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Derivatives in Vector Notation

  • #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]
 

Answers and Replies

  • #2
berkeman
Mentor
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Do you know the equation for the grad operator [tex]\nabla[/tex] in spherical coordinates?
 
  • #3
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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
3,763
8
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
3,763
8
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
berkeman
Mentor
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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
berkeman
Mentor
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7,463
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
HallsofIvy
Science Advisor
Homework Helper
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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
3,763
8
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 gonna 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
nrqed
Science Advisor
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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
nrqed
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marlon said:
What do you mean by this ?

Ok, i am gonna 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
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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]
 
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