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davefoc
9th April 2006, 11:08 PM
It seems that it is possible to tell the difference between a rotating frame of reference and a non-rotating frame of reference even if those frames of reference are in an area of space where no stars are visible for reference.

Objects fixed in the rotating frame of reference will experience centripetal force. Objects fixed in the non-rotating frame of reference won't experience centripetal force.

What's going on here? I understand this problem is addressed in general relativity but can the answer be boiled down so that even I could understand it? If space is a completely empty void then would it be possible to tell the difference between a rotating and a non-rotating frame of reference? What in the void would there be to interact with the matter within it to let the matter sense whether it is rotating or not?

I understand that I might be wildly confused here, but I am curious enough about what is going on to allow my ignorance to become more widely known. So please feel free to allude to my lack of insight or to mention it directly.

Yllanes
10th April 2006, 01:15 AM
Forget relativity for the moment. Newton's laws only work in inertial (non-accelerating) reference frames. Whenever you are in a non-inertial frame, pseudoforces appear (centrifugal force, Coriolis force, azimutal force, drag force). This means that in that frame F =/= m a, you have to add all these terms.

Now consider this. Imagine a group of people on a mery-go-round and take into account special relativity. The platform its measured, first at rest. We get, not surprisingly, 2*pi*r = l. Now it starts rotating at relavistic speed. A man in the centre measures it again. The value for its radius is the same as before, because the velocity is perpendicular to the radius, so there is no measured contraction. However, the circunference is tangential to velocity, so it is contracted. This means that, for this rotating platform, 2*pi*r =/= l. You could say that there is another metric at work, we are no longer in Euclidean space, but in a hyperbolic geometry (notice that we haven't talked about a graviation, just acceleration in special relativity). Incidentally, we know that the centripetal acceleration on the disk is not gravity, because it doesn't go to zero at infinity. Exercise: Would the angles of a triangle sum more or less than 180º on this platform?

CurtC
10th April 2006, 07:13 AM
I think davefoc's question has to do with what is "space"? Most of us have an intuitive idea that it's just how objects are arranged with respect to each other, but he's asking, if there are no other objects in our thought experiment, what then does space describe? It has to be more than that, because obviously enough you can still tell the difference between rotating and non-rotating frames.

I guess the answer is that space is more than the description of where stuff is. The laws of physics depend on space, independently of other bodies in the universe. Truly understanding this might require mind-altering substances.

Ririon
10th April 2006, 08:36 AM
One of these days I will get a good relativity book, a comfortable and undisturbed place to read and some tea and shortbread. :D

davefoc
10th April 2006, 09:04 AM
Forget relativity for the moment. Newton's laws only work in inertial (non-accelerating) reference frames. Whenever you are in a non-inertial frame, pseudoforces appear (centrifugal force, Coriolis force, azimutal force, drag force). This means that in that frame F =/= m a, you have to add all these terms.

Now consider this. Imagine a group of people on a mery-go-round and take into account special relativity. The platform its measured, first at rest. We get, not surprisingly, 2*pi*r = l. Now it starts rotating at relavistic speed. A man in the centre measures it again. The value for its radius is the same as before, because the velocity is perpendicular to the radius, so there is no measured contraction. However, the circunference is tangential to velocity, so it is contracted. This means that, for this rotating platform, 2*pi*r =/= l. You could say that there is another metric at work, we are no longer in Euclidean space, but in a hyperbolic geometry (notice that we haven't talked about a graviation, just acceleration in special relativity). Incidentally, we know that the centripetal acceleration on the disk is not gravity, because it doesn't go to zero at infinity. Exercise: Would the angles of a triangle sum more or less than 180º on this platform?

Ylanes, thank you for your answer. I am not sure I understood it well enough to know whether you have addressed the basic idea of my question. If the merry go round is floating in space completely devoid of all references and completely devoid of all external gravitational effects people on the merry go round can still tell if its spinning or not. It seems that there is something in the void they are floating in that interacts with the matter of the riders and the matter of the merry go round such that it is possible to tell the difference between spinning and not spinning. If this were not so they might just be able to change their frame of reference to one that is spinning at the same rate they are and then they would think they are not spinning.

kuroyume0161
10th April 2006, 09:07 AM
I guess the answer is that space is more than the description of where stuff is. The laws of physics depend on space, independently of other bodies in the universe. Truly understanding this might require mind-altering substances.

But don't cause confusion that 'space independent of other bodies in the universe' is an absolute frame of reference. :rolleyes:

davefoc
10th April 2006, 09:33 AM
I think davefoc's question has to do with what is "space"? Most of us have an intuitive idea that it's just how objects are arranged with respect to each other, but he's asking, if there are no other objects in our thought experiment, what then does space describe? It has to be more than that, because obviously enough you can still tell the difference between rotating and non-rotating frames.

I guess the answer is that space is more than the description of where stuff is. The laws of physics depend on space, independently of other bodies in the universe. Truly understanding this might require mind-altering substances.

CurtC, thanks for the answer. You seem to be suggesting that this is just an unresolved issue in physics. It is interesting that so much is understood, but this seemingly simple issue isn't.

I would not have independently recognized it as an issue. For me there was spinning and not spinning and I never thought about how the issue forces a non-rotating frame of reference on the problem before one can make calculations that predict the orbits of planets and the forces on the merry go round riders. I have been reading through an elementary book on relativity and they mentioned the issue without providing an explanation that I understood.

One obvious possibility that you allueded to is that a vacuum is not really devoid of some kind of structure. Would it be wrong to think of this as some sort of aether? You may remember that I mentioned that I thought that there must be some kind of aether because of the fact that a light beam travels at the same speed with respect to another light beam traveling in the same direction. That is a light beam neither gains nor loses distance with respect to another light beam traveling in the same direction. This suggests to me that an aether does exist even if it is not the simple aether originally envisioned that didn't explain time and distance contraction from observers in different inertial frames of reference.

It seems that modern physics instruction is pretty much devoted to shooting down the idea of an aether which tends to make me think that I have entered kook land with my thinking on this.

Yllanes
10th April 2006, 09:39 AM
What I meant is that rotating frames are easy to identify, because Newton's second law is not satisfied in them. Things accelerate that shouldn't, etc.

If this were not so they might just be able to change their frame of reference to one that is spinning at the same rate they are and then they would think they are not spinning.

Again, I'm not sure what you mean. If you are in a rapidly rotating room, even if there is nothing at all outside, you will stick to the walls. For you the room is not rotating.

roger
10th April 2006, 10:35 AM
Dave, it has nothing to do with any property of space. It's Newton's laws. An object in motion stays in motion, and an object at rest stays at rest. Recall the (non-relativistic) equation F=MA. This tells us that if there is acceleration, you will feel force,and if you apply force, there will be an acceleration.

So, adding all that up. Take a stick, and rotate it around it's center, like a helicopter blade. Without reference to any other object in the universe, you see that the tip is continuously changing position and direction of motion with relation to the stick's center point. Hence, the tip is undergoing acceleration. If you were sitting on that tip, you would feel that acceleration as a force (due to F=MA).

Now, if it was a big stick floating in empty space you would not be able to tell what that force was coming from. You wouldn't perceive the stick as rotating, but you would feel the force. Move around on the stick and you could probably surmise what was happening and where the rotation point was because the forces would change as you move closer to and away from the center of rotation.

Ziggurat
10th April 2006, 11:13 AM
Another way to think about it is to start with a SINGLE reference frame (which we index with some x,y,z,t coordinates), and assume Newton's laws hold in that frame (that is, force causes acceleration, no acceleration means no force), but without prior knowledge of what happens in any other reference frames. We can do experiments in this reference frame, we can throw balls around, play with masses on springs, etc. and confirm that Newton's laws hold in this single frame. Now the question becomes: what happens if we CHANGE reference frames? In other words, instead of (x,y,z,t), what happens if we use some other coordinates (x',y',z',t')?

Well, all you really need to do is plug in the transformations for (x,y,z,t) -> (x',y',z',t') into your equations of motion (Newton's laws), and you can find out what the equations of motion in your new reference frame look like. And it turns out that if Newton's laws hold in ONE reference frame, they will also hold in any other frame which differs from the first by some constant rate of translation (that is, a moving, non-accelerating frame). This is classical Galilean relativity. But there are SOME coordinate transformations under which your equations of motion DO change. Those include frames which are rotating and frames which are accelerating. But we can get the equations of motion for the new reference frame to LOOK like the original equations of motion by separating it into a component which is the same as before (acceleration = force/mass) plus whatever is different for this new frame (the so-called fictitious forces, such as centrifugal and corriolis forces).

You can do the same process with special relativity with a few adjustments. Classically, Newton's second law is often written as F = m d^2x/dt^2 (acceleration as the second derivative of position with respect to time). In SR, this is replaced with F = dp/dt (where p is the momentum) - this substitution makes no difference classically (where p = m dx/dt), but it does matter in relativity (since p is a little more complicated). But given that substitution, you can establish that only a certain class of coordinate transformations (Lorentzian boosts) will leave your equations of motion the same, and other transformations (such as rotations, or galilean boosts) will not.

jj
10th April 2006, 11:13 AM
Exercise: Would the angles of a triangle sum more or less than 180º on this platform?

Top or bottom? :p

Alkatran
10th April 2006, 12:46 PM
I'd like to point out another reason you can tell rotation from linear movement: it takes at least 2 particles to get rotation. It only takes 1 particle to move in a straight line.

If there was only one particle in the universe there would be no way to measure anything. With 2 particles you can measure things like relative distance, speed and acceleration. Being able to measure relative acceleration, along with a bit of knowledge of how things move around, is all you need to figure out if rotation is happening.

epepke
10th April 2006, 12:50 PM
It seems that it is possible to tell the difference between a rotating frame of reference and a non-rotating frame of reference even if those frames of reference are in an area of space where no stars are visible for reference.

Objects fixed in the rotating frame of reference will experience centripetal force. Objects fixed in the non-rotating frame of reference won't experience centripetal force.

What's going on here? I understand this problem is addressed in general relativity but can the answer be boiled down so that even I could understand it? If space is a completely empty void then would it be possible to tell the difference between a rotating and a non-rotating frame of reference? What in the void would there be to interact with the matter within it to let the matter sense whether it is rotating or not?

I understand that I might be wildly confused here, but I am curious enough about what is going on to allow my ignorance to become more widely known. So please feel free to allude to my lack of insight or to mention it directly.

It sounds as if your understanding is pretty good.

Note that this thought experiment is pretty old. Einstein himself thought about it and even vacillated. So you're in good company.

However, I think it has been resolved.

In Special Relativity, there's no problem. SR only applies to inertial frames. A rotating reference frame is undergoing acceleration. Rotation is not relative under SR. So there you have it.

In General Relativity, it's a bit trickier. The forces that you feel in a rotating frame (or, more precisely, inertial resistance to those forces), are not only like gravity, they are gravity. If you declare the universe as non-rotating and the frame rotating, then it reduces to the SR case. If you declare the frame non-rotating and the universe rotating, then the rotation of the universe creates a gravitational field.

This is, I think, why people are talking about space and spacetime here. This can be hard to think about, because space doesn't seem like it has stuff in it, if there's nothing else in the universe. There are no little wires or dots that you would be able to see, so there's no stuff that it's easy to see that would cause this. (If you get into quantum behavior, there's reason to believe that there's a lot of stuff in the vacuum, but let's not go there.)

However, spacetime does have some properties. There's a relationship between space and time given by c. There are also things called geodesics, which are inertial paths and correspond to what in flat spacetime (SR) would be called straight lines. Light travels along geodesics, too, and they're called null geodesics.

Now, why spacetime works like this, nobody knows, at least from a classical view. (QED, as I've mentioned, explains the geodesics rather nicely, but let's not go there.) What is important is that the mathematics of a rotating frame in a non-rotating universe, and the mathematics of a non-rotating frame in a rotating universe are exactly the same, and so there is no way at all to distinguish between them.

