B Stationary frames of reference

[name123]

What curves are you talking about? The worldlines of distant objects in the universe will be straight lines in non-rotating Cartesian coordinates, since those are just the coordinates of an inertial frame. See above.

Apologies, you are right of course. What about if the Cartesian coordinate system was rotating with the observer? The observed objects could then be observed to be traveling at a speed greater > $c$ within that coordinate system, and so using that coordinate system in the equations, would not gamma (for the observed distant stars) involve imaginary numbers?

 

[name123]

Proper acceleration, the kind that an accelerometer measures and that is the same no matter what coordinate system I choose, happens or not according to whether the object is in free fall or not. Freefall paths through flat spacetime appear in space like the straight lines of Euclidean geometry; freefall paths through curved spacetime do not. eiether way they are freefall paths and there is no proper acceleration.

Thanks.

Sorry one more thing. Am I correct in thinking that in a spacecraft if one were to change direction using a thruster, the initial velocity would have no effect on the how much the thruster changed the direction, or to the readings of an onboard accelerometer?

 

[name123]

Apologies, you are right of course. What about if the Cartesian coordinate system was rotating with the observer? The observed objects could then be observed to be traveling at a speed greater > $c$ within that coordinate system, and so using that coordinate system in the equations, would not gamma (for the observed distant stars) involve imaginary numbers?

Sorry I think I overlooked that if the frame was not inertial, then you would have to use equations which took the angular velocity of the frame into account.

 

[PeterDonis]

What about if the Cartesian coordinate system was rotating with the observer?

Then it would not be an inertial coordinate chart and things would work differently. Plus the metric would look a lot messier than the one I wrote down; I used cylindrical coordinates precisely because they make things look much simpler in a rotating chart.

The observed objects could then be observed to be traveling at a speed greater > $c$ within that coordinate system

At a coordinate speed greater than $c$, but coordinate speed is not physical speed. That is the point I was making in my previous long post. The objects would be traveling on timelike worldlines. See below.

would not gamma (for the observed distant stars) involve imaginary numbers?

No, because the analogue of $\gamma$ in non-inertial coordinates is not based on coordinate speed, it's based on the worldline being timelike.

Am I correct in thinking that in a spacecraft if one were to change direction using a thruster, the initial velocity would have no effect on the how much the thruster changed the direction

How would you measure how much the thruster changed the direction?

or to the readings of an onboard accelerometer?

This would be independent of initial velocity (which is coordinate dependent anyway), yes.

I think I overlooked that if the frame was not inertial, then you would have to use equations which took the angular velocity of the frame into account.

Yes, and that's what I wrote down in my previous long post. Note that there is an extra "cross term" involving $dt d\phi$ in the metric. That is a reflection of "the angular velocity of the frame". So the analogue of $\gamma$ for a given worldline now has to take into account, not just $dt$, but $d\phi$ as well. That's one way of describing how the worldline of an object "at rest relative to the universe" (and therefore "moving" in this rotating frame) can still be timelike even though $1 - \omega^2 r^2$ is zero or negative.

 

[PeroK]

Sorry I think I overlooked that if the frame was not inertial, then you would have to use equations which took the angular velocity of the frame into account.

A simple example of why you must remain in a single inertial reference frame is as follows, and I think this also answers your question about what happens if you accelerate in discrete steps with periods of constant velocity.

If you measure the distance to an object at rest, then you accelerate towards it, stop accelerating and take a new measurement, then the distance may have reduced greatly owing to length contraction.

If you take these measurements and divide the difference by the elapsed time on your watch, then you have a velocity of sorts, which may be far in excess of the speed of light. But, because the initial and final measurements were taken in different inertial reference frames, this is not a valid velocity in terms of SR equations. You cannot take that "velocity" and calculate a gamma factor with it.

In physics as well as mathematics, context is vital. In SR in particular you have to be careful that the assumptions under which the equations were derived hold good for your experiment.

