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I know that the Foucault pendulum rotation in relation to Earth is a proof that the object is inertial in relation to the distant stars. But what makes them more important than the Earth? Are they an absolute and universal inertial frame? How can we prove that? Please elaborate.

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Nobody claims that the distant stars are an inertial frame. But the center of mass of a sufficient number of distant stars is expected to be (the total momentum of the universe is 0). – Fabian Jan 26 '12 at 18:06
@Fabian why the center of mass of distant stars are more important than the mass of the Earth? AFAIK The gravitational effects of this small planet over the object are stronger. – Jader Dias Jan 26 '12 at 18:09
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@Vam'çá it's not gravity that is causing the apparent rotation. I say "apparent" rotation because that's what it is. We are seeing the effect of a combination of real and fictitious forces that, when observed from our viewpoint as stationary with respect to the accelerating Earth, looks like rotation. – Mark Beadles Jan 26 '12 at 19:02

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Actually the path of the Foucault Pendulum is not "fixed" (even approximately!) to the "fixed" stars. Unless the pendulum is installed at one of the Earth's poles (as someone has done), then the point of suspension is in constant rotation with the Earth itself. $\therefore$ the pendulum is really not in an intertial frame.

Consider a pendulum at the equator, swinging in a North South plane. It's obvious from symmetry that the plane of this pendulum doesn't rotate with respect to the earth and that, relative to an inertial frame, it rotates once every 24 hours. - UNSW, Austl.

A very good discussion of the forces (real and fictitious) on the pendulum can be found at this UNSW site. The vector that points from the suspension point toward the Earth is in constant acceleration and has a precession period that varies according to latitude.

This animation from the Wikipedia article on the Foucault pendulum may help show how the plane of the pendulum is rotating. enter image description here

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I thought it rotated in periods of 32 hours (the sidereal day) – Jader Dias Jan 27 '12 at 13:14
The sidereal day is 23.93447 hours. – Mark Beadles Jan 27 '12 at 14:23
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Maybe he got confused when he read "At the latitude of Paris a full precession cycle takes 32 hours" from the Wikipedia article – Jader Dias Jan 28 '12 at 16:21
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It is not true that the Foucault Pendulum is "inertial in relation to the distant stars". The distant stars are moving in various random directions at various random speeds and are certainly not in the same inertial frame as the pendulum. Our galaxy is rotating, so it can't be used as an inertial frame. The visible universe is expanding and probably accelerating, so it's certainly not an inertial frame.

In Foucault's day the movement of the stars wasn't visible because they are so far away, so it was common to assume that the stars were fixed, and therefore represented a fixed framework that you could use as an absolute frame of reference. Actually that's not a bad approximation, but it is only an approximation.

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If the distant stars aren't the inertial frame, why does the Foucault Pendulum rotates? I think it is inertial to something other than the Earth, what is that other thing? – Jader Dias Jan 26 '12 at 18:43
The Foucault's Pendulum doesn't rotate. It carries on swinging in the same plane while the earth rotates beneath it. – John Rennie Jan 26 '12 at 19:06
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@JohnRennie What you say is true only at the poles. Elsewhere the pendulum must rotate. – Mark Beadles Jan 26 '12 at 19:54
@JohnRennie What determines the "same plane"? What is this plane referece? – Jader Dias Jan 27 '12 at 13:05
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I suspect this is moving away from what Vam'çá will find helpful. I'm guess he's puzzled why what is apparently "locking" the pendulum in place. For the purposes of this discussion I suggest we assume the pendulum is at one of the poles - if you try the experiment at the equator you'll find there is no apparent rotation of the pendulum anyway. – John Rennie Jan 27 '12 at 15:15
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