Stationary frames of reference

[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).

 

[Dale]

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).

If you have two possible sources of an effect and you wish to determine which one causes it then you must vary them independently. Think of scenarios where the gravity is constant, but the normal force is different. Like a man standing on a glass floor which breaks. Gravity is the same before and after, but the normal force changes to zero after. What does the accelerometer record?

 

[Boing3000]

the reduction in normal force is due to the reduction in gravitational force is it not

If you want to think into GR framework you have to realize that gravity is not a force. "Gravitating" if what things do when no force apply to them (<= inertial), when they are in "free-fall", like (falling)apple, satellite, the Earth (the whole system you included). It is also said those inertial things follow straight line following (curved)spacetime.

On the ground your free-fall is stopped by the ground exerting a force on you and your accelerometer.

 

[name123]

If you have two possible sources of an effect and you wish to determine which one causes it then you must vary them independently. Think of scenarios where the gravity is constant, but the normal force is different. Like a man standing on a glass floor which breaks. Gravity is the same before and after, but the normal force changes to zero after. What does the accelerometer record?

So if you were in the jury at a court case where a person had attached a rope to a winch and placed it through a hole in a wall, and wrapped it around a person and then turned the winch on such that the person was killed, that you would accept the argument that the person turning on the winch was not the cause of the death? The defence argument being that the basis to determine whether the forces generated by the winch (which the defendant was responsible for) contributed to the death, you would have to consider whether there had been a paper wall, and their conclusion that the forces generated by the winch being unrelated to the effect (the death). Their argument that the source (the wall) of the effect (of the person dying) was one that the defendant had no causal connection to.

Could the gravitational force not influence the calculated normal force in a situation?

Why did the glass floor break, was there a force that prior to its breaking it was resisting?

 

[name123]

If you want to think into GR framework you have to realize that gravity is not a force. "Gravitating" if what things do when no force apply to them (<= inertial), when they are in "free-fall", like (falling)apple, satellite, the Earth (the whole system you included). It is also said those inertial things follow straight line following (curved)spacetime.

On the ground your free-fall is stopped by the ground exerting a force on you and your accelerometer.

I am using language pretty loosely here. I could change the term gravity for mass if you like, but one term seems to be useful in highlighting the point you are getting to. As in the influence of mass on "Gravitating".

Why does the ground differentiate on the force it exerts. To explain what I mean, imagine
Scenario 1) I touch a wall with my finger (the wall does not break)
Scenario 2) A car drives into a wall (the wall does not break)

With architecture is there no consideration of the force being placed on the wall? And a concept of the amount of resistance that the wall can give?

 

[PeterDonis]

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

The Earth's surface pushing up on you. Just as in the accelerating rocket in flat spacetime, the 1g proper acceleration you feel is due to the rocket's floor pushing up on you.

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.

This is true, but it still doesn't mean the acceleration you feel standing on the Earth's surface is "due to gravity". It just means that, if the Earth's mass were smaller so it curved spacetime less (more precisely, curved spacetime less at the same radius from the center), the Earth's surface wouldn't push up on you as hard. "Gravity" here is the curvature of spacetime, but it isn't curvature of spacetime that's pushing up on you, it's the Earth's surface.

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?

No. Once more: in GR, gravity is not a force. It's spacetime curvature. Spacetime curvature doesn't push on anything.

So if you were in the jury at a court case where a person had attached a rope to a winch and placed it through a hole in a wall, and wrapped it around a person and then turned the winch on such that the person was killed, that you would accept the argument that the person turning on the winch was not the cause of the death?

Of course not. The person is killed because the rope exerts a force on their neck and breaks it. But that doesn't contradict the fact that in GR, gravity is not a force. It isn't gravity that breaks the person's neck. Or spacetime curvature either.

I am using language pretty loosely here.

Yes, and you need to stop doing that. We are talking about physics, and you need to use language precisely in physics or you will not understand things properly.

Scenario 1) I touch a wall with my finger (the wall does not break)
Scenario 2) A car drives into a wall (the wall does not break)

You keep on giving examples of things that aren't gravity exerting force on something, and then trying to argue that somehow that means gravity is a force. Can you see the problem?

 

[name123]

This is true, but it still doesn't mean the acceleration you feel standing on the Earth's surface is "due to gravity". It just means that, if the Earth's mass were smaller so it curved spacetime less (more precisely, curved spacetime less at the same radius from the center), the Earth's surface wouldn't push up on you as hard. "Gravity" here is the curvature of spacetime, but it isn't curvature of spacetime that's pushing up on you, it's the Earth's surface.

I am thinking of Gravity as the measured influence of "mass" on gravitation. What is the theory behind why the Earth's surface, or a wall, or a table, pushes up more in some places than others? With a spring bed for example I would expect it to be that there was more force on some areas than others (but as I understand the respondents there is no difference in force on areas). I was assuming that the curvature created a force on a body which lead to motion of that body, which on contact with another could be resisted or not (if the resistance was not enough) by that body.

