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Interpretation of the fictitious force

by NoWorry
Tags: fictitious, force, interpretation
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WannabeNewton
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Mar18-13, 01:40 PM
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Quote Quote by A.T. View Post
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.
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, [itex]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} [/itex] picks up that extra negative term next to the [itex]F^{i}[/itex] 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?

Quote Quote by A.T. View Post
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.
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.
Jorriss
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Mar18-13, 06:28 PM
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"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?
AlephZero
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Mar18-13, 08:08 PM
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Quote Quote by Jorriss View Post
"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?
jtbell answered for the first term.

For the second term, see http://en.wikipedia.org/wiki/D'Alembert's_principle
NoWorry
#22
Mar18-13, 08:59 PM
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Quote Quote by vanhees71 View Post
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.
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.

Quote Quote by jtbell View Post
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
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=ZWoY...opage&q&f=true.

Quote Quote by WannabeNewton View Post
Just to drive home this point, I would like to quote Kleppner, from page 344 of his "An Introduction to Mechanics" text.
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?
NoWorry
#23
Mar18-13, 09:11 PM
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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),
[itex] m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0} [/itex]
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
[itex]-m\boldsymbol{\ddot{R}}_{0} [/itex]
is what makes the non-inertial reference frame to accelate with a translational acceleration [itex] \boldsymbol{\ddot{R}}_{0}[/itex] 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)
[itex] 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} [/itex]
(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)
[itex] 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}[/itex]
(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.

I appreciate your answers though.
stevendaryl
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Mar20-13, 12:14 AM
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Quote Quote by NoWorry View Post
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),
[itex] m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0} [/itex]
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
[itex]-m\boldsymbol{\ddot{R}}_{0} [/itex]
is what makes the non-inertial reference frame to accelate with a translational acceleration [itex] \boldsymbol{\ddot{R}}_{0}[/itex] 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)
[itex] 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} [/itex]
(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)
[itex] 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}[/itex]
(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.

I appreciate your answers though.
The following might help:

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

[itex]\boldsymbol{A} = A^x \boldsymbol{e'_x} + A^y \boldsymbol{e'_y} + A^z \boldsymbol{e'_z}[/itex]

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


[itex]\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}[/itex]

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

[itex]\dfrac{D}{Dt}\boldsymbol{e'_i} = \boldsymbol{\omega} \times e'_i[/itex]

where [itex]i[/itex] is either [itex]x[/itex], [itex]y[/itex], or [itex]z[/itex].

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

[itex]\dfrac{D}{Dt}\boldsymbol{A} = \dfrac{d}{dt} \boldsymbol{A} + \boldsymbol{\omega} \times \boldsymbol{A}[/itex]

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

[itex]\dfrac{D}{Dt}\boldsymbol{A} = \dot{\boldsymbol{A}} + \boldsymbol{\omega} \times \boldsymbol{A}[/itex]

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

[itex]\dfrac{D}{Dt}\boldsymbol{r} = \dot{\boldsymbol{r}} + \boldsymbol{\omega} \times \boldsymbol{r}[/itex]

Taking another derivative gives:
[itex]\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} [/itex]

[itex] = \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})[/itex]

This simplifies to:
[itex]\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})[/itex]

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

[itex]\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})[/itex]

Now, multiply through by the mass [itex]m[/itex] to get

[itex]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})[/itex]

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

[itex] m\dfrac{D^2}{Dt^2}\boldsymbol{r} = \boldsymbol{F}[/itex]

So in terms of the force [itex]\boldsymbol{F}[/itex] we have

[itex]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})[/itex]

At this point, people sometimes define the "inertial force" [itex]\boldsymbol{F}_{inertial}[/itex] to be the extra terms
[itex]\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})[/itex]

So the equations of motion look like

[itex]m \ddot{\boldsymbol{r}} = \boldsymbol{F} + \boldsymbol{F}_{inertial} [/itex]


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