# Interpretation of the fictitious force

1. Mar 17, 2013

### NoWorry

Hello. Before I ask a question I want you to remind that it is my first time to use this page. And I am not even good at speaking english. My question is about the fictitious forces. I have made a long proof for the formula of the force in the non-inertial reference frame with the second law of Newton, that is

$m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0} -m\left (\boldsymbol{\dot{\omega}} \times \boldsymbol{r} \right )- \underbrace{2m\left (\boldsymbol{\omega} \times \boldsymbol{\dot{r}} \right )}_\text{Coriolis kraft} -\underbrace{m\left [\boldsymbol{\omega} \times \left (\boldsymbol{\omega} \times \boldsymbol{r} \right ) \right ]}_\text{Centrifugal kraft}$

Note that "Kraft" in danish means "force". I use danish because I live in denmark. This equation where the left side is the force in the non-inertial reference frame. The first term to right is the total force in the inertial reference frame. The two last last terms to right is the rotation reference frame.
The question is: How do I interpret this equation when I don't know what (second and) third term to right say? I think (I am not sure) the second term to right is what that makes the non-inertial reference frame to accelerate from the origin of the inertial reference frame while the third term is to make rotate with the angular acceleration $\boldsymbol{\dot{\omega}}$. Could you please explain/verify?

Last edited: Mar 17, 2013
2. Mar 17, 2013

### vanhees71

First of all, this is the equation in the non-inertial frame, nothing else. More precisely it's the most general case of a reference frame, which is both tranlationally accelerated and rotating with a non-constant rotation axis and angular velocity. Your interpretations of the various inertial-force terms are correct.

Second, the name "fictitious force" is misleading. I'd call them "inertial forces", and they are as real as any other forces. If you call those "fictitious" then, within the General Theory of Relativity, gravitation would be a "fictitious force", and nobody would come to the idea to call gravitation fictitious.

3. Mar 17, 2013

### Staff: Mentor

The third term on the right side, $m(\dot {\vec \omega} \times \vec r)$, is often called the "transverse force" in English. Wikipedia calls it the "Euler force", which I've never seen before.

http://en.wikipedia.org/wiki/Rotating_reference_frame

4. Mar 17, 2013

### D H

Staff Emeritus
This is not the most general case; things can get even hairier. For example, the non-constant unit length some use to address the restricted elliptical three body problem.

As far as naming them, there's no real need is there? "A rose by any other names smells just as sweet." The centrifugal and coriolis forces have names because they are encountered so often. The fictitious forces due to the linear acceleration of the frame origin and the rotational acceleration of the frame axes don't come up so often, so they don't have standard names (as far as I know).

5. Mar 18, 2013

### stevendaryl

Staff Emeritus
This is a terminological argument that will never be resolved, but I consider that completely wrong. Fictitious forces are NOT as real as any other forces. That's why they're called "fictitious". That's not misleading, it's descriptive. A fictitious force is not a force. Bringing up GR doesn't make your case, because in GR, gravity is not a force.

6. Mar 18, 2013

### vanhees71

Well, if you call gravity not a force, then you are right. For me gravity is one of the fundamental interactions.

7. Mar 18, 2013

### stevendaryl

Staff Emeritus
The real issue is what it means to take the derivative of a vector-valued quantity. Newton's equations of motion can be written as:

$m \dfrac{d \vec{V}}{dt} = \vec{F}$

where $\vec{V}$ is the velocity. But what does it mean to take the derivative of a vector-valued quantity? "Fictitious forces" arise from making the naive assumption that

$(\frac{d \vec{V}}{dt})^i = \frac{d V^i}{dt}$

Why do I say that that's "naive"? Because that assumption is inconsistent. If it is true for inertial Cartesian coordinates, then it's provably false for curvilinear coordinates.

To see this, write $\vec{V}$ as a linear combination of basis vectors $e_x$ and $e_y$:

$\vec{V} = V^x e_x + V^y e_y$

Now, re-write it in terms of polar coordinates, using basis vectors $e_r$ and $e_\theta$:

$e_r = \frac{x}{r} e_x + \frac{y}{r} e_y$
$e_\theta = -\frac{y}{r} e_x + \frac{x}{r} e_y$

Note: the relationship between the basis vectors is NOT constant. So $\frac{d e_r}{dt}$ is NOT equal to zero, in general.

