30
$\begingroup$

I am constructing a thought experiment about a spinning object that is floating in intergalactic space. I assume that this object is about the size of a planet so that it will have enough gravity so that a Foucault pendulum will work, although I'm not sure that this is necessary for the thought experiment.

I can easily determine that this object is spinning if I stand on the object and observe the galaxies around me rise and set. Similarly, I can determine the axis of rotation.

For simplicity, I go to one of the poles of my object, and I set up my Foucault pendulum. What will I see? And why will I see it? I assume that in intergalactic space, the gravity is very small (galaxies are very far away, and the mass of my object is very small compared to the mass of a galaxy) so that there will be little coupling between the gravitational field of my object and the gravitational field of the galaxies around me.

$\endgroup$
20
$\begingroup$

This is indeed a Big Question; you have essentially stumbled into Mach's principle.

For an even more bewildering version: suppose that in that bit of intergalactic space, you have two spherical objects, which are rotating relative to each other about their separation axis, with the distant stars stationary with respect to object 1. Our current understanding of physics is very clear that a Foucault pendulum on object 1 will not precess, but if placed on a pole of object 2 it will precess relative to object 2 (and keep in plane with a pendulum on a pole of object 1). The reasons for this, however, are not as clear, and if I understand correctly they are still a matter of debate, but maybe someone closer to that field can clarify.

$\endgroup$
  • 4
    $\begingroup$ There were some fairly insightful comments here that a moderator just unceremoniously dumped. $\endgroup$ – Robert Harvey Feb 25 '16 at 21:55
  • $\begingroup$ @RobertHarvey yes, for two reasons: (1) the participants in the comment exchange indicated that it was concluded, and (2) some of the things said were slightly inappropriate and indicative of further inappropriate comments to come. If you think the insights brought up in the comments would constitute a good answer, or a followup question, feel free to post it. If you need to reference the comments to do so, I can undelete them for a short time so you can copy them into a text file. $\endgroup$ – David Z Feb 26 '16 at 13:11
  • 4
    $\begingroup$ @DavidZ: It just seems aggressive, that's all I'm sayin'. On Stack Overflow, I delete comments ruthlessly, but if they have anything at all to do with the actual question content (and are reasonably civil), I don't feel the need. It's not like we're going to run out of bits. Nor do I base my decisions on predictions of what might happen in future commentary. No, I don't want to post an answer; why would I do that? I'm not a subject matter expert. In short, I save my guns for the real battles, which we never seem to be short of. ♦ $\endgroup$ – Robert Harvey Feb 26 '16 at 15:48
  • $\begingroup$ @RobertHarvey probably a site culture difference. I've also posted a fair amount on SO, and I really do think we have more problems with comments here than you do there - though, admittedly, I don't see the junk that gets deleted on SO. Since it's a smaller site we get to know the frequent posters, and we notice a lot of the characteristic patterns that suggest an argument is about to break out, so it's easier to cut things off in advance. $\endgroup$ – David Z Feb 26 '16 at 16:50
  • 1
    $\begingroup$ @DavidZ That's fair enough. $\endgroup$ – Robert Harvey Feb 26 '16 at 16:53
16
$\begingroup$

While we may not be able to define a universal rest frame (Galilean invariance), we can still tell when frames are non-inertial. A spinning frame of reference is non-inertial, and thus there are non-inertial forces that arise, which we have ascribed to being "fictitious," which means that they are not fundamental, but rather a poor choice of reference. If we believe Newton's law to be what governs the universe, then we will always be able to tell a spinning frame. You can even tell how fast your frame is spinning just from local experiments, without needing an external reference frame such as the stars (e.g., the pendulum you mentioned). At either pole you will get a pendulum precession period equal to the rotation period of the planet.

Now the philosophical question about what is fundamental and what isn't, is essentially what I make of Mach's principle. And it's just that, a philosophical question.

$\endgroup$
5
$\begingroup$

How does a spinning object know it is spinning?

Let's step back. How does an object spin? First imagine a rod, if you stretch (strain) the rod to be longer than its natural rest length then like a spring there is a force (stress) on the parts trying to compress it.

An object spins when it has some velocity in one direction and yet it the orthogonal direction it is too long (strained) so it has a stress in the orthogonal direction. You could imagine a spring with masses on the two ends. At rest it has a particular length.

