.)

3. The force $\vec{F}_{kj}$ does not depend on the reference frame, but only of the nature of the interactions between the interacting bodies.

All that is valid when one describes dynamics as inertial reference frames. What happens if one deals with non-inertial frames?

A possibility is using again (1) where $\vec{a}_k = \vec{a}^{NI}_k + \vec{A}_k$. Above $\vec{a}^{NI}_k$ is the acceleration of the considered point in the non-inertial reference frame and $\vec{A}_k$ is the difference of these accelerations.

It is here assumed that the motion of the non-inertial reference frame is given with respect to the inertial one and thus $\vec{A}_k$ has a complicated functional form written below.

Eq. (1) can be re-written as $$m_k\vec{a}_k^{NI} = \sum_{j\neq k}\vec{F}_{kj}(\vec{x}^{NI}_k, \vec{x}^{NI}_j, \vec{v}^{NI}_k, \vec{v}^{NI}_j) - m_k \vec{A}_k \tag{2}$$

Notice that I used the coordinates of the points referred to the non-inertial frame. In particular, where the functions $\vec{a}_{O}(t)$ and $\vec{a}_{O}(t)$ are given, $$\vec{A}_k = \vec{a}_{O}(t) + 2\vec{\omega}(t) \wedge \vec{v}^{NI}_k + \vec{\omega}(t) \wedge(\vec{\omega}(t) \wedge \vec{x}_k^{NI}) + \dot{\vec{\omega}}(t) \wedge \vec{x}_k^{NI}$$

Using Eq. (2), it is natural to intepret $$\vec{F}_k(t,\vec{x}^{NI}_k, \vec{v}^{NI}_k) := - m_k \vec{A}_k$$ as a new further "force" acting on the point with position $\vec{x}_k^{NI}$, and which take place only in the considered non-inertial reference frame.

It is evident that these "forces" violates the conditions (1)-(3) valid for the "true" forces describing Newtonian interactions.

That is basically due to the fact that they are not generated by another material point $\vec{x}_{N+1}$, but they are here a simple mathematical expedient. (When passing to general relativity, this view completely changes.)

For this reason they are called fictitious forces.

The issue about the forces as the cause of the acceleration seems a bit of metaphysical flavour to me, and I consider it quite irrelevant and also a bit misleading in reference to the considered issue.

The "fictitious" nature of the inertial forces should be clear from the above discussion.

  • $\begingroup$ (foundered vs. floundered.) $\endgroup$ Nov 10, 2023 at 21:56
  • $\begingroup$ Many thanks for correcting my typos! Regarding foundered vs floundered my intention was to use the former, but I do not know if it can be used as in Italian in that context. $\endgroup$ Nov 10, 2023 at 22:07
  • $\begingroup$ Regarding the definition of inertial frames, it is necessary to postulate that bodies which are far away do not apply any forces, right? Bodies which are far away cannot influence each other in any conceivable way, that is an additional assumption, right? $\endgroup$ yesterday

I have never taught General Physics for engineers, but I applaud the mindset of the student that asked the question. He/she has a valid point: if there is no way to practically distinguish two phenomena they are same for all practical purposes.

Coming to the particular question, I second what @physicopath says: there is no such thing as a fictitious force, however (if I understood his/her answer correctly) this is not because it is not there, but because a "fictitious" force is as good as any other force, because it causes an acceleration (in the reference frame where it has appeared). And by the fact, this is precisely what Einstein's equivalence principle is about: there is no way to tell a "fictitious force" in an accelerated frame from gravity (locally). So, in this sense, the answer to @physicopath 's question to the doubting students would be gravity.


Yes, the student’s comment does make sense in most circumstances, given our most basic causal intuitions, but it’s impossible to see why without stepping back and figuring out where these intuitions come from in the first place. As you note, they’re certainly not evident in the “bare equations” like $F=ma$, which is just a statement of correlations. (And, as everyone knows by rote, correlation is not causation.)

So what is causation in the context of physics models? There’s been an enormous increase in our understanding of causation over the past 30 years, and it’s now widely understood in many technical fields. Unfortunately, it hasn’t spread much to physics. Still, physics students have causal intuitions, physics teachers have causal intuitions, and it’s well worth our time to examine and explain those intutions, so that we can tap into them. Our brains are excellent causal inference machines, as Judea Pearl puts it.

