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.