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Is curved space allowed to show discontinuities (steps)?

  1. Sep 16, 2011 #1
    Recall the gravitational effect of a hollow sphere upon a test mass inside ... no net attraction of the test mass anywhere within the sphere.

    Let this sphere be large and dense enough to have 1 G of attraction on a test mass resting on its outer surface.

    Plotting the measure of effect on the test mass positioned along a line from the center of the sphere to the sphere's surface and continuing radially outside the sphere, at first approximation it looks like the plot of effect against radius would show no effect from the center to the sphere's radius, then 1 G effect continuing from there and gradually diminishing.

    This plot would appear to represent a discontinuity at the radius where the effect steps from 0 to 1 G.

    In thinking about this, I imagine we make a hole in the sphere to see what happens as we pass a test mass through it...

    If the hole is made by removing material from the sphere, that changes the effect, especially when in very close proximity to the hole, because the subtended area of the plane of the hole with respect to the test mass inside approaches 1/2 at the hole plane itself. And there is the missing mass of the hole to account for...

    Maybe an improved version of making the hole might be to use a radially retracting system like the old fashioned aperture of a camera - where equal amounts of sphere mass would be moved and aligned around the hole uniformly. Since the plotting experiment is done along a single radial line from the sphere center through the hole center, there might be at least one particular redistribution of hole mass to the surface around it that results in the same original plots - 0 inside from center to radius, then 1 G from there diminishing with further distance. But it look like there will be an important error based on the diameter of the hole - as soon as the test mass passes through the plane of the hole from the center of the sphere, there will be a region between that plane and the radius of the sphere where the test mass will find that all of the sphere's mass is on one side before reaching the radius. Likewise the test mass approaching from outside will encounter the same thing after it has passed the radius but before passing through the plane of the hole center.

    But maybe for thought experiment purposes we can just imagine no actual hole is necessary and that the sphere is permeable to the test mass... maybe the sphere is comprised of a geodesic lattice? Anyway, the point is, what is going on at the radius when the attraction steps from 0 to 1? Is this thinking correct? Are there any other instances of gravity making what appears to be discontinuous step functions?

    What does this mean for curved space models? Can curved space "step"?
     
  2. jcsd
  3. Sep 16, 2011 #2
    Not really, because: how thick is the shell?
     
  4. Sep 17, 2011 #3
    I thought about the thickness of the shell...

    So you are thinking that the effect will be 0 inside the inner radius, then in passing through the shell, the parts of the shell at further radius continue to have 0 effect as the parts of the shell at lesser radius begin to develop a central attraction that rises with increasing passage through the shell to the outer radius...

    To me, what that does is change the plot so it has a zero level inside the inner radius, an increasing level through the radius of the shell thickness, and then a third level that starts with 1 G at the outer radius and diminishes from there...

    So then the question is about the transition points (inner and outer shell radii).

    I can see the inner transition changing from 0 to greater than zero in a smooth way.

    For some reason I think that the outer transition is not going to be smooth. Even if the effect at the outer radius is equal when approached from each side (no step), it seems like the inner side effect might be ramping up either straight or exponentially approaching the outer radius, and the outer effect side ramping up exponentially in its approach. That would make the transition point kind of sharp.
     
  5. Sep 17, 2011 #4

    A.T.

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    The function of gravitational acceleration along the radius is continuous at both points. It's first derivate is not, at both points.
     
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