Talk:Fictitious force

Notational suggestion

Einstein notation and Newton's notation for derivatives could be used to make the mathematical derivation section more concise. For example:

$\mathbf {F} _{\mbox{fictitious}}=-m(\mathbf {a} _{\mathrm {AB} }+2v_{j}{\dot {\mathbf {u} }}^{j}+x_{j}{\ddot {\mathbf {u} }}^{j}).$

compared to

$\mathbf {F} _{\mbox{fictitious}}=-m\mathbf {a} _{\mathrm {AB} }-2m\sum _{j=1}^{3}v_{j}{\frac {d\mathbf {u} _{j}}{dt}}-m\sum _{j=1}^{3}x_{j}{\frac {d^{2}\mathbf {u} _{j}}{dt^{2}}}.$

The rest of the section could be edited in a consistent manner. Or is the original one better? What do you guys think?

666th root of unity (talk) 03:27, 11 October 2012 (UTC)

Disagree. The original is much more accessible for general readers. —DIV (120.17.231.4 (talk) 04:55, 7 September 2016 (UTC))

Focault pendulum doesn't prove rotation of the Earth

In the end of the "Detection of non-inertial reference frame" section, the article incorrectly says that the focault pendulum proves the rotation of the Earth. Suppose the Earth is not rotating and that the pendulum is going along a really thin elliptical path that is slow slightly elliptical that we can't tell with the naked eye that that it's not passing through the bottom position, then the fact that acceleration varies as the sin of the angle from the bottom rather than linearly combined with the noneuclidian geometry of a sphere making the circumference of any circle on a sphere be less than π times its diameter means the pendulum will still precess anyway if it's swings aren't perfectly linear. An observed rotational period of 24 hours or more is probably so long that that distance away from the bottom position the pendulum must pass in order to have that rate of precession is so slight that we can't tell with the naked eye that it's not passing through the bottom position.Blackbombchu (talk) 18:26, 25 September 2013 (UTC)

Please change this entry to "apparent force"

Wouldn't it be better to encourage the use of the term "apparent force", which is well-used in literature, has the same definition, and is more apt?

"Fictitious" implies "doesn't exist" or "made-up". "Apparent" means "what you detect" without implying it's what's actually there.

"Apparent wind" is commonly used in sailing, and is perfectly apt. No sailer would ever talk about "pseudo-wind" or "fictitious wind" as he sets his sails to the apparent wind. Shouldn't physicists be at least as clear in use of terms as sailors? — Preceding unsigned comment added by Learjeff (talk • contribs) 17:46, 17 June 2014 (UTC)

Yes, I have been told many times that the forces I feel when I accelerate or break or turn when driving are fictitious - as in illusory - they are not, they are real. We have equivocation over the meaning of the word "fictitious" - in normal parlance it means illusory, not really there, a perceptual mistake... whereas in physics it means that the origin of a force is the accelerating inertial frame as opposed to mechanical interaction. People are thus misled by the word to believe that science tells us these forces are illusory. The use of "apparent" is less likely to mislead (as pointed out above) since it is (more) neutral with respect to ontology.Richwil (talk) 10:27, 17 August 2015 (UTC)

Name

Pseudo force is a much better name for this, and the page should be moved. Bumblebritches57 (talk) 17:51, 25 February 2015 (UTC)

Agree. "Pseudo", "inertial", "d'Alembert", etc. are all as equally technically correct as "fictitious", but "fictitious" too easily leads laity into thinking there's something not real about it all, which does them a disservice. 72.74.19.224 (talk) 14:02, 29 March 2015 (UTC)

Wrong information about fictitous forces

In § Detection of non-inertial reference frame, the article says an observer can tell whether or not they're in an accelerating frame of reference but that's not true. They would be unable to distinguish a rectilinear fictitious force from a gravitational field or a combination thereof. Since the gravitational constant is so small, any gravitational field the same strength as that at the surface of Earth would vary at an ever so slow rate with respect to position, to tiny a difference to distinguish it from a rectilinear fictitious force which would have a uniform force field according to Newtonian physics. Blackbombchu (talk) 15:54, 23 June 2015 (UTC)

Rotating coordinate systems

The result of the derivation in 'Rotating coordinate systems' currently reads:

$\mathbf {F} _{\mathrm {fict} }=-2m{\boldsymbol {\Omega }}\times \mathbf {v} _{\mathrm {B} }-m{\boldsymbol {\Omega }}\times ({\boldsymbol {\Omega }}\times \mathbf {x} _{\mathrm {B} })-m{\frac {d{\boldsymbol {\Omega }}}{dt}}\times \mathbf {x} _{\mathrm {B} }.$

in which $\mathbf {x} _{\mathrm {B} }$ is the position vector for the particle/object in the rotating reference frame B (not A!). It is stated that the first term is the Coriolis force, the second term is the centrifugal force, and the third term is the Euler force.
Contrast the formulæ in this article with those on 'Rotating reference frame':

• the Coriolis force: $\mathbf {F} _{\mathrm {Coriolis} }=-2m{\boldsymbol {\Omega }}\times \mathbf {v} _{\mathrm {r} }$

• the centrifugal force: $\mathbf {F} _{\mathrm {centrifugal} }=-m{\boldsymbol {\Omega }}\times ({\boldsymbol {\Omega }}\times \mathbf {r} )$

• and the Euler force: $\mathbf {F} _{\mathrm {Euler} }=-m{\frac {\mathrm {d} {\boldsymbol {\Omega }}}{\mathrm {d} t}}\times \mathbf {r}$

in which $\mathbf {r}$ is the position vector of the particle/object in the inertial reference frame A (not B!).
Although $\mathbf {r}$ and $\mathbf {x} _{\mathrm {B} }$ could be equal for some specific choices of the rotating reference frame B, it does not appear that they are generally equal, and I didn't see a clear restriction defined in this article.
—DIV (120.19.160.192 (talk) 14:23, 6 September 2016 (UTC))

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That is wrong, because of conservation of the angular momentum

"The rotating observer sees the walker travel a straight line from the center of the carousel to the periphery,"

the angular speed is, $\omega (r)=L/(m*r^{2})$ - a non linear function of radius, because the angular momentum L=const

Therefore the traveled path is not a straight line

Link request

The section "Orbiting and rotating" does not give a full solution to the problem it addresses. The article https://en.wikipedia.org/wiki/Clohessy-Wiltshire_equations is the proper and academically accepted solution to the problem. Can we please place a link there? -Theanphibian ^{(talk • contribs)} 17:51, 18 August 2019 (UTC)