At this point, if you still have emotional difficulties, you might want to go into the relativistic quantum field theories, such as QED, in which there is stuff everywhere, and you can't have an otherwise empty universe. Even if there is just a single electron in the universe, it has amplitudes everywhere.

Yllanes
10th April 2006, 02:25 PM
Davefoc, my connection broke when I was editing a previous post. I wanted to add that maybe I took your question at a more basic level than it was intended. Google for "Mach's Principle", maybe this is what you meant (the Wikipedia article on this is not very complete, and I don't have time right now to explain it in detail myself).

davefoc
10th April 2006, 02:41 PM
I have just read through several of the posts. I don't have time to understand them all right now, but my initial reaction is that only epepke has really understood what I am talking about (that is not to say that I understand exactly what epepke is talking about yet).

Let me say this.
There are two ways we know that something is rotating.
1. We assume some non-rotating frame of reference and notice that the object is moving with respect to the non-rotating frame of reference we assumed.
2. The forces and motions that we calculate and or sense for the object are consistent with Newton's laws of motion in the particular non-rotating frame we assumed.

For instance consider the earth rotating around the sun. How do we know that the earth is rotating around the sun rather than they are both fixed in some frame of reference?

The answer seems pretty obvious. We look at the background stars and make a pretty good guess as to how a non-rotating frame of reference is oriented and we notice the earth is moving a lot and the sun is moving a little bit too in that frame of reference.

Ahah we say, our guess as to what the non-rotating frame of reference must be pretty good because when we make calculations on the earth and sun's motions based on Newton's equations we come up with a pretty good description of how the sun and the earth are actually moving through the non-rotating reference frame we assumed. Further we might even refine our idea of exactly how the non-rotating reference frame is oriented by working backward from the motions of the earth and sun that are observed.

But the thing that is not obvious is that we have had to assume that there is such a thing as a non-rotating frames of reference in the vacuum of space to make all this work out. And that seems to suggest (to me at least) that there is something going on with respect to the interaction between mass and a vacuum that has some directional properties that allows us to detect whether we are rotating or not.

ceptimus
10th April 2006, 03:03 PM
...we have had to assume that there is such a thing as a non-rotating frames of reference in the vacuum of space to make all this work out.I don't think so. If we are able to observe centripetal forces at work, then we know a system is rotating. If we don't observe any such forces then we know it isn't. You could be placed inside a spaceship, with no windows and no means of observing anything outside, and with a few simple experiments you'd be able to tell if the spaceship was rotating or not.

davefoc
10th April 2006, 03:21 PM
I don't think so. If we are able to observe centripetal forces at work, then we know a system is rotating. If we don't observe any such forces then we know it isn't. You could be placed inside a spaceship, with no windows and no means of observing anything outside, and with a few simple experiments you'd be able to tell if the spaceship was rotating or not.
You would use one of the two ways of detecting rotation that I mentioned previously. You would detect the forces causes by rotation and deduce that you are rotating. You could also establish you rate of rotation with respect to a hypothetical non-rotating frame. So far I am with you I think. And if you looked outside your spaceship you'd probably feel like you did a pretty good job of deciding what the non-rotating frame was because you'd observe the stars and see that your hypothetical non-rotating frame wasn't rotating with respect to them.

OK, but my question goes to why there should be a special non-rotating frame at all if the space you are floating in consists of a complete nothingness that is incapable of interacting at all with your ship.

ceptimus
10th April 2006, 03:38 PM
Imagine you are in the spaceship in zero-G and objects are just hanging motionless (with respect to the ship and each other). We would normally conclude this was a non-rotating system. But if you say that there is nothing outside the ship for it to rotate 'with respect to' and therefore you might just as well consider that it is rotating, then you would have to come up with a whole new physics to explain why objects with no forces acting on them move in circles.

I think the apparent paradox of 'what is the system rotating with respect to?' is just a trick of our way of thinking. Objects don't care whether they're in a rotating frame of reference or not - they just obey their nature and move in straight lines unless a force causes them to deviate. The 'frame of reference' thing is just an analytical concept that exists in the minds of humans. It has no basis in reality.

Ziggurat
10th April 2006, 03:38 PM
OK, but my question goes to why there should be a special non-rotating frame at all if the space you are floating in consists of a complete nothingness that is incapable of interacting at all with your ship.

To a certain extent, I'm not sure a satisfactory answer exists. Take things to an elementary enough level, and I don't think you can answer the "why should they" questions, but only the "do they" questions.

epepke
10th April 2006, 03:44 PM
I have just read through several of the posts. I don't have time to understand them all right now, but my initial reaction is that only epepke has really understood what I am talking about (that is not to say that I understand exactly what epepke is talking about yet).

Let me say this.
There are two ways we know that something is rotating.
1. We assume some non-rotating frame of reference and notice that the object is moving with respect to the non-rotating frame of reference we assumed.
2. The forces and motions that we calculate and or sense for the object are consistent with Newton's laws of motion in the particular non-rotating frame we assumed.

For instance consider the earth rotating around the sun. How do we know that the earth is rotating around the sun rather than they are both fixed in some frame of reference?

Stick to the carousel in space, or a rotating space station. That's clearer. Something rotating, but held together with wires and girders and stuff so that its bits don't fly off into space.

The situation of a planet revolving around the sun or even rotating about its axis is a bit hairier, because you have to deal with gravity as keeping the system together as well as the gravitational effects of rotation. This gets tricky and confusing almost instantly.

Much better to have the thing kept together with wires. Assume the mass of the carousel or space station is small enough that the gravitational effects of that mass is negligible, so we can concentrate on the gravitational effects of the rotation.

Yllanes
11th April 2006, 01:46 AM
As I said before, maybe you are thinking something similar to Mach's principle, which says roughly (and the 'roughly' is the important part):

The inertia of a system is caused by its interaction with the rest of the universe, i.e., every particle in the universe eventually has an effect on every other particle.

According to Mach, if there were a single object in the universe, it would be impossible to determine whether it was rotating.

Einstein was inspired by these ideas when he developed GR, but the theory turned out to be quite anti-Mach. If we have an universe with a rotating bucket of water (Newton's thought experiment) it would feel centrifugal forces, even if it were otherwise completely empty (according to GR).

The real problem with Mach's principle is that it is so vague as to be be meaningless. You could try for a more precise formulation, but until you translate it to mathematical terms it means nothing. And translating it is not an easy problem. For the initiate, one book that discusses this is Gravitation and Inertia, by Ciufolini & Wheeer.

OK, but my question goes to why there should be a special non-rotating frame at all if the space you are floating in consists of a complete nothingness that is incapable of interacting at all with your ship.

You say, if the universe is otherwise completely empty and we can see no stars, what is the room rotating with respect to? The answer is that it is rotating with respect to the metric of spacetime. If you don't think that the metric is a 'real' concept, like a star, walk into a black hole... They are nothing but pure metric, no matter.

screw_dog
11th April 2006, 06:39 AM
If we are able to observe centripetal forces at work, then we know a system is rotating. If we don't observe any such forces then we know it isn't.

Now, why spacetime works like this, nobody knows, at least from a classical view. (QED, as I've mentioned, explains the geodesics rather nicely, but let's not go there.) What is important is that the mathematics of a rotating frame in a non-rotating universe, and the mathematics of a non-rotating frame in a rotating universe are exactly the same, and so there is no way at all to distinguish between them.
I'm sorry, but these two quotes appear to be in direct opposition.

Which is it? Can we tell if the windowless spaceship we are in is spinning?

Yllanes
11th April 2006, 07:40 AM
I'm sorry, but these two quotes appear to be in direct opposition.

He said a non rotating frame in a rotating universe is no different from a rotating frame in a non rotating universe. You can't tell between those two, so it makes no sense to speak of a rotating universe, but you can tell between a non rotating and a rotating frame in a non rotating universe.

Which is it? Can we tell if the windowless spaceship we are in is spinning?
Yes.

Budric
11th April 2006, 08:25 AM
I have a question about Newton's laws. Someone mentioned that you start getting ficticious forces in a rotating frame of reference. I keep thinking back to an old video in high school where they showed these two guys sitting at a table and pushing an object (ball or something) and it would not go in a straight line. Then the camera zoomed out and you see they're sitting on a platform that's rotating them and the table.

My question is why can I perform experiments at my table and things don't go in weird directions because the earth is rotating around the sun. Is it because the force is very small (large radius)? Or is it just my desk?

Hellbound
11th April 2006, 08:46 AM
Budric:

The force is small. You can see this effect with long pendulums (One of the museums here has one, I wanna say the Smithsonian but I could be wrong).

Actually, the effect you see is due to the Earth rotating on it's axis; the effect from it's rotation around the sun would be so small as to be neglible (I think, not sure here).

You see some of this curving effect in weather patterns on the Earth, as well. THe major movements of air masses are, in part, driven by these effects.

The term for this is Coriolis Force. A quick google should bring you all kinds of info on it :).

Yllanes
11th April 2006, 11:15 AM
Actually, the effect you see is due to the Earth rotating on it's axis; the effect from it's rotation around the sun would be so small as to be neglible (I think, not sure here).
Yes. The centrifugal acceleration due to the rotation is of the order of 0.01*cos d (in m/s/s, where d is the latitude). The effect of the rotation around the Sun is of the order of 0.0001 m/s/s. (Compare to gravity, ~10 m/s/s). It's an easy excercise to calculate the effect due to the rotation of the Solar System around the Galaxy (take 8 kpc as the radius of the orbit and 220 km/s as the speed).

My question is why can I perform experiments at my table and things don't go in weird directions because the earth is rotating around the sun. Is it because the force is very small (large radius)? Or is it just my desk?

To further illustrate the point Huntsman made, notice that the Earth needs a whole day to complete a full rotation, so the effect of the rotation will only be significative in processes that take at least this time. The ball across the table takes a second, a hurricane takes more time. Storms in the Northern (Southern) hemisphere rotate counterclockwise (clockwise) for this reason. The other famous example is Foucault's pendulum, which Huntsman also mentioned. If you have a very big pendulum and let it oscillate, the plane of its motion will slowly rotate (we call this effect 'precession'). This was demonstrated by Foucault 150 years ago and was the first clear proof of the rotation of the Earth.

epepke
11th April 2006, 03:00 PM
I'm sorry, but these two quotes appear to be in direct opposition.

Which is it? Can we tell if the windowless spaceship we are in is spinning?

We are, after all, two different people. You can tell us apart with the naked eye.

Ceptimus, I think, is speaking from a Newtonian + Special Relativity perspective. In that framework, rotation isn't relative, and you can declare (in that framework) that you are spinning.

I am coming from a General Relativity perspective. In that framework, there is no way at all to tell the difference between the forces that you feel from spinning and the forces that you would feel from a gravitational field of the same shape, which would be a configuration of spacetime exactly the same as if space itself (and the associated inertial paths) were spinning around you. So under GR, you can tell that either you are spinning, or the universe is spinning around you, but you cannot tell which. The point is, under GR, it is equally valid to say that spacetime is rotating as it is to say that you are rotating. The mathematics comes out exactly the same no matter which way you do it. So you cannot say that you are absolutely rotating, just that you're rotating relative to spacetime (or it is rotating relative to you).

