Is the Integral of 1/r^2 Over a Closed Loop Zero?

Click For Summary
SUMMARY

The integral of 1/r^2 over a closed loop is proven to be zero using Cauchy's theorem on residues. By selecting two points, A and B, and two curves, C_1 and C_2, the line integrals along these curves are shown to be equal, leading to the conclusion that the loop integral is zero. The function 1/τ^2 is analytic except for a pole of order 2 at τ=0, confirming that the residue at this pole is zero, thus validating the integral's value.

PREREQUISITES
  • Understanding of complex analysis, specifically Cauchy's theorem.
  • Familiarity with line integrals and contour integration.
  • Knowledge of residues and singularities in complex functions.
  • Basic concepts of analytic functions and their properties.
NEXT STEPS
  • Study Cauchy's theorem and its applications in complex analysis.
  • Learn about contour integration techniques and their significance.
  • Explore the concept of residues and how to calculate them for different functions.
  • Investigate analytic functions and their behavior near singularities.
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on complex analysis, as well as physicists dealing with loop integrals in theoretical contexts.

rocket
Messages
9
Reaction score
0
i want to prove that this integral along a closed loop:

[tex]\oint (1/r^2) dr[/tex]

is equal to zero. but I'm not sure how to prove it. i was wondering if someone can show me a rigid proof for this. I think I'm missing something here because I'm not really that familiar with loop integrals.
 
Physics news on Phys.org
Pick two points, A and B, and two curves, C_1 and C_2 where C_1 goes from A to B and C_2 from B to A.

Now calculate the line integral along the curve C_1 and C_2. They should both be equal, in which case, the loop integral will be 0.
 
Since [tex]1/\tau^2[/tex] is an analytic function with a singularity in 0 (a pole of order 2), the contour your want to use is closed, then you can use the Cauchy's theorem on residues.

[tex]\oint 1/\tau^2 \, d\tau = 2i\pi \cdot 0[/tex]

where 0 is the residue of [tex]1/\tau^2[/tex]
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
9
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
12
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K