When it spins it is longer and the two masses have a velocity orthogonal to the spring. It's literally longer and the parts are literally moving with respect to each other.

There is no way at all in which it is any way like the stationary spring. Just because it didn't stretch much doesn't mean it isn't stretched. A spinning object bulges at its equator, that's how it spins.

Now as for how you know. You could look at the parts and notice they are strained by measuring their separations between each other and considering the materials they are made of and how far apart their natural separations are you see that they are too far apart (that's measuring the strain). You can also measure the stress. You could also use the comoving coordinates of the parts as a reference frame and check to see if Newton's laws hold without fictional inertial forces (they won't). You could take something that moves through a vacuum at a steady speed such as light and send it around one way and then send it around another way and see if they get around in the same amount of time (they won't).

A rotating object and a not rotating object are different and there are thousands of ways to tell the difference. It's a bit absurd to even imagine they are similar in any way. Spin a spring and literally watch it get longer. What's confusing about that in the slightest?

$\endgroup$
  • 6
    $\begingroup$ You're ignoring his question and saying "duh", restating the question as an assertion. That doesn't explain anything. Hey, it bothered Newton! $\endgroup$ – JDługosz Feb 24 '16 at 21:21
  • 3
    $\begingroup$ @JDługosz What question am I ignoring? If you removed my ears and eyes and spun me, I'd still be able to feel it because I'd be stretched out. You are physically different when you are spinning. I disagreed with the OP that he could tell spin by looking at galaxies. Someone could have painted images on a large spherical shell and be spinning the shell. But you can tell when you spin. You can feel it when you are spinning because you are stressed when you spin. You can feel stress. $\endgroup$ – Timaeus Feb 24 '16 at 21:28
  • 2
    $\begingroup$ "How does stuff tell that it's spinning?" "It does!" $\endgroup$ – JDługosz Feb 24 '16 at 21:33
  • 3
    $\begingroup$ @JDługosz Did you read anything I wrote? If you had a cloud of dust, it can't have stress and it can't spin. It could orbit, but that's different (and the whole look at the sky wouldn't tell the difference), and the difference is essential. To spin, when the parts go in directions there is a stress. Stresses are literally what you feel. If you don't know how things feel, go learn, then you'll know what I'm talking about. Refuse to learn and you won't know. I can't make you learn. The stresses accelerate the parts, the acceleration of the parts is the spin. Feel the stress to know spin $\endgroup$ – Timaeus Feb 24 '16 at 21:39
  • 1
    $\begingroup$ @JDlugosz: When I was a child, we did an experiment in which I was blindfolded, sat in a chair, and spun around. I could tell that I was spinning, not because I was stressed, but because of the semicircular canals in my skull, near the ears. This is a real concern for airplane pilots - the human ear is easily fooled, and that can lead to fatal airplane crashes. In my original question, I assumed that I was the observer, however, in thinking about your comments, I think that a cybernetic observer would be better - less prone to illusion. $\endgroup$ – user1928764 Feb 25 '16 at 8:13
1
$\begingroup$

The basic rule is that space has no "origin", so only relative coordinates are possible. Thus, motion is relative and only meaningful with respect to other objects. Now we also have all directions being equivilent so you have no preferred axes, and orientation is only relative too.

But, starting with that, working out what are essentially Newton's laws of motion, you discover that angular velocity is not relative, as it links up with linear acceleration. That's the same thing: given no absolute position you find you also have no absolute motion (first derivitive) but do have absolute acceleration (second derivative).

Just start with the idea of no absolute position/direction and follow the math: when do absolute quantities pop out, and when do they not?

$\endgroup$
  • 2
    $\begingroup$ I suspect you may have hit the source of the confusion here. Given that, apparently, the laws of physics are intrinsically invariant with respect to position, orientation and velocity (= the time derivative of position), it's somewhat surprising that they're not invariant with respect to angular velocity (= the time derivative of orientation). But of course, an extended object with non-zero angular velocity necessarily (either breaks apart or) experiences non-zero centripetal acceleration, which is also absolute. $\endgroup$ – Ilmari Karonen Feb 25 '16 at 11:56
1
$\begingroup$

The obscure principle of angular energy will settle the matter.

The spinning object has angular momentum, which means that most of its particles have ordinary momentum about the center, which means they possess kinetic energy, and therefore by $E=mc^2$, we know they possess gravitational attraction what we can measure if we are precise enough.

