# B Stationary frames of reference

1. Jan 6, 2017

Thanks

2. Jan 6, 2017

### Battlemage!

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?

3. Jan 6, 2017

### Staff: Mentor

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

4. Jan 24, 2017

### name123

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?

5. Jan 24, 2017

### Staff: Mentor

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.

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.

6. Jan 24, 2017

### Staff: Mentor

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.

7. Jan 24, 2017

### Staff: Mentor

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.)

8. Jan 25, 2017

### Staff: Mentor

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.

9. Jan 25, 2017

### name123

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)?

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?

Last edited: Jan 25, 2017
10. Jan 25, 2017

### Staff: Mentor

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

They are the same thing.

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

11. Jan 25, 2017

### Staff: Mentor

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

12. Jan 25, 2017

### Staff: Mentor

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

As Dale said, it would not.

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.

13. Jan 25, 2017

### name123

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?

14. Jan 25, 2017

### Staff: Mentor

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

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.

What? I don't understand this question.

15. Jan 25, 2017

### Staff: Mentor

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.

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.

16. Jan 25, 2017

### name123

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?

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?

17. Jan 25, 2017

### Staff: Mentor

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

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.

18. Jan 25, 2017

### name123

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.

19. Jan 25, 2017

### Staff: Mentor

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.

20. Jan 25, 2017

### name123

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).

Last edited: Jan 25, 2017