Derivatives in Vector Notation

In summary, the equation for the electrostatic potential of an electric dipole, \Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}}), can be used to find the electric field by taking the gradient, \mathbf{E} = - \nabla \Phi. The derivative of the potential is given by \mathbf{E} = \frac {1} {4\pi\epsilon_0 r^3} \left(3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{
  • #1
dimensionless
462
1
I've got an equation (from Wikipedia)for the electrostatic potential of an electric dipole. It looks like this:

[tex]
\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})
[/tex]

E is the electric field
r, r, r\hat are as above
p is the (vector) dipole moment
e0 is the primitivity of free space

To find the electric field I have to take the derivative as follows.

[tex]
\mathbf{E} = - \nabla \Phi
[/tex]

The derivative looks like this:

[tex]
\mathbf{E} = \frac {1} {4\pi\epsilon_0 r^3} \left(3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}\right)
[/tex]

I'm confused by the vector notation. Why do I have r\hat multiplied by r\hat? Why is the electric field simply not

[tex]
\Phi (\mathbf{r}) = \frac {1} {2\pi\epsilon_0 r^3} (\mathbf{p}\cdot\hat{\mathbf{r}})
[/tex]
 
Physics news on Phys.org
  • #2
Do you know the equation for the grad operator [tex]\nabla[/tex] in spherical coordinates?
 
  • #3
dimensionless said:
The derivative looks like this:

[tex]
\mathbf{E} = \frac {1} {4\pi\epsilon_0 r^3} \left(3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}\right)
[/tex]

I'm confused by the vector notation. Why do I have r\hat multiplied by r\hat? Why is the electric field simply not

[tex]
\Phi (\mathbf{r}) = \frac {1} {2\pi\epsilon_0 r^3} (\mathbf{p}\cdot\hat{\mathbf{r}})
[/tex]

To answer your first question - it's not r\hat times r\hat, but the scalar product of p\hat and r\hat times r\hat (basically a scalar multiple of the unit vector)

As for the second question, it's not what you think it is because the expression is in spherical polar coordinates
 
Last edited:
  • #4
This is a classic one, especially because you need to calculate the gradient of a scalar:

First apply :
[tex]\nabla \frac {1} { r^2} ({p}\cdot\hat{r}) = \nabla ( \frac {1} { r^2}) ({p}\cdot\hat{r}) + \frac {1} { r^2} \nabla ({p}\cdot\hat{r}) [/tex]

To get the gradient of a scalar:
[tex]\nabla \frac {1} {r^2} = \frac {-2 \vect(r)}{r^4}[/tex] ; the r in the numerator is a vector !

To get this, just write the vectors in x and y coordinates for example.

The [tex]\frac{1}{r^2} = \frac{1}{x^2+y^2}[/tex]
Take the derivative to x and to y, then add up these two values. Keep in mind that each result will have a unitvector [tex]e_x[/tex] and [tex]e_y[/tex].

The p is easy, since it does not depend on r.

marlon
 
  • #5
Besides, shouldn't the [tex]\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})[/tex]

be

[tex]\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r} (\mathbf{p}\cdot\hat{\mathbf{r}})[/tex]


so no r^2 but r ?

Otherwise, your given potential will not yield the solution that you have been given, after calculating the gradient.

regards
marlon
 
  • #6
marlon said:
[tex]\nabla \frac {1} {r^2} = \frac {-2 \vect(r)}{r^4}[/tex] ; the r in the numerator is a vector !

Could the above be written as

[tex]
\nabla \frac {1} {r^2} = \frac{\partial \frac {1} {r^2}}{\partial r} \hat{\mathbf{r}} + \frac{1}{r} \frac{\partial \frac {1} {r^2}}{\partial \theta} \hat{\mathbf{\theta}} + \frac{1}{r sin(\theta)} \frac{\partial \frac {1} {r^2}}{\partial \psi} \hat{\mathbf{\psi}}
[/tex]

??
 
  • #7
marlon said:
Besides, shouldn't the [tex]\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})[/tex]

be

[tex]\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r} (\mathbf{p}\cdot\hat{\mathbf{r}})[/tex]


so no r^2 but r ?

Otherwise, your given potential will not yield the solution that you have been given, after calculating the gradient.

regards
marlon

I think it is right because/if [tex]\hat{\mathbf{r}} = \mathbf{r}/r[/tex] .
 