The point being that under the GR view, more is relative than in SR or Newtonian/Galilean mechanics. In N/G mechanics, velocity in a straight line is relative, and there is no way to build a local experiment that will tell you your velocity, but times and distances are not relative. SR adds optical experiments as well, and shows times and distances to be relative, though acceleration is not relative (with the cost that now the speed of light is now, in some sense, absolute). GR shows acceleration to be relative, in the sense that there is no way to distinguish acceleration from gravitational effects.

ceptimus
11th April 2006, 03:27 PM
Objects in the rotating spaceship will tend to move further apart. It's difficult to see how to arrange gravity to do this, unless you allow negative gravity or negative time.
Also the rotating ship will have an axis of rotation, and the apparent forces on the objects in the ship depend on their distance from the axis. This is difficult to mimic with gravity: gravitational forces tend to act in a 3d spherical manner, rather than the more 2d cylindrical distribution that results from rotation.

epepke
11th April 2006, 04:09 PM
Objects in the rotating spaceship will tend to move further apart. It's difficult to see how to arrange gravity to do this, unless you allow negative gravity or negative time.
Also the rotating ship will have an axis of rotation, and the apparent forces on the objects in the ship depend on their distance from the axis. This is difficult to mimic with gravity: gravitational forces tend to act in a 3d spherical manner, rather than the more 2d cylindrical distribution that results from rotation.

I love these discussions, because I learn more and more about the difficulties in understanding this stuff.

See, within GR, there is no way at all to tell the difference between a rotating spaceship and a non-rotating universe and a non-rotating spaceship in a rotating universe.

I want to be very clear about that: no way at all. I think forgetting that is where people get hung up.

So if you put a spaceship, and you set it spinning, after it's spun up, there is no way at all to tell the difference between two. No way at all. If some aliens with infinite amounts of exotic matter or a warp drive or whatever set spacetime spinning around you, there would be no way at all of telling the difference.

Because there's no way at all, you can't even decide that there is a difference to tell in the first place. You could as easily say that when you span up the spaceship, you were really spinning up spacetime. The cases are in all ways equivalent under GR.

It's whether universe really "is" rotating or not rotating; there is no God with the universe in a little room to tell you the "real" answer. The only answers that exist are measurements from your frame of reference. Nothing else even has any meaning.

Now, it may be emotionally troubling or hard to think about, and one might not be able to think about the universe rotating without drinking a lot of beer first, but GR provides no way whatsoever to tell the difference.

Dark Jaguar
11th April 2006, 04:23 PM
So the reason that model comes to that conclusion is because the model says a spinning universe would basically drag the stuff in our "non-spinning" space ship off to the sides of the ship, and not cause them to sort of veer clockwise/counterclockwise in relation to the "spin" of the universe? That seems like quite a strong gravitational effect for matter that's so far away it's gravity has no detectible effect on us at all (detectable as in every day experience, which being shoved up against the wall of a spinning spaceship would count as).

I'm fully willing to embrace the idea that centrifugal forces are essentially us being dragged around by the universe spinning around us (or whatever frame of reference you'd like, as they would in fact be identical), but it does seem like they are all a little far away to be having such a strong affect, but only in relation to spinning about. Or, is it more like the "straight paths" are determined by the net result of adding up all those gravitational effects, so the old classical "centrifugal force is caused by everything's straight direction constantly being redirected into a loop" is still in effect, but what is a straight path to take is being determined by the entire universe?

epepke
11th April 2006, 04:51 PM
Or, is it more like the "straight paths" are determined by the net result of adding up all those gravitational effects, so the old classical "centrifugal force is caused by everything's straight direction constantly being redirected into a loop" is still in effect, but what is a straight path to take is being determined by the entire universe?

That's exactly correct, as near as I can tell.

The "straight paths" are called geodesics, and they're determined by the geometry of spacetime itself.

Theres a minor wrinkle, though. Minor conceptually, though it makes the mathematics huge. The geodesics are in spacetime, not just in space. It's important, because you might think that something that isn't moving in your frame of reference isn't going to start moving, just because some lines get bent. But it is moving: it's moving through time. Its odometer might not be changing, but its clock is.

Dark Jaguar
11th April 2006, 06:28 PM
Ah, there's an interesting wrinkle! But I think I'll stick with fully wrapping myself around geodesics first. It's just great to finally get a handle on the GR equivilant of centrifugal force.

I'll add it's nice because this way centrifugal forces aren't any "special thing" that has to have it's own solution any more to me. It's now just a natural consequence and just a particular arrangement of things in GR as I had understood it before, but just hadn't properly been applying to that situation.

Art Vandelay
11th April 2006, 07:35 PM
You could say that there is another metric at work, we are no longer in Euclidean space, but in a hyperbolic geometry (notice that we haven't talked about a graviation, just acceleration in special relativity). Well, a metric is a function on the entire space, and here we'll talking about a three-dimensional subspace, so that's not exactly an appropriate term. Also, we're not in a different space, we're in a different frame of reference.

Exercise: Would the angles of a triangle sum more or less than 180º on this platform?The term "triangle" is not well-defined.

If space is a completely empty void then would it be possible to tell the difference between a rotating and a non-rotating frame of reference? Of course not. In order for you to tell the difference between different frames of reference, you must exist. And if you exist, then space is not void (it has you in it).

I am coming from a General Relativity perspective. In that framework, there is no way at all to tell the difference between the forces that you feel from spinning and the forces that you would feel from a gravitational field of the same shape, which would be a configuration of spacetime exactly the same as if space itself (and the associated inertial paths) were spinning around you.Wait- are you saying that gravity normally works like a rotating reference frame, or are saying that it's theoretically possible for there to be a configuration of gravity that acts like a rotating reference frame? I don't think you've been quite clear about this.

GR shows acceleration to be relative, in the sense that there is no way to distinguish acceleration from gravitational effects.Locally.

You say, if the universe is otherwise completely empty and we can see no stars, what is the room rotating with respect to? The answer is that it is rotating with respect to the metric of spacetime. I think that a more direct answer is that it's rotating with respect to itself. If you're in a room, then there isn't just one object in the universe. There's you, the ceiling, the floor, the four walls, etc. All of these are moving with respect to each other.

If you don't think that the metric is a 'real' concept, like a star, walk into a black hole... They are nothing but pure metric, no matter.Are you saying that blacks holes are not made up of matter?

My question is why can I perform experiments at my table and things don't go in weird directions because the earth is rotating around the sun.The Earth is not rotating around the sun. The Earth is revolving around the sun, which isn't quite the same thing. If you look at a rotating table, the ball's route is taking up a significant portion of the table, so the rotation of the table is significant. The table doesn't take up a significant portion of the Earth, so the Earth's rotation doesn't cause significant effects. Hurricanes do take up a significant portion of the Earth, so the Earth's rotation affects them, but the Earth's revolution doesn't. To be affected by the Earth's revolution, something would have to be not merely a significant portion of the Earth, but of the Earth's orbit.

Also, the orbit of the Earth involves gravity, which obviously counterbalances the centrifugal force (otherwise, it wouldn't be an orbit). Locally, the Earth's frame is actually inertial (ignoring the Earth's gravity). You only see the effect of the sun in tidal forces.

The force is small. You can see this effect with long pendulums (One of the museums here has one, I wanna say the Smithsonian but I could be wrong).Actually, I don't think you'd really see it at all. As I said, the rotation and the sun's gravity cancel each other out.

Actually, the effect you see is due to the Earth rotating on it's axis; the effect from it's rotation around the sun would be so small as to be neglible (I think, not sure here).Its.

epepke
11th April 2006, 10:05 PM
Wait- are you saying that gravity normally works like a rotating reference frame, or are saying that it's theoretically possible for there to be a configuration of gravity that acts like a rotating reference frame? I don't think you've been quite clear about this.

It doesn't matter, at all, how gravity normally works. GR is based on the equivalence principle. The effects of acceleration (or, as I said, more accurately, resistance to acceleration) are gravity.

And GR has solutions for every possible configuration. That's why the equations are so big and hard to solve.

Art Vandelay
11th April 2006, 11:59 PM
It doesn't matter, at all, how gravity normally works. GR is based on the equivalence principle. The effects of acceleration (or, as I said, more accurately, resistance to acceleration) are gravity.No, it's not. Gravity is the curvature of space time. Acceleration is the curvature of a path. The priciple of equivalence says that spacetime is locally indistinguishable from a flat spacetime. Come to think of it, it's a rather vacuous statement. The actual spacetime can be locally approximated by an uncurved spacetime. I.e., it's differentiable.

epepke
12th April 2006, 12:59 AM
No, it's not. Gravity is the curvature of space time. Acceleration is the curvature of a path. The priciple of equivalence says that spacetime is locally indistinguishable from a flat spacetime. Come to think of it, it's a rather vacuous statement. The actual spacetime can be locally approximated by an uncurved spacetime. I.e., it's differentiable.

Art, I can only answer questions. I cannot cure your mental disorder, whatever it may be.

Yllanes
12th April 2006, 01:11 AM
Well, a metric is a function on the entire space, and here we'll talking about a three-dimensional subspace, so that's not exactly an appropriate term. Also, we're not in a different space, we're in a different frame of reference.

That doesn't make sense. Of course you can talk about the metric of a rotating disk, or a cylinder, or a sphere, why wouldn't you? And you can talk about the spatial part of the metric of spacetime, what do you think we are doing when we talk about cosmology and the different universes (open,closed, flat)?

We can define a metric in a very general class of manifolds. And a manifold is a very general concept. For example, the configuration space (the set of all posible positions of a system of particles) in classical mechanics is a manifold. The space of all velocities at all points is its tangent bundle and the phase space (momenta and positions) its cotangent bundle. You can very well define a metric, through the mass tensor (kinetic energy), as a diffeomorphism between the two.

The term "triangle" is not well-defined.

Why not? It has three sides, which are straight lines. The straight lines are not the ones you would find on a plane, but that doesn't matter.

Are you saying that blacks holes are not made up of matter?

In a sense they aren't. Once they are formed, they are a singularity and a metric. They are not matter in the sense stars are. Black holes are pure gravitation, one of the reasons theoretical physicists like them so much. Once you have a star, you need an astrophysicist to tell you the equation of state. With black holes there's only geometry to worry about.

Actually, I don't think you'd really see it at all. As I said, the rotation and the sun's gravity cancel each other out.

I don't know what you are saying here. Foucault's pendulum is an observable non inertial effect of the rotation of the Earth around its axis. It doesn't take a significant part of the surface, by the way, it just takes time, which, as I said before, is the important thing.

Hellbound
12th April 2006, 07:02 AM
I don't know what you are saying here. Foucault's pendulum is an observable non inertial effect of the rotation of the Earth around its axis. It doesn't take a significant part of the surface, by the way, it just takes time, which, as I said before, is the important thing.

I think he responded before reading farther in my post, where I pointed out that the effect was due to the Earth's rotation, rather than it's orbit around the Sun.

Yllanes
12th April 2006, 10:25 AM
I think he responded before reading farther in my post, where I pointed out that the effect was due to the Earth's rotation, rather than it's orbit around the Sun.
That's what I thought, but he also said: 'Actually, I don't think you'd really see it at all. As I said, the rotation and the sun's gravity cancel each other out'. He made a point of distinguishing between 'revolving' and 'rotating'.

And, anyway, the effect of the orbit around the Sun is minute, but that doesn't mean it is cancelled by gravity.

davefoc
12th April 2006, 11:24 AM
Well, I am almost off on a little mini-vacation but I wanted to take a moment to post my incomplete thoughts about all this.

I thought Ceptimus did a good job of expressing the reason that if you just think about Newton's laws of motion then it becomes obvious that you don't need any special property of space to figure out when something is rotating and when it is not.

But then I go back to the notion of the merry-go-round rotating, but rotating relative to what? If the universe was rotating at the same rate as the rotating merry-go-round we would judge the merry-go-round as not rotating. It seems difficult to reconcile the two views.

I think now that the reason for the apparent conflict is that we tend to assume that the notion of a straight line is absolute. An object without any forces acting on it moves in a straight line.

But it is actually the universe that controls our ability to judge what is straight. If the universe rotated differently than however it happens to be rotating now we wouldn't know it because we would judge the path of an object without external forces on it as straight in that other hypothetical universe. But in fact the straight in the hypotheical universe would not be the same as straight in the current universe.

I was hoping to think this out a little better before I posted, but I am about to leave so I decided to post now. I think it is very possible that what I have said above as already been said, and I just didn't quite understand it. I also think that maybe I am just wrong here and still don't quite get it.