What your student is doing, mentally, is starting with some physical situation of some finite system, and then imagining how it might be different. Specifically, they are imagining the difference as originating from outside the system – some way they might “reach in” and “intervene” on some variable, changing it to some different value. Whatever they imagine changing, that’s the “cause”. All the other parts of the system which are correlated with that change are the “effect”. This is the interventionist account of causation in a nutshell: causes are the variables one imagines directly intervening on, from outside the system. (There are few more technical details concerning “arrow breaking”, preventing other parts of that system from also causing those variables, but that’s not really needed for this simple case.)

What your student is realizing is that for any conventional force on an object, that force itself is the natural point of external intervention. If I want to physically change a single-particle system on which F=ma is true, I notice that the force is caused by something outside that system, providing me a natural variable on which I can imagine intervening. Changing that external force, the rules of physics tell me that the acceleration will change accordingly. So we see the force as the cause and the acceleration as the effect.

But in most situations with fictitious forces, the system is now some object that happens to be accelerating or rotating, and that movement is correlated with the fictitious forces inside the object. There’s no longer any imagined external intervention on which I can imagine directly changing those forces. Instead, what I imagine is that I can change the acceleration of the object, from outside the system. (Perhaps, using conventional forces!). So the very same logic now tells your student that the acceleration is the point of external intervention, the cause, and the internal fictious forces are the effect, not a cause in their own right.

By changing the scope of the system, you could maybe find special exceptions to this pattern of real vs fictitious forces, but they would the exceptions that prove the rule: causes are the points where we imagine intervening from outside a system, making things different. This reasoning is system-specific, and cannot be found in the bare physics equations themselves, but that doesn’t mean that causation isn’t real. As long as you slow down to think through what our minds are doing when we quickly make causal inferences, we see that our causal assessments are not mere whims, but actually follow well-prescribed logical rules.


My recommendation is to start with emphasizing the central role of inertia.

I will first discuss the concept of Inertia, and from there I will move to the expression 'fictitious force'.

In daily life: just the act of walking around involves inertia. Our subconscious sense of balance is well trained to take Inertia into account. (That is a skill set that we are generally not conscious of, so I won't elaborate this example.)

Now a case of using inertia that is vivid: hammering in a nail. You swing the hammer, and at the instant the hammerhead makes contact with the head of the nail the nail opposes the velocity of the hammerhead and starts decelerating it. For a couple of milliseconds the force that decelerates the hammerhead is larger than the friction force of the nail against the wood that it is being hammered in.

I cannot emphasize this enough: We use inertia all the time. Example: there is a door you need to go through, but you notice it's somewhat stuck at one corner, just one corner. Pushing while standing still: you can't get the door to budge, but you can feel it's close to giving way. So you take a step back, and with a thud you put your weight against the door, and that overcomes the friction. We've all done that many times. You don't have to think about it, you just do it. (You only take a single step back; when you feel it will take more than that you're just not going to bother.)

Putting your weight against the door is analogous to hammering in a nail. When you put your weight against that door, with a hefty thud, then the door has to provide a force to decelerate you. That is how that thud overcomes the friction.

Inertia is in a category of its own. While Inertia is forceful indeed, we don't have the option of categorizing Inertia as a form of force.

We count a phenomenon as a force when it is an interaction between two objects. Example: the electrostatic repulsion in Rutherford's Alpha particles experiment.

The criterion to be categorized as a force: the interaction makes two moving objects exchange momentum.

In the case of gravity: example of gravitational interaction with exchange of momentum: gravitational assist.


Inertia acts in opposition to change of velocity, but for that opposition to be elicited there must be some agent that causes change of velocity.

That means that Inertia in and of itself cannot be a causal agent.

Inertia manifests itself in response to change of velocity; Inertia cannot be a causal agent.

Inertia as reference of acceleration

The fact that Inertia is the same everywhere makes it a reference for acceleration. For velocity, on the other hand, there is no absolute reference, hence the principle of relativity of inertial motion. But there is an omnipresent reference for acceleration: Inertia.

