Gradients of 1/r: Solutions from Griffiths' Electrodynamics

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In summary, Christianjb states that the gradient operator is differentiable with respect to a different co-ordinate system, which causes one to lose a minus sign. He then provides a derivation for the potential at a point given the position and dipole moment of the point.
  • #1
Flux = Rad
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Homework Statement



This is from Griffiths' Intro to Electrodynamics. He is discussing the field of a polarized object of dipole moment per unit volume [tex] \vec{P} [/tex] viewed at [tex] \vec{r} [/tex].

He then states:

[tex] \nabla ' \left( \frac{1}{r} \right) = \frac{ \hat{r}}{r^2} [/tex]

Where [tex] \nabla ' [/tex] denotes that the differentiation is with respect to the source co-ordinates [tex] \vec{r}' [/tex]


Homework Equations





The Attempt at a Solution



Following from the definition of the gradient,

[tex] \nabla ' \left( \frac{1}{r} \right) = \frac{-1}{r^3} \left[ x \frac{ \partial x}{\partial x'} \hat{x}' + y \frac{\partial y}{\partial y'} \hat{y}' + z \frac{\partial z}{\partial z'} \hat{z} \right]
[/tex]

So I guess all would be well as long as
[tex] \frac{\partial x}{\partial x'} \hat{x}' = - \hat{x} [/tex]
However, this isn't clear to me at the moment
 
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  • #2
The easiest method is to use the chain rule in r

d/dx = d r^2/dx * d/d r^2

r^2 = x^2+y^2+z^2

d r^2/dx = 2x

so d/dx= 2x d / dr^2 = 2x dr/dr^2 d/dr = (x/r) d/dr

d/dx (1/r) = -(x/r) 1/r^2
 
  • #3
Well, just to complete the above:

grad (1/r) = (d/dx xhat + d/dy yhat +d/dz zhat) (1/r)
= -x/r^3 xhat -y/r^3 yhat -z/r^3 zhat
= -(x,y,z)/r^3=-rhat/r^2

apologies for being too lazy to latex this.

(edit- corrected dumb mistakes)
(edit again- corrected dumb corrections. Hopefully this is right now. I had to write it down)
 
Last edited:
  • #4
The 'pro' method is to memorize the useful formula

grad (f(r))= rhat df(r)/dr

You can prove this using the methods above.
 
  • #5
Why not use the gradient operator for spherical polar coordinates, noting that f is a function of r alone?
[tex]\nabla f=\hat{r}\frac{\partial f}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial f}{\partial \theta}+\hat{\varphi}\frac{1}{r\sin\theta}\frac{\partial f}{\partial\varphi}[/tex]
 
  • #6
Thanks for the help so far.

Christianjb, what you've written is correct, but it proves
[tex] \nabla \left( \frac{1}{r} \right) = \frac{- \hat{r}}{r^2} [/tex]

I'm cool with this, but my problem is that Griffiths is differentiating with respect to a different co-ordinate system (hence the prime on the gradient operator), which seems to cause one to lose a minus sign. He calls this the source coordinates and they are integrated over since we are not dealing with a point charge.

Christo, so if I use the spherical gradient, ignoring angular parts,
[tex] \nabla f = \hat{r} \frac{\partial f}{\partial r} [/tex]
And presumably I can extend this to my case by changing the co-ordinate system so that
[tex] \nabla ' f = \hat{r}' \frac{\partial f}{\partial r'} [/tex]
Now substituting 1/r for f,
[tex] \nabla ' \left( \frac{1}{r} \right) = \hat{r}' \frac{\partial}{\partial r'} \left(\frac{1}{r} \right) = \frac{- \hat{r}'}{r^2} \frac{\partial r}{\partial r'} [/tex]

So that this time it appears that I require
[tex] \hat{r} = - \frac{\partial r}{\partial r'} \hat{r}' [/tex]
in order to be in agreement with Griffiths.

This is kinda neater than what I first posted with individual components, but I'm still not sure why it's true.
 
  • #7
Ok, I didn't notice the prime. What's the relationship between the primed and the unprimed coordinates? It may turn out that you cannot ignore the angular parts-- just because f is a function only of r it doesn't mean that f is a function of only r'.

(I don't have the text, so am relying solely on what you write here!)
 
  • #8
Yeah, I was beginning to realize that was the problem. The relationship between the co-ordinate systems isn't explicitly stated.

What we've got is the usual arbitrary blob in space, which has a dipole moment per unit volume [tex] \vec{P} [/tex]. We want to know what the potential is due to this blob.

For a simple dipole [tex] \vec{p} [/tex] we have
[tex]
V (\vec{r}) = \frac{1}{4 \pi \epsilon_0 } \frac{\hat{\mathcal{R}} \cdot \vec{p}}{\mathcal{R}^2}
[/tex]

Where [tex]\mathcal{R}[/tex] is the vector from the dipole to the point at which we are evaluating the potential.

So in our case we have a dipole moment
[tex]
\vec{p} = \vec{P} d \tau'
[/tex] in each volume element [tex] d \tau' [/tex] so the total potential is:

[tex]
V(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int_\mathcal{V} \frac{ \hat{\mathcal{R}} \cdot \vec{P}( \vec{r}') }{\mathcal{R}^2} d \tau '
[/tex]

Now he states that
[tex]
\nabla ' \left( \frac{1}{\mathcal{R}} \right) = \frac{\hat{\mathcal{R}}}{\mathcal{R}^2}
[/tex]
where the differentiation is with respect to the source coordinates [tex] (\vec{r}') [/tex]


To me, this isn't very clear but I reckon that we've got an origin. We want the potential at point [tex] \vec{r} [/tex] from the origin. Now we have a dipole at position [tex] \vec{r} ' [/tex]. And we are told that our point is at [tex] \vec{ \mathcal{R} } [/tex] from the dipole.
So that surely [tex] \vec{\mathcal{R}} = \vec{r} - \vec{r}' [/tex]
 
  • #9
It's a simple sign change when you diff wrt the axis coordinates.

In 1D- moving the origin 1 unit to the left has the effect of increasing all x values by 1. Thus the signs are reversed.
 

1. What is the Gradient of 1/r?

The Gradient of 1/r, also known as the inverse square law, is a mathematical formula used to describe the relationship between the strength of a force and the distance between two objects. It states that the force between two objects is inversely proportional to the square of the distance between them.

2. How is the Gradient of 1/r used in science?

The Gradient of 1/r is used in a variety of scientific fields, including physics, astronomy, and engineering. It is commonly used to describe the strength of gravitational and electric fields, as well as the intensity of light and sound waves.

3. How is the Gradient of 1/r calculated?

The Gradient of 1/r is calculated by taking the derivative of the inverse square law formula. This involves dividing the constant value by the distance between the two objects squared. The resulting value represents the strength of the force at a given distance.

4. What are some real-life examples of the Gradient of 1/r?

One real-life example of the Gradient of 1/r is the force of gravity between two objects, such as the Earth and the Moon. The force of gravity decreases as the distance between the two objects increases and follows the inverse square law. Another example is the intensity of light from a point source, which decreases as the distance from the source increases.

5. Are there any limitations to the Gradient of 1/r?

While the Gradient of 1/r is a useful mathematical tool, it does have some limitations. It assumes that the two objects involved are point sources with no other objects affecting the force or field. In reality, this is often not the case, and the force or field may be affected by other factors, such as the shape or density of the objects.

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