Confusion over what constitutes a uniform gravitational field in relativity

Question

Suppose we have some observer moving upwards with a constant proper acceleration, by the equivalence principle this is the same as the observer remaining stationary in a gravitational field, like standing on the surface of the earth.

The issue I'm having is that this situation is often called "a uniform gravitational field", let me explain why I find this problematic.

The uniformly accelerating observers coordinates are describable by the rindler coordinates, and in the rindler coordinates as you move further up from the observer it requires less and less acceleration to remain stationary with respect to them, so can we really call the gravitational field uniform?

If we instead force any stationary observer to accelerate at the same rate, by their individual rindler coordinates they would appear to move away from each other, which makes it hard to claim they're "stationary" at all.

So in one case it's hard to call it "uniform", and in the other it's hard to say anything is "stationary".

This confuses me greatly.

Any help with understanding it would be appreciated.

RE-EDIT:

a. I'm well aware that the acceleration to stay stationary in Rindler coordinates varies from place to place, for this I was asking why they're often referred to as a "uniform gravitational field" when the acceleration needed is non-uniform.

b. I was then asking if you could construct a gravitational field where the acceleration needed to stay stationary is the same everywhere, unlike the Rindler coordinates.

c. I've also seen it said that the coordinates of any observer stationary in a gravitational field are locally Rindler, however in rindler coordinates equivalently accelerating things appear to move away from each other, so how would this be avoided if said hypothetical uniform gravitational field existed?

Answer 1

a. The Rindler coordinates are sometimes referred to as a "uniform gravitational field" because they have the property of not exhibiting any tidal-forces, like a classical uniform gravitational field.

b. You can, the metric for one such field is $ds^{2}=e^{2gx}dt^{2}-dx^{2}$ for $c=1$ in $1+1$ dimensions.

c. Imagine a sphere, where the lines of latitude are time coordinates and the lines of longitude are spatial coordinates. It can be observed that any two lines of latitude remain equidistant from each other. Objects "dropped" from lines of latitude will not continue along them, but will instead travel downwards or upwards depending on if they were dropped above or below the equator, this is because they travel in a straight line along the surface. The rate that they drop depends on the elevation of the latitude line. Therefore an object which stays on a line of latitude will have to accelerate to maintain its position, and this acceleration will vary depending on the elevation. So we can see that objects that run along lines of latitude remain at a constant distance from each other, but accelerate at different rates. Even though a small subsection of the sphere looks like a flat plane, it will still exhibit this behavior. This is caused by the curvature of the sphere.

As far as I'm aware this is analogous with the problem of "local Rindler coordinates", in Rindler coordinates two objects experiencing the same acceleration will appear to move away from each other and any spacetime will look flat like the Rindler coordinates on a small scale. However, if the spacetime is curved, like a sphere, the rules of the truly flat Rindler coordinates do not necessarily apply, even though it looks like them locally.

Answer 2

In every position of an uniform gravity field a test mass receives the same amount of momentum in same time.

An observer in a low position of an uniform gravity field says that a test mass receives the same amount of momentum everywhere in the field in same time, and said amount is large.

An observer in a high position of the aforementioned uniform gravity field says that a test mass receives the same amount of momentum everywhere in the field in same time, and said amount is small.

Addition:

At the Rindler-horizon of an accelerating rocket the accelerometer of a hovering observer reads an infinite value.

At one micrometer from the Rindler-horizon of an accelerating rocket the accelerometer of a hovering observer reads a very large value. So one micrometer causes an infinite difference in measured accelerations.

At the position of the rocket the reading of an accelerometer is the acceleration of the rocket.

When an accelerometer is winched up along a mast attached to the rocket, the reading decreases. At infinite height the reading is infinitely close to zero.

If rinler-observer A sends a pulse of light with energy X to a rindler-observer B, and rindler-observer B says that the energy of the light pulse is 10*X, then also the reading of the accelerometer of rindler-observer B is ten times of the reading of the accelerometer of the rindler-observer A.

Addition2:

In classical mechanics "uniform gravity field" means a gravity field with no tidal forces.

In general relativity "uniform gravity field" means the same as in classical mechanics.

Answer 3

The equivalence is only valid locally, meaning over effectively infinitesimal time and space "distances". I suggest you're effectively pointing out that over larger distances, the equivalence breaks down.