What is the force on an electric dipole due to another electric dipole?

[aftershock]

Homework Statement

Two electric dipoles, oriented as shown in the figure, are separated by a distance r.

What is the force on p₁ due to p₂

Homework Equations

F = (p dot ∇)E

The Attempt at a Solution

I'm confused on the p dot ∇ part.

How is the divergence of p not just zero?

It seems p1 would just be some constant in the x hat direction.

 

[diazona]

Think about that carefully. $\mathbf{p}\cdot\nabla$ is not the divergence of $\mathbf{p}$!

 

[aftershock]

Think about that carefully. $\mathbf{p}\cdot\nabla$ is not the divergence of $\mathbf{p}$!

I'm not sure I understand that. ∇⋅p would be divergence? And dot products are commutative.

 

[diazona]

This isn't really a dot product, though, it's a dot "application." That is, it still means that $\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$, but you can't assume that e.g. $a_x b_x$ means "$a_x$ multiplied by $b_x$," because you are dealing with things that don't get multiplied. Think about it: does $x \frac{\partial}{\partial x}$ mean $x$ multiplied by $\frac{\partial}{\partial x}$? Does $\frac{\partial}{\partial x} x$ mean $\frac{\partial}{\partial x}$ multiplied by $x$?

 

[aftershock]

This isn't really a dot product, though, it's a dot "application." That is, it still means that $\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$, but you can't assume that e.g. $a_x b_x$ means "$a_x$ multiplied by $b_x$," because you are dealing with things that don't get multiplied. Think about it: does $x \frac{\partial}{\partial x}$ mean $x$ multiplied by $\frac{\partial}{\partial x}$? Does $\frac{\partial}{\partial x} x$ mean $\frac{\partial}{\partial x}$ multiplied by $x$?

Oh ok so if the vector is in front of del I'm basically taking the components and multiplying them by differential operators, where if the vector was to the right of del i would be taking derivatives of the components?

 

[diazona]

Yep, that's the idea.

A good way to think about it is that the components of $\mathbf{p}$ are multiplicative operators. Just like the differential operator $\frac{\partial}{\partial x}$ takes a function $f(x)$ and turns it into $f'(x)$, a multiplicative operator $x$ takes a function $f(x)$ and turns it into $xf(x)$. The composition of two operators is also an operator, so $p_x \frac{\partial}{\partial x}$ is an operator that means "take the derivative and then multiply by a number."

 

[aftershock]

Yep, that's the idea.

A good way to think about it is that the components of $\mathbf{p}$ are multiplicative operators. Just like the differential operator $\frac{\partial}{\partial x}$ takes a function $f(x)$ and turns it into $f'(x)$, a multiplicative operator $x$ takes a function $f(x)$ and turns it into $xf(x)$. The composition of two operators is also an operator, so $p_x \frac{\partial}{\partial x}$ is an operator that means "take the derivative and then multiply by a number."

Thanks, really helped me.