 

[pervect]

As a general overview, I should mention that General Relativity does have concept(s) that relate to "frames of reference", but there are some important differences. It's hard to be precise in lay language, but the biggest change is that what passes for "frames of reference" in GR are for the most part purely local. So to give an example, if one is in the MIR space-station in Earth orbit, one can construct a local frame of reference that "moves with" the space-station and covers the section of space-time within the space-station without significant problems. Trying to cover too much area (an entire orbit) with a single frame of reference tends to cause issues, though.

So if one is basing all of one's physical understanding on the existence of a frame of reference, one may feel at a bi of a lost, since they don't quite work the same way in GR. What does exist is the concept of a coordinate system. In general the properties of a coordinate system are pretty basic, the only requirement is a unique mapping from every physical event (a point in space at a specific time) in the region covered by the coordinate system (which often covers all of space-time, but this is not actually required) to a set of numbers, called "coordinates", that tell one when and where the event occured. This is a more basic, more primitive construct that represents the structure of space-time than a "frame of reference".

Getting things like distances out of these generalized coordinates is more involved, however, the mathematical tool that gets distances and/or time intervals out of coordinates is called a metric.

 

[Nugatory]

Sorry one more thing. Am I correct in thinking that in a spacecraft if one were to change direction using a thruster, the initial velocity would have no effect on the how much the thruster changed the direction, or to the readings of an onboard accelerometer?

It would have no effect on the reading of the onboard accelerometer (and note that "onboard" is the only kind of accelerometer there is).

"How much" it changes the direction is going to be coordinate-dependent, as is the initial and final velocity. Consider two coordinate systems: In both of them the spaceship is traveling along the x-axis, and in both of them the thruster is pointed along the y-axis (so is pushing the spaceship in a direction perpendicular to its initial direction. In both coordinate systems the thruster changes the y component of the spaceship's velocity from 0 m/sec to 100 m/sec over a period of one second - this is 10g, a high but not unrealistic acceleration.

However, in one coordinate system the spaceship is initially moving at 100 m/sec along the x-axis and in the other coordinate system the spaceship is initially moving at 1 m/sec along the x axis. You could, if you wished, consider these two coordinate systems to be the two "points of view" of two observers, one moving at 100 m/sec relative to spaceship the other moving at 1 m/sec relative to the ship (and therefore moving at 99 m/sec relative to one another).

One observer will say that the spaceship made an almost 90-degree change of course, and the other will say that it made a 45-degree change of course. In one frame the spaceship's final velocity vector will be $\vec{v}=\hat{x}+100\hat{y}$ and in the other it will be $\vec{v}=100\hat{x}+100\hat{y}$; these are clearly different directions. (Here, $\hat{x}$ and $\hat{y}$ are unit vectors in the x and y directions).

Note that although this is the relativity forum, the example above is classical - all the velocities and accelerations are so small that no relativistic corrections are needed.

 

[name123]

Yes, and that's what I wrote down in my previous long post. Note that there is an extra "cross term" involving $dt d\phi$ in the metric. That is a reflection of "the angular velocity of the frame". So the analogue of $\gamma$ for a given worldline now has to take into account, not just $dt$, but $d\phi$ as well. That's one way of describing how the worldline of an object "at rest relative to the universe" (and therefore "moving" in this rotating frame) can still be timelike even though $1 - \omega^2 r^2$ is zero or negative.

Yes I was a bit slow getting it. Thanks for the explanation and patience.

 

[name123]

It would have no effect on the reading of the onboard accelerometer (and note that "onboard" is the only kind of accelerometer there is).

"How much" it changes the direction is going to be coordinate-dependent, as is the initial and final velocity. Consider two coordinate systems: In both of them the spaceship is traveling along the x-axis, and in both of them the thruster is pointed along the y-axis (so is pushing the spaceship in a direction perpendicular to its initial direction. In both coordinate systems the thruster changes the y component of the spaceship's velocity from 0 m/sec to 100 m/sec over a period of one second - this is 10g, a high but not unrealistic acceleration.