Yes, and you need to stop doing that. We are talking about physics, and you need to use language precisely in physics or you will not understand things properly.

While I am grateful for your help, I had assumed that it was the burden of the more educated in physics (that were trying to help the less educated) to try to empathise with what the less educated were getting at and explain (as there could be loads of less educated that were thinking the same way). As opposed to them being pedantic until the less educated phrased the question in a way that gave the more educated no option to misunderstand what was being asked.

 

[name123]

No. Once more: in GR, gravity is not a force. It's spacetime curvature. Spacetime curvature doesn't push on anything.

So change in motion requires no force? Does not curvature perpetuate at the speed of light and does it not have an effect?

Of course not. The person is killed because the rope exerts a force on their neck and breaks it. But that doesn't contradict the fact that in GR, gravity is not a force. It isn't gravity that breaks the person's neck. Or spacetime curvature either.

But if you have two possible sources of an effect (the wall or the winch) and you wish to determine which one causes it then should you not (as Dale suggested) vary them independently. Think of scenarios where the winch is constant, but the normal force of the wall is different. Like a man being pulled through a paper wall which breaks. The winch force is the same before and after, but the normal force (of the paper wall) changes to zero after.

 

[PeterDonis]

I am thinking of Gravity as the measured influence of "mass" on gravitation.

In other words, you are thinking of gravity as the measured influence of mass on gravity? That doesn't even make sense.

In GR, the best way to think of "gravity" is as spacetime curvature. That is the "gravity" that does not depend on your choice of coordinates, and that appears in the laws of physics.

What is the theory behind why the Earth's surface, or a wall, or a table, pushes up more in some places than others?

General relativity, whose laws--the Einstein Field Equation--are based on spacetime curvature and stress-energy (which is how what you ordinarily think of as "mass", i.e., the source of "gravity", appears). You solve the EFE under particular conditions that you are interested in, and that tells you the geometry of spacetime. Then you look at different possible worldlines in that spacetime geometry, and the laws tell you what proper acceleration (how much "pushing up") will be experienced by an object following that worldline.

change in motion requires no force?

"Change in motion" depends on your choice of coordinates. So it can't appear in the laws of physics.

Does not curvature perpetuate at the speed of light

Changes in curvature propagate at the speed of light. But for a gravitating body like the Earth, the spacetime curvature is, to a very good approximation, static, i.e., unchanging, so there's nothing to propagate.

does it not have an effect?

Yes, spacetime curvature has an effect. See above.

if you have two possible sources of an effect (the wall or the winch)

I don't understand. How is the person in your scenario killed? I thought it was the rope attached to the winch breaking their neck. Is it the winch pulling them through the wall and killing them by contact with the wall?

Assuming the latter is the case, I still don't see the point. The person wouldn't have been pulled through the wall if the winch hadn't been turned on. So whoever turned the winch on is responsible for their death. The fact that a wall made of something less sturdy wouldn't have killed them is irrelevant. And this whole scenario seems to me like a quibble; we're not talking about legalistic reasoning, we're talking about physics.

Think of scenarios where the winch is constant, but the normal force of the wall is different. Like a man being pulled through a paper wall which breaks. The winch force is the same before and after, but the normal force (of the wall) changes to zero after

If the winch isn't turned on, the force of the wall on the person is zero regardless of what the wall is made of. So the causal factor that varies the force is the winch. But this whole argument is still, as I said above, a quibble.

 

[PeterDonis]

I had assumed that it was the burden of the more educated in physics (that were trying to help the less educated) to try to empathise with what the less educated were getting at

We can try, but if you are unable or unwilling to learn the correct language in which to describe what you are getting at, this is often not possible. Basically what we have been telling you is that you are using the wrong concepts, and your language reflects that. We are trying to get you to change the concepts you use, from ones that don't work to ones that work. It won't do any good for you to continue to insist on us answering your questions using your concepts, because they don't work.

In conclusion: this is a "B" level thread, and I think we have answered your questions as well as they can be answered in a "B" level thread. To briefly summarize:

(1) In GR, gravity is not a force. It's spacetime curvature. Only things that are actually felt as proper acceleration are forces in GR.

(2) The laws of physics don't include things that depend on your choice of coordinates. That includes "rest", velocity, and coordinate acceleration.

To go into more detail about these things would require a thread at a level higher than "B", which would require you to have the background for a discussion at that level. I would suggest working through a GR textbook: Sean Carroll's online lecture notes are a good choice to start with:

https://arxiv.org/abs/gr-qc/9712019

This thread is closed.

 

[Dale]

So if you were in the jury ...

You seem to be trolling now rather than looking for help learning GR. If you want to learn these concepts then you are going to have to pay attention with an open mind. And if you want us to help you then you will have to respond helpfully rather than sarcastically.

This thread is closed temporarily. PM me with a sincere analysis of the normal force, gravitational force, and accelerometer reading, and I will reopen it.