# Gravity candy wrapper test

1. May 3, 2014

### duordi134

I am in a zero mass space ship at rest in my chosen frame of reference.

A cylinder with a diameter equal to the sun and of infinite length is passing at 8 minutes light travel distance and has a velocity of 99% x the speed of light traveling with the length of the cylinder with respect to my preferred reference frame.

I drop a candy wrapper out the window (with zero velocity with respect to me) and record the time required to have the candy wrapper
fall to the cylinder by my clock.

The cylinder is stopped (has zero velocity with respect to my preferred reference frame) and is at an identical distance as it was for the previous test. I repeat the experiment by dropping a candy wrapper at zero velocity with respect to myself and record the time required for the candy wrapper to hit the cylinder.

Are the times recorded for the tests identical or different?

Duordi

2. May 3, 2014

### phinds

Just as a side comment, you should reword your thought experiment since as stated, it not only requires the spaceship to be zero mass (a GOOD trick!) but it also requires YOU to be zero mass (an even better trick).

3. May 3, 2014

### duordi134

Phinds,
I am from the planet Gargon and being composed entirely of light we are weightless, that is if we have not
eaten a candy bar.
We have a weakness for candy bars.

Duordi

4. May 3, 2014

### Staff: Mentor

Google around for "linear frame dragging".

To make sense of this problem you will want to get clear in your mind exactly what the world-line of the spaceship looks like when the ship is hovering at a constant distance from the tube and at rest relative to its surface; and when the ship is hovering at a constant distance from the tube and moving relative to it. They are different world-lines and neither world-line is a geodesic so there will be proper acceleration; the proper acceleration is different for the two.

5. May 3, 2014

### Staff: Mentor

Or, more rigorously, we can consider your mass and the mass of the candy wrapper to be negligible compared with the mass of the moving tube.

6. May 3, 2014

### PAllen

The times would be different, and the difference would be that when the cylinder is moving 99% speed of light the acceleration of the candy wrapper would be larger by a factor nearly 2γ compared to the stationary cylinder scenario. To a closer approximation, the factor would be (1+β)γ.

β = v/c
γ = √(1-v^2/c^2)

7. May 3, 2014

### duordi134

Y is the ratio of relativistic mass to rest mass for perpendicular motion.

This would mean that the acceleration would be about twice the correction one
might expect from replacing the rest mass with relativistic mass.

So there is a real measurable difference in acceleration that would be large
enough to be identified and proven.

I look forward to the time when I will be able to do this myself.

Duordi

P.S.
The discussion on linear frame dragging seem to indicate it would be a very miniscule affect.
There was also a static mass increase discussed which seem to indicate that a small increase
in inertia will be created when other masses are placed nearby.

I never heard this before.

Last edited: May 3, 2014
8. May 3, 2014

### Staff: Mentor

Then you can't conduct the experiments you describe, because those experiments require you to be following a timelike worldline. If you're made purely of light, you must be following a null worldline. That means you can't do things like be at rest relative to the cylinder, or drop a candy wrapper that starts out at rest relative to you, etc. To do those things, you have to be following a timelike worldline, which means you have to have nonzero invariant mass.

9. May 3, 2014

### Staff: Mentor

This is true for a spherically symmetric source, but is it true for an infinite cylinder?

10. May 4, 2014

### 256bits

Good point.
Candy wrappers from two seperate ships would not be falling towards a common centre.

Edit: Depending upon the orientation wrt the longtitudinal axis of the cylinder.

Last edited: May 4, 2014
11. May 4, 2014

### PAllen

I'm basing it on the same factor appearing in attraction of dust and light beams as in the classic paper on body deflection. Basically, I'm combining the results of this paper, on (non-relativistic) dust beams and light beams:

http://arxiv.org/abs/gr-qc/9811052

with the classic paper on body deflection. The factors seem to come out the same. Then, I am making a small leap of faith that a cylinder can't be different from dust beam of the same configuration.

12. May 4, 2014

### Staff: Mentor

But those factors are for parallel (and antiparallel) beams; i.e., in the scenario under discussion in this thread, if the cylinder lies along the $z$ axis, the particles being deflected would also need to be moving along the $z$ axis. I don't think these results generalize to a test object moving in a plane perpendicular to the $z$ axis.