However, in one coordinate system the spaceship is initially moving at 100 m/sec along the x-axis and in the other coordinate system the spaceship is initially moving at 1 m/sec along the x axis. You could, if you wished, consider these two coordinate systems to be the two "points of view" of two observers, one moving at 100 m/sec relative to spaceship the other moving at 1 m/sec relative to the ship (and therefore moving at 99 m/sec relative to one another).

One observer will say that the spaceship made an almost 90-degree change of course, and the other will say that it made a 45-degree change of course. In one frame the spaceship's final velocity vector will be $\vec{v}=\hat{x}+100\hat{y}$ and in the other it will be $\vec{v}=100\hat{x}+100\hat{y}$; these are clearly different directions. (Here, $\hat{x}$ and $\hat{y}$ are unit vectors in the x and y directions).

Note that although this is the relativity forum, the example above is classical - all the velocities and accelerations are so small that no relativistic corrections are needed.

Thanks

 

[name123]

As a general overview, I should mention that General Relativity does have concept(s) that relate to "frames of reference", but there are some important differences. It's hard to be precise in lay language, but the biggest change is that what passes for "frames of reference" in GR are for the most part purely local. So to give an example, if one is in the MIR space-station in Earth orbit, one can construct a local frame of reference that "moves with" the space-station and covers the section of space-time within the space-station without significant problems. Trying to cover too much area (an entire orbit) with a single frame of reference tends to cause issues, though.

So if one is basing all of one's physical understanding on the existence of a frame of reference, one may feel at a bi of a lost, since they don't quite work the same way in GR. What does exist is the concept of a coordinate system. In general the properties of a coordinate system are pretty basic, the only requirement is a unique mapping from every physical event (a point in space at a specific time) in the region covered by the coordinate system (which often covers all of space-time, but this is not actually required) to a set of numbers, called "coordinates", that tell one when and where the event occured. This is a more basic, more primitive construct that represents the structure of space-time than a "frame of reference".

Getting things like distances out of these generalized coordinates is more involved, however, the mathematical tool that gets distances and/or time intervals out of coordinates is called a metric.

Thanks

 

[name123]

A simple example of why you must remain in a single inertial reference frame is as follows, and I think this also answers your question about what happens if you accelerate in discrete steps with periods of constant velocity.

If you measure the distance to an object at rest, then you accelerate towards it, stop accelerating and take a new measurement, then the distance may have reduced greatly owing to length contraction.

If you take these measurements and divide the difference by the elapsed time on your watch, then you have a velocity of sorts, which may be far in excess of the speed of light. But, because the initial and final measurements were taken in different inertial reference frames, this is not a valid velocity in terms of SR equations. You cannot take that "velocity" and calculate a gamma factor with it.

In physics as well as mathematics, context is vital. In SR in particular you have to be careful that the assumptions under which the equations were derived hold good for your experiment.

Thanks

 

[Battlemage!]

The same way you, standing on the surface of the Earth, are at rest yet accelerating. "Accelerating" here means "feeling acceleration", i.e., feeling weight. You can feel weight and still be at rest.

The concept of "being at rest" is frame-dependent, and you can always choose a frame in which any chosen object is at rest, regardless of whether it is feeling weight or not.
There isn't one. "What you actually feel" is the direct observable; that's the best way of distinguishing it that you're going to get.

What about being local to the accelerating object? Wouldn't your distance from the object have to be zero in order to properly measure it's proper acceleration?

 

[PeterDonis]

Wouldn't your distance from the object have to be zero in order to properly measure it's proper acceleration?

Strictly speaking, only an accelerometer in contact with the object itself can measure its proper acceleration. Which I think means yes.

