Is it correct to call the Coriolis an pseudo-force or an illusion?

[watchwolf49]

I’ve been poking around Wikipedia this morning and found this pearl of wisdom, if I might paraphrase here a bit … when a physicist attempts to explain something, they like to use the simplest explanation … and for some phenomena this simplest explanation uses a rotating frame-of-reference and the Coriolis force that appears in this frame … it’s not that these phenomena can’t be explained in a stationary frame-of-reference … but it gets very complicated fairly quickly …

For example … if we wish to explain the motion of the Moon around the Sun … it’s easiest to invoke the pseudo-force of gravity … the “action from a distance” explains how the Moon orbits the Earth and how the Earth orbits the Sun … compare that to using the intersection of two surfaces of revolution in time/space … 99.9% of your audience will be lost as soon as we mention “field tensor” … and in this case the differences are so minute that Newton’s Law of Gravity is close enough …

Are pseudo-forces “real”, who knows … are they useful, you bet …

[scr4]

Magnetism is also a “pseudo force” in the same sense. It’s just electrostatic force acting differently in different reference frames.

[scr4]

@Whack-a-Mole

• Let go of the rope and the bucket does not fly straight away but on a tangent.

If the centrifugal force is real why doesn’t the bucket fly straight away?

Same reason a cannonball doesn’t drop straight down as soon as it leaves the cannon.

[JoseB]

I think this really cool educational film from 1960, courtesy of the University of Toronto, might help understanding the things related to frames of reference in physics. Link follows:

- YouTube

Enjoy!

[Andy_L]

@Andy_L

Because you’re switching frames of reference mid-description.

In the frame of reference rotating with the bucket, there’s an outward force, and when the rope is let go of, the bucket moves outward (from your still rotating arm or body) with the same acceleration you’d expect due to the outward force you felt (for short periods of time - which is all the Principle of Equivalence demands of the universe).

Meanwhile in the non-rotating frame of reference, the force you feel is the resistance of the bucket to your demand that it curve its path - and so in that frame of reference, when the rope is cut, the bucket gets its revenge by refusing to bend its path anymore - so it moves out on a tangent.

Just to run the numbers:

Consider a rock swinging around on a rope of length L, with rotational frequency of w radians per second.

While the rock is still tethered to the rope, the x coordinate will be Lcos(wt) and the y coordinate will be Lsin(wt) at all times. In coordinates rotating with the rock, r=L and theta is 0 for all times (the rotating frame has theta=wt relative to the non-rotating frame).

For convenience, we’ll cut the rope at time 0 when the rock is at x=L and y=0, moving in the y direction at a speed of wL.

After that time, x remains L for all time, and y=wLt.

What does this look like in the rotating frame?

Rnew=square root (L^2+w^2L^2t^2) and thetanew=arc tan (w*t)-wt

As long as the small angle approximation applies, Rnew is approximately L+L*(1+w^2L^2t^2/2) and thetanew = wt- (wt)^3/3 - and for very short times that means that it looks like the rock is moving away from where it was at an acceleration of w^2*L^2/2

@JoseB

I think this really cool educational film from 1960, courtesy of the University of Toronto, might help understanding the things related to frames of reference in physics. Link follows:

https://youtu.be/aRDOqiqBUQY

Enjoy!

I think I saw that one in High School.

[davidmich]

@Derleth

Gravity as Newton understood it was a force, yes, and gravity can be mathematically treated as a force in most circumstances, especially if you do all your math from the perspective of someone standing on a planet.

That last bit is the important piece of all of this: These pseudo-forces appear and disappear depending on where your observer is relative to other things. If you write all your equations from the perspective of an observer in a box which just got dropped out of a plane, there is no gravity*: Everything they observe moves in straight lines, balls hang in the air when they let go, and so on, just as if they were in a sealed box out beyond Pluto. From their perspective, the apparent force is the one dragging the ground towards them at an ever-increasing rate; they certainly don’t feel it, and it doesn’t impact them at all yet. In short, someone in freefall isn’t in an accelerating reference frame, and the only people who see them accelerating are the ones on the ground, who are in an accelerating reference frame.

*(This is human-scale objects immediately near Earth. Tidal effects are, like, a zillionth-order correction in this context.)

This gets into something very important in the study of General Relativity: The equivalence principle, which states that there’s no observable difference between accelerations due to things like rockets and accelerations due to being on the surface of a planet. The usual statement of this principle, however, is imprecise, and some introductions to the concept would have you believe that a scientist in a sealed room could under no circumstances tell whether they were in Hoboken or on a rocket to Jupiter and beyond the infinite accelerating at 9.8 m/s[sup]2[/sup]. This overly-naïve platitude is falsified by Foucault’s pendulum, which will precess at high and low latitudes on Earth and not on the rocket due to Coriolis forces. Full circle, complete with parallel transport.

Thanks Derleth. Thank you all. Very helpful.