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Why are some forces are considered pseudo-forces while some are considered real or physical forces?

The definition of pseudo-forces that I know of is that they exist in noninertial reference frames but don't exist in inertial reference frames. Among other arguments is that they don't have an identifiable source.

However, a person certainly "feels" the pseudo-force such as in a car that is rounding a corner. How can it be proved, then, that the centrifugal force is not a real force? Why isn't what is "real" determined by the non-inertial reference frame?

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  • $\begingroup$ Who are 'we'? Since you are asking this question, logically it does not include you yourself, which is a contradiction. Perhaps reformulate as: "What distinguishes physical and pseudo forces?" $\endgroup$
    – my2cts
    Feb 12 at 9:50
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    $\begingroup$ @my2cts, you're right, I stated it as "we", because it is a generally accepted fact. $\endgroup$
    – ten1o
    Feb 12 at 9:55
  • $\begingroup$ But you yourself don't know. There is no we except me. $\endgroup$
    – my2cts
    Feb 12 at 9:59
  • $\begingroup$ Your question touches on an interesting point. If you reformulate it I may write an answer. $\endgroup$
    – my2cts
    Feb 12 at 10:07
  • $\begingroup$ Because the inertial force can be more simply vewied and explained by the stationary frame. Kind of like earth being center point and travels in more complicated spiral graph pattern vs simpler ellipse. $\endgroup$ Feb 13 at 14:06
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Real forces satisfy two fundamental requisites

  1. They do not depend on the reference frame

  2. They satisfy the third law of classical dynamics

Pseudoforces violate both requirements. Indeed, they appear only in some reference frames called, non-inertial in addition to real forces and they are added just to impose the validity of $\vec{F}=m\vec{a}$ also in those reference frames. Secondly, all real forces always arise in pairs in the Newtonian formulation of mechanics: the body B exerts a force $\vec{F}$ on the body A and simultaneously the body A exerts the force $-\vec{F}$ on the body B. If $\vec{F}$ is a pseudoforce acting on $A$, there is no $B$ and there is no $-\vec{F}$ acting on it.

(I stress that the notion of pseudoforce is proper of Newton's formulation of classical mechanics. When passing to more general relativistic formulations, that distinction between forces and pseudoforces is not so sharp also because the language is different, and pseudoforces share the same geometric nature of the gravitational interaction: they are no longer "forces". So that it is difficult to directly compare the two scenarios.)

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  • $\begingroup$ But why are these the requisites of what is a real force, specifically the first one? $\endgroup$
    – ten1o
    Feb 12 at 10:21
  • $\begingroup$ That is just the formulation of classical mechanics invented by Newton. Why these choices? I do not know, the most honest answer should be because this formulation works very well in our every-day experience. The first requisite is entangled with another deeper requisite known as the Galileian invariance of classical mechanics. $\endgroup$ Feb 12 at 10:22
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    $\begingroup$ If I know the functional form of a force acting on a body $F=F(t,x,v)$ in an inertial reference frame and I assume that $F$ takes the same values in every inertial reference frames, then F=ma (used as a differential equation) produces the correct motion of the body in every other inertial reference frame. $\endgroup$ Feb 12 at 10:28
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This is a comment really.

Thinking of Newoton's force, F=ma as really F=dp/dt where p is the momentum vector gives the perspective where what is called a fictitious force and a force are the same thing. It also makes momentum important and makes force liable to the conservation of momentum law. This perspective leads naturally to the use of the term "force" in particle physics, interpreting Feynman diagrams, where there is a dp/dt at the vertices. It is the first order exchanges of gauge bosons that characterizes the "forces" of the standard model into electromagnetic , strong and weak.

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Because it is different in different reference frames. You would expect a universal force to be the same regardless of how you look at it.

Also, it is not easy to figure out the source of the force.* You feel like being pushed by something you can't identify, which appears to break with Newton's 3rd law. That is an indicator (although not certain proof) of it possibly being an effect we feel due to something else than forces.

Finally, you won't feel the centrifugal force if you close your eyes.** This indicates that it might be an illusion, a mind trick, due to the surroundings changing.


*Think centrifugal forces, Euler forces and the push you feel when standing in a braking bus, and then try to look for the source.