Art Vandelay
12th April 2006, 12:43 PM
And you can talk about the spatial part of the metric of spacetime,No, you can't. In relativity, there is no such thing as "the" spatial part. What constitutes "the spatial part" depends on your coordinate system.

what do you think we are doing when we talk about cosmology and the different universes (open,closed, flat)?They're talking about the entire spacetime. An open universe, for instance, keeps on expanding forever. "Forever" refers to time.

We can define a metric in a very general class of manifolds. Yes. But if you split spacetime into a bunch of subsets and define metrics on each, you no longer can discuss "the" metric; you'd have say something like "the metric of this subset" rather than just "the metric".

You can very well define a metric, through the mass tensor (kinetic energy), as a diffeomorphism between the two.Huh? Metrics always map to the real numbers (a one dimensional space). Diffeomorphisms always map between manifolds of the same dimension. How can a metric be a diffeomorphism?

Why not? It has three sides, which are straight lines. The straight lines are not the ones you would find on a plane, but that doesn't matter.Sure it does. If they're not the ones you would find on a plane, you would have to define what they are.

In a sense they aren't. Once they are formed, they are a singularity and a metric."Singularity" means that our coordinate system breaks down, and is unable to describe what's inside. That doesn't mean that nothing is inside.

They are not matter in the sense stars are. Black holes are pure gravitation, one of the reasons theoretical physicists like them so much. So black holes cannot hold charge? They don't emit radiation? Is there a requirement that something be more than "pure gravitation" to be matter? Is Dark "Matter" something more than pure gravitation?

I don't know what you are saying here. Foucault's pendulum is an observable non inertial effect of the rotation of the Earth around its axis. I'm saying that practically speaking, there are no observable effects of the revolution around the sun in the Earth's reference frame.

Art Vandelay
12th April 2006, 12:51 PM
That's what I thought, but he also said: 'Actually, I don't think you'd really see it at all. As I said, the rotation and the sun's gravity cancel each other out'. He made a point of distinguishing between 'revolving' and 'rotating'.Yeah, I messed up. That should be "revolution".

And, anyway, the effect of the orbit around the Sun is minute, but that doesn't mean it is cancelled by gravity.Locally. If you're displaced from the center of gravity of the system(which we are), then minute effects show up. If you were at the exact center of gravity, there would be no effects.

69dodge
12th April 2006, 01:29 PM
I think that a more direct answer is that it's rotating with respect to itself. If you're in a room, then there isn't just one object in the universe. There's you, the ceiling, the floor, the four walls, etc. All of these are moving with respect to each other.They are? For each pair of objects, the distance between the two remains the same as time passes. So in what sense is anything moving with respect to anything else? Only in the dynamical sense that centrifugal force is observed to exist, but not in any purely kinematical sense.

This is davefoc's question. How does the universe "know" that it should apply centrifugal force, if everything looks exactly the same as a nonrotating room where it shouldn't?

(Of course, we don't have a nearly empty universe to experiment with, so we don't actually know that centrifugal force could exist in that situation. Mach would say it wouldn't exist, I assume.)

Dark Jaguar
12th April 2006, 01:44 PM
I suppose in that hypothetical, you'd have to ask how you would get something to start spinning. I can only think of setting up two rockets on opposite sides aimed in opposite directions. Then however, you suddenly have something to be rotating relative to, the spent rocket fuel. If you decided to kick start it by literally kicking it at an angle to get it spinning, you are what it spins relative to. Essentially, the only way that thing could be spinning in this nearly null space is to toss something overboard for it to spin relative to, or set an engine inside of it that spins up itself, but again then you have two frames of reference. At any rate, from what I've been told here, if all you have is that single object, it is impossible for it to be considered "spinning" and centrifugal forces won't happen since that's the only object determining what is and is not a straight line.

69dodge
12th April 2006, 01:45 PM
Objects in the rotating spaceship will tend to move further apart. It's difficult to see how to arrange gravity to do this, unless you allow negative gravity or negative time.
Also the rotating ship will have an axis of rotation, and the apparent forces on the objects in the ship depend on their distance from the axis. This is difficult to mimic with gravity: gravitational forces tend to act in a 3d spherical manner, rather than the more 2d cylindrical distribution that results from rotation.Newtonian gravity couldn't do it, but "gravity" in general relativity can be very different from Newtonian gravity, although in common cases they're close.

Standing still while the entire universe rotates around you is very uncommon. :D

Although I'm not quite sure how to make sense of "the entire universe rotates around you" when there isn't anything in the universe other than you . . .

Ziggurat
12th April 2006, 01:47 PM
"Singularity" means that our coordinate system breaks down, and is unable to describe what's inside. That doesn't mean that nothing is inside.

To nitpick slightly, here's the relevant definition (from http://www.m-w.com/dictionary/singularity )

3 : a point at which the derivative of a given function of a complex variable does not exist but every neighborhood of which contains points for which the derivative exists

The event horizon for a black hole is a singularity for the Schwartzchild metric, but is not for other metrics used to describe the same kind of black hole. It is therefore only a coordinate singularity (as is the origin for Euclidean space using polar coordinates), since changing coordinate systems can make it (the singularity of the event horizon, not the event horizon itself) vanish.

The center of a black hole is a singularity because the curvature is infinite. Because this singularity in the curvature exists regardless of the metric you use, it is considered a physical singularity. I wouldn't quite say that our coordinate system "breaks down", nor would I say there's any "inside" to describe (that is, assuming GR is correct and doesn't need any modifications on these length scales).

Jekyll
12th April 2006, 02:49 PM
Sure it does. If they're not the ones you would find on a plane, you would have to define what they are.

A triangle always refers to the shortest lines connecting 3 separate points. It is well defined on every metric.

Yllanes
12th April 2006, 03:11 PM
No, you can't. In relativity, there is no such thing as "the" spatial part. What constitutes "the spatial part" depends on your coordinate system.

They're talking about the entire spacetime. An open universe, for instance, keeps on expanding forever. "Forever" refers to time.

You are wrong. A closed universe is spatially finite but may go on forever in time (with a cosmological constant). In fact, our universe is probably spatially finite (closed) but will expand forever. Are you familiar with the concept of submanifold?

Yes. But if you split spacetime into a bunch of subsets and define metrics on each, you no longer can discuss "the" metric; you'd have say something like "the metric of this subset" rather than just "the metric".
That doesn't make sense. Tell me why can't I talk of he metric of a 2-dimensional rotating disk?

Huh? Metrics always map to the real numbers (a one dimensional space). Diffeomorphisms always map between manifolds of the same dimension. How can a metric be a diffeomorphism?
A metric in a differentiable manifold automatically defines a diffeomorphism between the tangent and cotangent bundles. If you do not know this basic concept of differential geometry, I don't know why you keep arguing. If you plug in two vectors, it gives a real number. If you plug in one vector, it gives a 1-form. This is the same as saying, informally, that a metric serves to lower and raise indices. p_a = g_ab q'^b

Sure it does. If they're not the ones you would find on a plane, you would have to define what they are.
It was quite clear what they were in the example I gave.

"Singularity" means that our coordinate system breaks down, and is unable to describe what's inside. That doesn't mean that nothing is inside.
There are no magnetohyrodynamics, no fusion, no convection, etc. This is what I meant. And, as Ziggurat pointed out, the centre is a physical singularity, with our current understanding.

I'm saying that practically speaking, there are no observable effects of the revolution around the sun in the Earth's reference frame. This is correct, althought, as you point out in your next post, you were not clear at one point and I misunderstood you.

Yllanes
12th April 2006, 03:19 PM
Locally. If you're displaced from the center of gravity of the system(which we are), then minute effects show up. If you were at the exact center of gravity, there would be no effects.

But we are not, so this is not relevant to the inertial or non inertial character of the Earth's reference frame. And if you are at the CM, it is not a question of gravity 'cancelling' the inertial forces, there just aren't any.

Ziggurat
12th April 2006, 03:30 PM
A triangle always refers to the shortest lines connecting 3 separate points. It is well defined on every metric.

I agree that it's well defined, but your definition isn't actually quite the definition you want: anything that satisfies it is certainly a triangle, but I think you can get triangles which don't satisfy that. What you really want is a generalization of a straight line other than merely the shortest distance between two points, and that generalization is usually refered to as a geodesic. A geodesic is always "locally" the shortest path between two points, but not necessarily globally. A simple example to get this distinction across is that you can't define the shortest path between the north and south poles, but you can rather easily define geodesics between the two (basically every longitude line).

For the question of triangles, here's how to construct a triangle which meets the geodesic definition but not the definition as you stated it. Consider a sphere, with three non-colinear points (A, B and C) mostly on one side of the surface. The shortest lines connecting them (call them AB, BC, and CA) certainly form a triangle, and matches your definition. But consider taking the line from A to B (AB) and forming a great circle around the sphere from that line. Now this greater circle can be broken into two line segments, one of which is short (AB) and one of which is long (I'll now call BA). The three lines BA, BC, and CA also form an object that I think should be called a triangle - it fits our intuitive definition (three straight lines connecting three points) because a great circle is the equivalent to a straight line on a sphere, but it doesn't match your definition. It does, however, match the geodesic definition.

Art Vandelay
12th April 2006, 04:36 PM
They are? For each pair of objects, the distance between the two remains the same as time passes. So in what sense is anything moving with respect to anything else?So you are saying that "motion" only refers to change in distance between two points?

The event horizon for a black hole is a singularity for the Schwartzchild metric, but is not for other metrics used to describe the same kind of black hole. Well, according to your definition, it's a collection of singularities.

The center of a black hole is a singularity because the curvature is infinite.Of course, that's true of everything, if you consider particles to be point-masses.

I wouldn't quite say that our coordinate system "breaks down", nor would I say there's any "inside" to describe (that is, assuming GR is correct and doesn't need any modifications on these length scales).Well, a single point doesn't have an "inside", regardless of the curvature.

A triangle always refers to the shortest lines connecting 3 separate points. It is well defined on every metric.But now you have to give a metric. Furthermore, "line" has to be defined, and "shortest line" is not guaranteed either existence or uniqueness. In a non-convex space, there are pairs of points with no lines connecting them.

That doesn't make sense. Tell me why can't I talk of [the?] metric of a 2-dimensional rotating disk? You can. You just can't define a metric for a 2-dimensional rotating disk, then refer to it as "the" metric when discussing the 3-dimensional space in which the disk exists.

A metric in a differentiable manifold automatically defines a diffeomorphism between the tangent and cotangent bundles.There's a bit of a difference between saying that a metric defines a diffeomorphism, and a metric is a diffeomorphism. Furthermore, I may indeed be deficient in my understanding of the math, but I don't see how a metric defines a diffeomorphism.

First of all, a metric applies to one space, so if you're trying to set up a diffeomorphism, don't you need two metrics (one for each space)? Secondly, let's consider the following situation:

You have two spaces, X and Y, and diffeomorphisms between each and the real numbers:
f: X->R
g: Y->R

There are obvious metrics d1, d2:
d1(x1,x2)=|f(x1)-f(x2)|
d2(y1,y2)=|g(y1)-g(y2)|

So now it looks like this defines a diffeomorphism
h: X->Y
h(x)=g-1(f(x))

But isn't h(x)=g-1(-f(x))
also a diffeomorphism? What's wrong with my thinking?

It was quite clear what they were in the example I gave.No, it wasn't.

Jekyll
12th April 2006, 05:05 PM
For the question of triangles, here's how to construct a triangle which meets the geodesic definition but not the definition as you stated it. Consider a sphere, with three non-colinear points (A, B and C) mostly on one side of the surface. The shortest lines connecting them (call them AB, BC, and CA) certainly form a triangle, and matches your definition. But consider taking the line from A to B (AB) and forming a great circle around the sphere from that line. Now this greater circle can be broken into two line segments, one of which is short (AB) and one of which is long (I'll now call BA). The three lines BA, BC, and CA also form an object that I think should be called a triangle - it fits our intuitive definition (three straight lines connecting three points) because a great circle is the equivalent to a straight line on a sphere, but it doesn't match your definition. It does, however, match the geodesic definition.
This doesn't meet the geodesic definition because the large piece of the great circle is not at a mimima. In fact, it is the largest curve that doesn't 'wriggle'.
But I take your point about the purality of solutions and that you can define them in terms of local rather than global solutions. You'll just have to use a more complex example next time :D .

