What will be ##\int_{v'}\vec \nabla_{v'}(1/R)dv'##

  • Thread starter Apashanka
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In summary, the volume integral is written in terms of surface integral and reduces to the surface integral when the radius is large.
  • #1
Apashanka
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Homework Statement
What will be ##\int_{v'}\vec \nabla_{v'}(1/R)dv##
Relevant Equations
if the source be represented by the vector ##\vec r'## and another point of interest be ##\vec r## where ##\vec R=\vec r-\vec r'## with respect to a fixed origin,
Now from vector theorem ##\nabla_r'(1/R)=\frac{\hat R}{R^2}##
Now what will be ##\int_v'\nabla_r'(1/R)dv'## over a spherical volume....can anyone please help me out??
Problem Statement: What will be ##\int_{v'}\vec \nabla_{v'}(1/R)dv##
Relevant Equations: if the source be represented by the vector ##\vec r'## and another point of interest be ##\vec r## where ##\vec R=\vec r-\vec r'## with respect to a fixed origin,
Now from vector theorem ##\nabla_r'(1/R)=\frac{\hat R}{R^2}##
Now what will be ##\int_v'\nabla_r'(1/R)dv'## over a spherical volume...can anyone please help me out??

if the source be represented by the vector ##\vec r'## and another point of interest be ##\vec r## where ##\vec R=\vec r-\vec r'## with respect to a fixed origin,
Now from vector theorem ##\vec \nabla_{r'}(1/R)=\frac{\hat R}{R^2}##
Now what will be ##\int_{v'}\vec \nabla_{r'}(1/R)dv'## over a spherical volume...can anyone please help me out??
 
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  • #2
Apashanka said:
Problem Statement: What will be ##\int_{v'}\vec \nabla_{v'}(1/R)dv##
Relevant Equations: if the source be represented by the vector ##\vec r'## and another point of interest be ##\vec r## where ##\vec R=\vec r-\vec r'## with respect to a fixed origin,
Now from vector theorem ##\nabla_r'(1/R)=\frac{\hat R}{R^2}##
Now what will be ##\int_v'\nabla_r'(1/R)dv'## over a spherical volume...can anyone please help me out??

Problem Statement: What will be ##\int_{v'}\vec \nabla_{v'}(1/R)dv##
Relevant Equations: if the source be represented by the vector ##\vec r'## and another point of interest be ##\vec r## where ##\vec R=\vec r-\vec r'## with respect to a fixed origin,
Now from vector theorem ##\nabla_r'(1/R)=\frac{\hat R}{R^2}##
Now what will be ##\int_v'\nabla_r'(1/R)dv'## over a spherical volume...can anyone please help me out??

if the source be represented by the vector ##\vec r'## and another point of interest be ##\vec r## where ##\vec R=\vec r-\vec r'## with respect to a fixed origin,
Now from vector theorem ##\vec \nabla_{r'}(1/R)=\frac{\hat R}{R^2}##
Now what will be ##\int_{v'}\vec \nabla_{r'}(1/R)dv'## over a spherical volume...can anyone please help me out??
The volume integral is written in terms of surface integral ##\int_s(1/R)\vec ds'##
##R=√[(\vec r-\vec r')•(\vec r-\vec r')]=√(r^2+r'^2-2r•r')=r'√(1+\frac{r^2}{r'^2}-2r•r'/r'^2)## for large ##r' ,R=r'##
##\int_s(1/r')r'^2sin\theta d\theta d\phi\hat r'=4\pi r'##
Is it correct??
 
  • #3
Apashanka said:
The volume integral is written in terms of surface integral ##\int_s(1/R)\vec ds'##
##R=√[(\vec r-\vec r')•(\vec r-\vec r')]=√(r^2+r'^2-2r•r')=r'√(1+\frac{r^2}{r'^2}-2r•r'/r'^2)## for large ##r' ,R=r'##
##\int_s(1/r')r'^2sin\theta d\theta d\phi\hat r'=4\pi r'##
Is it correct??
The volume integral is having an ans. ##\int_{v'}\vec \nabla_{r'}(1/R)dv'=\frac{4\pi}{3}\vec r## over a large spherical volume,where ##\vec R=\vec r-\vec r'## is with respect to the origin coinciding with the centre is the sphere.
Can anyone please help me out in proving this...
 
