Physics/relativity question: Is centrifugal force "magic"?

Factual Questions
41

You’re talking about Earnshaw’s theorem, which does indeed hold for gravity (at least classical Newtonian gravity, which is mathematically similar to a classical electric force). On the other hand, Earnshaw’s theorem doesn’t hold for diamagnetics, which is one reason magnetic levitation is actually possible.

Now, whether we can use Earnshaw’s theorem to say something about creating a constant gravitational field is not so obvious to me. Earnshaw’s theorem isn’t talking about a constant field (e.g., one where everything is pulled down by the same amout no matter where you are in the volume of interest), it’s talking about a field that pulls everything towards some fixed point in empty space, making that point a stable equilibrium where objects will be inclined to just hover in place.

42

Well, you’re a lot less massive than your space station (I assume), so not much is going to happen. The point is angular momentum is conserved, not angular velocity. Your angular momentum depends, among other things, on your mass, so if the station is much, much more massive then you then it can spin much, much more slowly and still counterbalance your angular momentum.

43

I’m not worried about slowing the space station (though if I were running counter to it, I guess I’d be speeding it up).

I wanna fly! Can I fly?

44

Oh, I see what you’re saying. Off the top of my head I’d say yes, you could fly, at least until air resistance slowed you down. Just like on Earth if you move fast enough you can fly (i.e., you can orbit the earth). If the station matches Earth’s gravity g, then the speed you need to cancel out its rotation is v = sqrt(R g), which is the same formula that applies to trying to orbit the Earth at surface level. (I’m approximating the Earth as a sphere here, so your orbit is a perfect circle.) Except, the station presumably has a much smaller radius R than the Earth, which should make the necessary speed much easier to achieve.

With air resistance, though, I don’t think you’d stay up for long.

45

Chronos wrote:

With the infinite mass sheet is the field really uniform? Isn’t there a discontinuous direction shift near the surface? Or is this just a nitpick and we don’t care what happens near the surface?

46

Read Rendezvous With Rama by Arthur C. Clarke. In it a guy flies a human powered small plane, not unlike the Gossamer Albatross, across the central axis of a giant spinning cylinder drifting through our solar system. I never liked Clarke’s description of that flight though - he kept saying that “gravity” (expression of the centrifugal force due to the spinning cylinder) increased as he drifted away from the central axis during his flight from one end of the cylinder to the other. Since he was not in contact with the cylinder, no such thing would happen, however, there would be air resistance pushing him in a “downward” direction to contend with, but it seems to me that this would be very nonlinear and almost insignificant near the axis.

47

If I did the math right, I get a constant acceleration due to gravity of 2 pi G times the mass per area of the sheet.

–

You can treat the sheet as concentric rings around the point directly below you. Consider a ring of mass with radius R and mass dM = 2 pi sigma R dR (where sigma is the mass per area).

If you’re a distance z above the ring, the acceleration due to gravity from that ring is just:

da = G dM cos(theta) / (z^2 + R^2)

directed straight downward, since z^2 + R^2 is the distance from you to any point on the ring.

theta is the angle to the ring off the straight-downward axis. We have to include cos(theta) because the horizontal force of gravity for each segment of the ring is canceled by the segment on the opposite side of the ring, so we’re left with just the vertical component.

cos(theta) = z / sqrt(z^2 + R^2)

Plugging in this and the definition of dM, we have the acceleration due to a single ring of mass as:

da = 2 pi G sigma z R dR / (z^2 + R^2)^(3/2)

To get the full acceleration, integrate over all rings that make up the plane.

a = Int_{R=0…infinity} [ 2 pi G sigma z R dR / (z^2 + R^2)^(3/2) ]
= (2 pi G sigma z ) Int_{R=0…infinity} [ R dR / (z^2 + R^2)^(3/2) ]
= (2 pi G sigma z ) (1 / z) for z > 0
= 2 pi G sigma

Thus the acceleration is always 2 pi G sigma straight down, independent of z.

