# Reference frames

pmb
Hi Hurkyl

Switching into a rotating frame changes the geometry of space. The rotating observer will see the universe differently than the nonrotating observer because in his frame, space-time has a different geometry. Space-time, in both pictures, has zero curvature, but the geometry is different, thus switching frames will introduce a "space-time distortion".

I assumed that Carter was asking about curvature.

And I wonder about the rest of what you say. To me the "geometry" is an intrinsic property of a manifold once a metric is attached to it. When you change frames all you're really doing is chaning the coordinate system. You're not chaning the geometry of spacetime itself. The metric is a geometric object and when you change coordinates wha you're doing is merely changing the components - not altering the geometric entity itself. E.g. Think of the analogy with vectors since they too are geometric objects like the metric. If you have a Cartesian vector and you change coordinates then you change the components of the vector but you don't change the geometric object itself.

Pmb

pmb
Originally posted by mich
In a freefalling elevator, the air within the elevator
isn't accelerating, so there will be a measurable pressure difference
in the elevator proportional to the acceleration; if so why is it equivelant to being inertial if an experiment can prove it is not?

mich

Inside a freely falling elevator the air *is* accelerating. It's accelerating doward with the elevator - that's if the elevator is sealed such that it's isolated.

Pmb

Hurkyl
Staff Emeritus
Gold Member
I guess I should say the geometry of the reference frame rather than the geometry of space-time!

pmb
Originally posted by Hurkyl
I guess I should say the geometry of the reference frame rather than the geometry of space-time!

But what do you mean by this? Let's consider Euclinean geometry for a moment. Specifically E^2 - a plane for example. In cartesian coordinates the metric is given by

dL*2 = dx^2 + dy^2

I know change coordinates to polar coordinates

dL^2 = dr^2 + r^2 dtheta^2

If I understand you correctly this situation is analogous to the one in GR with regard to the rotating frame. I start with a flat space - I change coordinates - I still have a flat space. The metric looks different because of the curvilinear coordinates. What is the "distortion" you're refering to?

Perhaps you're refering to the curvilinear coordinates and its the curvilinear coordinates thats the distortion? Is so how is this different/same for a uniform g-field/accelerating frame of referance?

Thanks

Pete

Hurkyl
Staff Emeritus
Gold Member
The alternate coordinates don't even have to be that exotic. Spinning your reference frame amounts to a helical twist:

t' = t
x' = x cos &omega t - y sin &omega t
y' = x sin &omega t + y cos &omega t
z' = z

The twist preserves locally Minowski frames located on its axis, and is a smooth deformation from the identity map.

Is so how is this different/same for a uniform g-field/accelerating frame of referance?

Here the distortion is that (via a change of coordinates) one reference frame sees a uniform nonzero classical gravitational field and another reference frame sees a Minowski metric everywhere.

pmb
Originally posted by Hurkyl
The alternate coordinates don't even have to be that exotic. Spinning your reference frame amounts to a helical twist:

t' = t
x' = x cos &omega t - y sin &omega t
y' = x sin &omega t + y cos &omega t
z' = z

The twist preserves locally Minowski frames located on its axis, and is a smooth deformation from the identity map.

Here the distortion is that (via a change of coordinates) one reference frame sees a uniform nonzero classical gravitational field and another reference frame sees a Minowski metric everywhere.

But that's not a distortion in the geometry. That's a result of the coordinates being curvilinear.

Pete

Hurkyl
Staff Emeritus
Gold Member
The metric is the geometry; it is what defines lengths and angles.

And to top it off, the spinning observer using the helical coordinates is in a locally Minowski frame, so by the EEP, we can't consider the nonspinning coordinates any more special than the helical coordinates.

The geometry of the two coordinate charts are different; they have different metrics, different geodesics, et cetera. They, of course, refer to the same space-time, but that doesn't make them the same.

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I can't understand what you guys are talking about right now, but I can imagine this:

From the point of view of the rotating reference frame, by SR, the circumference of your rotation will be contracted. It therefore won't match up with the euclidian 2[pi]r. So I assume the geometry of the system is somehow warped, although I guess spacetime, which is somehow a compilation of all reference frames (?), isn't actually distorted.

--Carter

pmb
Hi Hurkyl

The metric is the geometry; it is what defines lengths and angles.
I agree. However when you change the coordinates you don't change the geometry. Lengths and anlges *do not change* under a coordinate transformation. In fact that is part of the defintion of the metric, i.e. the metric, g_ab, is defined such that the "length"

ds^2 = g_ab*dx^2 dx^b (summation implied over a and b)

such that it remains unchange when you change the coordinates.