  • #4
Okay I have made a schematic diagram
IMG_20190512_213054.jpg

##\vec r## is fixed,##\vec R=\vec r-\vec r'##
now ##\int_{v'}\vec \nabla_{r'}(1/R)dv'=\int_{s'}(1/R)\vec ds'## is done over a spherical surface having radius ##r_s## then it reduces to ##\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\phi)}r_s^2sin\theta d\theta d\omega\hat r_s=2πr_s\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\phi)}sin\theta d\theta(\vec r-\vec R) ## where ##\omega## is the azimuth 0 to 2π ,##\theta=0## to ##\pi##.
Can anyone will give any hints how to deal with such type of integrals??
Note:within the integral ##\vec R=\vec r-\vec r_s## has it's tail at the surface of the sphere
 
Last edited:
  • #5
Apashanka said:
Okay I have made a schematic diagram
View attachment 243456
##\vec r## is fixed,##\vec R=\vec r-\vec r'##
now ##\int_{v'}\vec \nabla_{r'}(1/R)dv'=\int_{s'}(1/R)\vec ds'## is done over a spherical surface having radius ##r_s## then it reduces to ##\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\phi)}r_s^2sin\theta d\theta d\omega\hat r_s=2πr_s\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\phi)}sin\theta d\theta(\vec r-\vec R) ## where ##\omega## is the azimuth 0 to 2π ,##\theta=0## to ##\pi##.
Can anyone will give any hints how to deal with such type of integrals??
Note:within the integral ##\vec R=\vec r-\vec r_s## has it's tail at the surface of the sphere

This is similar to the integration in the shell theorem, which you may have encountered. E.g:

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html#wtls
 
  • #6
@PeroK
Okk ,then in that case the integral reduces to
##\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\theta)}r_s^2sin\theta d\theta d\omega\hat r_s=2πr_s\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\theta)}sin\theta d\theta(\vec r-\vec R)=4\pi\vec r-\int_s2\pi r_s\frac{\vec R}{r_s^2+r^2-2rr_scos(\theta)}sin\theta d\theta##
The problem here is within the integral there is ##\hat r_s## which is not constt. throughout the integration
 
  • #7
Apashanka said:
@PeroK
Okk ,then in that case the integral reduces to
##\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\theta)}r_s^2sin\theta d\theta d\omega\hat r_s=2πr_s\int_{s'}\frac{1}{r_s^2+r^2-2rr_scos(\theta)}sin\theta d\theta(\vec r-\vec R)=4\pi\vec r-\int_s2\pi r_s\frac{\vec R}{r_s^2+r^2-2rr_scos(\theta)}sin\theta d\theta##
The problem here is within the integral there is ##\hat r_s## which is not constt. throughout the integration
The link I gave you shows a technique for tackling that.
 
  • #8
PS if you have studied electromagnetism then a quick way is to find a charged sphere that generates the same integral for the electric field.
 
  • #9
PeroK said:
PS if you have studied electromagnetism then a quick way is to find a charged sphere that generates the same integral for the electric field.
I understood this but the problem is how to tackle the unit vector ##\hat r_s## within the integral which is not constt. ,##r_s## is the radius of the sphere of integration
 
  • #10
Apashanka said:
I understood this but the problem is how to tackle the unit vector ##\hat r_s## within the integral which is not constt. ,##r_s## is the radius of the sphere of integration
IMG_20190513_114747.jpg

Our integration is ##\int_s(1/R)\vec ds=\int_s(1/R)r_s^2sin\theta d\theta d\phi \hat r_s,##
##R=√(r^2+r_s^2-2rr_scos\theta)##
##\hat r_s=(\vec r-\vec R)/r_s##
Putting these it reduces to ##2\pi r_s \int_s[\frac{\vec r}{R}-\hat R]sin\theta d\theta##
Now since ##\vec r## is constt. the first part is solvable but how to do the second part ??
 
  • #11
Okay I am asking for this type of integrals is due to ,I have came across this in the following snap of a paper by @Nick Kaiser
IMG_20190513_205158.jpg

In deriving this equation 8 from eq. 6 I think this integral will come in way for the term in the RHS containing ##\bar n##.Will anyone please help me out??or any idea in deriving eq.8 from eq.6...
 

1. What does the integral ##\int_{v'}\vec \nabla_{v'}(1/R)dv'## represent?

The integral represents the average change in the reciprocal of the distance from a point in space to a given reference point, over a specified volume.

2. How is the integral ##\int_{v'}\vec \nabla_{v'}(1/R)dv'## calculated?

The integral is calculated by taking the gradient of the reciprocal of the distance function, and then integrating it over the specified volume.

3. What is the significance of the integral ##\int_{v'}\vec \nabla_{v'}(1/R)dv'## in scientific research?

The integral is commonly used in fields such as physics and engineering to calculate the potential or force at a point due to a distribution of sources. It is also used in fluid dynamics to calculate the velocity field of a fluid.

4. Can the integral ##\int_{v'}\vec \nabla_{v'}(1/R)dv'## be negative?

Yes, the integral can be negative if the distance function is decreasing over the specified volume. This indicates a decrease in potential or force at a point due to the distribution of sources.

5. How does the integral ##\int_{v'}\vec \nabla_{v'}(1/R)dv'## relate to the concept of a point charge in physics?

The integral is related to the concept of a point charge in physics through Coulomb's Law, which states that the electric field at a point is equal to the gradient of the electric potential at that point. The integral represents the average change in the electric potential over a specified volume, and can be used to calculate the electric field due to a distribution of point charges.

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