Art Vandelay
12th April 2006, 05:09 PM
This doesn't meet the geodesic definition because the large piece of the great circle is not at a mimima. But being a minimum (minima is the plural) is not part of the definition of a geodesic. In fact, geodesics can be (local) maxima.

Jekyll
12th April 2006, 05:27 PM
But being a minimum (minima is the plural) is not part of the definition of a geodesic. In fact, geodesics can be (local) maxima.
Not according to what I'm reading:
http://mathworld.wolfram.com/Geodesic.html
"A geodesic is a locally length-minimizing curve."
http://en.wikipedia.org/wiki/Geodesic#Metric_geometry
"In metric geometry, a geodesic is a curve which is everywhere locally a distance minimizer"

What definition are you working from?

Edit: Oh, if the tangental component of the second derivitive of the path is zero. Never mind.

69dodge
12th April 2006, 07:18 PM
So you are saying that "motion" only refers to change in distance between two points?I don't see what else it could mean.

How can something move relative to itself? If it is taken as the frame of reference relative to which motion is measured, then by definition it's not moving.

Art Vandelay
13th April 2006, 12:14 AM
I don't see what else it could mean.It means changing location. Suppose ABC is an equilateral triangle with sides of 10 miles. If a cop is at point A, and sees a car go from point B to point C in 5 minutes, do you think he's going to say "well, he's not moving at all, because his distance from me didn't change"?

Not according to what I'm reading:
http://mathworld.wolfram.com/Geodesic.html
"A geodesic is a locally length-minimizing curve."
http://en.wikipedia.org/wiki/Geodesic#Metric_geometry
"In metric geometry, a geodesic is a curve which is everywhere locally a distance minimizer"Depends on what you mean by "locally". Also, spacetime actually doesn't have a metric; it has a psuedometric, and because of this, all geodesics have maximum subjective time. Certainly a great circle is a geodesic. Do you have any definition that excludes it?

Yllanes
13th April 2006, 02:26 AM
defines[/i] a diffeomorphism, and a metric is a diffeomorphism. Furthermore, I may indeed be deficient in my understanding of the math, but I don't see how a metric defines a diffeomorphism.
Do you know about dual spaces? We have V, the space of vectors and V*, the space of linear applications V -> R. We say V* is the dual space to V and we can define a dual basis. But these spaces are not canonically isomorphic, the isomorphisms you can define between them depend on the basis. If you are in a riemannian manifold, the metric automatically is a base independent isomorphism. It is also a diffeomorphism if everything is smooth (a diffeomorphism is an application such that both the application and its inverse are differentiable).

There's no point arguing that a metric is a diffeomorphism between the tangent and cotangent bundles, because this is basic geometry, not something I have just invented. How much do you know about geometry and analysis? I can't explain these things unless I know where to start.

First of all, a metric applies to one space, so if you're trying to set up a diffeomorphism, don't you need two metrics (one for each space)?
No. A metric has two slots g(·,·) if you fill both of them you get a number. If you fill one of them you get a covariant vector. Its inverse, g-1, has also two slots, to be filled with covariant vectors. If you only fill one you get a contravariant vector. Haven't you seen this kind of expressions:

gabva=vb (TM -> T*M)
gcdvc=vd (T*M -> TM)

Whe I say the metric is an isomorphism I mean the formulae above. The metric raises and lowers indice, and the relation va -> va is unambiguous once you have a metric. That's the second thing you learn about a metric (the first being the bit about the scalar product). Of course, gac is the inverse application.

Secondly, let's consider the following situation:

You have two spaces, X and Y, and diffeomorphisms between each and the real numbers:
f: X->R
g: Y->R

There are obvious metrics d1, d2:
d1(x1,x2)=|f(x1)-f(x2)|
d2(y1,y2)=|g(y1)-g(y2)|
Those seem to be distances, rather than metrics. d1(x1,x1) = 0, for all x1. This is not a metric, every vector would have length 0.

So now it looks like this defines a diffeomorphism
h: X->Y
h(x)=g-1(f(x))

But isn't h(x)=g-1(-f(x))
also a diffeomorphism? What's wrong with my thinking?
I'm sorry, I don't understand what you are saying here, or how it is relevant to the metric.

That doesn't make sense. Tell me why can't I talk of [the?] metric of a 2-dimensional rotating disk?

You can. You just can't define a metric for a 2-dimensional rotating disk, then refer to it as "the" metric when discussing the 3-dimensional space in which the disk exists.

The 3-d space doesn't matter. The third coordinate doesn't add anything special, we can concentrate on the other two. If you think of the 3D metric as a 3x3 matrix, we can talk about the 2x2 submatrix, which is the interesting part. The disk could be embedded in a 6-dimensional space, all we wanted to talk about is the 2 relevant dimensions. And I can very well talk about *THE* metric.
http://www.randi.org/latexrender/latex.php?%3Cbr%20/%3E%0A%5Cleft%28%5Cbegin%7Barray%7D%7Bcccccc%7D%3C br%20/%3E%0A1&0&0&0&0&0%5C%5C%3Cbr%20/%3E%0A0&1&0&0&0&0%5C%5C%3Cbr%20/%3E%0A0&0&1&0&0&0%5C%5C%3Cbr%20/%3E%0A0&0&0&1&0&0%5C%5C%3Cbr%20/%3E%0A0&0&0&0&a&b%5C%5C%3Cbr%20/%3E%0A0&0&0&0&c&d%3Cbr%20/%3E%0A%5Cright%29%5Cend%7Barray%7D%3Cbr%20/%3E

Why do I need to carry all first 4 trivial rows? If the only part that can change is the 2x2 submatrix a,b,c,d, then I can refer to it as the metric. And I don't know how I have been dragged into this. If you are on a rotating disk, you don't have an Euclidean metric, you have a different thing. That's all I said.

69dodge
13th April 2006, 12:18 PM
[Motion] means changing location.That's fine, but how do you know what an object's location is at various times, so that you can decide whether it has changed location? Only by seeing how far away it is, at those times, from other things that are assumed not to be moving.

Suppose ABC is an equilateral triangle with sides of 10 miles. If a cop is at point A, and sees a car go from point B to point C in 5 minutes, do you think he's going to say "well, he's not moving at all, because his distance from me didn't change"?Don't you want BC to be an arc of a circle centered at A, instead of a side of a triangle? But, anyway, the cop will say that the car moved because its distance from point A changed. Looking only at its distance from a single point (the cop) isn't enough, except in one dimension. In two dimensions, you need two points; in three, three.

69dodge
14th April 2006, 06:54 AM
But, anyway, the cop will say that the car moved because its distance from point A changed.Oops. I meant point B, of course.

Art Vandelay
14th April 2006, 02:31 PM
There's no point arguing that a metric is a diffeomorphism between the tangent and cotangent bundles, because this is basic geometry, not something I have just invented.Well, actually, there is a point arguing it. You are thinking of a metric tensor. A metric is a function d: (X cross X)->R.

Those seem to be distances, rather than metrics. d1(x1,x1) = 0, for all x1. This is not a metric, every vector would have length 0.Yes, that's what a metric is, a distance function. The distance between any vector and itself is indeed 0.

The 3-d space doesn't matter. The third coordinate doesn't add anything special, we can concentrate on the other two. If you think of the 3D metric as a 3x3 matrix, we can talk about the 2x2 submatrix, which is the interesting part. So is g(a,b) defined if a is in this 2D subspace, but b is not?

The disk could be embedded in a 6-dimensional space, all we wanted to talk about is the 2 relevant dimensions. And I can very well talk about *THE* metric.The example you give is a metric tensor for the whole space, and therefore isn't a counterexample to my claim that if you have a metric (I said metric, but it also applies to metric tensors) for just the subspace, then you cannot call it "the" metric of the space.

Why do I need to carry all first 4 trivial rows?You don't have to. You could, for, instance, express it as:
I4 0 0
0 a b
0 c d

You just have to remember that those other rows and columns exist.

If you are on a rotating disk, you don't have an Euclidean metric, you have a different thing. That's all I said.Well, more precisely, if you define your reference frame so that the disk is at rest, you have a different metric tensor.

My real problem with what you said is that you implied that it is some special feature of relativity that results in such wacky things as C !=2 pi r. But there's no need to involve relativity. Suppose you look at normal 3D space, with the normal, Euclidean metric tensor. No, let's look at a "circle" of radius r. If we define a "circle" to be the set of all points a constant distance from the center, then this "circle" is actually a sphere. Okay, now let's suppose that we're only going to discuss the set of points comprising a cone with an apex which coincides with the center of that sphere. The points on that cone are the only points in our new space. Now, I'm not doing anything with the metric tensor. It's still the old Euclidean metric tensor, just restricted to the cone. If you look at all the points in our new space that are a distance of r from the apex, this now will be a circle (or two, depending on how you define "cone"). But the circumfrance of the circle will not be 2 pi r. Once you play around with what "circle" and "radius" mean, it's rather easy to "prove" that C!=2 pi r, even without relativity.

That's fine, but how do you know what an object's location is at various times, so that you can decide whether it has changed location? Only by seeing how far away it is, at those times, from other things that are assumed not to be moving.In 3D space, the x coordinate in determined by the distance to the yz plane. The yz plane is not a "thing" in the sense of a material object.

But, anyway, the cop will say that the car moved because its distance from point A changed. "point A' is a completely abstract concept. If you're going to allow it as a reference point, why not allow abstract coordinates?

Yllanes
14th April 2006, 03:11 PM
Well, actually, there is a point arguing it. You are thinking of a metric tensor. A metric is a function d: (X cross X)->R.

Yes, that's what a metric is, a distance function. The distance between any vector and itself is indeed 0. We are talking about relativity. In GR and differential geometry metric <-> g. There is no ambiguity. The point is that you don't seem to know what g is. Look at my original post:

You can very well define a metric, through the mass tensor (kinetic energy), as a diffeomorphism between the two.

And you said:
Huh? Metrics always map to the real numbers (a one dimensional space). Diffeomorphisms always map between manifolds of the same dimension. How can a metric be a diffeomorphism?
If you knew what I was talking about, there would have been no ambiguity. I said 'You can define a metric, through the mass tensor'. This is in every book about the mathematics of theoretical mechanics or differential geometry. It is basic that the phase space is a symplectic manifold and so on... That paragraph was not of course aimed at explaining all those concepts, I just wanted to make clear that spacetime is not the only useful manifold in physics.

The distance d makes absolutely no sense here. Can you substract two vectors on different points on a curved manifold? Answer: no you cannot.

Everything goes through g. What is the distance between two points on a curved surface? Hint: if we define the distance as the arc length of a geodesic joining them, the result involves g.

I'm going to drop the whole 'metric of the subspace' thing because it is waste of time. The point is that you can talk about the metric of a rotating disk and it makes no difference in how many dimensions the disk is embedded.

My real problem with what you said is that you implied that it is some special feature of relativity that results in such wacky things as C !=2 pi r. But there's no need to involve relativity.
[example snipped]

Yes, but you are living on a cone, I don't have any need for that. The only thing I said is that on a rotating frame, our space is no longer Euclidean. What you are saying is that other spaces are not Euclidean.

69dodge
14th April 2006, 04:09 PM
In 3D space, the x coordinate in determined by the distance to the yz plane. The yz plane is not a "thing" in the sense of a material object.Yes, exactly. So, in the real world, how do you experimentally determine what the distance of an object to the yz plane is?