Pmb

pmb
Originally posted by CJames
I can't understand what you guys are talking about right now, but I can imagine this:

From the point of view of the rotating reference frame, by SR, the circumference of your rotation will be contracted. It therefore won't match up with the euclidian 2[pi]r. So I assume the geometry of the system is somehow warped, although I guess spacetime, which is somehow a compilation of all reference frames (?), isn't actually distorted.

--Carter

First off - I disagree with that conclusion. However this is one of those things which are still being debated in physics. In fact this appeared in the Am. J. Phys. back in 2000 as I recall.

Pmb

take pride in the fact that you can even begin to understand it.
Thanks Pete.

The derivatives are with respect to proper-time but the curves are plotted in a spacetime diagram with respect to coordinate time.
Proper time being the time within a given reference frame and coordinate time being what?

Also, I was wondering if my example with two worldlines being the top and bottom of an elevator was correct?

Thanks for all this discussion Hurkle and Pete.

pmb
Howdy Carter

You're most welcome!

Proper time being the time within a given reference frame and coordinte time being what?

That is not what proper time is. Proper time is the time which passes on a particluar clock. Coordinate time is just plain good ole time. Hard to explain without knowing how much you know about special relativity. Suppose there are two flashes of light. One flash here and one flash far away but they do not happen at the same time. But it just so happens that a spaceship passes through both places where the flashes occured and when they occured. The time amount of time between these events as recorded in my frame of referance is just "time". The time as recorded on a clock inside the rocketship is called the proper time between the events. So the proper time between two events along a given worldline is the amount of time recorded on a clock whose wordline is that worldline and which passed through both events.

Whew!

re - "Also, I was wondering if my example with two worldlines being the top and bottom of an elevator was correct?"

Actually I don't recall the question. Sorry I didn't see it. I'll try to dig it up later. I'm rather pooped now.

Pete

Hurkyl
Staff Emeritus
Gold Member
I see why we're in disagreement. I'm talking about the coordinate geometry, you're not.

Specifically, for an example, I'm talking about the fact that in the global Minowski frame, dx is orthogonal to dt, but in the twisted frame, dx is no longer orthogonal to dt. Of course, dx in the twisted frame is not the same as the dx in the global Minowski frame; that's the distortion caused by changing reference frames.

I consider this perspective important because it relates the abstraction of a curved space-time back to the Euclidean approximation into which we stuff it for our observations.

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pmb
Hi Hurkyl
I see why we're in disagreement. I'm talking about the coordinate geometry, you're not.
Ah! Okay. I thought that was the case. Yes. And that's what I meant by curvilinear coordinates. Ever see a straight line plotted in polar coordinates? Buy that I mean if you take two straight axes one of which has the "r" values and the other has the "theta" values then plot the (r,theta) points y = ax + b. You will get what appears to be a "warped" line. plot a bunch of these and it will look like a "distortion" in a way.

Pete

mich
Originally posted by pmb
Inside a freely falling elevator the air *is* accelerating. It's accelerating doward with the elevator - that's if the elevator is sealed such that it's isolated.

Pmb

Sorry pmb, this is not what I meant.
I meant the air molecules inside the elevator aren't freefalling; therefore it's cause for acceleration must be due to the force inside the accelerating elevator. This will cause a greater air pressure on top of the elevator than on the floor....that's what it seems to me anyway.

mich

pmb
Originally posted by mich
Sorry pmb, this is not what I meant.
I meant the air molecules inside the elevator aren't freefalling; therefore it's cause for acceleration must be due to the force inside the accelerating elevator. This will cause a greater air pressure on top of the elevator than on the floor....that's what it seems to me anyway.

mich

Why don't you think the air particles are in free-fall?

Pete

mich
Originally posted by pmb
Why don't you think the air particles are in free-fall?

Pete

I was thinking in the line that air excerts an equal pressure
everywhere inside the elevator, which would have caused a positive pressure on the ceiling of the elevator while it was freefalling...
It just came to mind that when the elevator is at rest, the air pressure is greater on the floor due to the gravitational force, equalizing the pressure inside the elevator while it is freefalling...
which is the same as saying it is inertial.

Thanks for your help.
mich

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Actually I don't recall the question. Sorry I didn't see it. I'll try to dig it up later. I'm rather pooped now.

I'm going to add this. Imagine two parallel worldlines following two geodesics. The lines in curved space will separate away from each other. Hence, from the perspective of one worldline, the other is curved and is therefore accelerating away. If these two worldlines correspond to the top and bottom of an elevator, we can see that a tidal force is taking place. (Of course, in reality, the top and bottom of the elevator are attatched too rigidly to accelerate away from each other, and will in reality only "try" to accelerate away from each other. But is the basic consept correct?)

My mind works in examples and I appologize for that.

Hi again Pete.

That is not what proper time is. Proper time is the time which passes on a particluar clock.
Doesn't a particular clock have to remain within a particular reference frame (the reference frame of that clock)?

