I do not need to define the gravity well.
Gravity well - Wikipedia, the free encyclopedia
I have been looking for what happens to this in a massive hollow sphere.
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I do not need to define the gravity well.
Gravity well - Wikipedia, the free encyclopedia
I have been looking for what happens to this in a massive hollow sphere.
Do you mean - what gravity is like in the interior of a hollow shell of matter ?
I'm thinking you would be weightless inside the hollow sphere, pulled equally by all particles of the spherical-shell in collective no matter what one's location within it. The gravitational potential inside the hollow sphere would be essentially that that exists at the surface of the shell, and that gravitational potential would be the same everywhere inside the hollow sphere. Spacetime would be flat inside, and the gravity field inside should have no gradient. That's been my understanding of it.
Thank You,
SinceYouAsked
The g = zero inside a hollow sphere. That's early undergrad classical physics and calculus.
But the depiction from a GR standpoint that piques my curiosity.
The force of gravity is zero inside the hollow sphere, but the gravitational potential is not zero. Far as GR goes, Imagine a heavy hollow bowling ball at rest on a very large trampoline. It puts a dent into the 2d surface of the tramp. However, where the trampoline touches the bowling ball, imagine the trampoline being perfectly flat (level) inside that area. A level trampoline surface corresponds to flat spacetime, no medium curvature, and so no force of gravity can be experience there. The flat spacetime inside the sphere is lower than the flat spacetime far from the sphere, which is essentially the height of the trampoline had no ball been placed in it. I've never taken GR courses, and so I defer the math to the experts here.
Thank You,
SinceYouAsked
SYA's explanation of it is essentially correct. Space-time inside the cavity would be completely flat, however, the metric tensor would differ from the Minkowski metric at infinity outside the sphere ( though it will be isometric to it ). Basically this means that there are no tidal forces inside the cavity, but the region is nonetheless at a different ( uniform ) gravitational potential than a reference clock at infinity, so clocks placed inside the cavity will be gravitationally dilated.
This is physically identical to the Newtonian description of the scenario ( sans time dilation, of course ).
OK, I am confused. Wouldn't the gravity inside the sphere behave pretty much like the gravity outside the sphere by decreasing as you left the interior surface and approached the center of the sphere to only become zero at the spheres center of mass?
No, the gravity in the interior is perfectly uniform, even in the good old Newtonian case : Gravity Force Inside a Spherical Shell
OK, I am still confused Markus.
Other than how massive you assume the shell to be what is the difference between the situation in your link
Gravity Force Inside a Spherical Shell
and this one
Hole Through the Earth Example
From Gravity Force Inside a Spherical Shell The net gravitational force on a point mass inside a spherical shell of mass is identically zero! Physically, this is a very important result because any spherically symmetric mass distribution outside the position of the test mass m can be build up as a series of such shells.
so in effect a sphere can be thought of as a set of shells.
Note also that this argument assumes the shells to be of infinitesimal thickness.
As for the hole in the sphere you could assume it to be of infinitesimal diameter too.
Like I said, I find this idea of the net gravity being uniformly zero even though you should at least have the gforce from the particle inside the sphere increasing as it approaches the mass of the sphere's wall (even if you don't consider the force of gravity caused by the sphere itself) while the force on a body falling through what is effectively a thick walled sphere decreases as it falls very confusing.
The only way it makes any sense to me is if we assume the center of the earth to be the limiting case of the spheres. If I think of it that way the force on the body falling through the earth is the sum of the forces on the body from the exterior of the spheres it is falling through.
However this link seems to muddy that up again with wantin to treat the spheres as having a point mass at the center.
Gravity Force of a Spherical Shell
Yes, I know they made that point in http://hyperphysics.phy-astr.gsu.edu...rthole.html#c1 but it is not really as clear as it should be.
The zero gravity for a point mass anywhere within empty sphere can be determined by calculus. What happens is that the elements of the closest wall have a stronger influence but the quantity of the elements on the opposite farther side make up for their weaker influence. Its a college physics problem given as an assignment or taught by the prof.
One must be familiar with basic calculus to do these calculations, and doing them is perhaps the best way to understand. Reading texts has never been a good way to learn mathematical intensive sciences.
http://hyperphysics.phy-astr.gsu.edu...sphshell2.html is the way I remember it.
As for falling into a tunnel straight to a solid sphere's center. Only the mass of the sphere inside (below) the body will exert a force. The shell outside the radius of the body's distance from the sphere center will exert no net force on the falling body. This is or the same reasons given in the hollow sphere above. That outer shell is thin as the object starts falling and it thickens as the object falls until it is the entire sphere when the object is at the center. Similarly the part of sphere that exerts gravity is the whole sphere when the object starts falling and it becomes progressively smaller as the object falls to the center until it is zero at the center. If the tunnel goes all the way through the sphere, then the object will oscillate back and forth with a period. Again a question given on assignments and tests in physics classes.
Just to add to what pikpobedy said - in General Relativity, the relevant concept is called Birkhoff's Theorem. It implies, among other things, that the space-time in the interior cavity of a thin shell must be flat and isometric to Minkowski space-time.