We have that the celestial objects of the Solar System are orbiting the center of mass of the solar system. (The next level up is that our solar system is, together with billions of other stars, orbiting the center of mass of the Galaxy.)

As seen from the Earth the planets go into retrograde motion from time to time. That acceleration is recognized as apparent acceleration. Within the context of the solar system as a gravitationally bound system: for each planet its true acceleration is its orbiting motion relative to the center of mass of the solar system.

Fictitious force

The term "fictitious force" is used in two different contexts. Authors should point out that distinction, but many authors don't, which tends to make their narrative incoherent.

There are oodles of videos with a person riding on a merry-go-round, and throwing a ball, with the camera co-rotating with the merry-go-round. The true acceleration of the ball is zero. We have that as seen from the rotating point of view there is an apparent acceleration.

A vivid example of the other context is the amusement ride called 'Rotor'.

Rotor ride: you are on the inside of a vertical cylinder, with your back against the wall. As the rotor spins up the G-load that you experience increases.

The special property of the G-load during a Rotor ride is that the magnitude of the G-load is constant.

We experience G-load in many different situations, but in all of those other situations the magnitude of the G-load is non-constant. Example: when sitting in a car, and the car pulls up hard you feel you are being pressed into the seat, because of your inertia.

In the case of the rotor ride: the faster the rotation rate, the stronger the required centripetal force.

As we know, inertial mass is equivalent to gravitational mass. Our sense of balance is tuned to perceiving the 1 G of G-load of gravity as a force towards the Earth, which it is. In a rotor ride the sensation of G-load is indistinguishable from the sensation of G-load of gravity. Our brain automatically infers the presence of a centrifugal force, just as we infer the presence of the Earth's gravity. Again, in the case of rotation (at constant rotation rate, and constant radial distance) the automatic inference that the brain does is much more vivid than in other cases because the magnitude of the G-load is constant.

Inertia as reference for acceleration

The only way to formulate theory of motion at all is to make Inertia the central organizing principle. If you don't do that you don't have a theory of motion.

In the equation of motion for motion relative to a rotating coordinate system there are the centrifugal term and the coriolis term. (You can refer to them as 'centrifugal force' and 'coriolis force', but you have to keep in mind they are about manifestation of Inertia.)

The following property is key:
The centrifugal term and the coriolis term are both proportional to the rotation rate of the rotating coordinate system. That rotation rate is the rotation relative to the inertial coordinate system.

That is: the equation of motion for motion relative to a rotating coordinate system is dependent on the inertial coordinate system as its reference. If you don't reference the inertial coordinate system then you don't have the means to set up a working equation of motion.

That underlines:
The only way to formulate theory of motion at all is to make Inertia the central organizing principle.


I was giving them an explanation using Reference Frames (real forces are visible in every frame while fictitious ones are not), but they weren't convinced

Your explanation is correct. Them not being convinced does not imply your explanation is wrong.

Does his reasoning make sense? Is there a rigorous way to include a cause-effect relationship in dynamics (and in fictitious forces in particular)?

There is no rigorous way to treat Newton’s laws as cause-effect relationships. The issue is time. Causes, by definition, precede effects. None of Newton’s laws involves quantities that occur at different times.

So a causal equation must relate something at an earlier time (the cause) to something that occurs later (the effect). A good example is the retarded potential: $$\phi(\vec r, t)=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec r’, t_r)}{|\vec r -\vec r’|}d\vec r’$$ where $t_r=t-\frac{|\vec r - \vec r’|}{c}<t$. The retarded time is earlier than the time. So this equation states that the charge density causes the scalar potential. There is no ambiguity about which is the cause and which is the effect because the form of the equation is a correct causal form.