"point A' is a completely abstract concept. If you're going to allow it as a reference point, why not allow abstract coordinates?Sorry, I wasn't clear. The cop doesn't think about the point abstractly. He notices that the car moves away from some concrete object at the point.

But nothing in a rotating room moves away from anything else, if the universe is otherwise empty.

Art Vandelay
14th April 2006, 08:06 PM
Yes, exactly. So, in the real world, how do you experimentally determine what the distance of an object to the yz plane is?You don't. How do you experimentally determine that something is at rest?

Sorry, I wasn't clear. The cop doesn't think about the point abstractly. He notices that the car moves away from some concrete object at the point.Strictly speaking, he notices that the distance between them increases. That the car is moving an assumption. For that matter, how does he determine that the distance between them increases? What if this takes place in the middle of the desert, and there are no landmarks? Will the cop be able to know how fast the car's going?

But nothing in a rotating room moves away from anything else, if the universe is otherwise empty.So then there's no paradox. Nothing is moving away from each other, there's no measureable difference between it and a room at rest, and everything makes sense.

This is in every book about the mathematics of theoretical mechanics or differential geometry.Where can I find it in Do Carmo's Differential Geometry of Curves and Surfaces?

The distance d makes absolutely no sense here. Can you substract two vectors on different points on a curved manifold? Answer: no you cannot.Of course not. They're not in the same vector spaces. I already explained this. For d(x,y) to make sense, x and y have to be from the same metric space. You can define d(x,y) where x and y are different points on a manifold, or you can define a different d(x,y) where x and y are both members of the tangent bundle of a particular point.

Is g defined if the two vectors are from different points?

What is the distance between two points on a curved surface? Hint: if we define the distance as the arc length of a geodesic joining them, the result involves g.How do you define "geodesic"? What if there are no geodesic joining them? Or more than one?

The point is that you can talk about the metric of a rotating disk and it makes no difference in how many dimensions the disk is embedded.I've already addressed this.

What you are saying is that other spaces are not Euclidean.No, I'm saying you don't have to go to a non-Euclidean space. The cone is Euclidean.

Yllanes
15th April 2006, 03:53 AM
Where can I find it in Do Carmo's Differential Geometry of Curves and Surfaces?
You can't, I'm afraid. That book is only about curves and surfaces and I was talking about more general manifolds. You would need a book on theoretical mechanics (Arnold's Mathematical Methods of Classical Mechanics is the classic, there are also several, quite advanced, books by Marsden) or a more advanced book on differential geometry (Spivak's, Choquet-Bruhat's, the list is endless). A very good book on differential geometry and applications in physics (a lot of applications) is Frankel's The Geometry of Physics. Great book, doesn't assume any previous knowledge of differential geometry and is not as painstakingly abstract as others, but goes very far and touches many, many topics. I heartily recommend this to anyone with some mathematical and physical knowledge who wants to go further.

Of course not. They're not in the same vector spaces. I already explained this. For d(x,y) to make sense, x and y have to be from the same metric space. You can define d(x,y) where x and y are different points on a manifold, or you can define a different d(x,y) where x and y are both members of the tangent bundle of a particular point.
Yes. But the tangent space at each point is isomorphic to R^n, so the distance between two vectors there is not very interesting. By the way, so there's no misunderstanding, the tangent bundle is the collection of pairs (p,v), where p is a point and v a vector on that point. In other words, the tangent bundle is roughly the collection of all tangent spaces. Vector fields live on the tangent bundle and 1-form fields live on the cotangent bundle. On a riemannian manifold, there is a diffeomorphism between the two bundles, the metric. But even if we don't have a metric, those spaces still exist (and are manifold of dimension 2n, if n is the dimension of the original manifold).

Is g defined if the two vectors are from different points?
No, in the sense that you cannot do g(u,v), if u and v are on different points, because g is a function of the point. But g is a tensor field, so it is defined everywhere. If u(x) and v(x) are also vector fields, then g(u(x),v(x)) is defined everywhere.

How do you define "geodesic"? What if there are no geodesic joining them? Or more than one?

This is not important for what I was saying. I just wanted to illustrate that the metric is a field, defined continously and differentiably on all the manifold. The distance is no such thing. Sometimes it is easy to talk about the distance between two points, as the arc length of a geodesic joining them.

For example, let's talk about surfaces. You know about that if you have read Do Carmo's. There are several (equivalent) definitions of geodesic on a surface. For instance, a curve is a geodesic if it can be parameterised with acceleration always perpendicular to the tangent plane at each point. Or we can say that a curve is a geodesic if the tangent plane to M is always perpendicular to its osculating plane. Or a curve with extremal arc length (not always a local minimum, as you pointed out earlier). You can also talk about covariant derivatives, Christoffel symbols and geodesic curvature, though I don't remember if that book covered these things. Anyway you define it, if x is a point of M and v a vector on the tangent plane at x, there's always one (and only one) geodesic that passes through x with velocity v. So the concept is well defined.

To summarise, because we have hijacked this thread (sorry, davefoc). The fundamental notion of metric in differential geometry is g, the metric tensor, not the distance. If we want to define some kind of distance, chances are we will do it through g. The metric tensor g is 'useful' for many other things. It is a tensor field, provides a diffemorphism between the tangent and cotangent bundles (=it serves to raise and lower indices), determines the curvature, etc. Einstein's equations say G_mv = k T_mv (k is a unit-dependant constant). T is the amount of energy-momentum and G is a tensor defined through the Riemann tensor which is in turn a function only of the metric tensor. I probably should have been more careful when talking about 'metric', I didn't think it would give raise to misunderstandings. I will say always 'metric tensor' in the future.

We were talking about different things with our 'metric'. Can we give it a rest now? Of course, if you want to ask something about differential geometry, feel free to do it, but let's not keep on arguing about the difference between d <-> g.

Art Vandelay
15th April 2006, 11:25 AM
This is not important for what I was saying. I just wanted to illustrate that the metric is a field, defined continously and differentiably on all the manifold. The distance is no such thing. Actually, you're now using another term with more than one meaning: field. There's the physics meaning, and the math meaning. It looks like you're using the physics meaning. Anyway, you can't talk about continuity without there being some topology defined. For a metric space, the open ball topology is canonical, which makes d continuous by definition.

For example, let's talk about surfaces. You know about that if you have read Do Carmo's. There are several (equivalent) definitions of geodesic on a surface. For instance, a curve is a geodesic if it can be parameterised with acceleration always perpendicular to the tangent plane at each point. Or we can say that a curve is a geodesic if the tangent plane to M is always perpendicular to its osculating plane. Or a curve with extremal arc length (not always a local minimum, as you pointed out earlier). You can also talk about covariant derivatives, Christoffel symbols and geodesic curvature, though I don't remember if that book covered these things.Do Carmo takes d to be given, and bases the definitions on that. For instance, the term "extremal arc length" is defined in terms of arc length, and arc length is defined in terms of distance. Acceleration and orthogonality are also tied to distance. So if you want to define distance in terms of geodesics, you must find a way to define geodesics that doesn't rely on a preexisting d, otherwise it's a circular definition.

Anyway you define it, if x is a point of M and v a vector on the tangent plane at x, there's always one (and only one) geodesic that passes through x with velocity v. So the concept is well defined.Yes, but if you want the distance between point x and point y, you are guaranteed neither existance nor uniqueness of a geodesic passing through both. "The distance between x and y is the arc length of a geodesic from one to the other" therefore is not a valid definition.

Yllanes
15th April 2006, 12:08 PM
Actually, you're now using another term with more than one meaning: field. There's the physics meaning, and the math meaning. It looks like you're using the physics meaning.
No, I'm not. Field means several things in mathematics. It can be field as in 'the complex field' or it can be a vector/tensor field. Check a book on differential geometry by a pure mathematician and you'll see that they also use it. It's not my fault that sometimes the same word has more than one meaning.

Anyway, you can't talk about continuity without there being some topology defined.
A topological manifold is a space locally homeomorphic to R^n. Of course a topology is defined. A differentiable manifold has even more structure, and a riemannian manifold even more.

For a metric space, the open ball topology is canonical, which makes d continuous by definition As I said I'm not going to discuss this further, because it's pointless. The distance you are talking about is not a tensor field.

So if you want to define distance in terms of geodesics, you must find a way to define geodesics that doesn't rely on a preexisting d, otherwise it's a circular definition. Neither of my definitions were circular. Take the first one 'a curve is a geodesic if it can be parameterised with acceleration always perpendicular to the tangent plane at each point'. How does this rely on a distance? I don't recall how Do Carmo does all this. I didn't like that book when I was studying classical differential geometry, so I'm not familiar with it.

Yes, but if you want the distance between point x and point y, you are guaranteed neither existance nor uniqueness of a geodesic passing through both. "The distance between x and y is the arc length of a geodesic from one to the other" therefore is not a valid definition. Sometimes it is, that's all I said. It was just an example. Take a cylinder. It is isometric to the plane. It's geodesics are the isometric images of straight lines (circles, helices and straight lines). The only geodesic joining two points on a cylinder (and there always is one) is the image of the straight lines joining their inverse images on the plane. Its arc length is just the length of the line, the Euclidean distance between the inverse images of the points. We know they are isometric because they have the same metric tensor. And, again, I don't care for the definition of distance between two points on a manifold, it was just an example.

If you want, I can define for you the intrinsic distance between two points. Maybe this is the concept you are always talking about. Let a be a differentiable map a: I=[0,1] -> R^3. We define its length as the integral L(a) = \int_0^1 |a'(t)| dt. Notice that this definition corresponds to the arc length of a curve. Now, if M is a surface and p,q two points on M, we call S_{pq} to the set of all curves a: I -> R^3 such that a(0)=p, a(1)=q and a(I) is contained in M. With all this, we can define the intrinsic distance between two points as the infimum of the L(a) : a in S_{pq}. Is this the definition you are using? The cylinder example works with this definition. If it is, you will agree with me that this distance is not a function of the point, not a field and so it is not the metric we talk about in GR. Notice further that it doesn't act on vectors, but is a function of two points.

SkepDick
15th April 2006, 02:31 PM
Not to change the subject, but I have a question. Mach's Principle suggests to me that in the early universe, when all the matter was closer togeather than it is today, inertial forces were either greater or smaller. If the strength of inertial forces is, in fact, changing with time, would this have a significant impact on the large-scale evolution of the universe?

Yllanes
15th April 2006, 03:30 PM
Not to change the subject, but I have a question. Mach's Principle suggests to me that in the early universe, when all the matter was closer togeather than it is today, inertial forces were either greater or smaller. If the strength of inertial forces is, in fact, changing with time, would this have a significant impact on the large-scale evolution of the universe?
Take into account that Mach's principle, in its original form, is too vague to really mean anything solid. Furthermore, if you try to formulate it precisely you may get false results. You can try to get around this by saying, as Einstein did, that it only holds in closed universes. Here's an excerpt from the introduction of Wald's General Relativity, the favourite book of relativity workers:

The second much less precise set of ideas which motivated the formulation of general relativity goes under the name of Mach's principle. In special relativity as in prerelativity notions of spacetime, the structure of spacetime is given once and for all and is unaffected by the material bodies that may be present. In particular,"inertial motion" and "nonrotating" are not influenced by matter in the universe. Mach [...] found this idea unsatisfactory. Rather, Mach felt that all matter in the universe should contribute to the local definition of "nonaccelerating" and "nonrotating"; that in a universe devoid of matter there should be no meaning of these concepts. Einstein accepted this idea and was strongly motivated to seek a theory where, unlike special relativity, the structure of spacetime is influenced by the presence of matter.

The new theory of space, time, and gravitation -general relativity- proposed by Einstein states the following: The intrinsic, observer-independant, properties of spacetime are described by a spacetime metric, as in special relativity. However, the spacetime metric need not have the (flat) form it has in special relativity. [...] Furthermore, the curvature of spacetime is related to the stress-energy-momentum tensor of the matter in spacetime via an equation postulated by Einstein. In this way, the structure of spacetime [...] is related to the matter content of spacetime, in accordance with some (but not all!) of Mach's ideas.