 

[name123]

Proper acceleration, the kind that an accelerometer measures and that is the same no matter what coordinate system I choose, happens or not according to whether the object is in free fall or not. Freefall paths through flat spacetime appear in space like the straight lines of Euclidean geometry; freefall paths through curved spacetime do not. eiether way they are freefall paths and there is no proper acceleration.

Hi sorry for bothering you again, but I just had some more questions about the accelerometer.

1) With the Earth spinning around its axis, how with an accelerometer can you tell the proportion of the reading from the Earth spinning around its axis and the proportion from gravity ( I am assuming the effect from the spin will produce an effect equivalent to some acceleration towards the Earth's core)?

2) With the Earth rotating around the Sun I imagine that there is an increased pull from the Sun in midday due to the gravitational pull of the Sun (as you would be slightly closer), but a pull in the other direction from the acceleration due to the Earth orbiting the axis of the Sun (which I assume is equivalent to some acceleration towards the Sun's core, ignoring that the orbit is not circular (does that make a difference?)), which is greater, and is there a name for the effect?

 

[PeterDonis]

With the Earth spinning around its axis, how with an accelerometer can you tell the proportion of the reading from the Earth spinning around its axis and the proportion from gravity

First, there is no "proportion from gravity"; gravity does not cause any proper acceleration. What you are thinking as "the proportion from gravity" is the effect of the Earth pushing on the accelerometer.

With that correction, the answer to your question is that you can't tell. The accelerometer just tells you the magnitude and direction of the total proper acceleration. It doesn't tell you how to break that total up into parts.

which I assume is equivalent to some acceleration towards the Sun's core

No, it isn't, because we are talking about proper acceleration, and as above, gravity does not cause any proper acceleration. The proper acceleration of the Earth as a whole, for example, is zero; it is in free fall orbiting the Sun.

 

[Dale]

With that correction, the answer to your question is that you can't tell.

I disagree a little here. A 6-degree-of-freedom accelerometer can separate out the rotational and linear accelerations. The rotational acceleration can be measured with gyroscopes and the linear acceleration can be measured with a mass on some springs.

 

[PeterDonis]

A 6-degree-of-freedom accelerometer can separate out the rotational and linear accelerations.

Locally, yes. But it can't tell you how the local accelerations are related to global properties such as the rotation of the Earth. For that you need global information.

In other words, if your local accelerometers tell you that you are in a frame with linear acceleration g pointing in some direction, plus rotational acceleration (a better term would be "vorticity", but I don't think we need to get into too much detail here) w, that could be because you are sitting at rest on a rotating planet, but it could also be because you are traversing a particular circular trajectory in flat spacetime, or it could be because you are "hovering" motionless (with respect to infinity) above a rotating neutron star or black hole--and there could be other possibilities as well that I haven't thought of. The only way to distinguish these possibilities is by looking at global information; you can't tell just from your local accelerometer measurements. (This is basically a consequence of the equivalence principle.)

 

[Dale]

Locally, yes. But it can't tell you how the local accelerations are related to global properties such as the rotation of the Earth. For that you need global information.

Yes, I agree. And with a system of accelerometers located throughout spacetime you can gather the global information also. Basically, I want to emphasize that these are measurable effects, although you may need to piece together lots of measurements.

 

[name123]

First, there is no "proportion from gravity"; gravity does not cause any proper acceleration. What you are thinking as "the proportion from gravity" is the effect of the Earth pushing on the accelerometer.

With that correction, the answer to your question is that you can't tell. The accelerometer just tells you the magnitude and direction of the total proper acceleration. It doesn't tell you how to break that total up into parts.

I had meant by the term "proportion from gravity" the proportion of the reading on the accelerometer (the question under (1) was about the reading on an accelerometer, the term "proper acceleration" was not mentioned). Just to confirm (I think by your mention of the Earth pushing on the accelerometer you were suggesting that it would) the accelerometer would register a reading due to gravity also would it not? The reason I thought it would is that I have read that Einstein assumed an equivalence "the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system".