So the same vector $\vec{V}$ can be expressed in terms of the basis $e_r$ and $e_\theta$ as:

$\vec{V} = (\frac{x}{r} V^x + \frac{y}{r} V^y) e_r + (-\frac{y}{r} V^x + \frac{x}{r} V^y) e_\theta$
$= V^r e_r + V^\theta e_\theta$

Now, take the time derivative of both sides:
$\frac{d\vec{V}}{dt} = \frac{dV^r}{dt} e_r + \frac{dV^\theta}{dt} e_\theta + V^r \frac{d e_r}{dt} + V^\theta \frac{d e_\theta}{dt}$

So, it's NOT true, in polar coordinates that
$(\frac{d\vec{V}}{dt})^r = \frac{dV^r}{dt}$ and
$(\frac{d\vec{V}}{dt})^\theta = \frac{dV^\theta}{dt}$

There are the "correction terms"
$V^r \frac{d e_r}{dt} + V^\theta \frac{d e_\theta}{dt}$

"Fictitious forces" arise from treating these correction terms as forces, but they don't actually have anything to do with forces; they are just terms in the computation of the derivative of a vector valued quantity.

8. Mar 18, 2013

### stevendaryl

Staff Emeritus
Not all interactions are forces.

9. Mar 18, 2013

### stevendaryl

Staff Emeritus
The extra terms are technically known as "connection coefficients".

Let me simplify from 3 dimensions to just 1.

If $x$ is an inertial Cartesian coordinate, then Newton's laws look like this:

$m \dfrac{d^2 x}{dt^2} = F$

Now, if we switch to a new coordinate $x'$, then the regular rules of calculus tell us:

$\dfrac{dx'}{dt} = \dfrac{\partial x'}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial x'}{\partial t}$

$\dfrac{d^2 x'}{dt^2} = \dfrac{\partial^2 x'}{\partial x^2} (\dfrac{dx}{dt})^2 + 2 \dfrac{\partial^2 x'}{\partial x \partial t} \dfrac{dx}{dt} + \dfrac{\partial^2 x'}{\partial t^2}$

So in addition to the term $\dfrac{\partial^2 x'}{\partial t^2}$, the second time derivative of $x'$ has the additional terms:

$\dfrac{\partial^2 x'}{\partial x^2} (\dfrac{dx}{dt})^2 + 2 \dfrac{\partial^2 x'}{\partial x \partial t} \dfrac{dx}{dt}$

These additional terms are due to the nonlinear relationship between the coordinate $x'$ and the coordinate $x$. In 3 dimensions, the first term generalizes to a "centrifugal force" and the second term generalizes to a "coriolis force" term. I don't think of them as "forces" at all, they are just terms in the derivative.

10. Mar 18, 2013

### vanhees71

Ok, then we can agree on the fact that there are no forces in nature, because everything is relativistic and thus there are only interactions. Of course, here I suppose, you mean by "force" an instantaneous interaction at a distance as is natural in Newtonian physics but not possible (as far as we know) in relativity due to causality constraints inherent in the structure of relativistic space-time.

11. Mar 18, 2013

### D H

Staff Emeritus

Forget about relativity, forget about quantum mechanics. Just looking at things from the perspective of Newtonian mechanics, there are a couple of very solid reason for calling those forces "fictitious". One is that unlike real forces, those fictitious forces are not subject to Newton's third law. There is no equal and opposite reaction to these fictitious forces. Another is that there are a set of reference frames, the inertial frames of reference, in which those fictitious forces vanish. You can't truly attribute cause and effect to those fictitious forces because this attribution vanishes in the inertial frames. A real force, on the other hand, is a frame invariant quantity. You can attribute cause and effect to real forces (in context of Newtonian mechanics, of course).

12. Mar 18, 2013

### A.T.

But since it is a "terminological argument", then "I consider that completely wrong" merely means "I don't like that terminology".

That is the point. Something that was a real force (In Newtons theory) becomes a fictitious force (in GR), by redefining which frame is inertial, and which is not.

13. Mar 18, 2013

### stevendaryl

Staff Emeritus
The fact that not all interactions are forces does not imply that no interactions are forces. A force is a very specific kind of interaction. In contrast, a "fictitious force" is not an interaction, at all. Unless you consider it to be an interaction with spacetime itself, I suppose.

14. Mar 18, 2013

### stevendaryl

Staff Emeritus
Not all choices of terminology are equal. The terminology of considering the extra terms to be "forces" is actually inconsistent. The extra terms appear whenever one takes derivatives of vector-valued quantities, whether or not there are any "forces" involved. If you have a vector field $\vec{A}(x,y,z,t)$, then its time derivative will look different in different coordinate systems. The transformation from one coordinate system to another will in general involve extra terms. To call these extra terms "forces" is just nonsensical. Forces are only relevant in the specific case that $\vec{A}$ is a velocity. In other cases, the same extra terms appear, but they can't possibly be interpreted as "forces". So you have to come with yet another term for these extra terms in this case. But then why call them "forces" when $\vec{A}$ happens to be a velocity?

