Calculate Nabla Operator for Potential Function with Distance r

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Homework Help Overview

The discussion revolves around calculating the nabla operator for a potential function defined in terms of distance r, specifically for the function \(\varphi (r) = \frac{1}{4\pi\epsilon_{0}}\frac{1}{r}\). The context involves vector calculus and the interpretation of scalar and vector fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct application of the gradient operator versus the divergence operator, with some attempting to simplify expressions involving Cartesian and spherical coordinates. Questions arise about the relationships between the vector \(\vec{r}\), the unit vector \(\hat{r}\), and the scalar r.

Discussion Status

The discussion is active, with participants providing hints and guidance on vector notation and simplification. There is recognition of the need to clarify the differences between various vector representations and their magnitudes. Some participants express confusion about vector notation and the relationships between components.

Contextual Notes

Participants are navigating the transition between Cartesian and spherical coordinates, with specific attention to the definitions and properties of vectors in these systems. There is an acknowledgment of potential misunderstandings regarding vector notation and the implications of the calculations being discussed.

bumclouds
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Does my solution look correct to you guys?

Homework Statement



Calculate:
\nabla \varphi (r)

If:
\varphi (r) = \frac{1}{4\pi\epsilon_{0}}\frac{1}{r}

with: r = \sqrt{x^{2}+y^{2}+z^{2}}

Homework Equations



n/a

The Attempt at a Solution



hw1.jpg

hw2.jpg
 
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You used the divergence not the gradient. Remember taking the gradient of a scalar field "creates" a vector.
 
Is this better? I can't figure out how to simplify it further.

hw4.jpg
 
Last edited:
That is correct, but you can simplify it much further. What is x^2+y^2+z^2? What is x \hat{e_1}+y\hat{e_2}+z \hat{e_3}?

Hint: \vec{r}=?
 
It is the derivative w.r.t. r times r-hat, because if you change r by dr, the step length in the r-hat direction is dr.


By definition, you have for a differentiable function f(x1,x2,...) that:

df = nabla f dot ds (1)

where ds is the displacement vector.

You also have that:

df = df/dx1 dx1 + df/dx2 dx2 + df/dx3 dx3 + ... (2)

In case of Cartesian coordinates, ds = (dx1, dx2,...), but in general this is not the case.
 
Cyosis said:
What is x^2+y^2+z^2

r^{2} = x^{2}+y^{2}+z^{2}?

Cyosis said:
What is x \hat{e_1}+y\hat{e_2}+z \hat{e_3}
- \hat{r} = -x \hat{e_1}-y\hat{e_2}-z \hat{e_3}?


So final answer:
\frac{-\hat{r}}{r^{5/2}} ?
 
Last edited:
No, |\hat{r}|=1 yet we know that |x \hat{e_1}+y\hat{e_2}+z \hat{e_3}| \neq 1. Again, do you know what \vec{r} looks like in Cartesian coordinates?
 
I'm thinking hard but I'm not sure..

I'm thinking something along the lines of.. the x-component of \hat{r}.. the y-component. or something? >_<
 
You're getting close. We are given that |\vec{r}|=\sqrt{x^2+y^2+z^2}=r. So we want to find s an expression for \vec{r} whose length is r and points in the \hat{r} direction.
 
  • #10
I don't understand the differences between:

\vec{r},\hat{r} and r!

=(
 
  • #11
A vector has a direction and a magnitude. A unit vector is a vector with magnitude 1. For example \hat{e_1} is the unit vector pointing in the x-direction with |\hat{e_1}|=1. These are used in Cartesian coordinates and I am sure you're familiar with them. In your problem however we are working with spherical coordinates. the unit vectors for spherical coordinates are \hat{r},\hat{\theta}, \hat{\phi}. So whereas you express a vector in Cartesian coordinates as a \hat{e_1}+b\hat{e_2}+c \hat{e_3} in spherical coordinates you would express it as \alpha \hat{r}+\beta \hat{\theta}+\gamma \hat{\phi}.

In spherical coordinates the length of a vector \vec{r} is given by |\vec{r}|=\sqrt{x^2+y^2+z^2}=r. Therefore r is the magnitude of \vec{r}. A vector also has a direction and the direction of \vec{r} is given by\hat{r}.

It is given in the problem statement that |\vec{r}|=\sqrt{x^2+y^2+z^2} and we have a vector,x \hat{e_1}+y\hat{e_2}+z \hat{e_3} in Cartesian coordinate which we would like to write as a vector in spherical coordinates. If you draw a Cartesian coordinate system and draw a radial vector on the axes, then how can you express this radial vector in terms of x,y,z? Compare it to the vector we're trying to write in spherical coordinates, also compare the length of x \hat{e_1}+y\hat{e_2}+z \hat{e_3} with the length of \vec{r}.

Lastly you also need to write (x^2+y^2+z^2)^{-3/2} in terms of r correctly, it is not r^{-5/2}.
 
Last edited:
  • #12
bumclouds said:
So final answer:
\frac{-\hat{r}}{r^{5/2}} ?

This is almost there... if you fix your confusion in how to notate each kind of vector, you only have to fix a minor arithmetical error
 
  • #13
=\frac{-r\hat{r}}{r^{3}}

??

so it's the negative r unit vector pointing in the r direction.. over r cubed.. right? =(

I seem to have a huge knowledge hole about vectors
 
  • #14
It is the correct answer, but how did you get to it?

so it's the negative r unit vector pointing in the r direction.. over r cubed.. right?

No it is the unit vector pointing in the -\hat{r} direction multiplied by the scalar r. That means it is a vector pointing radially towards the origin of the coordinate system and has a length r.
 
Last edited:

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