Moving faster but taking longer

[gonegahgah]

Cool Jesse that's actually what I would have thought; every bit.

I noticed in one of the links supplied in another thread that they compared black holes with acceleration and came to similar things such as event horizons. The only difference they noted was that you don't get matter torn apart at the event horizon for acceleration that you get for black holes.
As you mentioned, under normal conditions the stretching is so tiny to be ignored.
I would guess that black holes are not normal conditions so for even relatively small distances the gravitational difference is large. So much so that we would probably be torn apart at some point in our fall into a black hole; whereas we would not be torn apart at the event horizon of something accelerating away.
Is this correct?

I wondered to myself if we could simulate freefall but in deep space. Even the part where you eventually ka-splat with the Earth. I realized that if you accelerated something towards the person in the box then it could be made to seem equivalent to someone free-falling in the box on Earth. ie. They couldn't tell the difference; they would both feel weightless until they struck; in a similar experience.

That would give:
In gravitation field feeling weight (on surface) | In gravitational field weightless (freefall)
------------------------------------------------------------------------------------
Accelerating in deepest space feeling weight | Steady in deepest space accelerated at

So steady in deepest space being accelerated at would be equivalent to being in a gravitational field in free fall & accelerating in deepest space would be equivalent to being on the surface of a planet.

I'm pretty sure that this is basically just agreeing with what you are saying.
Is it okay?

 

[JesseM]

Cool Jesse that's actually what I would have thought; every bit.

I noticed in one of the links supplied in another thread that they compared black holes with acceleration and came to similar things such as event horizons. The only difference they noted was that you don't get matter torn apart at the event horizon for acceleration that you get for black holes.

Yes, an observer who accelerates at a constant rate forever will have a Rindler horizon beyond which events will never have their light catch up to him, although of course if he stops accelerating this horizon will disappear and the light from these events will be able to catch up with him.

As you mentioned, under normal conditions the stretching is so tiny to be ignored.
I would guess that black holes are not normal conditions so for even relatively small distances the gravitational difference is large. So much so that we would probably be torn apart at some point in our fall into a black hole; whereas we would not be torn apart at the event horizon of something accelerating away.
Is this correct?

That's right, the tidal forces actually go to infinity when you get to the singularity of a black hole. For a smaller black hole formed by the collapse of an ordinary star I don't think you could even get to the event horizon without being torn apart, but for much larger black holes like those found at the center of galaxies you could make it past the event horizon, you wouldn't get torn apart until you were closer to the singularity. And of course if you're an inertial observer crossing the Rindler horizon of another observer accelerating away from you, you don't feel any tidal forces at all.

I wondered to myself if we could simulate freefall but in deep space. Even the part where you eventually ka-splat with the Earth. I realized that if you accelerated something towards the person in the box then it could be made to seem equivalent to someone free-falling in the box on Earth. ie. They couldn't tell the difference; they would both feel weightless until they struck; in a similar experience.

That's true, although like I said the equivalence principle only works in a small enough window of space and time that the curvature of spacetime doesn't make any difference; so you'd have to pick a small time window for the freefalling observer, starting soon before he hit the ground and ending soon after, if you wanted to treat this as equivalent to a situation where an inertial observer is hit by an accelerating patch of ground in deep space.

That would give:
In gravitation field feeling weight (on surface) | In gravitational field weightless (freefall)
------------------------------------------------------------------------------------
Accelerating in deepest space feeling weight | Steady in deepest space accelerated at

So steady in deepest space being accelerated at would be equivalent to being in a gravitational field in free fall & accelerating in deepest space would be equivalent to being on the surface of a planet.

Yes, although the part where the inertial observer is being "accelerated at" by a piece of ground is only necessary if you want to look at a time window for the freefalling observer that includes that observer hitting the ground (and thus no longer being in freefall); normally physicists would just talk about the equivalence between a small window of spacetime where the observer is in freefall throughout with the laws of physics as seen by an inertial observer in deep space, so there'd be no need to worry about the ground hitting either of them.