  • #8
berkeman said:
Do you know the equation for the grad operator [tex]\nabla[/tex] in spherical coordinates?
I do if it's

[tex]
\nabla = \frac{\partial }{\partial r} + \frac{1}{r} \frac{\partial }{\partial \theta} + \frac{1}{r sin(\theta)} \frac{\partial}{\partial \psi}
[/tex]
 
  • #9
Why is [tex]\mathbf{p}[/tex] not treated as a constant?
 
  • #10
dimensionless said:
I do if it's

[tex]
\nabla = \frac{\partial }{\partial r} + \frac{1}{r} \frac{\partial }{\partial \theta} + \frac{1}{r sin(\theta)} \frac{\partial}{\partial \psi}
[/tex]
Correct. When you do the differentiations, you will end up with r-hat and theta-hat components, where the polarization p is a scalar constant inside each term. I'm not used to seeing the form of the answer that you list for the E field of a dipole. Do you have another textbook that might show you a different form, with the r-hat and theta-hat terms?
 
  • #11
dimensionless said:
I do if it's

[tex]
\nabla = \frac{\partial }{\partial r} + \frac{1}{r} \frac{\partial }{\partial \theta} + \frac{1}{r sin(\theta)} \frac{\partial}{\partial \psi}
[/tex]
Oh, and you omitted the r-hat, theta-hat, and phi-hat unit vectors from the three terms...
 
  • #12
dimensionless said:
Why is [tex]\mathbf{p}[/tex] not treated as a constant?
Because it is not in general a constant! [itex]\mathbf{p}[/itex] is, as you said initially, "the (vector) dipole moment" which may vary with position.
 
  • #13
dimensionless said:
I think it is right because/if [tex]\hat{\mathbf{r}} = \mathbf{r}/r[/tex] .
What do you mean by this ?

Ok, i am going to assume that r hat is just the vector r, ok ? If not, please define what it is interms of vector r and the scalar r. The definition that you give is one that i don't get ? What does it mean ?

According to me, the r squared in the denominator should be r. Just to be clear, the r is the magnitude of the vector r , right ? At least, that is what it should be.

I get this :
[tex]\vec {\nabla} \frac {1} { r} ( \vec {p} \cdot \vec {r}) = \vec {\nabla} ( \frac {1} { r}) ( \vec {p} \cdot \vec {r}) + \frac {1} { r} \vec {\nabla} ( \vec {p} \cdot \vec {r}) [/tex]


This first part of the sum is :
[tex] \vec{\nabla} \frac{1}{r} = \frac{- \vec{r}}{r^3}[/tex]

The second part :
[tex] \frac {1} { r} \vec {\nabla} ({\vec {p}} \cdot \vec {r}) = \frac {1} { r} {\vec {p}} \cdot \vec {\nabla} ( \vec {r}) = 3 \frac {1} { r} {\vec {p}} [/tex]

The 3 comes from the gradient of the r vector. To check this, just write down the expression for both nabla and r in x y and z coordinates.

Now fill all of this into the sum and multiply by -1

To get :
[tex]\vec {E} = \frac {1} {4\pi\epsilon_0 r^3} \left((\vec {p} \cdot \vec {r}) \vec {r}- 3 {\vec {p} r^2}\right)[/tex]


marlon
 
Last edited:
  • #14
Use a vector identity and you can indeed treat vector P as a constant since it is a vector of constant length.
 
Last edited:
  • #15
dimensionless said:
I've got an equation (from Wikipedia)for the electrostatic potential of an electric dipole. It looks like this:

[tex]
\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})
[/tex]

E is the electric field
r, r, r\hat are as above
p is the (vector) dipole moment
e0 is the primitivity of free space

To find the electric field I have to take the derivative as follows.

[tex]
\mathbf{E} = - \nabla \Phi
[/tex]

The derivative looks like this:

[tex]
\mathbf{E} = \frac {1} {4\pi\epsilon_0 r^3} \left(3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}\right)
[/tex]

I'm confused by the vector notation. Why do I have r\hat multiplied by r\hat? Why is the electric field simply not

[tex]
\Phi (\mathbf{r}) = \frac {1} {2\pi\epsilon_0 r^3} (\mathbf{p}\cdot\hat{\mathbf{r}})
[/tex]

Sounds like some replies have been given partial help and some other have made things more confusing. Let me add the following:

First, to answer your question, the key point is that, the derivative of the unit vector r hat is not zero! (but you may treat the dipole vector p as a constant vector here). That is the reason why you get two terms.