The point is that GR seems to agree sometimes with this principle and sometimes it doesn't. But GR is a much better tool to make predictions than Mach's principle ever was, so the best thing is to concentrate on what GR says. The mass-energy-momentum content is represented by a tensor T, which determines the structure of spacetime. This is what GR says, and it has worked perfectly thus far.

SkepDick
15th April 2006, 04:00 PM
I was really thinking about the main idea in Mach's Principle, i.e., that inertia is a property of spacetime (and the distribution of matter) rather than a property of matter. This is a facinating idea because it leaves open the possibility that inertial forces can change with time. If this were the case, there might be (I can't say how) effects on the evolution of the universe which have left traces that are observable today, ruling this idea in or out. I would advise readers that I am approaching the limits of my understanding of these concepts but would like to hear the thoughts of anyone more familiar with them than I.

Art Vandelay
15th April 2006, 06:19 PM
It's not my fault that sometimes the same word has more than one meaning.Wasn't saying it was. Just something to be aware of.

Neither of my definitions were circular. Take the first one 'a curve is a geodesic if it can be parameterised with acceleration always perpendicular to the tangent plane at each point'. How does this rely on a distance? I don't recall how Do Carmo does all this. Do Carmo mainly considers surfaces to be embedded in R^3, and pretty much everything depends on the inherent Euclidean metric of R^3. If R^3 didn't already have a metric, most of the definitions would be meaningless.

If you want, I can define for you the intrinsic distance between two points. Maybe this is the concept you are always talking about. Let a be a differentiable map a: I=[0,1] -> R^3. This is particular type of metric. It takes the space, embeds it in a space that already has a metric, and then derives a new metric from the parent space. It reminds me of the time Homer Simpsons says "I carved this spoon myself. From a bigger spoon." If you're talking about a completely abstract manifold, then this option is not available to you. (And, of course, it assumes that the space is path connected.)

Is this the definition you are using?It's an example of such a metric. It's not the only possibility. In general, a metric is any function d(x,y):(X cross X)->R where
d(x,y) = 0 iff x=y
d(x,y) > 0 otherwise
d(x,y) = d(y,x)
d(x,y) + d(y,z) >= d(x,z)

The properties of a manifold are largely determined by this metric. You can put a bunch of points on an open hemisphere and another bunch on a plane, for instance, and define metrics on each such that there's a homeomorphism that preserves the metrics. So it's not really how you label the points, but how you define the metric that determines the surface. You can label the points as if they're on a hemisphere, but if you define a flat metric, the set's going to be a plane, for all intents and purposes. Conversely, you can (locally) parameterize any two dimensional manifold with two variables, making it seem like a plane, but it's the metric that's going to differentiate it from an actual plane. If, as in your example, you give a cylinder a flat metric, then it's going to be (locally) identical to a plane, for all practical purposes. You could take the exact same set of points, give it a different metric, and have it (locally) be, for all practical purposes, the same as a sphere. Or a torus. Or a Klein bottle. Or a completely abstract space.

69dodge
15th April 2006, 06:26 PM
So then there's no paradox. Nothing is moving away from each other, there's no measureable difference between it and a room at rest, and everything makes sense.If centrifugal force is present, you can measure it with, e.g., a spring scale, even though nothing is moving relative to anything else.

So the question is, if this room is the only thing in the universe, and nothing in the room is moving relative to anything else, might centrifugal force still be present?

"Paradox" is probably too strong a word---if centrifugal force is present, then it just is, and that's that---but it is an interesting question, I think. The naive view is that rotation causes centrifugal force, rather than "rotation" just being another name for the existence of centrifugal force. On that view, we need a definition for "rotation" other than "the existence of centrifugal force", and we can take "rotation with respect to everything else in the universe" as that definition. But what if there isn't anything else in the universe to rotate with respect to?

Art Vandelay
15th April 2006, 06:56 PM
How can a spring scale measure force?

Art Vandelay
15th April 2006, 07:05 PM
Another question: suppose we were to that since everything is relative, we can consider the Earth to be at rest and the stars to be orbiting around us. Then a star 4 light years away is moving about 25 light years each day, or more than 9000 times the speed of light. Does this contradict the principle that nothing can travel faster than light?

davefoc
15th April 2006, 09:06 PM
If centrifugal force is present, you can measure it with, e.g., a spring scale, even though nothing is moving relative to anything else.

So the question is, if this room is the only thing in the universe, and nothing in the room is moving relative to anything else, might centrifugal force still be present?

My own cut at this which I put forth a bit in the post before I left on a little vacation to Death Valley is that space controls the motion of an object moving without forces applied to it. It is the line drawn by an object moving without external forces that serves as our fundamental method for distinguishing a rotating frame from a non-rotating frame.

If the universe is rotating and we are synchronized with that rotation of course we will not sense any rotation. But if one part of the universe was rotating with respect to the space we are in we will see the path of an object moving without external forces in the part of the universe we aren't in as moving in a curve in the universe we are in.

This is true not only of objects but of light beams as well. It seems like space not only controls the path of objects but of the light that passes through it.

I have also considered how a line that is formed by stretching a string between two points in the part of the universe which is rotating differently than the part we are in. Would it be curved like the light beam or does the method of stretching a string to define a line produce a line which is straight even if it is viewed from an alternate universe which is rotating at a different rate from the observing section of the universe?

At first I thought it would be curved like the light beam, but now I'm not so sure.

When I first started thinking along this line, I thought that I might have begun to understand what is meant by the GR term, bending of space. It seems that what is meant by the bending of space is exactly this idea that space is modified in such a way that light and objects moving without external forces take a different path than they would in another part of space.

Of course, all of this leans in the direction of something that seems like it must be true to me. Space has structure.

Yllanes
16th April 2006, 01:23 AM
Do Carmo mainly considers surfaces to be embedded in R^3, and pretty much everything depends on the inherent Euclidean metric of R^3. If R^3 didn't already have a metric, most of the definitions would be meaningless.
Which is why all you are saying doesn't make sense for general manifolds. We want intrinsic properties, not something that depends on the embedding.

This is particular type of metric. It takes the space, embeds it in a space that already has a metric, and then derives a new metric from the parent space. It reminds me of the time Homer Simpsons says "I carved this spoon myself. From a bigger spoon." If you're talking about a completely abstract manifold, then this option is not available to you. (And, of course, it assumes that the space is path connected.)

This is not a metric in the relativity sense, it is a distance. In relativity we don't call the distance a 'metric'. Of course it is not for any manifold, it is valid for surfaces, which is what Do Carmo was talking about. I didn't claim it was available for general manifolds, I didn't use it at all and I don't use it in my everyday life (hint: my everyday life has very much to do with differential geometry).

It's an example of such a metric. It's not the only possibility.

You are really tiresome. I was asking you if that's what you are using, I was not claiming it is the only possible distance.

In general, a metric is any function d(x,y):(X cross X)->R where
d(x,y) = 0 iff x=y
d(x,y) > 0 otherwise
d(x,y) = d(y,x)
d(x,y) + d(y,z) >= d(x,z)

Do you think I do not know that? But the metric we talk about in relativity is not this. This time I'm really finished with this. When we talk about metric in relativity we mean g. Have you read any relativity books? The same goes
for differential geometry in general manifolds.

Yllanes
16th April 2006, 01:43 AM
"Paradox" is probably too strong a word---if centrifugal force is present, then it just is, and that's that---but it is an interesting question, I think. The naive view is that rotation causes centrifugal force, rather than "rotation" just being another name for the existence of centrifugal force. On that view, we need a definition for "rotation" other than "the existence of centrifugal force", and we can take "rotation with respect to everything else in the universe" as that definition. But what if there isn't anything else in the universe to rotate with respect to?
As I said before, there always is a metric. The room rotates with respect to the metric. In GR rotating with respect to the metric is as real as rotating with respect to the stars.

Let's consider a simpler experiment: a glass of water spining. On Earth, we know that the water will be pulled to the sides of the glass. Imagine the glass is now located in an otherwise empty universe. GR predicts that the water will be pulled to the sides (which seems anti-Machian) just the same. And again, the glass rotates with respect to the metric.

Another different concept, to show once more that GR makes predictions and is not just vague philosophy. Consider the same glass of water, without rotation this time. Now we are in our universe. Imagine we rotate stars, galaxies, planets... around the bucket. Newtonian physics says that the water would remain flat. In GR the water would get curved. This is calling 'frame-dragging' and it has been experimentally verified (Google for Gravity Probe B).

When I first started thinking along this line, I thought that I might have begun to understand what is meant by the GR term, bending of space. It seems that what is meant by the bending of space is exactly this idea that space is modified in such a way that light and objects moving without external forces take a different path than they would in another part of space.
Yes, that's pretty much the basic meaning.

69dodge
16th April 2006, 11:46 AM
As I said before, there always is a metric. The room rotates with respect to the metric. In GR rotating with respect to the metric is as real as rotating with respect to the stars.It seems to me there is some difference: I can tell whether I'm rotating with respect to the stars by looking at the stars, and then based on that, I can predict whether I will measure a nonzero centrifugal force. But if there are no stars in the universe, I can't tell whether I'm rotating with respect to the metric except by measuring the centrifugal force and seeing whether it's zero. So GR doesn't really make any prediction in that case. All it says is, if centrifugal force is present, then centrifugal force is present. Yes, of course, but will centrifugal force be present?

I guess we could say, GR says that centrifugal force might be present and Mach says that it definitely won't be. Does that sound basically right to you?

Let's consider a simpler experiment: a glass of water spining. On Earth, we know that the water will be pulled to the sides of the glass. Imagine the glass is now located in an otherwise empty universe. GR predicts that the water will be pulled to the sides (which seems anti-Machian) just the same. And again, the glass rotates with respect to the metric.Is there any way to tell whether the glass is rotating, in order to predict whether the water will be pulled to the sides, besides just seeing whether the water is in fact pulled to the sides?

Another different concept, to show once more that GR makes predictions and is not just vague philosophy. Consider the same glass of water, without rotation this time. Now we are in our universe. Imagine we rotate stars, galaxies, planets... around the bucket. Newtonian physics says that the water would remain flat. In GR the water would get curved. This is calling 'frame-dragging' and it has been experimentally verified (Google for Gravity Probe B).They're not finished with the data analysis yet for Gravity Probe B. But I think there have been previous experiments that verified frame dragging?

I'm still not sure exactly what GR predicts. According to GR, is there any reason why all the matter in our universe isn't rotating with respect to the metric, just as a single room in an empty universe might be rotating? That is, would it be consistent with GR if we did experience centrifugal force while not rotating with respect to the stars? I don't see why not. But in fact we don't. So maybe Mach was right.

69dodge
16th April 2006, 12:00 PM
How can a spring scale measure force?Huh? Force is exactly what spring scales measure.

Just use it in the normal manner to weigh something. If it shows that the weight is nonzero, then your room in an empty universe is not at rest.

Yllanes
16th April 2006, 12:46 PM
I guess we could say, GR says that centrifugal force might be present and Mach says that it definitely won't be. Does that sound basically right to you?

Actually, GR says that the centrifugal force will be present. It makes a definite prediction. The scenario is a glass of water in an otherwise empty (asymptotically Minkowskian) spacetime. As I said, it rotates with respect to the metric. If you don't believe me, notice that the metric is a field ijust like the electromagnetic field. Only that now, instead of a vector, we have a rank (0,2) tensor at each point. Gravitational waves (provided they exist) are a good indication that the geometry is as good a field as any.

I'm still not sure exactly what GR predicts. According to GR, is there any reason why all the matter in our universe isn't rotating with respect to the metric, just as a single room in an empty universe might be rotating?
Yes, there are observational bounds on the rotation of the universe.