13. May 4, 2014

### PAllen

The deflection between a non-relativistic (could be almost stationary) dust beam and a light beam seems relevant. Then the factor of 2 for such a case becomes (1+beta) for a relativistic beam, rather than light. The 'slow beam' could be a beam of candy wrappers.

14. May 4, 2014

### pervect

Staff Emeritus
Several things are making me nervous here. First is the tension in the cylinder walls and it's gravitational effects. This isn't included in the model Pallen describes. But I think the tension is potentially an important modifier to the gravitation here.

If we have to deal with what the stress energy tensor of the rotating cylinder is, to me this implies some sort of material model like the hyper-elastic one used in Greg Egan's relativistic hoops. Unfortunately, we know that that model can fail if you push it to hard - the Lagrangian can become singular even. Unless we can somehow use primarily gravity to keep the hyper-rotating cylinder from tearing itself apart by centrifugal force, without having it collapse to a black hole when we stop it - which seems doubtful, but I haven't run any calculations.

Second is the well-known "Rotating cylinders and the possibility of global causality violation", that makes me think that the behavior of infinite cylinders could be ill-behaved. Tipler uses the "Van Stockum dust" for the metric in that paper, I'm not sure if the van stockum dust is really a good model for a rotating cylinder, in spite of said usage. The other issue I see with using the dust solution (besides it's ill behavior in terms of potentially being a time machine) is s the fact that there's no good way to stop the cylinder from rotating using that solution.

Sorry for all the fear, uncertanity, and doubt, but I'm not sure the problem is well-defied enough to have a really good solution at this point. The infinite nature of the cylinder is one issue of concern, the fact that we don't really know what the cylinder is made of is another (the idea that we can spin up a cylider to 99% of the speed of light is really sort of a relic from the idea of a rigid body, we don't actually know of a material that allows this).

All in all, I'd be happier with comparing a rotating black hole to a non-rotating one if the idea is to ask what the effects of angular momentum are on "mass" and gravitation. Of course this isn't what the OP asked, and I haven't done any calculations. I'm not sure if the OP's motiviation is what I described, so I'm not sure the calculations would be of interest to the OP. There are well known formula for spinning up black holes, the effect of spinning them up on their mass, and the metric associated with them. So it's more of a mater of working through the details if we use black holes, without any of the issues I see with the cylinder.

15. May 4, 2014

### Jonathan Scott

I don't believe that "linear frame dragging" is relevant here as such; that term relates to the dragging effect of linear acceleration.

There is however a possible component of "gravitomagnetism" and rotational frame dragging relating to the motion of the cylinder. I don't think that causes any deviation of the line of falling towards the cylinder, but it would cause something to rotate (in a plane through the axis of the cylinder) as it falls.

I'd expect the magnitude of the moving cylinder's field to be increased not only by the effect of the additional kinetic energy due to the motion, but also due to the increased density per unit length due to the Lorentz contraction, which means that overall it would increase as the square of the gamma factor.

16. May 4, 2014

### Staff: Mentor

I'm not reading the OP as hypothesizing a relativistically spinning cylinder; I'm reading the OP as hypothesizing a non-spinning cylinder that is moving (relative to the test object) at relativistic speed along its own axis: i.e., if the cylinder's axis is the $z$ axis, the cylinder itself is also moving in the $z$ direction, without any spin in the $x - y$ plane.

17. May 4, 2014

### phinds

Yeah, that's how I read it too, since he said "traveling with the length of the cylinder"

18. May 4, 2014

### duordi134

The comic book edition

You are correct PeterDonis but I have learned a lot about relativity by this discussion.
It sounds like it takes a teem of experts to solve a "simple" problem.

Is there a location which discusses the GR problems which have been worked out and what the results are?

The "comic book" edition is preferred.

Thanks

Duordi

19. May 4, 2014

### PAllen

That sounds reasonable, then my amended estimate would be:

(1+β)γ^2

20. May 6, 2014

### duordi134

Just a thought

Is it possible that you are double dipping and the Lorentz contraction density increase and the kinetic energy increase are two views of the same thing?

Just a thought.

Duordi

21. May 6, 2014

### PAllen

I don't think so. Think about each atom in the cylinder. In a frame in which it moving rapidly along its axis, as compared to one where it isn't moving, the number density of atoms is increased by gamma. Further, each atom has energy increased by gamma. The 1+beta deals with a further correction to reach that for light, the effect is doubled for a given energy density. IOW, I think Jonathan Scott's argument is correct.