Is tidal acceleration frame dependent?
How does it transform?

(This is just a guess, so don't quote me.)

I don't know if this is a good question to ask because a good frame should have no tidal forces. If you cannot define a good frame, I don't know that you can define a Lorentz transformation.

Here's another shot.

Suppose you were standing on the surface of the Earth. The force on your head, due to the Earth, is less than the force on your feet. (Actually, the force acts at your center of mass, or center of gravity, but work with me.) No, for a Lorentz Transform, you can boost either along the surface of the Earth, or perpendicular to it.

If you boost along the surface of the Earth, you will always feel the same difference in force between your head and your feet.

If you boost perpendicular to the surface of the Earth, then you have to worry about SR effects. First, if the boost velocity is small, the difference between the force on your head and your feet falls to zero like
\Delta F \sim \frac{M_F}{R_F^2}-\frac{M_F}{R_H^2},
where R_F and R_H are the distance between the center of the Earth and your feet, and the center of the Earth and your head. (Another contradiction---why do we use the center of the Earth, and NOT the center of you???)

The thing that I'll have to think about it what happens if you boost with a relativistic velocity? The Lorentz length factors would act to contract your total length in the Earth's frame, but you don't notice anything in your frame.

Ahh yes. The tidal force is defined in your frame. So it doesn't make sense to ask what it is in the Earth's frame. So we get back to the first answer:)

I think all of this can be avoided by choosing a logical reference frame. This is the problem that Zanket has---one cannot define arbitrarily large reference frames.

I don't know if this is a good question to ask because a good frame should have no tidal forces.
I don't understand. How do you choose a frame with no tidal forces?

I don't understand. How do you choose a frame with no tidal forces?

Make it pointlike.

That's not what I'd call a good frame!

I think the term "reference frame" may be misunderstood.

The way one works on curved space time is to use something called "vierbeins", a German word for "many legs". The vierbeins tell you how to make a plane tangent to the space-time at a point. Basically, the vierbeins tell you how to do calculus on the curved manifold by giving you a little flat place to work on, where it is easy to do calculus. Because the manifold must be non-singular, you can always do this---think of it like zooming in on a surface untill all you see is a flat space-time. At the end of the day, you stitch all of the little flat pieces together.

The vierbeins are essentially basis vectors in the tangent space. The reference frame is defined by the vierbeins, which are defined only at one point in space-time. The coordinate system that the vierbeins define are good inasmuch as the size of the coordinate system (i.e. the reference frame) is much bigger than the curvature of the manifold.

In this case you will always have some approximation, that the experiment is done in an approximately flat space-time.

When I say "reference frame", I'm thinking of a coordinate system covering some manifold, curved or flat. The integration of the flat elements defined by the verbeins, I guess.

Is that not right?

When I say "reference frame", I'm thinking of a coordinate system covering some manifold, curved or flat. The integration of the flat elements defined by the verbeins, I guess.

Well, you have to do everything in flat space. Or, maybe it's easier to do everything in flat space (I don't know). So you have to go through the business of defining vielbeins, which tell you how to cut up your curved manifold into little flat patches.

Also, I think there is some theorem in differential geometry that says you can never have one good coordinate system for any curved manifold. The classical example is the sphere. The basic idea is that you will always have a coordinate singularity somewhere. I will check on the details tomorrow when I get into my office. If physicsmonkey is running around, he/she seems to remember all of these things much better than I do:)

The easy example of a coordinate singularity is the Schwarzchild metric, which is singular at the horizon. This is just a coordinate singularity because, with valid choice of coordinate systems, one can show that the horizon is smooth. I can't remember the name of the coordinates, but they exist.

When an isolated body of mass is rotating in space, does it physically grip and twist the space around it? ie: would it twist the rubber sheet?

Thanks Ben, but I'm not getting what I'm after.
Let me put it like this...

In a free-fall lab, two test particles some distance apart which are momentarily at rest are found to be accelerating relative to each other.
Say that during some small time frame, the particle positions follow the paths:
(x1,y1,z1) = (1 - k.t^2, 0, 0)
(x2,y2,z2) = (k.t^2 - 1, 0, 0)

A similar lab with passes the first lab at high speed (velocity = (v, 0, 0)), and examines the same particles, using a coordinate system with the same alignment and origin.

Can I naively use the Lorentz transform to get the particles' motion in the second lab's frame, and what should I expect the result to be?

Interesting, eh Pete?

In a free-fall lab, two test particles some distance apart which are momentarily at rest are found to be accelerating relative to each other.
Say that during some small time frame, the particle positions follow the paths:
(x1,y1,z1) = (1 - k.t^2, 0, 0)
(x2,y2,z2) = (k.t^2 - 1, 0, 0)

A similar lab with passes the first lab at high speed (velocity = (v, 0, 0)), and examines the same particles, using a coordinate system with the same alignment and origin.