Imagine only SR at play. A rotation of the Universe could be detected by measuring redshift of light from distant sources, looking in a direction perpendicular to the axis of rotation and comparing it with sources on the axis. The redshift would not be the usual (since there is no radial velocity) but a cuadrupolar redshift, due to time dilation at the sources. Roughly the same happens in GR.

In short: a rotating universe is theoretically possible in GR, well defined, and different from a nonrotating one. However, there are strong observational bounds (isotropy) on the rotation velocity.

davefoc
16th April 2006, 12:54 PM
Is there any way to tell whether the glass is rotating, in order to predict whether the water will be pulled to the sides, besides just seeing whether the water is in fact pulled to the sides?

That is a very good way of putting the question that I was trying to pose at the beginning of this thread.

I don't know if it is a scientifically accepted answer, but right now I think my thoughts about space having a structure that allows a body to move through it with a constant speed on a particular path without external forces is a plausible answer.

In the case of your pail, it is only necessary to determine a frame of reference in which light beams trace a straight path (determined say by a stretched string) and fix your pail with respect to that frame of reference to know that the water in your pail will remain flat. Alternatively to the light beams floating objects could also be used.

I don't know that I am not stating something that is so obvious that it isn't worth stating and if that is the case then I apologize.

Here is an article in Wikipedia on a somewhat related subject:
http://en.wikipedia.org/wiki/Aether_and_general_relativity

69dodge
16th April 2006, 03:23 PM
Another question: suppose we were to that since everything is relative, we can consider the Earth to be at rest and the stars to be orbiting around us. Then a star 4 light years away is moving about 25 light years each day, or more than 9000 times the speed of light. Does this contradict the principle that nothing can travel faster than light?I'm pretty sure it doesn't, because the basic idea of general relativity is that any reference frame is as good as any other, even reference frames that would usually be called "accelerated". But I don't know enough about the details of GR to say exactly why it doesn't. Probably the reason has something to do with the fact that you're here and the star is there, and, as you discussed with Yllanes, one can't compare a vector here with a vector there. (The vectors in this case being velocities.) My understanding is, the relative velocity of two objects close to each other can't exceed the speed of light, but the relative velocity of two objects far from each other isn't well-defined.

Yllanes
16th April 2006, 03:50 PM
I'm pretty sure it doesn't, because the basic idea of general relativity is that any reference frame is as good as any other, even reference frames that would usually be called "accelerated". But I don't know enough about the details of GR to say exactly why it doesn't. Probably the reason has something to do with the fact that you're here and the star is there, and, as you discussed with Yllanes, one can't compare a vector here with a vector there. (The vectors in this case being velocities.) My understanding is, the relative velocity of two objects close to each other can't exceed the speed of light, but the relative velocity of two objects far from each other isn't well-defined.
I didn't see this originally. Of course, it is not in contradiction with relativity. The fact is that we can define lots of things that are sort of like a velocity, but can be greater than c. However, information is never propagated faster than c.

Let us consider a simpler example. Assume you have a laser pointer in your hand and aim it to a distant wall. You move your hand to the left and to the right and look at the point on the wall. It's obvious that the point is moving faster than your hand. In fact, if the laser point reached the Moon, the dot on the Moon would move faster than c (check this, it is easy). Now, does this mean that we have broken the 'light barrier'? No, because the dot cannot carry any information (or energy) faster than the speed of light. Imagine two observers on the Moon, A and B. The dot from the laser pointer sweeps the distance between them faster than a light signal would. Can we use this to communicate them at superluminal speeds? Think it yourself: We give them the following instructions:

A: When you see the dot, fire a bullet.
B: When you see the dot, take cover, as A has shot at you.

Has the warning message travelled from A to B faster than light? Once you work out this example, you will understand the one about the stars. But I think it is important that I do not give all the details, because the way to 'grok' these things is by doing them.

Another excercise (from Taylor & Wheeler): The maker of an oscilloscope claims a writing speed (the speed with which the spot moves across the screen) in excess of the speed of light. Is this possible, or a claim for the Million Dollar Challenge?

Edited to fix typos, my keyboard is dying...

69dodge
16th April 2006, 04:32 PM
Actually, GR says that the centrifugal force will be present. It makes a definite prediction. The scenario is a glass of water in an otherwise empty (asymptotically Minkowskian) spacetime. As I said, it rotates with respect to the metric. If you don't believe me, notice that the metric is a field ijust like the electromagnetic field. Only that now, instead of a vector, we have a rank (0,2) tensor at each point. Gravitational waves (provided they exist) are a good indication that the geometry is as good a field as any.I think we're kind of talking past each other here.

GR predicts that centrifugal force will be present, if the glass of water is rotating. If it's not rotating, GR predicts that centrifugal force won't be present. Correct?

So now the question is, how can we find out whether the glass is rotating, so that we can decide what GR predicts? The metric is considered real in GR, but it's not something that we can see, like we can see stars. Is there some way to distinguish a glass that's rotating with respect to the metric from a glass that's not rotating with respect to the metric, except by the presence or absence of centrifugal force? In other words, is there any way that we could decide in advance whether centrifugal force will be present, rather than just checking to see whether it is or it isn't? I don't see how we could. That's why I said, "GR says that centrifugal force might be present."

If we check and we find that centrifugal force is present, then we can say, "oh, I guess the glass must be rotating with respect to the metric." And if we find that centrifugal force is not present, then we can say, "oh, I guess the glass is not rotating with respect to the metric." But there's no way, beforehand, to look at the metric "directly" and see whether or not the glass is rotating with respect to it.

Do you agree?

Yes, there are observational bounds on the rotation of the universe.

Imagine only SR at play. A rotation of the Universe could be detected by measuring redshift of light from distant sources, looking in a direction perpendicular to the axis of rotation and comparing it with sources on the axis. The redshift would not be the usual (since there is no radial velocity) but a cuadrupolar redshift, due to time dilation at the sources. Roughly the same happens in GR.

In short: a rotating universe is theoretically possible in GR, well defined, and different from a nonrotating one. However, there are strong observational bounds (isotropy) on the rotation velocity.Yes, I agree with all this.

Lots of different possible universes are consistent with GR. We, of course, live in just one universe. This one universe is not rotating, as far as we can tell (correct?), which is just what Mach's principle says should be the case. So why should we hold against Mach's principle the fact that GR says other universes are possible which contradict it, when the one universe we know about doesn't contradict it?

Art Vandelay
16th April 2006, 04:47 PM
Has the warning message travelled from A to B faster than light?No, it hasn't traveled from A to B at all. Are you saying that if the stars are rotating around us, they are not going from one point to another?

Just use it in the normal manner to weigh something. What is the normal manner?

In the case of your pail, it is only necessary to determine a frame of reference in which light beams trace a straight path (determined say by a stretched string) and fix your pail with respect to that frame of reference to know that the water in your pail will remain flat.You're confusing space and spacetime. Light travels in a straight line in spacetime, not space. The string is measuring distance in space. How would rotation affect the string?

Yllanes
16th April 2006, 05:01 PM
Do you agree?
I don't seem to understand what you mean, I'm sorry. I think we are talking about different things. On the one hand, you can mathematically define the universe with the rotating glass and nothing else and get the definite prediction that there will be centrifugal forces. On the other hand (and this may be what you are saying), we cannot, experimentally, decide whether the universe with the glass is rotating except for the existence or absence of centrifugal forces.

So, GR says that there will be forces. We don't need to see anything to predict this because, once Einstein's equations are established, it is just a mathematical result, as the concept of a rotating glass is well defined in a mathematical sense.

Knowing this, if we are in the universe, we can look at the centrifugal forces. If there aren't any we can say the glass is not rotating. If there are, we can say it is. You seem to think this way of determining whether something is rotating is 'weaker' than just looking at the stars. But I disagree. Example: Hubble measured redshift and deduced that the galaxies are moving away from us. He didn't see them moving, just measured a consequence of their movement. This is the same thing, which is why I say that there will be forces if the glass is rotating.

So why should we hold against Mach's principle the fact that GR says other universes are possible which contradict it, when the one universe we know about doesn't contradict it? I wasn't explicitly holding it against it. If and when we measure gravitational waves, however, I will hold that against Mach. My real issue with the principle is that it is too vague to be useful for predictions, i.e., science. If you formulate it on an unambiguous way, inconsistencies may arise. It is also dangerous for the layman for several reasons, not the least of them being that it may give rise to thoughts that GR is not a very definite and rigourous theory. And also it may make you think that the geometry is a result of the mass configuration alone, when in fact is a dynamical thing.

However, to be fair, the case may be made also for it. Misner, Thorne and Wheeler have several pages about it in their book and the book I quoted on an earlier post by Ciufolini and Wheeler has an extended discussion of it, including one precise formulation via gravitomagnetic effects. Einstein loved it, and wrote to Mach to congratulate him on it. Stephen Hawking said:

The observed isotropy of the microwave background indicates that the universe is rotating very little if at all [...] This could possibly be regarded as an experimental verification of Mach's Principle.

And now for something completely different:

No, it hasn't traveled from A to B at all. Are you saying that if the stars are rotating around us, they are not going from one point to another?

The light beam has passed through both points. This is a very close analogy to the stars. I don't know what you are saying here. Are you saying that relativity is wrong because stars move faster than c? I assume you are not, but believe me, the light beam and the galaxies rotating are really the same example.

Art Vandelay
16th April 2006, 08:10 PM
The light beam has passed through both points.One set of photons went through point A. Another set of photons went through point B. No photon went from one to the other.

I don't know what you are saying here. Are you saying that relativity is wrong because stars move faster than c?I am not saying anything. I am asking questions.

I assume you are not, but believe me, the light beam and the galaxies rotating are really the same example.How so? And how is it not the same as going from the Earth to Mars in a minute?

davefoc
16th April 2006, 09:57 PM
Originally Posted by davefoc :
In the case of your pail, it is only necessary to determine a frame of reference in which light beams trace a straight path (determined say by a stretched string) and fix your pail with respect to that frame of reference to know that the water in your pail will remain flat.

You're confusing space and spacetime. Light travels in a straight line in spacetime, not space. The string is measuring distance in space. How would rotation affect the string?
I may be confusing space and spacetime but I also don't think I made my suggestion quite clear.

I propose three ways of determining if something is straight while floating in space.
1. Stretch a string between two points the string will be straight.
2. propel an object. The line formed by the center of gravity of the object will be straight if you are not measuring it relative to a rotating frame.
3. Fire a laser beam. The line formed by the beam will be straight if you are not measuring it relative to a rotating frame.

So my thought was to stretch a string and look to see if the propelled object or the laser beam moved parallel to it. If they are, at least one axis of the frame you are in is non-rotating. Do the same experiment in a direction that neither intersects the original line nor is parallel to it and you can determine if your frame is non-rotating in all three axies.

You could skip the string if you just have a good measurement system so you could set up way points in the frame of reference you are in that are on a line. If you can propel your object so that they go past each of the way points in your frame of reference then your frame of reference is not rotating in at least one axis.

Of course, it is much simpler to just instrument some sort of merry go round that is fixed to your reference frame and look for centrifugal forces. If there aren't any you are in a non-rotating frame.

If I was going to restate the question that I originally started this thread with it would be, "What is the property of space that allows us to detect whether we are in a non-rotating or rotating frame of reference?"

The problem with putting the question quite like that was that when this thread was started I didn't know whether it was a property of space or something else that allowed us to distinguish between rotating and non-rotating.

It now seems like it probably is a property of space that allows us to determine rotating from non-rotating. And that property seems to be the path that space imposes on an object that is moving through it without external forces.

ETA: Another roughtly equivalent statement that applies to the merry go round is that when something is moving in our frame of reference in a curved path a force will be required on the object to cause it to move in that curved path. Hence when the merry go round rotates in our frame of reference and we detect centrifigal forces we know that the merry go round is rotating relative to a non-rotating reference frame of the space we are in.