The best way to do the calculation is not to use r hat but the vector r. Then you have to take the gradient of


[tex]
\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^3} (\mathbf{p}\cdot{\mathbf{r}})
[/tex]

Using the spherical coordinate representation of the gradient, you should convince yourself easily that [itex] {\vec{ \nabla}} {1 \over r^3 } = -3 {{\hat r} \over r^4} [/itex].

Then, you also have to take the gradient of the dot product [itex] {{\vec p}} \cdot {\vec{r}}[/itex]. Here, the best way is to go to Cartesian coordinates in which case you simply have to compute

[tex] ({\vec i} {\partial \over \partial x} +{\vec j} {\partial \over \partial y}+{\vec k} {\partial \over \partial z}) ( p_x x + p_y y + p_z z) [/tex]

(where I have used that [itex] {\vec {r} } = x {\vec i} + y {\vec j} + z {\vec k} [/itex] in Cartesian coordinates). You should easily convince yourself that the result is simply the vector [itex] {\vec {p}}[/itex]!

Putting all the pieces together gives the desired result.

Patrick
 
  • #16
marlon said:
What do you mean by this ?

Ok, i am going to assume that r hat is just the vector r, ok ? If not, please define what it is interms of vector r and the scalar r. The definition that you give is one that i don't get ? What does it mean ?
But he/she *did* define it in terms of the vector r and scalar r! He/She wrote explicitly [itex] {\hat{ r}} = { {\vec r} \over r}[/itex]. There is no other way to define it in terms of the vector r and scalar r! It means that r hat is the vector r divided by its magnitude. It's a very common notation!

The second part :
[tex] \frac {1} { r} \vec {\nabla} ({\vec {p}} \cdot \vec {r}) = \frac {1} { r} {\vec {p}} \cdot \vec {\nabla} ( \vec {r}) = 3 \frac {1} { r} {\vec {p}} [/tex]
The first step is wrong. One cannot move in the divergence to the right like this. My previous post shows that the result is the vector p.

Regards

Patrick

EDIT: One way to see that it is wrong is that the vector p is dotted into something which is what? a scalar? That is clearly inconsistent.
 
Last edited:
  • #17
nrqed said:
like this. My previous post shows that the result is the vector p.

Regards

Patrick

EDIT: One way to see that it is wrong is that the vector p is dotted into something which is what? a scalar? That is clearly inconsistent.

Indeed, i made a mistake there. Thanks for the correction

regards

marlon
 
  • #18
berkeman said:
Do you have another textbook that might show you a different form, with the r-hat and theta-hat terms?

[tex]
\mathbf{E} = \frac {p} {4\pi\epsilon_0 r^3}( 2\mathbf{cos}\theta\mathbf{a}_{r} +\mathbf{sin}\theta\mathbf{a}_{\theta} )
[/tex]
 
Last edited:

1. What is a derivative in vector notation?

A derivative in vector notation is a mathematical operation that describes the rate of change of a vector quantity with respect to a given variable. It is represented by the symbol ∏ or d/dx.

2. How is a derivative in vector notation different from a traditional derivative?

A derivative in vector notation takes into account not only the magnitude of the change, but also the direction. This is represented by a vector instead of a scalar value, as in traditional derivatives.

3. What are some common applications of derivatives in vector notation?

Derivatives in vector notation are often used in physics, engineering, and other fields to describe the motion and changes of objects in three-dimensional space. They are also used in vector calculus to solve complex mathematical problems.

4. How is a derivative in vector notation calculated?

To calculate a derivative in vector notation, you can use the same rules as traditional derivatives, but with vectors instead of scalar values. This includes the product rule, quotient rule, and chain rule.

5. Are there any limitations to using derivatives in vector notation?

One limitation of using derivatives in vector notation is that they cannot be applied to non-continuous functions, as the concept of direction does not apply. Additionally, some problems may be more easily solved using traditional derivatives instead of vector derivatives.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
945
Replies
33
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
838
  • Calculus and Beyond Homework Help
Replies
9
Views
699
Replies
1
Views
243
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
269
  • Special and General Relativity
Replies
5
Views
269
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
6
Views
409
Back
Top