I The Equivalence Principle as a Starting point of GR

[cianfa72]

Either or both. What difference do you imagine?

Sorry, I've not a precise idea about as an accelerometer is made. My point was just that to measure the proper acceleration felt by an object you have to attach the accelerometer case to it

 

[jbriggs444]

Sorry, I've not a precise idea about as an accelerometer is made. My point was just that to measure the proper acceleration felt by an object you have to attach the accelerometer case to it

The case around an accelerometer is not essential. In principle, a [single axis] accelerometer is nothing more than a known mass on a spring and a scale against which the spring's length can be assessed. You attach the measuring scale and the free end of the spring to the object whose acceleration you wish to measure. You then see how long the spring is when the mass has settled into a position.

Putting a case around the whole assembly keeps contaminants out and keeps anyone from "putting their fingers on the scale"

 

[cianfa72]

The idea is that if you bolt the case to the ceiling, it does not matter whether you attach the spring to the case or to the ceiling.

Thus basically the accelerometer bolted to the elevator ceiling and the object attached to the spring 'share' the same proper acceleration measured by the accelerometer (actually the spring inside it) against its measuring scale

 

[Herman Trivilino]

Please, could you elaborate a bit around about what you mean with "response of objects inside the elevator" in both cases (no-gravity, gravity) ? thanks

With gravity, you release a ball and it accelerates towards the floor. Without gravity, you release a ball and the floor accelerates towards the ball. This is an "equivalence".

 

[epovo]

No. What you are describing is coordinate acceleration, which is not the same in all inertial frames.

Do you mean that each inertial frame may have its spatial coordinates rotated in its own way, giving different coordinate values but the same magnitude to the observed acceleration vector?

 

[jbriggs444]

Do you mean that each inertial frame may have its spatial coordinates rotated in its own way, giving different coordinate values but the same magnitude to the observed acceleration vector?

If I understand your point, it is that rotating a coordinate system 90 degrees to the right would mean that an "eastward" coordinate acceleration becomes a "northward" coordinate acceleration -- different coordinate values but the same magnitude.

No, that is not what @PeterDonis has in mind. Peter's point applies to "boosts" of a coordinate system.

Consider, for instance a "rest" coordinate system where an object with a 1 gee proper acceleration has a 9.8 meter/sec^2 coordinate acceleration. Now boost to a coordinate system where the object is moving at c minus 1 meter per second. The object still has a proper acceleration of one gee. Wait for one coordinate second and look at the velocity again. It cannot be as much as c. The coordinate acceleration in this frame is sure to be less than 1 meters/second^2 (actually, a lot less).

 

[epovo]

I understand now. I never thought of that! Thank you

 

[alantheastronomer]

Thanks PeterDonis. I am processing.

So gravity is a force in Newtonian physics but according to GR, which is the best theory we have about gravity, it is not a force anymore (but originates from the curvature of spacetime). I know physicists are trying to unify the four forces of nature (weak nuclear force, strong nuclear force, gravity, electromagnetic force) with the weak nuclear and electromagnetic one already being unified...But what happens with gravity now since it is not a force anymore and is explained/tied to this geometrical framework?

Thanks for any insight.

I thought this question was a very insightful observation. Gravity is a manifestation of mass in the same way that electromagnetism is a manifestation of electric charge and the strong force a manifestation of color charge. So even though it is not a force, it is the result of a fundamental property of nature.

 

[cianfa72]

Consider, for instance a "rest" coordinate system where an object with a 1 gee proper acceleration has a 9.8 meter/sec^2 coordinate acceleration. Now boost to a coordinate system where the object is moving at c minus 1 meter per second. The object still has a proper acceleration of one gee. Wait for one coordinate second and look at the velocity again. It cannot be as much as c. The coordinate acceleration in this frame is sure to be less than 1 meters/second^2 (actually, a lot less).

The boost you are talking about should be basically a change of inertial system from the "rest" coordinate system. Which are the corresponding transformations for the coordinate acceleration of the object ?

 

[Nugatory]

The boost you are talking about should be basically a change of inertial system from the "rest" coordinate system. Which are the corresponding transformations for the coordinate acceleration of the object ?

The relativistic velocity addition rule.

 

[cianfa72]

The relativistic velocity addition rule.

Using that does it result the coordinate acceleration of the 1 gee proper acceleration object is different in the two system of reference ?

 

[Nugatory]

Using that does it result the coordinate acceleration of the 1 gee proper acceleration object is different in the two system of reference ?

Yes.

 

[vanhees71]

I thought this question was a very insightful observation. Gravity is a manifestation of mass in the same way that electromagnetism is a manifestation of electric charge and the strong force a manifestation of color charge. So even though it is not a force, it is the result of a fundamental property of nature.

The flaw in this argument is that in GR the source of gravity is not only mass but any kind of energy, momentum and stress. Also in electromagnetism it's not only electric charge but also electric current distributions (on a fundamental level also the magnetic moments of particles due to spin).

 

[epovo]

Using that does it result the coordinate acceleration of the 1 gee proper acceleration object is different in the two system of reference ?

I think it's
$$ a = g ( 1 - v^2/c^2) ^ {3/2} $$

 

[pervect]

I think it's
$$ a = g ( 1 - v^2/c^2) ^ {3/2} $$

This is true in one dimension, when the acceleration is parallel to the velocity.

If we let $\alpha$ be the magnitude of the proper acceleration, $\vec{v}$ be the coordinate velocity, $\vec{a}$ be the coordinate acceleration, $\gamma = 1/\sqrt{1-v^2/c^2}$ be the gamma factor, and let $a = ||\vec{a}||$, we can state:

In the parallel case $\alpha = \gamma^3 \, a$

In the perpend caase $\alpha = \gamma^2\,a$

In the general case we can write:

$$\alpha^2 = \gamma^4 \, (\vec{a}\cdot \vec{a}) + \frac{\gamma^6}{c^2}\,(\vec{v}\cdot \vec{a})^2$$

see [link]

I've added a "missing" factor of c^2, the original assumed c=1

In one spatial dimension, $\vec{a}$ and $\vec{v}$ are parallel, so we can write $\vec{v} = v \, \vec{x}$ and $\vec{a} = a \, \vec{x}$ . Thus we have $\vec{a} \cdot \vec{a} = a^2 (\vec{x} \cdot \vec{x}) = a^2$ and $\vec{v} \cdot \vec{a} = v \, a$, and the above simplifies to

$$\alpha^2 = \frac{a^2}{(1-v^2/c^2)^3}$$

which implies $\alpha = \gamma^3 \,a$

When $\vec{a} \cdot \vec{v}=0$, i.e. the acceleration is perpendicular to the velocity, we have instead

$$\alpha^2 = \frac{a^2}{(1-v^2/c^2)^2}$$

which implies $\alpha = \gamma^2 \, a$