Expert: Courtney Seligman - 4/21/2011

Question
QUESTION: Hello Professor Seligman,
Almost every reference explains the apparent rotation of the Foulcault Pendulum on the coriolis force/effect.  I am having a hard time accepting this.  It seems that it is too small a force over too small of a distance to really cause this.  

I suspect that it is a red herring caused by the change in rotation depending on which side of the equator it is on.  
Though I don't have the math/physics/dymamics background to explain or understand it, I suspect that it is a complex orbital mechanics problem linked to the motion of the pivot of the pendulum when it is not at 90 degrees North or South.

Full discloser...  I am a docent at the California Academy of Sciences in San Francisco and I often present at the Foucault Pendulum there.  I have to dance around this a bit, but I do explain the sine calculation and spend a lot of time talking about gravity, Galileo, and science history.

Thanks for your thoughts and time.

ANSWER: I take it from the way you phrased your question that you understand the situation at the Poles -- namely, the pendulum just swings back and forth in a constant plane while the Earth rotates beneath it, and it only looks like it is turning to the west because the observer is moving around it while the Earth rotates. In any event, you may want to refer to my web page on the Coriolis Effect, at http://cseligman.com/text/planets/coriolis.htm, while or before going over the material below.

Continuing with that assumption, I presume what you're confused about is why the pendulum takes longer to turn at lower latitudes, according to the inverse sine of the latitude. The answer doesn't have anything to do with orbital mechanics, but with basic torques. I haven't looked up the math in a while, but am pretty certain it doesn't involve anything more complex than vector cross products; still, that's a lot more complicated (in the minds of non-mathematicians) than just stating the result. So you aren't likely to find a "simple" answer anywhere. However, things can be explained in a simple way if you don't mind ignoring the math and just accepting the fact that the inverse sine of the latitude does turn out to be the result.

At the Poles, of course, the pendulum maintains the same plane of motion, as the basic laws of physics would suggest it ought to, while the Earth rotates to the East underneath it, making it look like the pendulum turns to the West (as noted above, there is a discussion of this on my web site). But this requires a special mechanism for keeping the pendulum swinging back and forth (otherwise, friction with the air and the support would cause it to gradually slow down and stop). For the Foucault pendulum at your site, there is presumably a support that gives it a little push in the exact direction of its motion, so that it doesn't slow down, and just keeps going and going (the last I heard, the usual mechanism is a ring magnet centered on the support, which due to symmetry would give a tug in exactly the same direction the pendulum is moving, each time it nears the end of its swing; but there may be other methods of accomplishing the task that have been devised since I last looked into the matter).

I should also note that there is an absolutely terrific movie, "Frames of Reference", available on the Internet Archive, which has a nice discussion of the Coriolis effect near the end, and an explanation of why the Coriolis force is called a "fictitious force". In that movie, a pendulum is set to moving back and forth at the start, without any method of keeping it swinging back and forth, so at the end of the movie, the pendulum isn't swinging as far; but the fact that the plane of swing has changed is still obvious. As noted in that movie, any pendulum used to demonstrate this effect is referred to as a Foucault pendulum; but usually, they are set up to just keep going and going, so that over a few hours' time, the effect becomes very obvious.

Now, at the Poles, setting up a pendulum like this would be relatively simple (ignoring the conditions there), save for one thing. Namely, the building holding the support would gradually be turning around, with the rest of the Earth. So the support has to be designed so that it can rotate freely about the vertical (at the Poles) axis of the Earth's rotation. That way, the building can turn around the pendulum, but the support stays fixed in place, there are no forces on the pendulum save for the little tug that keeps it going in the same direction it was already going in, and its plane of swing is fixed, relative to the distant stars.

However, at any other latitude, such as San Francisco, there is another complication. Namely, not only is the building turning with the Earth's rotation, it is also moving around the Earth, along with everything else at its parallel of latitude. This means that the vertical and horizontal planes which define the motion of the pendulum are slowly twisting to the East, relative to the way they started off. As a result, even if the support can freely rotate around its vertical axis, it is slowly twisting to the East, along with those planes. This makes the pendulum also slowly twist to the East instead of staying in a truly constant plane, so its apparent westward motion is reduced by the amount that it twists to the East. At the Equator, the eastward twisting of the plane of its motion exactly equals the eastward rotation of the Earth, so the pendulum just swings back and forth in a straight line relative to the building, and doesn't appear to change its orientation at all. At mid-latitudes, the difference between the eastward rotation of the Earth and the eastward twisting of the plane of rotation allows the pendulum to appear to keep moving to the West, but at a slower rate -- namely, the rate given by the inverse sine of the latitude.