48

If you’re asking what happens if we actually pass throught the sheet:
if the sheet has some thickness the force of gravity drops linearly to zero as we go to the middle of the sheet, and then increases linearly until we exit the other side. If the sheet has zero thickness, then you’ve got a step discontinuity between the two sides.

49

tim314 I agree with your math, but at the same time the gradient has to be in the opposite direction on the other side of the sheet. I’m too lazy to pull out my Electrodynamics text but isn’t this also the way it works in EM?

WRT to centrifugal force, GR and Einstein here’s what the old boy himself says:

ETA on review I see the first part has already been answered

50

Okay - centrifugal force is as real as any other force when the reference frame is rotating. If you have a mass on a string connected to a spring scale attached at the axis of rotation, the scale measures a force.

But when you cut the string, the mass doesn’t go straight out, as you might think “centrifugal” means. It starts perpendicular to the spring, then spirals out. Also, the speed of the mass relative to the axis of rotation varies (due to the frame’s rotation). Is this just attributed to some Coriolis force?

51

  1. Einstein’s elevator example (which is different from CookingWithGas’s non-accelerating elevator) is only strictly true for single points, and he knew and said as much at the time.

  2. Continuing a hijack, does the math work in the 2001 scene? ? I saw the scene a little while ago, and from what I remember, the chamber/track looked to be about 10m in radius, and it took him about 10 seconds to do a circle. That implies running about 60m in 10 sec, which is more or less fast jogging speed, so that’s consistent.

Now doing that requires a centripetal acceleration of v^2/r = 36/10 m/sec^2 = 3.6 m/sec^2. Which means the centrifugal force that the rotating astronaut feels is about 1/3 of earth gravity.

I wonder if that’s enough downward force to give enough traction to actually run at 6 m/s, or would the jogger be either slipping or bouncing to the ceiling at every step?

52

To extend your post, and my earlier one about running counter to the rotation, once you leave the floor in a space station, what is there to drag you back down? It’s not like there’s any force acting on you to push you back to the floor.

Once you get off the floor, shouldn’t you stay there?

53

Suppose at a particular instant, the astronaut is standing on the (rotating) space station floor. He is experiencing a force due to the rotation, and consequently has an acceleration and a velocity. Once he jumps, he maintains his initial velocity vector due to his own inertia. From his perspective, this inertia is creating a fictitious force outward. This fictitious force is the centrifugal force, which brings him back into contact with the floor.

I think I’ve got that right.

54

Belrix, if we’re expressing things in the frame that rotates with the station, then there is a force pulling you back down. This arises from the fact that we’re doing our calculation in a non-inertial frame.

If you want to understand where this force comes from in the context of an inertial frame (say, the center of mass frame of the station), then consider this:
Suppose we’re standing still on the floor of the station, and facing against the direction of the station’s rotation. This means that we actually have a velocity directed towards our back, as the station pulls us along with it. So when we jump straight up off the floor, we continue to move backwards until we hit the floor of the station behind us. Of course, the station continued to rotate underneath us, so relative to the station it looks like we went straight up and down. It’s with reference to the intertial frame that I saw we have a velocity towards our back.

If we pick up some speed running against the station’s rotation before jumping, then our backwards velocity (in the inertial frame) is reduced to a smaller amount. But we still move backwards when we jump and slam into the floor. The difference is the station has now rotated more than us, so that relative to the station it seems like we jumped forward and fell to the floor.

If we pick up even more speed before jumping, so that we have zero velocity in the inertial frame (this is what you suggested above), then when we jump we don’t have any backwards velocity to slam us into the floor. (In priniciple I suppose we would have to worry about crashing into the ceiling, but I think it would be possible in principle to just kind of tuck our legs without giving us an upward velocity, so we’re floating in place.) Now, we’re floating in place relative to the inertial frame, but the station continues to rotate beneath us, so relative to the station it seems like we made a jump that covers infinite distance. Note that this isn’t some magical property of rotating space stations. In fact, you can make an infinite jump on Earth too if you can get going fast enough. This is just what it means to go into orbit. The advantage of the station is it has the same effective gravity at a smaller radius, which makes the speed you need to achieve less. [v = sqrt(R g) ]

Now, all this is neglecting air resistance. The air won’t be at rest relative to the inertial frame – rather, drag forces will cause the air to pick up velocity from the station’s rotation. So, in the inertial frame we’ll feel a backwards wind, which does give us a backwards velocity that causes us to hit the floor. (In the station’s frame, we can see it as us jumping forward and being slowed by air resistance).