Is there an experiment that established that the Earth is spinning (thrusting an object in space in the direction of the proposed spin and thrusting one against it or something)?

No, it isn't, because we are talking about proper acceleration, and as above, gravity does not cause any proper acceleration. The proper acceleration of the Earth as a whole, for example, is zero; it is in free fall orbiting the Sun.

I thought if there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration which would be a measurable effect (using an accelerometer). Would the Earth not be considered to be changing direction when orbiting the Sun, or was it that what I thought was wrong?

 

[Dale]

had meant by the term "proportion from gravity" the proportion of the reading on the accelerometer

0%. Accelerometers measure only proper acceleration which is due entirely to non gravitational forces. The gravitational acceleration is not detected by an accelerometer.

1) was about the reading on an accelerometer, the term "proper acceleration" was not mentioned)

They are the same thing.

I think by your mention of the Earth pushing on the accelerometer you were suggesting that it would) the accelerometer would register a reading due to gravity also would it not?

It would not. The acceleration only registers the proper acceleration, which comes entirely from the contact force pushing upwards.

 

[Nugatory]

Is there an experiment that established that the Earth is spinning (thrusting an object in space in the direction of the proposed spin and thrusting one against it or something)?

Yes. Google for "Foucault's pendulum".

Would the Earth not be considered to be changing direction when orbiting the Sun, or was it that what I thought was wrong?

The orbiting Earth is changing its direction in space, but not spacetime - it's following a straight line through spacetime.

 

[PeterDonis]

I had meant by the term "proportion from gravity" the proportion of the reading on the accelerometer

There isn't any "proportion" in the reading on the accelerometer; it's just one reading.

the accelerometer would register a reading due to gravity also would it not?

As Dale said, it would not.

I have read that Einstein assumed an equivalence "the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system"

Yes, but this doesn't mean what you appear to think it means.

Consider an accelerating rocket in flat spacetime (no gravity) with 1 g proper acceleration. You can stand on the floor of this rocket just as you would stand on the surface of the Earth, and if the only information you have is what you can measure inside the rocket, you have no way of telling whether it is in fact at rest on the Earth's surface or accelerating through free space at 1 g. That is the equivalence that Einstein was talking about.

But if we now consider what this tells us about a "gravitational field", it tells us that the acceleration you feel standing on the surface of the Earth is not due to "gravity"; it's due to the surface of the Earth pushing up on you, just as the acceleration you feel in the rocket in free space is due to the rocket pushing up on you. In other words, what the equivalence told Einstein was that "gravity", at least in the sense of "acceleration due to gravity", is not a force at all. It's just an artifact of the way you choose your reference frame.

 

[name123]

Yes, but this doesn't mean what you appear to think it means.

Consider an accelerating rocket in flat spacetime (no gravity) with 1 g proper acceleration. You can stand on the floor of this rocket just as you would stand on the surface of the Earth, and if the only information you have is what you can measure inside the rocket, you have no way of telling whether it is in fact at rest on the Earth's surface or accelerating through free space at 1 g. That is the equivalence that Einstein was talking about.

But if we now consider what this tells us about a "gravitational field", it tells us that the acceleration you feel standing on the surface of the Earth is not due to "gravity"; it's due to the surface of the Earth pushing up on you, just as the acceleration you feel in the rocket in free space is due to the rocket pushing up on you. In other words, what the equivalence told Einstein was that "gravity", at least in the sense of "acceleration due to gravity", is not a force at all. It's just an artifact of the way you choose your reference frame.

But if you had an accelerometer in the rocket, could you not just take a reading, because were you not all saying that accelerometers do not measure gravity? I was assuming that the 1g was supposed to be equivalent to the gravity on Earth, rather than a measurement of its spin. Was that assumption wrong?

 

[Dale]

But if you had an accelerometer in the rocket, could you not just take a reading

Yes, you can take an accelerometer reading in a rocket. It will measure the proper acceleration, as always.

were you not all saying that accelerometers do not measure gravity?