Can I naively use the Lorentz transform to get the particles' motion in the second lab's frame, and what should I expect the result to be?

I certainly can't give a definitive answer, but I think we can get some insight by supposing that neither lab "knows" that the acceleration is caused by tidal effects. As far as they know, the acceleration of the particles could be caused by tiny propellors on each test particle. That is to say we can approximate the situation as occuring in flat spacetime, with some unknown force pushing on the particles. Then, we'd expect the second lab to see a length-contracted, time-dialated version of the same events as the first lab. The causal structure would be the same (suppose the particles start out some distance apart and eventually crash into eachother; these events should still happen in sequence). So the frame force would still be there from the second lab's point of view (since the particles still accelerate), but its magnitude should change depending on how the length contraction and time dilation. I'm not sure quite what it would look like in the limit as v approaches c, but it seems that the particles would start out infintesimally close to one another due to length contraction, and accelerate infintesimally slowly, due to time dilation.

Hi Pete.

Then you can define proper acceleration---i.e. the acceleration the particle feels in its rest frame---by alpha:

\alpha = \gamma^3 \frac{du}{dt}.

where gamma is the relativistic thing:

\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}

Take the particles in a rest frame and v is the velocity of the moving frame, or vice versa.

u is the proper velocity.

Is tidal acceleration frame dependent?
How does it transform?I am not sure what you are asking with first question, and not qualified to attempt answer to the second.

If I transform the first question by replacing the word "acceleration" with the word "force" or better, both words "tidal acceleration" by "gravity gradient" I think I can answer that question with "No."

My reasoning is quite simple (the best type, until some wise guy shows it is not a simple problem):

Imagine a homogenious pure-iron ball that is in a slowing increasing gravity gradient field. (Perhaps achieved by it slowly moving towards a black hole.) Also imagine that as it slowing moves it slides in contact with a marked tape that records it location (and all prior locations, perhaps by a erassing the marks on the tape as it slides over them.) At some point which is a property of iron only it is ripped apart and ceases to erase the marks on the tape.

All frames have the same strength for iron and all must then agree that at the still marked (end of the erased) the gravity gradient was that which iron will not support. They may disagree on how far the iron ball was from the black hole, how fast it was moving etc. but not on the value of the gradient at the point the ball was ripped apart. I am strongly betting* that if the gradient is the same at that point for all frames, it must be the same for all points the ball could be (or was).

PS WTH is "tidal acceleration" ?

-----------------------
*For another point, use a copper ball, etc.

Billy---I think you and I both made complex answers to a simple question.

The acceleration is frame dependant, as is the force, as is the velocity as is...

The acceleration transforms in an easy way. See above:)

PS WTH is "tidal acceleration" ?

"The tidal acceleration is the differential acceleration between two points due to the difference in gravitational acceleration caused by their differing distances from a body." (scienceworld.wolfram.com (http://scienceworld.wolfram.com/physics/TidalAcceleration.html))

Observe the motion of two test particles in freefall. Tidal acceleration is the difference between the acceleration of the two particles.

That's a bit vague... which is why I'm asking about frame dependence. I think that both the difference between acceleration in the chosen frame and the difference between proper acceleration will be frame dependent, but I'm not sure on the details, and the maths looks like hard work.

Thanks Ben... sorry, I'd missed your earlier post.

...Observe the motion of two test particles in freefall. Tidal acceleration is the difference between the acceleration of the two particles. ...OK if there is a difference between "tidal acceleration" and "gravity gradient" it is only:
(1) The units differ by a mass term
and
(2) "gravity gradient" is a function well defined AT EVERY POINT in field (expect for singularities) whereas "tidal acceleration" is not as it depends upon the separation of your "two particles"

Thus it seems to me to be a stupid substituion to speak of "tidal acceleration" - It is too many special cases, and any one can be found by integrating the well defined "gravity gradient" field between the two points that happen to be of interest.

Also a dumb thing to use as the value is constantly changing as the separation of the two particles change.

Do you agree?

It depends on the context of the problem, I think.
I'm happy to talk about gravity gradient instead.

...I'm happy to talk about gravity gradient instead.Should be a short conversation. After noting it falls off as the inverse cube, what else is there to say?

I do like to point out that this means the gradient of small black holes at their event horizon is greater than the gradient of a big BH at its EH.

Do you have some favorate observation which also follows from the basic inverse cube law?

Perhaps the fact that big rocks, called moons, can not exist where the rings of Saturn are is worth mentioning?

I can not think of anyway to continue this conversation. - Ball in your court.:D