Wednesday, July 25, 2012

Appendix: Doing the Integration for a Hollow Sphere

Sadly, even most 1st year math students will not really understand integrating the hollow sphere, because of subtleties in the integration process. A 3rd year undergrad will do it easily, but still have little grasp of its application in physics.

The 1st year physics student will be able to do the application of the integral, for a simple case like a point particle and a ring, but will simply copy the textbook, because its all he'll have time for. He will have no idea when and where the formula will fail. The 3rd year undergrad will have a good grasp of the formulas and their applications, and perhaps struggle through the integration if he has to, but will have no real idea how the integrals were originally derived.

So to hold the hand of 1st and 2nd year college students in both physics and math, I've prepared a walk-through of the integration of the summation formula.

5. Appendix: Integrating the Hollow Sphere Formula

Let Q = ( D2 + 1 – 2Dx ) 3/2 then we break up the integral into two parts:
Let R = the right hand integral as a place-holder.
Integrating in parts, we handle the first (left hand) part using the following definitions:
We insert a constant to prepare for the substitution of dU, and bring the leftover outside the integral.
  We solve the indefinite integral, and make a note of the interval, but because the new function U has undefined constants, it is simpler to return to the variable x and the original limits of integration. Meantime we keep the form U as a handy shorthand.

We are now ready to unpack the right hand integral, R.
In this case, we choose a slightly different substitution, due to the complication of the x in the numerator having the same power as the x in the denominator.

Let a = ( D2 + 1) , b = ( -2D) , Let V = ( a + bx ) ½ , ( denominator = V3 )
V2 = ( a + bx ) , bx = ( V2 – a ) , x = (V2 – a)/b … dx = 2 / b * V * dV
Substituting and simplifying gives:

This is perhaps the trickiest part of the integration. We can’t simply accept the constant 1 for the left integral, because although we proceeded as if they were indefinite integrals, in fact they are definite integrals. The 1 here is really a placeholder for V0 which we must preserve to follow through:

Back-substituting V = ( a + bx ) ½ allows us to restore the limits of integration,
and we can group the terms by finding a common denominator:

More simplification can be achieved by substituting back in b = - 2D :

Now we are ready to combine left and right parts of the original integral.

 Restoring denominators, and taking 1/ D2 to the outside of the bracket gives:
 Grouping common denominators and substituting in a = (D2 + 1) gives:
  The final formula simplifies nicely without any x term left. Both denominators are now easily factored, and squaring a term then extracting its root is equivalent to performing the Absolute Value function (!):

This should now be recognizable as simply Newton’s inverse square Formula with a toggle factor that switches off the force when the test-mass at D is inside the radius ( = 1 ) of the hollow sphere.
Finally, we can generalize the formula for a sphere of any radius R :

All done!

We know the integration has been executed properly, because the result gives us precisely what Newton postulated, aside from the asymptotes at the surface points.

However, this is not a proof in any significant sense of Newton's Sphere theorem.
Its not the integration itself or its results for distances smaller than the radius which is actually under dispute, although the integration steps we have just followed require a lot more mathematical theoretical support and theorems than we have presented here.

Students should be forewarned that successful completion of the mathematical task is not any kind of proof either in regard to the formula's application to a physical problem, or in regard to gravitational theory in any way. Its simply some mathematical juggling.

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