Derivatives in Vector Notation

[dimensionless]

I've got an equation (from Wikipedia)for the electrostatic potential of an electric dipole. It looks like this:

$$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

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.

$$\mathbf{E} = - \nabla \Phi$$

The derivative looks like this:

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

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

$$\Phi (\mathbf{r}) = \frac {1} {2\pi\epsilon_0 r^3} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

 

[berkeman]

Do you know the equation for the grad operator $$\nabla$$ in spherical coordinates?

 

[neutrino]

The derivative looks like this:

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

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

$$\Phi (\mathbf{r}) = \frac {1} {2\pi\epsilon_0 r^3} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

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

 

[marlon]

This is a classic one, especially because you need to calculate the gradient of a scalar:

First apply :
$$\nabla \frac {1} { r^2} ({p}\cdot\hat{r}) = \nabla ( \frac {1} { r^2}) ({p}\cdot\hat{r}) + \frac {1} { r^2} \nabla ({p}\cdot\hat{r})$$

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

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

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

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

marlon

 

[marlon]

Besides, shouldn't the $$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

be

$$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r} (\mathbf{p}\cdot\hat{\mathbf{r}})$$


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

 

[dimensionless]

$$\nabla \frac {1} {r^2} = \frac {-2 \vect(r)}{r^4}$$ ; the r in the numerator is a vector !

Could the above be written as

$$\nabla \frac {1} {r^2} = \frac{\partial \frac {1} {r^2}}{\partial r} \hat{\mathbf{r}} + \frac{1}{r} \frac{\partial \frac {1} {r^2}}{\partial \theta} \hat{\mathbf{\theta}} + \frac{1}{r sin(\theta)} \frac{\partial \frac {1} {r^2}}{\partial \psi} \hat{\mathbf{\psi}}$$

??

 

[dimensionless]

Besides, shouldn't the $$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

be

$$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r} (\mathbf{p}\cdot\hat{\mathbf{r}})$$


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 $$\hat{\mathbf{r}} = \mathbf{r}/r$$ .

 

[dimensionless]

Do you know the equation for the grad operator $$\nabla$$ in spherical coordinates?

I do if it's

$$\nabla = \frac{\partial }{\partial r} + \frac{1}{r} \frac{\partial }{\partial \theta} + \frac{1}{r sin(\theta)} \frac{\partial}{\partial \psi}$$

 

[dimensionless]

Why is $$\mathbf{p}$$ not treated as a constant?

 

[berkeman]

I do if it's

$$\nabla = \frac{\partial }{\partial r} + \frac{1}{r} \frac{\partial }{\partial \theta} + \frac{1}{r sin(\theta)} \frac{\partial}{\partial \psi}$$

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?

 

[berkeman]

I do if it's

$$\nabla = \frac{\partial }{\partial r} + \frac{1}{r} \frac{\partial }{\partial \theta} + \frac{1}{r sin(\theta)} \frac{\partial}{\partial \psi}$$

Oh, and you omitted the r-hat, theta-hat, and phi-hat unit vectors from the three terms...

 

[HallsofIvy]

Why is $$\mathbf{p}$$ not treated as a constant?

Because it is not in general a constant! $\mathbf{p}$ is, as you said initially, "the (vector) dipole moment" which may vary with position.

 

[marlon]

I think it is right because/if $$\hat{\mathbf{r}} = \mathbf{r}/r$$ .

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 :
$$\vec {\nabla} \frac {1} { r} ( \vec {p} \cdot \vec {r}) = \vec {\nabla} ( \frac {1} { r}) ( \vec {p} \cdot \vec {r}) + \frac {1} { r} \vec {\nabla} ( \vec {p} \cdot \vec {r})$$


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

The second part :
$$\frac {1} { r} \vec {\nabla} ({\vec {p}} \cdot \vec {r}) = \frac {1} { r} {\vec {p}} \cdot \vec {\nabla} ( \vec {r}) = 3 \frac {1} { r} {\vec {p}}$$

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 :
$$\vec {E} = \frac {1} {4\pi\epsilon_0 r^3} \left((\vec {p} \cdot \vec {r}) \vec {r}- 3 {\vec {p} r^2}\right)$$


marlon

 

[Random Variable]

Use a vector identity and you can indeed treat vector P as a constant since it is a vector of constant length.

 

[nrqed]

I've got an equation (from Wikipedia)for the electrostatic potential of an electric dipole. It looks like this:

$$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^2} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

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.

$$\mathbf{E} = - \nabla \Phi$$

The derivative looks like this:

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

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

$$\Phi (\mathbf{r}) = \frac {1} {2\pi\epsilon_0 r^3} (\mathbf{p}\cdot\hat{\mathbf{r}})$$

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


$$\Phi (\mathbf{r}) = \frac {1} {4\pi\epsilon_0 r^3} (\mathbf{p}\cdot{\mathbf{r}})$$

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

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

$$({\vec i} {\partial \over \partial x} +{\vec j} {\partial \over \partial y}+{\vec k} {\partial \over \partial z}) ( p_x x + p_y y + p_z z)$$

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

Putting all the pieces together gives the desired result.

Patrick

 

[nrqed]

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 ${\hat{ r}} = { {\vec r} \over r}$. 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 :
$$\frac {1} { r} \vec {\nabla} ({\vec {p}} \cdot \vec {r}) = \frac {1} { r} {\vec {p}} \cdot \vec {\nabla} ( \vec {r}) = 3 \frac {1} { r} {\vec {p}}$$

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.

 

[marlon]

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

 

[dimensionless]

Do you have another textbook that might show you a different form, with the r-hat and theta-hat terms?

$$\mathbf{E} = \frac {p} {4\pi\epsilon_0 r^3}( 2\mathbf{cos}\theta\mathbf{a}_{r} +\mathbf{sin}\theta\mathbf{a}_{\theta} )$$