15. Mar 18, 2013

### WannabeNewton

Just to drive home this point, I would like to quote Kleppner, from page 344 of his "An Introduction to Mechanics" text:

If one is in a frame where a fictitious force is present, one can always find a coordinate transformation to an inertial frame where said force vanishes, and as steven noted this is akin to being able to make the christoffel symbols vanish identically for a given coordinate system on flat space. This is reason enough to claim they are artifacts of the coordinate system and not due to true physical interacts between bodies because such interactions are non vanishing in inertial frames so one cannot simply transform them away with a coordinate transformation like one can with fictitious forces. Of course I speak, and steven speaks, in the framework of Newtonian mechanics so there is no need to bring GR into this.

16. Mar 18, 2013

### stevendaryl

Staff Emeritus

Fundamentally, Newton's $F = m A$ is a vector equation:

$\vec{F} = m \vec{A} = m \dfrac{d \vec{V}}{dt}$

Since it is a vector equation, then you can choose a coordinate system, and write down the corresponding equation for components:

$(\vec{F})^i = m (\vec{A})^i = m (\frac{d \vec{V}}{dt})^i$

Now the move that some people take next is to assume the following:

$(\frac{d \vec{V}}{dt})^i = \frac{d}{dt} V^i$

If it's not clear what this is saying, we can break it down into two operations on vectors:
1. Take the $i$th component.
2. Take the time derivative.

The equation above assumes that these two operations can be done in either order, and you get the same answer. In the expression:

$(\frac{d \vec{V}}{dt})^i$

you first take the time derivative, to get $\frac{d \vec{V}}{dt}$, and then take component $i$ to get $(\frac{d \vec{V}}{dt})^i$. In the expression:

$\frac{d}{dt} V^i$

it's done in the opposite order: first you take component $i$ to get $V^i$, then you take the time derivative, to get $\frac{d}{dt} V^i$. In general, the order matters! The difference between these two expressions is the so-called "fictitious forces":

$F^i_{fict} = \frac{1}{m} [\frac{d}{dt} V^i - (\frac{d \vec{V}}{dt})^i]$

The right-hand side doesn't have anything to do with forces, particularly. It's just terms arising from vector calculus involving non-Cartesian coordinates.

17. Mar 18, 2013

### A.T.

How do you know which frames are inertial? There are no fictitious force in them.
How do you know which forces are fictitious? They don't exist in inertial frames.
GR vs. Newton demonstrates nicely how you to make a fictitious force into a real force, by changing the convention about which frames are considered inertial.

18. Mar 18, 2013

### stevendaryl

Staff Emeritus
That's true. If the only information you have is the paths of test particles, then you don't know whether coordinate acceleration is due to forces, or due to the use of noninertial, curvilinear coordinates. In order to identify inertial coordinates, you need test particles in freefall (not acted upon by any forces), and there is no way to know for sure whether your test particles have forces acting on them, or not.

But this ambiguity about how to interpret the motions of test particles doesn't really help the case of viewing centrifugal and coriolis terms in the equations of motion as forces. It's not that someone observed the motion of objects and hypothesized these forces to explain the motion. In this case, you start with the equations of motion in inertial cartesian coordinates, and derive the corresponding equations of motion in rotating, polar coordinates. The extra terms are not are not a surprise that requires a physical explanation, they are a mathematical consequence of the equations of motion in the original coordinate system.

19. Mar 18, 2013

### WannabeNewton

Inertial frames all lie in the equivalence class of frames related by galilean transformations. Can you find an element of this equivalence class that has "force" terms purely due to the coordinate dependence of the basis vectors used to define the frame and not due to physical interactions between bodies? Sure if we are in the frame of an observer, in some small enough confined experimental setup, there is no way we could tell or even care if said frame is "accelerating" or "inertial" but we are talking about operational definitions given by Newtonian theory for a class of preferred frames. In arbitrary coordinate systems, $F^{i} - \Gamma ^{i}_{jk}\dot{x}^{j}\dot{x}^{k} = (\frac{dA}{dt})^{i} - \Gamma ^{i}_{jk}\dot{x}^{j}\dot{x}^{k} = V^{j}\triangledown_{j}V^{i} - \Gamma ^{i}_{jk}\dot{x}^{j}\dot{x}^{k} = \ddot{x}^{i}$ picks up that extra negative term next to the $F^{i}$ term, which in an inertial frame of course vanishes because it is simply a coordinate artifact; I agree that an observer in a frame defined by this coordinate system will not KNOW or CARE that these terms are coordinate artifacts but we are speaking with respect to the preferred class of frames are we not?