Accelerometers do not measure the acceleration due to gravity nor to inertial forces. They only measure proper acceleration, which is never due to gravity or inertial forces.

I was assuming that the 1g was supposed to be equivalent to the gravity on Earth, rather than a measurement of its spin

What? I don't understand this question.

 

[PeterDonis]

I was assuming that the 1g was supposed to be equivalent to the gravity on Earth

1 g is a proper acceleration; it is the acceleration you feel standing on the surface of the Earth. But just from that measurement alone, you can't tell whether you are feeling 1 g proper acceleration because you are standing on the surface of the Earth, or because you are accelerating in flat spacetime. You need other information to distinguish those two cases.

rather than a measurement of its spin

The 1 g proper acceleration by itself tells you nothing about spin. Other measurements (such as the gyroscopes Dale mentioned) are needed if you want to know about spin.

 

[name123]

1 g is a proper acceleration; it is the acceleration you feel standing on the surface of the Earth. But just from that measurement alone, you can't tell whether you are feeling 1 g proper acceleration because you are standing on the surface of the Earth, or because you are accelerating in flat spacetime. You need other information to distinguish those two cases.

What would be causing the 1g proper acceleration if you were standing on the surface of the Earth? We might just be going around in circles here, and you meaning that if there was no Earth but you were in free fall due to gravity that there would be no acceleration, and therefore claiming that gravity does not cause proper acceleration. Whereas I have been discussing taking measurements with an accelerometer while standing on Earth, and talking about the acceleration due to gravity. The reason I was considering a proportion of the reading, in the situation, to be due to gravity, was because if the Earth had been mainly hollow for example, it would have curved spacetime less, and therefore there would be less gravity, and therefore the measured acceleration standing on the surface of the Earth would have been less. The greater reading being due to greater gravity. Do you think that perhaps that is the issue. For example would there be less measured acceleration using an accelerometer when standing on the surface of a large sphere if the sphere had less mass?

The 1 g proper acceleration by itself tells you nothing about spin. Other measurements (such as the gyroscopes Dale mentioned) are needed if you want to know about spin.

But part of the accelerometer reading would be due to the Earth spinning though would it not?

Also regarding the Earth orbiting the Sun, Nugatory mentioned:

"The orbiting Earth is changing its direction in space, but not spacetime - it's following a straight line through spacetime."

Can I assume that straight lines in spacetime depend on velocity?

 

[Dale]

What would be causing the 1g proper acceleration if you were standing on the surface of the Earth?

The normal force of the ground pushing up on the bottom of your feet

Whereas I have been discussing taking measurements with an accelerometer while standing on Earth, and talking about the acceleration due to gravity.

I understood that and I am pretty sure that @PeterDonis did also. None of the proper acceleration is due to gravity, it is all due to the normal force. Note that the proper acceleration is upwards, which is the direction of the normal force, not the gravitational force.

 

[name123]

The normal force of the ground pushing up on the bottom of your feet

But would that not be linked to gravity. For example if you were standing on a sphere with less mass you'd experience less acceleration because there would be less gravitational force.

 

[Dale]

But would that not be linked to gravity. For example if you were standing on a sphere with less mass you'd experience less acceleration because there would be less gravitational force.

The reduced proper acceleration is due to the reduced normal force. In this scenario both the normal and the gravitational forces changed together. Consider instead situations where the normal force is different but the gravitational force is the same, or vice versa.

 

[name123]

The reduced proper acceleration is due to the reduced normal force. In this scenario both the normal and the gravitational forces changed together. Consider instead situations where the normal force is different but the gravitational force is the same, or vice versa.

In the scenario I gave, with large spheres of differing mass, the reduction in normal force is due to the reduction in gravitational force is it not? So if the gravitational force was increased the normal force would be also would it not (given the context where there would be a normal force)? Making the gravitational force an indirect cause if not a direct cause (in that scenario).