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    $\begingroup$ I don't really agree with this answer. In the relativistic context, causes must precede effects that occur at a different point in space. But in the Newtonian context, forces are transmitted instantaneously (e.g. in Newton's law of universal gravitation), sometimes even over a distance. It's true that Newton's laws are only an approximation to our world, but they are fully conceptually self-consistent - and in this limiting regime, causes do not necessarily precede effects. $\endgroup$
    – tparker
    Nov 8, 2023 at 4:36
  • $\begingroup$ @tparker you are certainly free to disagree. The standard definitions of cause and effect has causes occurring before effects, but you are free to use non-standard definitions. Without that temporal ordering, however, it becomes difficult or impossible to distinguish the cause from the effect. So I am comfortable basing my argument on the standard definitions. $\endgroup$
    – Dale
    Nov 8, 2023 at 12:58
  • $\begingroup$ Do you have a citation? I was not aware that there were “standard definitions” for “cause and effect”. $\endgroup$
    – tparker
    Nov 8, 2023 at 13:18
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    $\begingroup$ Sure, Wikipedia is good for standard definitions: en.wikipedia.org/wiki/Causality#Necessary_and_sufficient_causes "If x is a necessary cause of y, then the presence of y necessarily implies the prior occurrence of x. The presence of x, however, does not imply that y will occur" and "If x is a sufficient cause of y, then the presence of x necessarily implies the subsequent occurrence of y. However, another cause z may alternatively cause y. Thus the presence of y does not imply the prior occurrence of x." $\endgroup$
    – Dale
    Nov 8, 2023 at 13:55
  • $\begingroup$ We'll have to agree to disagree about whether "Wikipedia is good for standard definitions," or whether in a Newtonian framework, the position of the Sun has no causal influence on the motion of the Earth or vice versa. But to me, it's more conceptually useful to flip the definitions around: start with the nontrivial observation that causal influences can be totally ordered, and then use that total ordering to define the distinction between the past and the future. $\endgroup$
    – tparker
    Nov 9, 2023 at 2:09

Fictious forces do not exist in reality,as the name suggests, They are "Fictious", When we write equations in non-inertial reference frame, We add an extra 'Mathematical term having dimensions of force" to make newton's laws work even when observed from non inertial frame, and in most of the cases, this mathematical term Is indeed of form "$m \cdot a$", Which makes us to imagine and call these terms "Imaginary/Fictious forces", Hence no causation can be attached to such force as these force(real interaction) don't actually(in physical reality) exist, The existence of these "fictious forces" is merely due to the fact that we "coined" the terms for our simplicity, or else we always have to think out of non-inertial frames and apply equations, which are sometimes very tedious..

for instance, The "existence" of such thought of "fictious force" can be attached to causation that , analysing motions in non-inertial motions caused us to "think/imagine" of such terms as "Fictious forces"


Newton's first law states that "A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force."

Maybe you could expand on this using a similar choice of words in something like:

"If you are using a reference frame that is not moving at a constant speed, that could be zero, in a constant direction or whose axes do not always, in space and in time, remain pointing in the same directions a body who is really at rest or set in motion at a constant speed in a straigh line may appear to accelerate. This apparent acceleration is sometimes explained as being due to fictional forces."

To answer you question more specifically I would like to think of it something like:

"Real forces cause objects to change speed or direction while "fictitious forces" can explain why objects appear to change speed or direction when you use a coordinate system that is undergoing linear acceleration or whose axes to not at all places in time and space point in the same direction."

As a typical example of fictional forces due to using coordinate axes that points in different directions at different places in space I would write something like:

Lets say we have a reference system attached to a (non-spinning) spherically symmetric planet with no gravity where you have a coordinate "r", which is the distance "upwards" from the centre of the planet but in reality points in different directions depending on where you are.

If you are standing on the surface and shooting an object with velocity v horizontally the ball will appear to accelarate as $a_r=m\frac{v_t^2}{r}$ where $a_r$ represents the acceleration in the radial "r" direction and $v_t$ is the velocity in the plane perpendicular to the radial direction.

This acceleration is a "centrifugal acceleration" and is totally fictious as the object you are shooting is really moving in a straight line.

However, if you attach a stone of mass "m" to one end of a rope of length "r" and grab the other end of the rope and start swinging it around in a circle with peripheral velocity of "v" you will feel a "real centrifugal force" of $F=mv^2/r$ pulling your hand.

I think this is a confusing part, that sometimes the centrifugal force is more like a real force but sometimes it is more like an artifact of the choice of coordinate system.


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