But it's a different description of reality. The very definition of what is inertial is different in SR as opposed to Newtonian mechanics. The inertial frames in Newtonian mechanics are the preferred frames related by galilean transformations and in SR those related by lorentz transformations. Within the confines of Newtonian mechanics, said things hold true.

Last edited: Mar 18, 2013
20. Mar 18, 2013

### Jorriss

"How do I interpret this equation when I don't know what (second and) third term to right say?"

Has anyone even attempted to answer his question?

21. Mar 18, 2013

### AlephZero

jtbell answered for the first term.

For the second term, see http://en.wikipedia.org/wiki/D'Alembert's_principle

22. Mar 18, 2013

### NoWorry

So you mean that my "guessing" interpretations about the second and third terms were right? I understand you very well about the calling the forces as "fictitious" forces. Since the laws of Newtons belong to the inertial reference frame, then socalled “forces” outside the inertial reference frame still can not consider as "true" forces. Therefore they are the fictitious forces. There were only very few lines in some books that call an accelerated reference frame (without rotation reference frame) that call it as an inertial force - only this simple reference frame, but most of the long equations the authors call them as fictitious forces - in the advanced reference frame.

Me netiher. I don't like reading something on Wikipedia even though it has very good and simple explaination. But I am just seeking for the explaination in depth. In reality there is no an acceptable name for it. See page 103 on http://books.google.dk/books?id=ZWoYYr8wk2IC&lpg=PR1&hl=da&pg=PA103#v=twopage&q&f=true.

Thanks. I needed it as a reference. I am also starting to think about a better explaination why we can't claim that the third law of Newton is valid in the non-inertial reference frame?

Last edited: Mar 18, 2013
23. Mar 18, 2013

### NoWorry

Firsly, if there has been no rotation in the non-reference frame the equation would have been like this, I call it as an accelerated reference frame (remember no rotating),
$m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0}$
which is very easy to interprete this equation. This is the equation for the 2. law in the non-inertial frame reference. The first term to right is the total force in the inertial reference frame while the last term, namely
$-m\boldsymbol{\ddot{R}}_{0}$
is what makes the non-inertial reference frame to accelate with a translational acceleration $\boldsymbol{\ddot{R}}_{0}$ in one direction straightly. If it transforms to the inertial reference frame, there would be no fictitious force this time.

Secondly, the rotation reference frame without the acceleration would have been like this (i)
$m\boldsymbol{\ddot{r}}=\boldsymbol{F}+ \underbrace{2m\left (\boldsymbol{\omega} \times \boldsymbol{\dot{r}} \right )}_\text{Coriolis Force}+\underbrace{m\left [\boldsymbol{\omega} \times \left (\boldsymbol{\omega} \times \boldsymbol{r} \right ) \right ]}_\text{Centrifugal Force}$
(This equation is according to the Classical Mechanics by John R. Taylor p. 343)

Then IF this rotation reference frame is about to accelerate, there should have been like this (ii)
$m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0} - \underbrace{m\left (\boldsymbol{\dot{\omega}} \times \boldsymbol{r} \right ) }_\text{“Whatever” Force} - \underbrace{2m\left (\boldsymbol{\omega} \times \boldsymbol{\dot{r}} \right )}_\text{Coriolis Force} -\underbrace{m\left [\boldsymbol{\omega} \times \left (\boldsymbol{\omega} \times \boldsymbol{r} \right ) \right ]}_\text{Centrifugal Force}$
(This equation is according to the Classical Mechanics by R. Douglas Gregory p. 477).

Now we can see that the third term is added "mysteriously". I don't even understand why there are positive terms in (i) while there are negative terms in (ii). That’s why I wanted to know how to interprete this (third) or other terms more specific without doubting.