** Sure, in the turning car or braking bus you feel the seat underneath you pulling sideways or backwards. You feel the car door or bus rail slamming into you, pushing you. But try jumping just before the turn or the braking - you won't feel anything and with your eyes closed you won't notice any changes at all. Not until you touch something again, that is.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Feb 12 at 13:41
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When you spin the ball attached to a string as you stated, what we call centrifugal force arises from the ball's inertial tendency to move away in a straight line per Newton's first law of motion. The string provides the centripetal force that keeps the ball from doing so. So we see the centrifugal and centripetal forces as resulting from the ball's inertial resistance to a change in acceleration rather than as direct forces alone.

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A pseudoforce indicates a force that would be absent in a different reference frame. A simple example is a linearly accelerated frame. If you weigh yourself in an accelerating elevator, your weight will deviate from its value in Earth frame. The spring inside the scale obeys Hooke's law, F=kx, tells you that x and thus the force exerted on it is different upon acceleration. This is a real force change, not a mathematical device. Why do we call this a pseudo force? Because we know that there is a simpler reference frame in which it is absent. By this reasoning gravity itself can be considered a pseudoforce. The simpler frame then is free fall.

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(This is a simplistic answer,that may be loose in language, and not universal. But may help a lot)

Newtonian mechanics is primarily formulated (in a simplostic sense) for inertial frames.

In a non-inertial (non relativistic) reference frame, the frame itself inherently incorporates its own acceleration/rotation. But you don't see it, really. When you walk on the rotating earth, or corner in a car, you feel some things but you don't necessarily directly perceive with your senses the non-inertial nature of your own reference frame.

So the effects you do perceive - that the car door pushes you inward, or the cars motion seems to throw you outward, or the non Newtonian precision measurement on a rotating Earth - in a way,the pseudo forces are manifestations, or (misleading!) perceptions, that either account for or offset the less easily perceived non Newtonian aspect of the frame.

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Why are some forces considered pseudo-forces while some are considered real or physical forces?

There are a variety of theoretical distinctions as listed in the other answers. However, since physics is an experimental science I prefer the practical/operational distinction:

Real forces can be detected with accelerometers and pseudo forces cannot.

However, a person certainly "feels" the pseudo-force such as in a car that is rounding a corner. How can it be proved, then, that the centrifugal force is not a real force?

Actually, this is incorrect. Removing all of the complicated neural processing and psychology, just stick with accelerometers as indicators of what is actually “felt”.

When you are in a car rounding a corner your accelerometer reads an inward acceleration. Thus the only force that is actually felt is the centripetal force. The centrifugal force is not detected. If it were then the accelerometer would read 0. So both the direction and the magnitude of the accelerometer reading contradict the claim that the centrifugal force is a real force that is “felt”.

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  • $\begingroup$ I think in actual practice, like engineering with accelerometers it is completely opposite actually. It won't detect gravity/free fall acceleration but will detect the said centrifugal forces, haha. $\endgroup$ Feb 13 at 14:10
  • $\begingroup$ @marshal craft if you have a smartphone then you can simply do the experiment and see. Turn on your accelerometer app and have a friend drive while you look at the accelerometer. When you turn left the real force is left and the centrifugal force is right. So a left reading is real force only, a right reading is centrifugal only, and a zero reading is both. You will find it reads the real force only, as I said $\endgroup$
    – Dale
    Feb 13 at 14:38
  • $\begingroup$ Don't need to, I've worked with lis3 mems accel from stm, they are virtually the same ones employed, except they don't have mems gyro. The idea is, when in free fall its accel at 9.8 m/s^2 as we all know yet the mems will read zero. Additionally when they are stationary they will read a default accel of 9.8 m/s^2, they are usually trimmed to zero this out, via software. $\endgroup$ Feb 14 at 11:18
  • $\begingroup$ @marshal craft the centrifugal force is the same way. In a rotating reference frame in free fall the accelerometer reads 0 even if it is accelerating outward by the centrifugal force. It does not detect the centrifugal force. If it is not in free fall, but rounding the bend, then it will sense only the centripetal force. It does not detect the centrifugal force. Given your experience with them, you really do need to do the experiment if you believe otherwise. My statements are demonstrably correct. $\endgroup$
    – Dale
    Feb 14 at 11:30
  • $\begingroup$ But it is true that it would not read the ficticous force i think. $\endgroup$ Feb 14 at 11:48
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In Newtonian physics, force is equal to the rate of change of momentum. In Lagrangian mechanics, we assert a function of position and velocity that governs the physics called the 'Lagrangian', and then 'force' is the partial derivative of the Langrangian with respect to position, and 'momentum' the partial derivative with respect to velocity. It looks a bit like:

