What is the Limiting Value of Integral in Newton's Shell Theorem for r=R?

In summary, the problem is that you are taking the limit of a discontinuous function. Depending on the way you do this, you could get different results.
  • #1
parshyaa
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Homework Statement
While deriving newton shell theorem for a hollow sphere of mass M , radius R and a point particle of mass m at a distance r, i am getting wrong answer while taking r=R (ie particle on the surface of spherical shell) Please tell me where i am going wrong.
Relevant Equations
F = ##\ (GmM/4r^2R)\int_{r-R}^{r+R}((r'^2+r^2-R^2)/r'^2)dr'##
?hash=e926ad216734d271bd1b3e05e80b4f1e.jpg

First i tried proving Newton shell theorem directly for r=R and solved the integral as above but still got the wrong solution.
?hash=e926ad216734d271bd1b3e05e80b4f1e.jpg

Here i tried using general case:
Here r' is the distance of a small ring from the point particle of mass m
So my doubt is when we take r=R and then evaluate this equation, limit goes from 0 to 2R and integral gives the value (GmM/2r^2)
Which is wrong, so where am i going wrong?
 

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  • #2
I believe the "relevant equation" is dimensionally incorrect ?
 
  • #3
hutchphd said:
I believe the "relevant equation" is dimensionally incorrect ?
Yep i had written r'instead of r'^2 and dr instead of dr', now its been edited
 
  • #4
The result is O.K. The "paradox" arises from considering an infinitely thin layer of mass. Inside the shell (r<R) the field is zero; just outside the the shell is GM/R^2. There is "jump" from zero to a finite value. On the surface, the result is the average of the values inside and outside.
 
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  • #5
Gordianus said:
The result is O.K. The "paradox" arises from considering an infinitely thin layer of mass. Inside the shell (r<R) the field is zero; just outside the the shell is GM/R^2. There is "jump" from zero to a finite value. On the surface, the result is the average of the values inside and outside.
Wow insight is good and satisfying, is their a mathematical theory behind this?, Some kind of theorem regarding integration ambiguity of this kind?
 
  • #6
parshyaa said:
Wow insight is good and satisfying, is their a mathematical theory behind this?, Some kind of theorem regarding integration ambiguity of this kind?
You have a function that is discontinuous at ##r = R##. Depending on the way you take the limit, you could end up with:
$$\lim_{r \rightarrow R^+} f(r), \ \ \lim_{r \rightarrow R^-} f(r)$$
Or something else. In your case it was something else, because you had an integral straddling the discontinuity, hence you got half of the right-hand limit and half of the left-hand limit.

The moral is to be careful taking a limit if the function is discontinuous. This idea will come up again in electromagnetism when looking at surface charges and surface currents.

One way round this problem (and an easier way to prove the shell theorem) is to use the potential, which is continuous at the shell boundary.
 
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  • #7
parshyaa said:
Wow insight is good and satisfying, is their a mathematical theory behind this?, Some kind of theorem regarding integration ambiguity of this kind?
Then you have some subtleties in vector calculus. You find the corresponding mathematical techniques in textbooks of electrodynamics, where one discusses the conditions for the fields at singular boundaries like surfaces. Then the differential operations like the divergence of a vector field have to be handled with some care, leading to the idea of surface divergences etc.
 
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  • #8
@PeroK ,@vanhees71, it seems this is the case of jump discontinuity
PicsArt_06-05-04.55.04.png

And still our solution manages to give average value, how cool is this?
 

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  • #9
I suggest reading a similar topic in Feynman's Lectures. Somewhere, he computed the force between the plates of a parallel plate capacitor. In this case, again, the field changes from zero (inside the plate) to a finite value (between the plates). Feynman used the average value and obtained the right value of the force. Perhaps he didn't follow the purest way, but he knew what he was doing.
 
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1. What is Newton's shell theorem?

Newton's shell theorem is a principle in classical mechanics that states that the gravitational force exerted by a spherically symmetric body on a particle outside of the body is the same as the force that would be exerted by a point mass located at the center of the body.

2. What is the significance of r=R in Newton's shell theorem?

The variable r represents the distance between the particle and the center of the body, while R represents the radius of the body. The theorem applies specifically to particles located at a distance r from the center equal to the radius R of the body.

3. How does Newton's shell theorem apply to real-life situations?

Newton's shell theorem is commonly used in astrophysics and celestial mechanics to calculate the gravitational force between celestial bodies, such as planets and stars. It is also used in engineering and physics to analyze the forces exerted on objects with spherically symmetric mass distributions, such as spheres and shells.

4. What are the limitations of Newton's shell theorem?

Newton's shell theorem assumes that the body in question has a spherically symmetric mass distribution, meaning that its mass is evenly distributed around its center. It also only applies to particles located outside of the body, and does not take into account the effects of other nearby bodies.

5. How does Newton's shell theorem relate to other laws and principles in physics?

Newton's shell theorem is closely related to Newton's law of universal gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. It also has connections to Gauss's law in electrostatics, which states that the electric field outside of a charged spherical shell is the same as the field that would be produced by a point charge located at the center of the shell.

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