1

I can't remember exactly what it was, but I remember going through a problem in physics related to gravity on and inside a sphere, and found that inside, gravity acts linearly as a result of some triple integral cancellation with an assumption on uniform density.

Suppose Earth itself was a perfect sphere and you could pass through it. Does gravity actually spontaneously transition from inverse quadratically to linearly? Would it really be some magic spontaneous switch in forces? Or is there something in the math that can explain a gradual transition from a quadratic factor to a linear factor?

CheeseMongoose
  • 87

2 Answers2

0

Note: Throughout the answer, I am treating the scenario from the point of view of Newtonian gravity.

Uniform solid sphere

The force function of a uniform sphere is continuous. Thus, the force never changes abruptly. Below is the graph between gravity and the distance from the sphere's center:

graph

Image source

As you can see, the function is a continuous one, and doesn't change it's value suddenly when you cross the surface.

Uniform spherical shell

However, in case of a uniform spherical shell, you will get to see a discontinuous force vs distance graph, something like this:

image

Image source

Here, the value of gravitational field/force inside the shell is zero, whereas just outside the shell, it's non-zero. This results in an abrupt change in force.

Physical feasibility

This discontinuity of a vector field is not at all unphysical or infeasible. There's nothing wrong to have an abrupt change in a force field. However, if a field has a potential associated to it (in other words, if the field is conservative), then the potential must be continuous. Because a conservative field is defined as the negative of the gradient of potential, and the gradient of the potential cannot be defined if the gradient is discontinuous. So, for a field to exist everywhere, it's potential must be continuous. The field doesn't need to satisfy continuity.

  • There is a very clear non-differentiable point in the solid sphere, this implies the speed of the transition is undefined. I didn't dispute that the curve was continuous, I asked about analyzing the speed of the change at such a non-differentiable point, if there is a more realistic model or explanation. It doesn't quite make sense that I could be one milimeter above the surface where gravity is completely and perfectly quadratic, then suddenly it's not just because I move 2 milimeters. There are functions capable of producing almost-cusp points but that are smooth nonetheless. – CheeseMongoose Jun 28 '20 at 14:05
  • 2
    Why does this bother you? Dig a 6 foot hole, jump in. Did you feel the discontinuity as you transitioned from inverse quadratic to linear? – JEB Jun 28 '20 at 14:22
  • That's the point. If you can't measure the effects of the discontinuity or non-differentiability, then maybe it's not really there. When an airplane breaks the sound barrier, it doesn't create waves at literally infinite frequency, that artifact is simply a failure of the classical model to predict the interference of proximal sound waves. Similarly, the acceleration due to gravity can't transition at literally infinite speed. Perhaps in a quantum model you could maybe argue "at random speed" which I haven't seen, but not infinite speed. – CheeseMongoose Jun 28 '20 at 14:38
  • 1
    @CheeseMongoose I believe this example was that of a spherical shell with infintessimal thickness. For such a shell to have non-zero mass, it must be infinitely dense, which is one of those cases where discontinuities appear. If your shell has some thickness, then the discontinuity starts to distort to look more and more like the uniform sphere example. – Cort Ammon Jun 28 '20 at 15:32
  • The uniform sphere example contains a point of non-differentiability. Do you think the transition happens at infinite speed? – CheeseMongoose Jun 28 '20 at 17:24
  • @CheeseMongoose There is no speed here. The field was always there, it never needed to transition or change. It is we, who are passing through the field, we are experiencing a different kind of radius dependence, and this is in no way as unphysical as you think it to be. –  Jun 28 '20 at 17:28
  • That radius dependence depends on time, so if you examine a falling object passing through the sphere, you're saying it experiences an infinitely fast change in the acceleration due to gravity. – CheeseMongoose Jun 28 '20 at 17:29
  • The object experiences the discontinuous (completely legit, nothing unphysical) but the field in the space never changes. –  Jun 28 '20 at 17:35
0

For a spherical body of mass $M$ and radius $R$, assuming the density is constant you can calculate the gravitational field strength at any distance $r$ from the centre of the body using Gauss's law. If you work through the algebra (which is a good exercise) you'll eventually find

$$g = \begin{cases} -\frac{MGr}{R^3}, &\qquad r \le R, \\ -\frac{GM}{r^2}, &\qquad r > R. \end{cases}$$

If you look at both expressions for $r = R$ you'll see that they agree. This means that the gravitational potential is continuous at $r = R$. If you take the first derivative with respect to $r$ of both expressions, however, you'll find that they are not the same. This means that the potential is not smooth in the mathematical sense of a smooth function.

DavidH
  • 3,275