$\frac{\partial L}{\partial x}=\frac{d}{dt}\frac{\partial L}{\partial \dot x}$ where $\frac{\partial L}{\partial x}=F$ and $\frac{\partial L}{\partial \dot x}=mv$

But Lagrangian mechanics allows positions $x$ to be expressed in non-inertial and curved coordinate systems (called 'generalised coordinates'), in which the relationship of the partial derivative to velocities in the flat, inertial coordinate system involves some extra terms. In an inertial reference frame (and assuming constant mass), these extra terms disappear and you just differentiate velocity to get acceleration (i.e. $F = ma$). In non-inertial or curved coordinates, you get extra terms on the right hand side, because you're differentiating with respect to something that itself varies. The trick with 'inertial forces' is just to transfer these extra terms over to the left hand side and call them part of the force. We're used to the form of the force being arbitrary, more so than the formula for acceleration.

Newtonian physics asserts as a postulate the existence of a set of inertial reference frames where these curvy extra terms vanish. It's not the only viewpoint on the matter, though.

In General Relativity, we allow much more freedom with regard to coordinate systems. If we consider a rock spinning in a static universe full of distant stars, Mach's principle suggests that this should be equivalent to a static rock in a universe with the distant stars spinning around it. And indeed, General Relativity predicts that there should be a gravitomagnetic force (a component of gravity analogous to the magnetic force in electromagnetism) that arises when there is a flow or current of mass, and in the case of the spinning universe turns out to match the centrifugal force. (See here.) Thus, the supposedly fictitious centrifugal force is just the gravitational force exerted by a spinning universe, and not fictitious at all! (Or at least, no more fictitious than gravity.)

In a sense, General Relativity pulls the same trick in reverse - it transfers all the forces over to the right-hand side of the equation and calls them curvature, leaving no 'real' forces. All particles follow free-fall paths through curved spacetime, and all forces are just aspects of curvature.

Of course, that doesn't apply in Newtonian mechanics, that asserts a simple inverse-square law for gravity, and a God-given set of inertial frames. Forces that can't be made to vanish and remain in the inertial frames are classified as 'real'.

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  • $\begingroup$ It is not correct to say that in GR there are no real forces. Only gravity is treated as curvature of spacetime and thus no longer a force, in the traditional sense. All other forces, in particular elecromagnetic forces are real forces which "force" bodies to follow non-geodesics paths. These non-geodesics paths are described as "accelerated" and thus we recover the usual $F= dp/d\tau$ written covariantly $\endgroup$
    – magma
    Sep 3 at 8:21
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Pseudo forces aren't physical forces because we, humans introduced them in order for Newton's 2nd law to work properly in non-inertial frame of references. Basically it's mathematical correctional term,very similar to cosmological constant introduced by Einstein in general relativity equations (if you know/heard of about that constant). For more details, you can check my other answer at : If an object is free falling under gravity, then where is the pseudo force applied?

Real forces like gravity, are 'real' because we see and are influenced by the effects like by gravity. So much so, that our evolution is based on to counter gravity of our earth and thrive on land.

Note : cosmological constant and pseudo forces are not related or similar things. They are merely mathematical correctional terms in their respective theories in order for theory to work properly.

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    $\begingroup$ Which part of physics is 'merely mathematical correction's and which part isn't? Which part of physics was not introduced by us humans? $\endgroup$
    – my2cts
    Feb 12 at 10:06
  • $\begingroup$ But they are for explaining real world stuff, not some mathematical fantasy. $\endgroup$ Feb 12 at 10:10
  • $\begingroup$ @my2cts Also, where can you feel the pseudo force? It is not a fundamental force or something arising from fundamental forces' interaction. Its the definition of the pseudo force that is a mathematical correctional term $\endgroup$ Feb 12 at 10:13
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    $\begingroup$ @Steeven exactly well answered $\endgroup$ Feb 12 at 13:03
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    $\begingroup$ @my2cts yes gravity is 'considered' pseudo force because it unlike other fundamental forces' which exchange force particles to 'conduct' force, gravity according to general relativity, gravity is curvature of spacetime. So if new particle for gravity is not discovered, gravity would be pseudo force. But according to classical mechanics point of view, gravity is as real like other force fields $\endgroup$ Feb 12 at 14:35

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