# B Box in gravitational field

1. Jul 18, 2016

### Einstein's Cat

Let's say there's a massive and uniform sphere of radius, r, and there's a box in its gravitational field and the sphere is so massive that the box is accelerated to a velocity near the speed of light; could the classical coordinate system still be used or is the Schwarzschild coordinate system needed?

Furthermore say there's an observers further from the gravitational field than the box. Is the affect of time dilation on different aspects of the box negotiable from the observers perspective and what is the radial coordinate of the observer? Wikipedia says that it is analogous to the classical distance; could the classical distance and classical coordinate system be used instead where the classical coordinate system is a three dimensional version of Cartesian coordinates?

I've researched this but do not understand.

Also is it suitible to you use classical physics to describe the acceleration of the box?

Last edited by a moderator: Jul 20, 2016
2. Jul 18, 2016

### Mr-R

What do you mean by "Classical coordinate system"? If you mean the Cartesian $(x,y,z)$ then it would not work because this is a relativistic situation(Acceleration $\approx c$ !!) . So you need Minkowski Coordinates $(t,x,y,z)$. But since we are dealing with gravitational fields and noninertial frames we need to take account of curvature in space-time. So we use the Schwarzchild metric.

Last edited by a moderator: Jul 18, 2016
3. Jul 18, 2016

### Staff: Mentor

Relative to what? As far as the box is concerned it is floating in empty space, although if you're in the box and look out in the right direction, you'll see the massive spherical body rushing towards you at relativistic velocities.
In any sufficiently small (and singularity-free) region every spacetime is locally flat, so you can use the Minkowski metric with ordinary Cartesian coordinates if you wish. You need to use the Schwarzschild metric when you're considering a region of space large enough that the curvature effects matter. There's an analogy with the curved surface of the earth here: If you're laying out the foundations of a house you can use ordinary x/y Cartesian coordinates as if the earth is flat; but if you're plotting a course across the ocean you'll have to pay attention to the curvature to calculate the correct great-circle route. Same thing around a gravitating object - you can ignore the curvature in small regions but not large regions. So if the box is whizzing by just under your nose, you can use ordinary coordinates at least for the few moments before it's moved far away again.

Again, that depends on the size of the box relative to the strength of the gravitational field. If the box is small enough that the effects of curvature from one side to the other is negligible, then the difference in time dilation effects from one side to the other will be also be negligible. The time dilation may be very large if the observer is very far from the box, but it will be about the same for both sides.

Only within a region small enough that curvature effects can be ignored (and even then you'll find classical polar coordinates more convenient). Consider two spheres around the object, one with Schwarzschild $r$ coordinate equal to $r_1$ and the second with Schwarzschild $r$ coordinate equal to $r_2$. the area of the two spheres will be $4\pi{r}_1^2$ and $4\pi{r}_1^2$ - but the distance between them will not be $r_2-r_1$. There is no way of making this work using flat-space Cartesian coordinates (or polar coordinates).
Not really. The box is accelerating, and classical physics says that means there a force on it (in GR, there's no force) but there's no way of calculating that force to see what the acceleration is - the classical $F=Gm_1m_2/r^2$ doesn't work under relativistic conditions.

4. Jul 18, 2016

### pervect

Staff Emeritus

If you are using coordinates in which the sphere is at rest, and you also assume the box was initially at rest in these coordinates and "dropped", allowing it to accelerate towards the sphere, aquiring near light-speed velocity in it's fall due to the gravity of the sphere, , then yes, it's a problem that needs the Schwarzschild coordinate system. (Actually, of course, you can use other coordinates, but they'd all be related to the Schwarzwschild coordinate by a 1:1 map, i.e. a diffeomorphism).

5. Jul 18, 2016

### Yukterez

With Newton the particle can be accelerated beyond c relative to the central mass, while under Einstein it can't, see Plot 3 in this post.

The distance from r1 to r2 is not r2-r1 but ∫(1/√(1-rs/r), r=r1..r2), see Flamm's paraboloid in this post.

