Understanding the Radial Attraction Force between Two Electric Dipoles

[LCSphysicist]

Homework Statement

.

Relevant Equations

.

I am extremelly confused with this question: Basically suppose we have two electric dipoles, parallel to each other, as follows in the figure:

I am supposed to show that the force is attractive and radial. But I am not getting how it can be radial (pointing toward each other)!

See, the force acting on a dipole is $$\vec F = (\vec p \cdot \vec \nabla ) \vec E$$

So the force is parallel to the electric field. Let's consider the force acting on $P1$ due to $P2$

But, here, $$\vec E = k\frac{3(\vec p . \vec r) \vec r - \vec p}{r^5} = \frac{-p_2}{4 \pi \epsilon r^3} \hat z$$

And, since $\hat z$ is constant, $(\vec p \cdot \vec \nabla)$ will not alter its direction, so the force is also in z direction !?

I am almost sure i am misinterpretating the electric field equation and i have not noticed that, since this formula was right in all the ways i have used it

 

[kuruman]

I think it is simpler to find an expression for the potential energy $U=-\vec p_2\cdot \vec E_1$ and then find the components of the gradient. Of course, $\vec E_1$ is the electric field due to dipole 1 at the location of dipole 2.

 

[LCSphysicist]

I think it is simpler to find an expression for the potential energy $U=-\vec p_2\cdot \vec E_1$ and then find the components of the gradient. Of course, $\vec E_1$ is the electric field due to dipole 1 at the location of dipole 2.

Indeed, it is the way it is generally solved. But, i am extremelly uncomfortable with my answer getting wrong using the well known electric field equation for dipoles and the force on a dipole.

What i mean is, until yesterday i would apply the equation everywhere i could, but now all of suden applying it here gives me the wrong answer. I would like to know why. More preciselly, i would really like to know that i have made some mistake, and the problem is not that we can't apply the formula here.

 

[kuruman]

I have to leave now so I cannot reply. If nobody has replied by the time I return, I will edit this post with my reply.

On edit: Several people replied by the time I got back so I posted my reply below, #14.

 

[Steve4Physics]

View attachment 300646
I am supposed to show that the force is attractive and radial.

Could there be a mistake in the question?

Consider each dipole to be a pair of separated charges, -q and +q. The direction of the dipole moment is from -q to +q.

The diagram in Post #1 is then effectively: +q₁ +q₂ -q₁ -q₂
The distance between +q₁ and +q₂ is smaller than the distance between +q₁ and -q₂. The net force on +q₁ due to dipole-2 is therefore away from dipole-2.

A similar argument applies to the other charges. This means the net force between the two dipoles (in the position shown shown in the diagram) is repulsive.

 

[Orodruin]

Homework Statement:: .
Relevant Equations:: .

See, the force acting on a dipole is F→=(p→⋅∇→)E→

So the force is parallel to the electric field.

This is not what the equation says. It says the force is the directional derivative of the field. There is no guarantee for the directional derivative to be parallel to the field itself.

 

[LCSphysicist]

Could there be a mistake in the question?

Consider each dipole to be a pair of separated charges, -q and +q. The direction of the dipole moment is from -q to +q.

The diagram in Post #1 is then effectively: +q₁ +q₂ -q₁ -q₂
The distance between +q₁ and +q₂ is smaller than the distance between +q₁ and -q₂. The net force on +q₁ due to dipole-2 is therefore away from dipole-2.

A similar argument applies to the other charges. This means the net force between the two dipoles (in the position shown shown in the diagram) is repulsive.

Oh Jesus. Sorry. Indeed, it is repulsive, not attractive, i wrote it wrong. BUT since the main problem here is the direction (radial), the question remains.

This is not what the equation says. It says the force is the directional derivative of the field. There is no guarantee for the directional derivative to be parallel to the field itself.

Oh well, i think the problem is that i have assumed the components are zero and then i derivated it >:( i should not put Ey=Ex=0 before deriving.

 

[Orodruin]

Oh well, i think the problem is that i have assumed the components are zero and then i derivated it >:( i should not put Ey=Ex=0 before derivative.