 

[gonegahgah]

Cool, sounds good. Can the duration of equivalence be made larger than a small window?

eg. If you had a peephole in the bottom of your box and we had a built base that we will call 'ground' and the only thing you can see through the peephole is this 'ground', could we make the observer in the box unable to know which situation - accelerated/gravity - they are in?

ie. If they are on this 'ground' on a planet's surface then they would feel no different to if they were on this 'ground' in deep space and the 'ground' were being accelerated - with them on it - at a decreasing rate with an equivalent proportional rate of 1/r² for both.

ie. Also if this 'ground' were on a planet's surface and the observer freefell towards it they would see and feel the same thing through the peephole as an observer in a steady box in deep space and the 'ground' were accelerated at them at an equivalent increasing rate; until it hit them and then the 'ground' would have to maintain a constant acceleration from that point.

Would these increase the window of equivalence Jesse?

 

[JesseM]

Cool, sounds good. Can the duration of equivalence be made larger than a small window?

eg. If you had a peephole in the bottom of your box and we had a built base that we will call 'ground' and the only thing you can see through the peephole is this 'ground', could we make the observer in the box unable to know which situation - accelerated/gravity - they are in?

ie. If they are on this 'ground' on a planet's surface then they would feel no different to if they were on this 'ground' in deep space and the 'ground' were being accelerated - with them on it - at a decreasing rate with an equivalent proportional rate of 1/r² for both.

ie. Also if this 'ground' were on a planet's surface and the observer freefell towards it they would see and feel the same thing through the peephole as an observer in a steady box in deep space and the 'ground' were accelerated at them at an equivalent increasing rate; until it hit them and then the 'ground' would have to maintain a constant acceleration from that point.

Would these increase the window of equivalence Jesse?

The problem with increasing the time window is that it allows tidal forces to become apparent for the observer freefalling in a gravitational field, which marks his experience as different from the one of the observer accelerating in deep space. The vertical tidal force due to gravity being stronger at your feet than your head is just one type of tidal force; there is also a horizontal form of tidal force due to the fact that objects on opposite sides of the room won't be pulled straight down but instead slightly towards the center of the room since they would both eventually have to converge on the center of the planet that's the source of gravity (see the final illustration on this page). So, if you're falling for an extended time, two small objects placed on either side of the room will eventually be observed to drift towards the center, demonstrating to the observer that he is not just accelerating in deep space. Tidal forces only go to zero in the limit as you pick a region of space and time whose size is approaching zero, and it's only in that limit that the equivalence principle becomes exact.

 

[gonegahgah]

I see what you mean.

The other insurmountable challenge I see is creating an equivalent observer & satellite scenario either; if we wanted to try.
Say we replace the person and box with a GPS and place an observer on the 'ground' in the examples.
Unfortunately we can not create an equivalent scenario between gravity and acceleration for this because we will not be able to create a constant equivalent for the observer on the 'ground' between the two types (ie acceleration vs gravity).

The other problem of course is that it is probably impossible to simulate an orbit scenario using acceleration anyway? That is probably correct; is it?

So it would be difficult (impossible?) to model the GPS using acceleration instead of gravity? So unfortunately we couldn't compare them this way?

 

[DrGreg]

As a curiousity, if the flat-earthists were right, i.e. if the Earth were not a sphere but an infinite flat plane, I think I'm right to say that the equivalence principle would be exactly true and not just locally true. Someone freely falling towards an infinite flat Earth would have no way of detecting gravity at all, there would be no tidal effects and technically the spacetime curvature would be zero!

 

[gonegahgah]

That sounds okay to me. How does it sound to everyone else?

Again - based on the dispersing wave analogy - I don't think gravity would decrease at all for any distance from the infinite flat surface giving - as Dr Greg says - zero space time curvature. ie. All things would fall in a straight co-ordinate line; rather than approaching quicker and quicker like ours does.

That is because the gravity wave would be generated from all points of the infinite flat plane and spreads onwards in tandem. Even if it were to sub-spread as an expanding part sphere arc square the surrounding sub-spread would reinforce it back up to full strength.

How curious? All imaginary of course.

 

[gonegahgah]

Actually you would still fall with a curve. The force acting on you for the flat Earth is always proportional to 1/r⁰ but it acts none-the-less.

 

[DrGreg]

Actually you would still fall with a curve. The force acting on you for the flat Earth is always proportional to 1/r⁰ but it acts none-the-less.

