what's the deal with rotating frames of reference

[Art Vandelay]

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]

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]

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]

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]

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.

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]

"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]

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]

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]

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]

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]

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]

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 .

 

[Art Vandelay]

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]

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]

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]

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]

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:

g_abv^a=v_b (TM -> T*M)
g^cdv_c=v^d (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 v^a -> v_a 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, g^ac 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.

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]

[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]

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]

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?