In some ways, an exact mathematical proof would be much simpler and more obvious than the preceding paragraph; but since you claim to be relatively unable to follow the math, perhaps this explanation, though less clear than a direct mathematical proof, will make more sense.

There is another way to explain the situation, which does not rely on either the twisting of planes or vector algebra. That is to think about the speed of the eastward motion of the Earth at different latitudes, and imagine that the pendulum is temporarily aligned exactly north and south. The support for the pendulum is moving to the east at a speed determined by its latitude. As the pendulum moves toward the pole, it passes over ground which is at a higher latitude, and therefore has a smaller circumference for its parallel of latitude. In other words, the part of the building which is closer to the Pole than the support is moving eastward more slowly than the support. Similarly, the part of the building which is closer to the Equator is moving eastward more rapidly than the support. Since the building is all one piece, this doesn't seem possible; but it isn't a problem if you remember that the vertical and horizontal planes which pass through the building are gradually turning in space, around the axis of the Earth's rotation. As the planes turn, the part of the horizontal plane closer to the Pole doesn't move around as big a circle as the part of the horizontal plane closer to the Equator, so even though the building moves as a unit, the more polar part is moving more slowly to the east than the more equatorial part.

Now, consider the pendulum bob as it moves toward the pole. Since it is attached to the support, it should be moving to the east at the same rate as the support; and as it passes below the support, the ground beneath it is moving to the east at the same rate as well, so there is no reason for the pendulum to seem to twist. But as it moves toward the Pole, the ground beneath it is moving eastward slower than the support, so the pendulum will move eastward relative to the ground beneath it. Conversely, as it moves toward the Equator, the ground beneath it is moving eastward faster than the support, so the ground will move eastward relative to the pendulum, which makes the pendulum appear to move westward relative to the ground. (This is shown in the diagrams on my web page, though not with exactly the same discussion.)

In the previous paragraph, it doesn't make any difference which hemisphere you are in. Movement toward the Poles results in an apparent eastward motion of the pendulum, while motion toward the Equator results in an apparent westward motion of the pendulum (relative to the ground beneath it). The result is that in the Northern hemisphere, the pendulum moves a little to the right, regardless of which way it is swinging. If moving toward the Pole, which is towards the North, the Eastward motion of the pendulum is to the right, relative to its original path. If moving toward the Equator, which is towards the South, the Westward motion of the pendulum is still to the right, relative to its original path. In the Southern hemisphere the directions are reversed, so the direction of swing appears reversed. Eastward motion toward the Pole becomes a leftward movement, and Westward motion toward the Equator becomes a leftward movement. So in the Northern hemisphere, the pendulum turns a little toward the right every time it swings, while in the Southern hemisphere, the pendulum turns a little toward the left every time it swings.

Now, how fast is this little swing to the right or left? That depends upon the difference in speed of the ground beneath the support, and the ground to the north or south of the support. Near the Poles, the ground is (if nearly horizontal) nearly perpendicular to the axis of rotation of the Earth, so moving closer to the Pole or further away from it increases the size of the parallel of latitude by quite a bit. As a result, the pendulum seems to swing to the right (in the northern hemisphere) or the left (in the southern hemisphere) pretty quickly -- in fact, once around to the West, each time the Earth turns once around to the East.

Near the Equator, things are very different. There, the ground is nearly parallel to the axis of rotation. As a result, the speed of the ground to the North or South of the support is almost exactly the same as the speed of the support. So the rate of swing of the pendulum is very slow. In fact, if you were exactly at the Equator, as the pendulum swung to the North, it would turn a little to the right, but when it swung back to the South, it would turn a little to the left, exactly canceling out the previous swing to the right, so it wouldn't seem to swing one way or the other, at all.

In other words, you can think of the problem as one of three-dimensional planes rotating around the Earth, which is the usual way of doing the math, or you can think in terms of the local conditions. Either way, near the Poles the rate of swing is as fast as the eastward rotation of the Earth, while near the Equator the rate of swing drops to zero.