So the proper time between two events along a given worldline is the amount of time recorded on a clock whose wordline is that worldline and which passed through both events.
So in other words, proper time is the time measured within that reference frame, not the time measured from another reference frame. I thought I said that, or maybe I still don't get it.

I'm still not clear on what coordinate time is, though. Is it the time recorded w/respect to the source of spacetime curvature? Or just the time recorded w/respect to some defined coordinate system that could be anywhere?

Sorry pmb, this is not what I meant.
I meant the air molecules inside the elevator aren't freefalling; therefore it's cause for acceleration must be due to the force inside the accelerating elevator. This will cause a greater air pressure on top of the elevator than on the floor....that's what it seems to me anyway.
Actually, the air particles will exert an equal pressure on all parts of the elevator (more or less, not accounting for tidal forces.)

pmb
Hi Clark

re - "Doesn't a particular clock have to remain within a particular reference frame (the reference frame of that clock)?"

Yes. Let me explain by giving an example of what proper time *is not*.

Suppose there are two clocks seperated by 1 light second in space (i.e. 1 light second is the distance traveled by light in 1 second) and are at rest with respect to each other in an inertial frame of referance - call this frame O. The clocks are synchronized so that they read zero at the same time in O. Each clock has a light in it. The lights are off before t = 0. Clock 1's light comes on at t = 0 sec. Clock 2's light comes on at t = 0.5 sec.

The time between these two events is dt = 0.5

That is an example of what I called "coordinate time". Just think of it as time. However it's meaningless to ask what the proper time between these events are since no clock can be present at each event along any worldline since it would mean that the clock traveled faster than light.

Pete

pmb
Clark - You can read more about this in a book a friend of mine wrote. See "Exploring Black Holes"

Front Matter (pdf file, ~95K)
Introduces the approach and describes the content of the book.

http://www.eftaylor.com/pub/front_matter.pdf

Chapter 1: Speeding (pdf file, ~422K)
Background in special relativity needed for general relativity.

http://www.eftaylor.com/pub/chapter1.pdf

Chapter 2: Curving (pdf file, ~332K)
Introduces curved spacetime and the Schwarzschild metric.

http://www.eftaylor.com/pub/chapter2.pdf

Student Project on the Global Positioning System (pdf file, ~62K)
The global positioning system is useless without general relativity.

http://www.eftaylor.com/pub/projecta.pdf

Revised Project F, The Spinning Black Hole (pdf file ~200k)
Includes fourth-printing correction of an error described in Errata file.

http://www.eftaylor.com/pub/SpinNEW.pdf

Errata for Printed Textbook (text file)

http://www.eftaylor.com/pub/errata.txt

Pete

Thanks for all the info Pete. Looks like I'm going to be busy for a while.

Err...name's Carter. LOL.

Hurkyl
Staff Emeritus
Gold Member
Ah! Okay. I thought that was the case. Yes. And that's what I meant by curvilinear coordinates.

I objected because you sounded like you're implying that the non-rotating reference frame is special, and the rotating frame is just a deformation of the "right" thing through a change of coordinates rather than a reference frame with equal standing to the non-rotating one.

dr-dock
Originally posted by CJames
Just a little question.

I'm a little curious. In the classical sense, a reference frame in freefall would, of course, be said to accelerate. However, no forces are felt within a freefalling reference frame. Am I correct, then, in assuming a freefalling reference frame to be an inertial reference frame? Am I also correct in stating that a reference frame on the surface of Earth is an accelerating reference frame, since a force of gravity is felt toward the ground? (Even though in the classical sense, this reference frame is "at rest.")

Don't limit this thread to an answer to my question. I would like a continuous discussion. Please? LOL.

Good day everybody.
no man that's all wrong.
if a object is in freefall then ask your self why it is moving constantly if you make a tunnel thru earth.for mua constant displacement suggest constant presence of nonzero force. the force and the equilibrium distance never actually drop down to zero but they(those vectors) only rotate around the constant energy vector. therefore freefall freame is inertial frame.i guess it's hard to understand me at first but i'll try to explain my self prety soon after i finish the particle simulator i'm developing right now.

Err... it isn't moving constantly. You have then a case of simple harmonic motion.

pmb
Originally posted by CJames
Thanks for all the info Pete. Looks like I'm going to be busy for a while.

Err...name's Carter. LOL.

Oh MAN! I keep forgetting that. Sorry Clark!

Pete

pmb
Originally posted by Hurkyl
I objected because you sounded like you're implying that the non-rotating reference frame is special, and the rotating frame is just a deformation of the "right" thing through a change of coordinates rather than a reference frame with equal standing to the non-rotating one.

Why did I sound like the non-rotating frame was special and where did sound like that? When I say "change coordinates" I'm talking changing from one system of spacetime coordinates to another system of spacetime coordinates.

pete