Last edited: Mar 18, 2013
24. Mar 20, 2013

### stevendaryl

Staff Emeritus
The following might help:

In general, a vector $\boldsymbol{A}$ can be written as a linear combination of basis vectors. Let $\boldsymbol{e'_x}$, $\boldsymbol{e'_y}$, $\boldsymbol{e'_z}$ be the basis vectors of a rotating coordinate system. Then we can write:

$\boldsymbol{A} = A^x \boldsymbol{e'_x} + A^y \boldsymbol{e'_y} + A^z \boldsymbol{e'_z}$

Now, if we take the time derivative of $\boldsymbol{A}$, there will be two sources of time-dependency: (1) The components $A^x$, $A^y$, and $A^z$ may change with time, and (2) The basis vectors $\boldsymbol{e'_x}$, $\boldsymbol{e'_y}$ and $\boldsymbol{e'_z}$ can change with time. So let the operator $\dfrac{D}{Dt}$ represent the complete time derivative, taking into account both effects, and let $\dfrac{d}{dt}$ be the operator that only takes into account the changes in the components. Then we can write:

$\dfrac{D}{Dt}\boldsymbol{A} = \dfrac{d}{dt} \boldsymbol{A} + A^x \dfrac{D}{Dt}\boldsymbol{e'_x} + A^y \dfrac{D}{Dt}\boldsymbol{e'_y} + A^z \dfrac{D}{Dt}\boldsymbol{e'_z}$

In the special case of a rotating coordinate system, we can express the time dependence of the basis vectors as follows:

$\dfrac{D}{Dt}\boldsymbol{e'_i} = \boldsymbol{\omega} \times e'_i$

where $i$ is either $x$, $y$, or $z$.

For this special case, we can summarize the total time derivative this way:

$\dfrac{D}{Dt}\boldsymbol{A} = \dfrac{d}{dt} \boldsymbol{A} + \boldsymbol{\omega} \times \boldsymbol{A}$

Let's let $\dot{\boldsymbol{A}} = \dfrac{d}{dt} \boldsymbol{A}$. Then we can write:

$\dfrac{D}{Dt}\boldsymbol{A} = \dot{\boldsymbol{A}} + \boldsymbol{\omega} \times \boldsymbol{A}$

Now, let's specialize to the case of the position vector:

$\dfrac{D}{Dt}\boldsymbol{r} = \dot{\boldsymbol{r}} + \boldsymbol{\omega} \times \boldsymbol{r}$

Taking another derivative gives:
$\dfrac{D^2}{Dt^2}\boldsymbol{r} = \dfrac{D}{Dt}\dot{\boldsymbol{r}} + (\dfrac{D}{Dt} \boldsymbol{\omega}) \times \boldsymbol{r} + \boldsymbol{\omega} \times \dfrac{D}{Dt} \boldsymbol{r}$

$= \ddot{\boldsymbol{r}} + \boldsymbol{\omega} \times \dot{\boldsymbol{r}} + \dot{\boldsymbol{\omega}} \times \boldsymbol{r} + (\boldsymbol{\omega} \times \boldsymbol{\omega}) \times \boldsymbol{r} + \boldsymbol{\omega} \times \dot{ \boldsymbol{r}} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})$

This simplifies to:
$\dfrac{D^2}{Dt^2}\boldsymbol{r} = \ddot{\boldsymbol{r}} + 2 \boldsymbol{\omega} \times \dot{\boldsymbol{r}} + \dot{\boldsymbol{\omega}} \times \boldsymbol{r} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})$

Now, here's where the negative signs come in: Solve the above for $\ddot{\boldsymbol{r}}$ to get:
This simplifies to:

$\ddot{\boldsymbol{r}} = \dfrac{D^2}{Dt^2}\boldsymbol{r} - 2 \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - \dot{\boldsymbol{\omega}} \times \boldsymbol{r} - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})$

Now, multiply through by the mass $m$ to get

$m \ddot{\boldsymbol{r}} = m\dfrac{D^2}{Dt^2}\boldsymbol{r} - 2m \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - m\dot{\boldsymbol{\omega}} \times \boldsymbol{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})$

The last move is to use Newton's law to rewrite:

$m\dfrac{D^2}{Dt^2}\boldsymbol{r} = \boldsymbol{F}$

So in terms of the force $\boldsymbol{F}$ we have

$m \ddot{\boldsymbol{r}} = \boldsymbol{F} - 2m \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - m\dot{\boldsymbol{\omega}} \times \boldsymbol{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})$

At this point, people sometimes define the "inertial force" $\boldsymbol{F}_{inertial}$ to be the extra terms
$\boldsymbol{F}_{inertial} = - 2m \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - m\dot{\boldsymbol{\omega}} \times \boldsymbol{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})$

So the equations of motion look like

$m \ddot{\boldsymbol{r}} = \boldsymbol{F} + \boldsymbol{F}_{inertial}$