Please note that we must use special relativity here only because we need mass-energy conversion. With pure Newtons mechanics we cannot easily prove mass-energy equivalence (I can get so far as $E=mk$ but it doesn't help because due to the way it is reached the formula no longer implies that $m$ further induces gravitational forces).

$\endgroup$
1
$\begingroup$

I wouldn't even get so complicated as the other answers and would just consider the Coriolis effect https://en.wikipedia.org/wiki/Coriolis_force on a pendulum. Or if you are trying to hit something with artillery.

$\endgroup$
0
$\begingroup$

There are already several excellent answers to this question, but I think there's a bit of detail missing from all of them. The fact that rotation stretches a spring doesn't necessarily tell us everything we need to know, as Timaeus suggests. (For my reasoning, see below.) And shouldn't it be possible to show that spinning objects know they are spinning without needing them to have a measurable gravitational field, as answers involving a pendulum require? It seems to me that the final answer must have something to do with basic facts about the local geometry of the universe, and the arrangement of objects in a rotating assemblage.

I am not 100% certain that the below works; I am open to feedback, and I'll edit or delete this answer if it turns out I've made a fundamental mistake. Please let me know!

To get at these issues, I'll consider four scenarios, involving the following experimental apparatus:

  • A long, narrow spring of unknown stiffness.
  • A set of thrusters attached to the spring, oriented to produce either (perfect) linear or (perfect) rotational acceleration.
  • A device attached to the spring that can measure its length when the thrusters aren't firing.
  • A memory bank and a simple calculator that can store and perform calculations on those measurements.

Two scenarios add another component:

  • A dead-reckoning device that can make continuous measurements of the spring's length during thruster activation.

I won't rely on any precise mathematical formulas, instead paying attention to basic patterns and correlations that we know from experience will certainly hold.

Scenario One

In this scenario, we imagine the spring initially moving linearly, in the direction of its long axis, at some unknown velocity. The thrusters are arranged to produce linear acceleration in either direction. In the first experiment, we do the following:

  1. Measure the length of the spring ($L_0$).
  2. Turn on the thrusters for a while in one direction, and then turn them off.
  3. Measure the length of the spring again ($L_1$).

We can repeat steps 2 and 3, firing in either direction, to get $L_2$, $L_3$, and so on. In this scenario we don't have a dead-reckoner, so we don't know anything about what the spring was doing while the thrusters were firing. And what do we find?

We find that every length measurement is the same—$L_0 = L_1 = L_2$, and so on. So with this set-up, we can learn nothing at all about the spring's linear velocity.

Scenario Two

In the second scenario, we make the following changes. Now, in addition to its unknown linear velocity, we imagine the spring rotating about its center of mass at some unknown rotational velocity. The thrusters are arranged to produce clockwise or counterclockwise acceleration around the current axis of rotation, and cause no linear acceleration.

Now we completely ignore linear movement, but follow roughly the same series of steps as above:

  1. Measure the length of the spring ($L^R_0$).
  2. Turn on the thrusters for a while in one direction (clockwise or counterclockwise), and turn them off.
  3. Measure the length of the spring again ($L^R_1$).

Again, we can repeat steps 2 and 3 to get $L^R_2$, $L^R_3$, and so on.

Now things look quite different. Pretty much every time, the length of the spring is different. We don't have any information about what happened while the thrusters fired, but now we can see the change in rotational velocity they caused because every time, the spring has become longer or shorter. Furthermore, we pretty quickly notice that every time they fire in one direction, the spring gets longer, and every time they fire in the other direction, the spring gets shorter.

First Interlude

At this point, it initially appears clear that rotational velocity is somehow different from linear velocity, because one stretches the spring, while the other doesn't. But it's still not totally clear why this should be. And there are still a lot of unknowns. We don't know how fast the spring was moving linearly, in the first case, or rotationally, in the second. And we don't yet seem to be any closer to being able to answer that question in either case. We know that the spring gets longer and shorter in the second case, but what if that's just a kind of "memory effect"? Does it really tell us anything we couldn't know in the first case?

Let's find out! Rather than only measuring the spring's length when acceleration is zero, let's use a dead-reckoner to measure and remember its length during acceleration and see what we can learn.