Yes if you differentiate coordinates by proper time (then you get the rapidity instead of the velocity) and no if you differentiate by coordinate time.

6. Jul 18, 2016

### Einstein's Cat

Could the spacetime be treated flat if it was say 4m in length and 3m from a sphere of 1m in radius? And also how would one calculate the acceleration of the box and thus it's displacement

7. Jul 18, 2016

### Staff: Mentor

That depends on how strong the gravitational field is. There's an easy way to find out: look for tidal effects. The box is in free fall, so if we release a weight inside the box it will just float. So we release six weights in the box - one just above the floor, one just below the ceiling, and one against each wall. In a flat-enough spacetime they will stay put. Any detectable curvature will show up as follows:
- The weights near the walls will drift closer to one another, because their paths aren't quite parallel - they point towards the center of the gravitating mass.
- The weight at the top and the weight at the bottom will move apart and towards the ceiling and the floor, because the bottom weight is a bit closer to the mass so is feeling a slightly stronger force.

The smaller the box and the weaker the gravitational field, the smaller these effects will be. When they're small enough to be irrelevant we say that the spacetime is flat enough to ignore the curvature. You'll notice that when that happens all of Newton's laws work perfectly; the objects started at rest (as far we can tell from inside the box) and because there is no force on them they stay that way.

Last edited: Jul 18, 2016
8. Jul 19, 2016

### Einstein's Cat

Does that mean when spacetime is flat enough Newtonian mechanics could be used to describe the box at relativistic velocities?

9. Jul 19, 2016

### A.T.

Newton or SR, depending on what happens inside the box and which frame you choose.

10. Jul 19, 2016

### Staff: Mentor

At relativistic velocities relative to what? If you're inside the box and free-falling with it, the only deviation from Newtonian mechanics you'll see will be the tidal effects I mentioned above, and these can become arbitrarily small as the box becomes smaller. If you're outside the box and it is moving at relativistic speeds relative to you, you'll need to use relativity. SR will work if you're only considering a very short period of time during which the box is right under your nose; as soon as the distances and times become large enough that curvature effects appear you'll have to use GR.

In all of these discussions, "large enough" and "small enough" depends on the strength of the gravitational field; the more the curvature the more effect it has and the shorter the distances over which it can be ignored.

11. Jul 19, 2016

### Einstein's Cat

The relativistic velocities would be relative to an observer further from the gravitational field than the box; and if I was to use GR then how would the box be affected by a sphere of g of x, and where it's accelerated to near the speed of light, relative to the observer?

12. Jul 19, 2016

### Einstein's Cat

Say there is a box of height, h, width, w, and length, l, with a distance of d from the centre of a sphere of radius, r, and mass,m. The box is accelerated by the gravitational field to near the speed of light.

How could this box be described in the gravitational field?
My knowledge of GR is very limited so I am entirely ignorant to this.

13. Jul 19, 2016

### A.T.

14. Jul 19, 2016

### Einstein's Cat

15. Jul 20, 2016

### Staff: Mentor

It's not clear what you mean by "a description". Do you want to know its trajectory? What it's like to be inside the box?

16. Jul 20, 2016

### Einstein's Cat

By description I mean what would happen to the box and the box would be stationary when time (t) is zero. Would it be compressed? To what extent?

17. Jul 20, 2016

### Staff: Mentor

Start by solving that problem using ordinary Newtonian gravity. You can model the box as eight point masses, one at each corner of the box, connected together by twelve rigid and massless sticks. Assuming just ordinary classical physics, what is the tension or compression in each stick when the box is free-falling in Newtonian $F=Gm_1m_2/r^2$ gravity?

That will give you some understanding into the various factors that influence what happens to the box. After you thoroughly understand that problem, you can start to think about the effect of corrections due to GR.

18. Jul 20, 2016

### Einstein's Cat

I've done the first part of the task; it is just the GR aspect that I am confused with