Integral equation for large surfaces

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Discussion Overview

The discussion revolves around the conditions under which terms of a surface integral can be neglected, particularly in the context of large surfaces and the behavior of a vector field at infinity. Participants explore the implications of the vector field's decay rate relative to the growth rate of the surface area.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that the vector field ##\vec A## is assumed to go to zero at infinity faster than the surface grows, which justifies neglecting certain terms in the integral.
  • One participant questions whether the integral could still be zero if the decay rate of the vector field is slower than the growth rate of the surface, suggesting a potential misunderstanding of the conditions.
  • Another participant asserts that the assumption of the field going to zero fast enough is necessary, referencing the example of a point charge's field.
  • It is mentioned that the field decreases as ##\frac{1}{r^2}## while the surface element grows as ##r^2 \sin\theta d\theta d\phi##, indicating that the integral is not zero under these conditions.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the decay rate of the vector field relative to the surface growth, indicating that the discussion remains unresolved regarding the conditions under which terms can be neglected.

Contextual Notes

The discussion highlights the dependence on specific assumptions about the decay rates of the vector field and the growth of the surface, which are not fully resolved.

Apashanka
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We often neglect the terms of a surface integral ##\int_v(\nabla•A)dv=\int_s(A•ds)## for ##s## to be very large or ##v## to be very large,
What is actually the reason behind this to neglect??
 
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The field ##\vec A## is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.
 
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Orodruin said:
The field ##\vec A## is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.
Okk got it ...thanks @Orodruin
 
Orodruin said:
The field ##\vec A## is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.
But if the rate at which the vector field ##A## goes to zero at infinity is slower than the rate at which the surface grows ,then also it would be zero ,isn't it??
 
No. You need the assumption of the field going to zero fast enough. Take the example of the field of a single point charge.
 
Orodruin said:
No. You need the assumption of the field going to zero fast enough. Take the example of the field of a single point charge.
Ok the field decreses as ##\frac{1}{r^2}## and the spherical surface element centered at the point charge grows as ##r^2sin\theta d\theta d\phi##
 
Apashanka said:
Ok the field decreses as ##\frac{1}{r^2}## and the spherical surface element centered at the point charge grows as ##r^2sin\theta d\theta d\phi##
And so the integral is not zero.
 
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