Of course, in the "local" example I assumed that the pendulum was swinging exactly North and South; and as its plane of motion changes, that would no longer be true. So the "local" method, although an easy way to explain what is going on, isn't as satisfactory as the three-dimensional method, because it doesn't allow for an explanation of why the pendulum continues to turn to the right (or left) as its motion becomes more East and West. And of course it does continue to turn, at exactly the same rate, regardless of which way it is swinging, so the vector algebra method is the only one you're likely to see explained mathematically.

I've rambled on for quite a while, so your eyes may have glazed over by this point. But hopefully, this gives you enough different ways of approaching the problem that you can see that it is strictly due to the rotation of the Earth under the pendulum, and has nothing to do with orbital dynamics, or for that matter, any real force. However, as stated in "Frames of Reference", if you DO ignore the rotation of the Earth, and try to explain the rotation as being due to the "fictitious" Coriolis force, the size of the force, though small, IS able to completely explain the motion. It is admittedly a very small force, being only a third of a percent of the weight of the pendulum bob at the Poles, and smaller and smaller as you approach the Equator. But the distance involved isn't as small as you think, as I'm pretty sure you're thinking in terms of the distance the pendulum swings back and forth in one swing. That is a small distance, and during one swing, the effect of the Earth's rotation, and the supposed (equal) effect of the Coriolis force, would be negligible. But the pendulum doesn't swing round all in one swing. It goes back and forth, over and over, thousands or even hundreds of thousands of times during one "rotation" of its plane. So the actual distance traveled by the bob is very, very large by the time it has completed one rotation around its support.

It's getting pretty late here, so I don't have the time or energy to check this for errors of omission or commission. Hopefully it's reasonably accurate and clear, or at least clear enough for you to follow the argument with the help of the diagrams on my web page, and/or the recommended movie. But if not, please feel free to ask for a clarification of any part of this muddle.

---------- FOLLOW-UP ----------

QUESTION: Hello Professor Seligman,

That was an amazingly quick response to my question!  Thank you for making such an effort.  

I remain not completely convinced that Coriolis effect or force is what is at work here.  I suspect that the red herring is that the pendulum changes direction of rotation as it crosses the equator.  I understand the effect and visualizations when discussing the north south travel of an airplane or a canon ball.  The rotational effect that we see in large fluid systems (ocean and atmosphere) is, to my understanding largely influenced by the nature of the fluid.  As the fluid is displaced more fluid flows in to the lower pressure area, or void, and this creates the rotation.
What I don't understand is why a pendulum continues to rotate when it is swinging parallel to the equator or lines of latitude.  It seems to me that the rate of rotation should appear to be maximum when swinging north/south and then decrease to no rotation if swinging east/west if it is fact coriolis.  

When you add the absolute minimal effect that coriolis has on something that moves such a small distance I just can't see it.  

To illustrate my question...  
Imagine a pendulum at the north or south pole with the pivot point half the distance of the swing of the pendulum away from the axis of the earth's rotation.  Say it's total swing is 100 ft.  Move the pivot point 50 ft from the axis.  The change in angular momentum from the point at the pole to the point 100 ft from the pole would go from zero to some value.  Now take this experiment to the equator and the change in momentum would be nill.  The change in distance of the bob from the axis would be almost nill.  
Now if you were to swing the pendulum parallel to the equator at the equator it is easy to understand why it doesn't rotate.  What I don't understand is why it rotates past this orientation when it is swinging somewhere more distant from the equator,if the rotation is due to coriolis.

Here are a couple of links that you may be interested in.
http://www.calacademy.org/products/pendulum/
http://www.calacademy.org/products/pendulum/pendulum_sales.html
http://www.calacademy.org/products/pendulum/pendspec.pdf  This one is the spec sheet for the pendulums that can be purchased through the California Academy of Sciences.

Thank you for your time and expertise.  I hope my questions make sense!