As long as we don’t mess up our calculations, we should get the same effect whether we treat the station as stationary and add the appropriate centrifugal force, or whether we treat the center-off-mass frame as stationary and say “there is no centrifugal force.”

55

(bolding mine)

That’s essentially correct, although I think it’s good to talk about one reference frame at a time. The sentence I bolded is a little imprecise, because (to my ear, at least) it makes it sound like there’s some frame (“his perspective”) in which he has both a sideways velocity (maintained by “inertia”) and experiences centrifugal force (the so-called “fictitious force”). Really what you’re trying to communicate is that from one perspective you’re just seeing the effects of inertia, but from another perspective you’re seeing the effects of a centrifugal force.

Specifically, from the “inertial frame” perspective, he has a velocity (maintained by inertia), and that brings him back to the floor. From the “rotating frame” perspective, there is a centrifugal force that brings him to the floor. (I won’t call it “fictitious”, because in that frame it’s as real as any force. Saying “fictitious” suggests that the inertial frame is somehow more legitimate than the rotating frame, in contrast to Einstein’s thinking on the matter.)

56

[Nitpick] As long as the jump height is very small compared to the radius of the station, that is. Otherwise, you need to start adding in Coriolis forces and other crazy things, and at some point the rotating reference frame will just not work at all.[/Nitpick]

57

Good point, the centrifugal force isn’t the only change that results in going from an inertial frame to a rotating one . . .

58

Thanks for breaking that down for me. Slight hijack here to satisfy my own curiosity: Isn’t it best to work out of an inertial frame, since things can get quite complicated in non-inertial frames. In this case, the system is very simple, and we can easily move from talking about the inertia frame to talking about the non-inertial frame (in this case, the rotating space station). My understanding is that for more complicated scenarios, trying to apply classical mechanics to a non-inertial frame would case all kinds of havoc.

59

Sorry to fly off the handle-- I take comfort in the hope that I’m not the first person to get hair-trigger testy on a message board.

I didn’t provide the cite initially because I didn’t have it then. And I didn’t know how to look it up because I didn’t know what the problem was called. The cite was from the link MikeS provided.

I was pretty sure that if the problem made sense, somebody smarter than me would have already thought about it-- I’m not so egotistical as to think my musings on physics are original, but I do tend to assume that if I come up with something interesting, some real physicist has probably written about the problem. (Which is still pretty egotistical, actually.)

Sorry I didn’t express the question more clearly. I have a bad habit of jumping two or three steps ahead in an argument without doing what my high-school math teachers called “showing your work,” and assuming that my listener somehow knows what I’m thinking anyway. (My girlfriend calls it “expecting people to read your mind,” and she doesn’t like it either. :D)

60

Sometimes things can get quite complicated no matter what we do, unfortunately.

In particular, sometimes what you want to calculate is actually the motion in a rotating frame. Let’s say you’re trying to figure out the force you’d need to, say, hit Moscow with an ICBM launched from the U.S. (although hopefully this isn’t something we need to worry about so much anymore). If you use a coordinate system that rotates with the Earth, you have the benefit that the coordinates of your launch site and you target are fixed. The downside is you have to include additional forces such as the centrifugal and coriolis forces.

The alternative is to work in an inertial frame (which doesn’t rotate with the earth). Then we don’t have these extra forces. But in that case the coordinates of your target change as a function of time, and we don’t know how long it will take the missile to get there. So now you have the more complicated problem of finding the force needed to make the path of the missile intersect the path of Moscow, and to have them both arrive at the point of intersection simultaneously.

So you can see why sometimes it’s worth going to a non-inertial frame even if it means dealing with a few extra forces.