Box in gravitational field

In summary, according to the speaker, the box would still work if it was in a gravity field, but you would need to use the Schwarzschild coordinate system.
  • #1
Einstein's Cat
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2
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?
 
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  • #2
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?

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 Schwarzschild metric.
 
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  • #3
Einstein's Cat said:
the box is accelerated to a velocity near the speed of light
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.
could the classical coordinate system still be used or is the Schwarzschild coordinate system needed]
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.

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?
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.

could the classical distance and classical coordinate system be used instead where the classical coordinate system is a three dimensional version of Cartesian coordinates?
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).
Also is it suitable to use classical physics to describe the acceleration of the box?
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.
 
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  • #4
Einstein's Cat said:
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?


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
Einstein's Cat said:
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
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.

Einstein's Cat said:
and what is the radial coordinate of the observer?
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.

Einstein's Cat said:
Also is it suitible to you use classical physics to describe the acceleration of the box?
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
Nugatory said:
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.
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
Einstein's Cat said:
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

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.
 
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  • #8
Nugatory said:
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.
Does that mean when spacetime is flat enough Newtonian mechanics could be used to describe the box at relativistic velocities?
 
  • #9
Einstein's Cat said:
Does that mean when spacetime is flat enough Newtonian mechanics could be used to describe the box at relativistic velocities?
Newton or SR, depending on what happens inside the box and which frame you choose.
 
  • #10
Einstein's Cat said:
Does that mean when spacetime is flat enough Newtonian mechanics could be used to describe the box at relativistic velocities?
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
Nugatory said:
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.
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
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.
 
  • #15
Einstein's Cat said:
It seems to me however that there's no description of a box in the gravitational field
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
Nugatory said:
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?
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
Einstein's Cat said:
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?
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
Nugatory said:
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.
I've done the first part of the task; it is just the GR aspect that I am confused with
 

Related to Box in gravitational field

What is a box in a gravitational field?

A box in a gravitational field refers to a physical object, such as a box, that is subject to the force of gravity. This means that the box will experience a downward acceleration towards the center of the Earth, as dictated by Newton's law of gravity.

How does the gravitational field affect the box?

The gravitational field exerts a force on the box, causing it to accelerate towards the center of the Earth. The magnitude of the force is directly proportional to the mass of the box and the strength of the gravitational field.

What factors affect the acceleration of the box in a gravitational field?

The acceleration of the box in a gravitational field is affected by two factors: the mass of the box and the strength of the gravitational field. The greater the mass of the box, the greater the force of gravity and thus the greater the acceleration. Similarly, the stronger the gravitational field, the greater the acceleration of the box.

Why does the box fall towards the center of the Earth in a gravitational field?

The box falls towards the center of the Earth in a gravitational field due to the force of gravity. According to Newton's law of gravity, every object with mass exerts a force on every other object with mass, and this force is directly proportional to the mass of the objects and inversely proportional to the square of the distance between them. As the Earth has a much larger mass than the box, it exerts a strong gravitational force on the box, causing it to accelerate towards the center of the Earth.

How does the gravitational field affect the shape of the box?

The gravitational field does not directly affect the shape of the box. However, the force of gravity can cause the box to deform or break if it is not strong enough to withstand the force. Additionally, the shape of the box can affect its acceleration in a gravitational field, as objects with greater surface area experience more air resistance and thus may have a lower acceleration compared to more aerodynamic objects.

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