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Answer
From my earlier followup (slightly modified):

I took a quick look at the first of the three sites you listed. The situation at the Poles and Equator is pretty clearly and simply discussed (much as in my original answer, but with the added benefit of diagrams), but at mid-latitudes, the discussion is too brief to be clear, and it's left to the reader to trust that it's correct. I'll have to think about how to explain the effect at mid-latitudes other than by just saying "the twist is the same, no matter what direction it's moving; it's just easier to explain if you imagine it's moving north and south".

I should point out one error in the discussion on the first site. Although many people may have thought the Earth was flat until a few hundred years ago, educated ancient Greeks (Hellenes) knew it was a sphere. In fact, it was old news that the Earth was spherical when Aristarchus proposed that it went around the Sun. It was only the question of whether the Earth or the Sun moved that was controversial; the fact that both were spheres was considered an obvious and simple result of natural philosophy more than 2500 years ago.
--------------------------

Now to finish the above (sort of):

After considerable thought, I don't believe I can provide a simple, clear answer to your question at this time. The three examples I used -- at the Poles, at the Equator, and at mid-latitudes but in a North-South direction -- are easy to explain and follow; but a correct answer for mid-latitude motions which are not exactly North and South requires a far more complex (and mostly mathematical) answer than I can conjure up in any reasonable amount of time (or words). So all I can do is summarize a few critical points:

(1) The distance the pendulum moves is not small -- it moves back and forth many times before there is enough rotation of its apparent plane of motion to notice -- so there is plenty of distance involved; it just doesn't seem like it, because what you notice is the distance it moves to one side or another, not the long-term motion. (If we wanted a straight-line demonstration of Coriolis forces, we would need distances of hundreds or thousands of feet; the pendulum doesn't swing that far, but it "moves" that far, back and forth and back and forth, over a period of time.)

(2) The Coriolis "force", though seemingly very small, is not too small to produce the (apparent) turning of the plane of motion. In fact, the point of using the concept of a Coriolis force, instead of thinking in terms of the turning of the Earth under the pendulum, is that (the following is set in caps to set the thought apart, not to yell at the audience) IN THE FRAME OF REFERENCE ATTACHED TO THE MOVING OBSERVER, THE TURNING OF THE PENDULUM APPEARS PERFECTLY REAL, AND EXACTLY WHAT IT WOULD BE IF THE CORIOLIS FORCE WAS PERFECTLY REAL. The reason the Coriolis force is so small is that any larger force would turn the plane of motion faster than what we observe. When we invent (that is, rely on) a "fictitious" force, its size is determined by its effects. The force required to turn the plane of the pendulum's swing is only 1/3 of 1 percent of the weight of the pendulum, or less (depending upon your latitude, and the corresponding rate of rotation of the plane); so that's the size of the Coriolis force. It may seem small, but it has to be, to be in line with the (slow) turning of the pendulum.

(3) The reason for even thinking about a Coriolis force is that if it was real, it would be possible (and relatively easy) to explain the motion of the pendulum without thinking about the actual geometry of the situation -- namely, that the horizontal and vertical planes defined by the floor of the building housing the pendulum, and the swinging back and forth motion of the pendulum, are slowly turning around the Earth's axis of rotation (at an angle which depends on your latitude), as the Earth rotates to the east. Describing the effects of this changing geometry is far more difficult than describing the effects of the imaginary (read "fictitious") Coriolis force; or at least, it is far more difficult in the general situation, where you can be at any latitude, with the pendulum swinging in any direction. It is simple in the three special cases already described. Otherwise, it requires vector cross-products and other (what would seem, to most non-mathematicians) horrific math. That's what you see in physics textbooks, and web pages devoted to a detailed discussion of things like the Coriolis force/effect; but since you are questioning that sort of discussion, I presume it is not the sort of answer you want to see. And I can't think of any way to avoid that kind of math, save for the special circumstances and general hand-waving I've done above, and here. It may not be a very satisfactory answer, but it's the best I can do at the moment.

In passing: I do plan to eventually discuss this topic in more detail on my website, both on the current page about Coriolis effects (which relies on a strictly non-mathematical approach, and is far briefer than this discussion), and on an "attached" page intended for individuals who would welcome a chance to see the actual math involved, or at least a more precise mathematical argument. I don't know if the result will include a presentation which would meet your current requirements; but you might check back in a year or two, to see if I've made any progress on the matter. (My time is divided between my regular work, the novels I write, and my website; so progress on either of the latter is slow and sporadic.)