Scenario Three

We return to the linear motion scenario, but we turn on our dead-reckoner, which tracks the length of the spring during thruster activation. We follow the same three steps as before, but taking continuous length measurements. And we find that every time we begin firing a thruster, the spring gets shorter! Then, when we turn the thruster off again, it gets longer again. (Assume that we start and end thruster activations gently, so that the spring doesn't bounce too much.)

Based on this, we have a lot more information. Assuming we know enough about springs and calculus, we can use the length measurements to determine how much force is being generated (at least relative to the unknown stiffness of the spring). If we know the mass of the device as well, we can use that to calculate information about acceleration, and from there, to determine information about how much faster or slower the spring is moving than it was when we first started taking measurements. So even though we don't know how fast the spring was moving at first, we can at least tell the difference between then and now.

Second Interlude

After doing this second experiment, the difference between rotational velocity and linear velocity is less obvious. In both Scenarios Two and Three, we can't tell what our initial rotational or linear velocity was. In both scenarios, we can tell when it increases or decreases. In Scenario Two we can tell because the spring gets shorter or longer. In Scenario Three we can tell because we keep track of the forces applied. These are superficially different methods of measurement, but does that superficial difference indicate any deeper difference between these two kinds of velocity? It's harder to tell.

To see the real difference between these two kinds of motion, we have to do one more experiment.

Scenario Four

In this experiment, we do the same thing in both the linear and rotational set-ups, with dead-reckoning turned on in the linear case.

  1. Measure the length of the spring.
  2. Turn on the thrusters in one direction—call it "to the right" in the linear case, and "clockwise" in the rotational case—for $T$ seconds, keeping track of the velocity change in the linear case.
  3. Turn off the thrusters and turn them on in the other direction—"to the left" / "counterclockwise"—for $2T$ seconds, keeping track of the velocity change in the linear case.

Now, suppose the first time we do this, we set $T$ to be pretty small, and here's what we see:

  • In the linear case, our final change in velocity is to the left.
  • In the rotational case, the spring is shorter than it was at first.

Then suppose we repeat step 3, doubling the value of $T$ each time, but continuing to alternate direction between "right" and "left," "clockwise" and "counterclockwise." We quickly see a correlation: each time the final change in linear velocity is to the left, the spring is shorter; each time the final change in linear velocity is to the right, the spring is longer.

Third Interlude

Imagine a world in which these correlations held forever. It would be a world in which any spring, when spun around its center of mass in a clockwise direction, stretches. But when spun around its center of mass in a counterclockwise direction, the same spring compresses.

This is clearly not the world we live in. Experience shows that in Scenario Four, eventually, once T gets large enough, the spring will stop compressing when we accelerate it in a counterclockwise direction, and will start stretching again. The only reason it was compressing when we spun it counterclockwise was that at first, it had been spinning at quite a high rate in the clockwise direction, and the resulting centripetal force had stretched it out quite a bit. The point at which the spring stops compressing and starts stretching again is the zero point—the fixed origin against which all other degrees of rotational velocity can be compared. No matter where we are in the universe—no matter what else is nearby—we can always discover that zero point. That's how a spinning object can tell that it's spinning.

Timaeus's answer follows a similar line of reasoning, but imagines that we know a lot of detail about the internal structure of the spring, so that we can tell what length it "should" have. The above shows that we don't even need that knowledge. Any system that behaves roughly like a spring (and many that don't behave like springs at all) will allow us to discover a rotational zero point using the above method.

But... Why?

This series of thought experiments hasn't directly answered the fundamental question—why can objects tell when they are rotating? Instead, it has allowed us to translate it into a different question: Why does spinning a spring faster around its center of mass always cause it to stretch, and never to compress? I'm honestly not sure I can answer that question either, but it sure seems easier!

If I had to venture a guess, it would be this: "because two points inside a circle are always a finite distance apart." You could spin a spring around in a way that would cause it to compress at first—imagine squeezing it very quickly, and spinning it at the same time, so that its ends move closer to the spring's center of mass as it rotates. But eventually, the ends of the spring would stop getting closer, and start getting farther apart again. They would move outside the circle defined by the spring's initial length, and the spring would stretch. The confining force of the spring would then begin to convert the remaining linear momentum into rotational momentum.

I don't know if that's the best way to answer the question. But at the very least, I hope this line of reasoning shows that the best answer relies only on local geometric conditions, and not on anything to do with the state of the rest of the universe.

$\endgroup$

protected by Qmechanic Feb 24 '16 at 20:39

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.