# Uniform Acceleration and Gravity (again)

1. Dec 22, 2007

### Jorrie

Same topic, new question, I think, so please suppress the Arghh!!!s. This is about detection of 'tidal gravity' effects inside a uniformly accelerated lab in free space.

As I understand it, if you release a spherical test particle cloud to 'free-fall from rest' inside a uniformly accelerated lab (where the acceleration is due to a force applied to one point, i.e., not Born-rigid acceleration), the cloud will display 'tidal stretching' in the acceleration direction relative to the lab. This can be deduced from the Nikolic paper that was discussed before.

My question: will one also be able to detect transverse (normal to the acceleration direction), 'tidal squeeze' effects on the particle cloud relative to the lab? I suppose "relative to the lab" is a bit loose, so let me try to tighten that: will an observer riding in (i.e. fixed to) the center of the lab be able to detect both a 'vertical' and a 'horizontal' deviation from the original spherical size of the 'free falling' particle cloud, where an observer 'hanging on to the ceiling' of the lab released the cloud and measured it to be spherical at release.

 I suspect that there is no transverse effect, because the situation is not the same as in a spherically symmetrical gravitational field around a central mass, but accelerated frames of reference are difficult beasts, so I'm not sure...[/edit]

Last edited: Dec 22, 2007
2. Dec 22, 2007

### pervect

Staff Emeritus
An observer in an accelerated ("Born rigid") spaceship will indeed measure different accelerations at the nose and tail of the space-ship. I would and have in the past described this as being due to a "tidal force", but it's on my to-do list to see how textbooks describe this situation.

There won't be any difference in the readings of the accelerometer between the port and starboard sides of the spaceship, however. This should be fairly obvious by symmetry.

3. Dec 22, 2007

### Jorrie

Agreed, but the question was actually about 'transverse tidal forces'. While sitting on Earth's surface, vertical accelerometers on the port and starboard side (at the same height) will also show no difference in reading, but there will be a slight horizontal 'tidal squeeze' towards the vertical centerline.

Can I take it that there will be no such horizontal tidal effect in a 'Born rigid' accelerating ship in free space? (I think the symmetry also answers this anyway...)

4. Dec 22, 2007

### publius

Again, forgive the new poster from jumping in, but I don't think I'd call the Rindler frame gradients "tides". As I understand it, "tides" are reserved for that experienced by free-fallers/floaters due to real curvature of space-time.

If you're in Rindler frame, any free-fallers you drop are simply inertial in flat space-time. They feel no tides. Now, what you, in your Rindler coordinates see of that free fall is a different matter. I think you will see initial "stretching" along the direction of your acceleration.

But at the end, that free-falling group is going to "freeze" at your Rindler Horizon (just like with Scharzschild, some last moment of their proper time gets stretched out to infinty according to your clock, and you see no events beyond that horizon), and so become very contracted as they asymptotically "freeze" and redshift to nothing.

But the free fallers themselves are globally inertial and see no tides whatsoever.

That's how I understand it.

-Richard

5. Dec 23, 2007

### yuiop

In my understanding a cloud free falling in a stationary lab in a gravitational field will experience tranverse squeezing. A cloud falling in an artificially accelerated lab (rocket) will not experience tranverse squeezing. The equivelence principle sets the condition that the measurements are local. In other words, the lab has to be very small for observers not to work out if they are in an accelerating rocket or stationary in a natural gravitational field.

If they are in a very long rocket and measure the acceleration at different points of the rocket from nose to tail they would probably work out the acceleration is proportional to 1/d and not to 1/$$d^2$$ as they would expect in a genuine gravitational field.

6. Dec 23, 2007

### Jorrie

Good point, thanks!

7. Dec 23, 2007

### Jorrie

I think you're right, because the geometry is different to, say, Schwarzschild geometry.

8. Dec 23, 2007

### pervect

Staff Emeritus
Free fallers will certainly feel no tidal effects, I agree.

The point is that someone in an accelerated frame will measure the amount of force required to hold him stationary in that frame as varying with position. Thus, as you approach the Rindler horizon, it requires infinite force to hold you there. A point right on the Rindler horizon requires infinite force to hold it there, a point close to but not yet at it requires less than infinite force.

So the observer who is co-moving with the Rindler coordinates feels not only an infinite acceleration, but an infinite rate of change of acceleration with "height", as he appraoches the Rindler horizon.

I think it is therefore best to say that the Riemann curvature tensor is equivalent to tidal forces experienced by a free-falling and non-rotating observer - because the tidal forces (naively defined, at least) depend on the acceleration.

At worst, one has only added unnecessary specifications to the measurement process of tidal forces. I think it's more cautions to overspecify than it is to ignore the issue.

Last edited: Dec 23, 2007
9. Dec 23, 2007

### publius

Pervect,

I was corrected once, either by someone directly or by something I read, I can't remember which, that the term "tides/tidal forces" were absolutely reserved for that experienced by inertial observers, and so "tides" were a feature only of curved space-time, etc, etc. That goes so far as the viewpoint that gravity = curvature = tides. No tides, no curvature, no "real" gravity. In that view, a Newtonian 'g' is just a coordiante thing, and one should view the only real gravity as the tidal g a free faller would see.

And so, in that view, one should not call gradients in proper acceleration felt or coordinate accelerations one sees in a general coordinate system as tides. In this view, So I've been calling that "pseudo tides" ever since, as tides were absolutely "locked" (pun intended) to Riemann curvature. :)

I got the impression that viewpoint was absolute gospel handed down from on high. But from you what say, it's not, and we can use "tides" to mean gradients for any observer.

Either way seems fine to me, though.

-Richard

10. Dec 23, 2007

### pervect

Staff Emeritus
If you could remember the details (who, where) it would be helpful. The reason that I make a point of talking about the issue is that the textbooks I read did talk about the equivalence of tidal forces and the curvature tensor, but did not make a point of saying that the tidal forces were those measured by an inertial, unaccelerated, and non-rotating observer.

I eventually came to realize that the geodesic equation describes tidal forces only for unaccelerated observers. But I wasted (or spent, at any rate) a lot of time wondering about the tidal forces experienced by observers, especially observers in rotating frames of reference. The Riemann tensor doesn't give you this - what the Riemann tensor gives you the tidal force measured by a non-rotating observer.

To make life more interesting, defining non-rotation in GR can be non-trivial. I've snipped my earlier discussion in the name of brevity and accuracy, but see for instance http://arxiv.org/PS_cache/gr-qc/pdf/0011/0011094v1.pdf

If you can find any such gospel in the literature or a textbook, I'd be interested. (Most of the literature, though, just talks about the Riemann, and assumes that people know what it is, except perhaps for a few pedagogical ones). It sounds to me like this might be one persons opinion rather than "gospel" but I haven't seen a lot written on the issue.

Last edited: Dec 23, 2007
11. Dec 25, 2007

### publius

You're most likely right, it is someone's "partisan" opinion, but it sure got put in my head. I think it stems from the real gravity = curvature school, and trying to ordain that tides to inertial (and zero Fermi-Walker derivatives) observers are the defining characteristic of curvature. Therefore, tides shouldn't be used for "coordinate gradients".

-Richard