Certainly free-falling objects accelerate towards an infinite flat Earth (as measured by someone on Earth). But they accelerate in such a way that objects that are initially stationary relative to each other will remain stationary -- i.e. no tidal forces. (And in general, all free-falling objects move at constant velocities relative to each other. That's pretty much the definition of flat spacetime.)

Note, however, from the point of view of someone on Earth, the free-falling objects are not a fixed distance apart, due to continuously-changing length contraction, which means that the acceleration of each object relative to Earth differs. In other words, the "acceleration due to gravity" still varies with height despite the fact that spacetime is flat.

 

[Ich]

That thing could be described by mirrored Rindler coordinates?

 

[DrGreg]

That thing could be described by mirrored Rindler coordinates?

Yes.

However, when I claimed that an infinite flat slab would result in flat spacetime I was relying on my memory. A little googling has left me confused.

Post #14 of this thread[/color] in this forum seems to suggest I might be wrong, but flat spacetime might instead be produced inside an off-centre spherical hole inside a uniform spherical planet!

On the other hand, I think equation (1) of this page[/color] of mathpages.com indicates I was right after all?

Everything I said in previous posts is true in the relativistic version of a "uniform gravitational field". What is no longer clear to me is whether there is a hypothetical distribution of matter that would be capable of generating such a field.

 

[JesseM]

I see what you mean.

The other insurmountable challenge I see is creating an equivalent observer & satellite scenario either; if we wanted to try.
Say we replace the person and box with a GPS and place an observer on the 'ground' in the examples.
Unfortunately we can not create an equivalent scenario between gravity and acceleration for this because we will not be able to create a constant equivalent for the observer on the 'ground' between the two types (ie acceleration vs gravity).

The other problem of course is that it is probably impossible to simulate an orbit scenario using acceleration anyway? That is probably correct; is it?

So it would be difficult (impossible?) to model the GPS using acceleration instead of gravity? So unfortunately we couldn't compare them this way?

Thanksgiving distracted me from this thread, but I wanted to reply to this before I forget. Assuming there is a significant distance between the GPS satellite and the person on the ground, so that you can't pick a window of spacetime containing both of them where spacetime curvature can be treated as negligible, then you're right that there's no equivalent to this in flat spacetime. From what I've read, all spacetime curvature is associated with some sort of measurable tidal effects; so this is why the equivalence principle only works in limits where curvature goes to zero, normally by zooming in on a small patch of spacetime, although I think you could also imagine letting the patch of spacetime have a fixed size but taking the limit as the size of the gravitating body goes to infinity.

 

[gonegahgah]

Cool.

I was wondering something. As you say you can have small windows of eqivalence.
Basically for a GPS satellite you have a weightless satellite and you have an observer with weight on the ground.
Would this be somehow equivalent in someway to someone being accelerated equivalent to surface g in deep space while a satellite traveled at the GPS speed?

I've just read what I wrote here and I can see the problem is that the person being accelerated would only have one instant where the satellite was traveling at GPS speed relative to them; the acceleration changing the relative speed constantly. So again there would only be a single window instant where this would be equivalent to the Earth scenario.
That's correct hey?

 

[JesseM]

Cool.

I was wondering something. As you say you can have small windows of eqivalence.
Basically for a GPS satellite you have a weightless satellite and you have an observer with weight on the ground.
Would this be somehow equivalent in someway to someone being accelerated equivalent to surface g in deep space while a satellite traveled at the GPS speed?

I've just read what I wrote here and I can see the problem is that the person being accelerated would only have one instant where the satellite was traveling at GPS speed relative to them; the acceleration changing the relative speed constantly. So again there would only be a single window instant where this would be equivalent to the Earth scenario.
That's correct hey?

I'm not sure there's any good way of defining "equivalence" such that a situation involving clocks with a large separation in curved spacetime (like the Earth clock and the GPS clock) can be called "equivalent" to any situation in SR, even for a moment. Normally "equivalence" refers to the idea that any experiment done in a certain patch of spacetime in the GR scenario will have the same result as the same experiment done in a certain patch of spacetime in an SR scenario, and even if the difference between the clock rates might be the same for the Earth clock and the GPS clock accelerating in flat spacetime as you envision, there could be other experiments done in this window which would not have the same result as they would in the curved spacetime around the Earth (perhaps one could take a large spring with one end near the Earth clock and one end near the GPS clock and see how it is pulled by tidal forces, for instance).