If they were equidistant from the observer, and both had surface detail, I think they would both appear flat.
I dunno. I was thinking this too, that they’d just look like they were slanted planes, but they converge on all sides. In our everyday experience that can only happen with curved surfaces.
Clearly someone needs to make a very very large and short building to test this out.
Would an infinite plane have enough gravity to hold the observer to the surface? Would an infinite plane be able to maintain an atmosphere? My WAG is you’d see infinity, as you suffocated to death floating away from the plane.
There’s an earth sized planet directly under the plane where you are standing. It does muck up the weather though.
A plane in its strictest sense has no thickness, and hence would have no mass. If you assume a flat massive object infinitely large in length and breadth but with finite thickness, say 1km, then it will also have infinite mass… but spread over an infinite volume.
I’m not sure how to work out the gravitational attraction in this case - presumably the attraction from the infinite mass in all directions would cancel out horizontally, but all that mass is below your centre of gravity, so I don’t think you’d float off. Wouldn’t it effectively cancel out to an infinite mass pulling you directly downwards?
As I recall, a plane of infinite size and finite uniform mass/surface area exerts a gravitational field that is constant regardless of your distance from the plane; if it pulls 1g when you are standing on it, it pulls 1g when you are a light-year away from it. None of this /r^2 nonsense with an infinite plane.
Huh? I’m hoping my misunderestimation of gravity or yours gets cleared up. It seems to my simplistic mind that as you move vertically away from a horizonatal plane, you are moving further away from all of the mass.
ETA: I guess there is no ‘r’ though.
http://www.pgccphy.net/ref/gravity.pdf
If you search for infinite plane you get the derivation.
It’s going to take me a while to fail to comprehend this. Is the ‘acceleration’ the key point here? I may be misunderstanding the concept of a ‘uniform gravitational field’. (I’m sure I’m misunderstanding it, now it’s a matter of how much).
The intuitive way to think about gravity above and infinite plane is to think about how much mass is below you. When you are standing the plane there is not a lot of mass the is below you to say 45 degrees of vertical but you are close. If you move 10 miles away from the plane there is a lot more mass below you to a 45 degree angle. You will notice that this mass goes up with the square of the distance just as the force of gravity per unit mass is falling with the square of the distance.
I shouldda known someone here would know the answer to my bullshit question.
The easiest way to derive the gravitational field of a uniform plane is via Gauss’s Law (which is usually used for electric fields, but the exact same principles apply). To get an intuitive feel for it, think of field lines. The fields can only begin or end on masses (charges) or at infinity. The greater the mass (charge), the more field lines there are. And the closer together the field lines are, the stronger the field. With an infinite plane, the field lines are all straight vertical, and since the only end at the mass, they all extend off to infinity. So they never get any further apart, and so the field strength is the same, no matter what your distance.
But isn’t there some problem in making gravity calculations using an object with infinite mass? (and therefore infinite gravity) How would the result be different if we consider a sphere with infinite mass?
Speaking of infinite gravity, at some point would the plane not exceed its own Schwarzchild radius since the SR of a plane would increase as the square of the radius, thus eventually catching up to the radius?
You are showing a fine definition of infinite john.
Divide your vision into 100 sections, and assume the “horizon” is at the height of 50. (all numbers are made up)
From height 0 to 1, you see about 1 foot of earth.
From height 1 to 2, you see about 1.5 feet of earth
from height 25 to 26 you see 50 feet of earth
from height 48 to 49 you see 3.7x10^22 feet of earth
from height 49.1 to height 49.2 you see 3.7x10^24 feet of earth
from height 49.98 to 49.99 you see 3.7x10^296 feet of earth
Do you understand this example? This is basically what all of calculus is based on.
As x (your vision) approaches lim a (the horizon) the amount of earth you see approaches infinity. You can keep splitting hairs until 49.999999 of 50, and see more and more earth in smaller and smaller amounts of space, essentially to an uncountable value, which we deem to be infinite.
I hope that made sense, calculus isn’t my strength. Carry on with the other talks now.
The Schwarzchild radius is relevant only for static spherically symetric solutions in general relativity.The bigger the depature from spherical symmetry the less relevant it is. An infinite plane is not spherically symmetric even as a rough approximation so the Schwarzchild radius is pretty much irrelevant.
Now that makes perfect sense. You should be writing the textbooks. Of course it would probably have been obvious if I understood much about gravity. And maybe it explains the r^2 stuff because all the lines would be diverging as they emanate from the surface of a sphere.
Yes, infinite planes of mass are problematic in general relativity, since they would indeed collapse into black holes. Whenever you have a mass contained within a sphere of radius that mass’s Schwarzschild radius, no matter how the mass is arranged, it’ll collapse (of course, after the collapse it will be spherically symmetric, no matter what it was before). Thus, any discussion of an infinite plane of mass must necessarily just be working in the Newtonian approximation.
And by the same reasoning, the gravity field of an infinite uniform cylinder declines as 1/r (vice 1/r/r)
That’s not strictly true, take for example a homogenous and isotropic universe. Clearly the matter contained in any sufficently large sphere will fall within it’s own Schwarzchild radius, but whilst there’s an inherent non-staticity in such a solution (which may or may not result in a big crunch singularity), no black holes form.
What has been shown in physically realistic nonspherical collapse, i.e. peturbations of the Kerr solution, will settle down to Kerr-like black holes. But that’s not the same as giving some overarching importance to the Scwarzchild radius (which may be difficlt to even define in many situatuons) or saying that the physically unrealistic infinite plane will collapse in on itself.
Infinite planes have been studied in general relativity and whilst, as is common in general relativity, the solutions won’t be static unless you make them static you need pressure/tensions/non-zero cosmological constant, they can be made so they do not collapse in on themselves.