That would indeed lead to issues since the derivative is taken in the z-direction and those components are zero only in the xy plane.

Anyway, your problems book of relativity is cool.

Thank you

 

[PhDeezNutz]

I think you misunderstand what $\left(\vec{p} \cdot \nabla \right)$ means.

in your case it would be $p_2 \frac{\partial }{\partial z}$ applied to each component of $\vec{E_{p1}}$

 

[PhDeezNutz]

Homework Statement:: .
Relevant Equations:: .

I am extremelly confused with this question: Basically suppose we have two electric dipoles, parallel to each other, as follows in the figure:
View attachment 300646
I am supposed to show that the force is attractive and radial. But I am not getting how it can be radial (pointing toward each other)!

See, the force acting on a dipole is $$\vec F = (\vec p \cdot \vec \nabla ) \vec E$$

So the force is parallel to the electric field. Let's consider the force acting on $P1$ due to $P2$

But, here, $$\vec E = k\frac{3(\vec p . \vec r) \vec r - \vec p}{r^5} = \frac{-p_2}{4 \pi \epsilon r^3} \hat z$$

And, since $\hat z$ is constant, $(\vec p \cdot \vec \nabla)$ will not alter its direction, so the force is also in z direction !?

I am almost sure i am misinterpretating the electric field equation and i have not noticed that, since this formula was right in all the ways i have used it

Whoa Whoa Whoa

$$\vec E = k\frac{3(\vec p . \vec r) \vec r - \vec p}{r^5} $$

This is unfortunate notation but there's not much we can do to get around it

$r$ in this equation is not the separation between the dipoles but literally the position vector from the origin.

The author or professor should have labeled the separation $\ell$ or something like that.

 

[Orodruin]

Whoa Whoa Whoa

$$\vec E = k\frac{3(\vec p . \vec r) \vec r - \vec p}{r^5} $$

This is unfortunate notation but there's not much we can do to get around it

$r$ in this equation is not the separation between the dipoles but literally the position vector from the origin.

The author or professor should have labeled the separation $\ell$ or something like that.

Considering one of the dipoles is located at the origin, the separation vector between the dipoles is the position vector of the second.

 

[PhDeezNutz]

Considering one of the dipoles is located at the origin, the separation vector between the dipoles is the position vector of the second.

you're right I jumped the gun. Sorry.

Edit: but at the same time if we're employing the $\vec{F} = \left( \vec{p} \cdot \nabla \right) \vec{E}$ approach shouldn't we use an expression for $\vec{E}$ in general before taking derivatives?

 

[Orodruin]

you're right I jumped the gun. Sorry.

Edit: but at the same time if we're employing the $\vec{F} = \left( \vec{p} \cdot \nabla \right) \vec{E}$ approach shouldn't we use an expression for $\vec{E}$ in general before taking derivatives?

As long as you insert the correct $\vec E$ (in this case the dipole field from p1) you will be fine. The problem here is simplifying the expression to the point that it is no longer valid when you move out of the xy plane (which is necessary if one wants to take derivatives in the z-direction).

 

[kuruman]

This is unfortunate notation but there's not much we can do to get around it

It's not only unfortunate, it's dimensionally incorrect and must be fixed. The second term in the numerator should be $r^2\vec p$. Here's what I would do with the corrected equation.

Write the problem in cylindrical coordinates to simplify it to two-dimensions. The electric field from source dipole 1 at the origin is (note the unit vectors in the numerator) $$\vec E = k\frac{3(\vec p \cdot \vec {r}) \vec r }{r^5}-\frac{\vec p}{r^3}.$$ WIth $\vec r=\rho~\hat {\rho}+z~\hat z$ and $\vec p=p~\hat z$ this becomes $$\vec E = k\frac{3pz(\rho \hat {\rho}+z \hat z) }{(\rho^2+z^2)^{5/2}}-\frac{p\hat z}{(\rho^2+z^2)^{3/2}}\implies
\begin{cases}
E_{\rho}=kp\frac{3\rho z}{(\rho^2+z^2)^{5/2}} \\
E_{z}=kp\left(\frac{3z^2}{(\rho^2+z^2)^{5/2}}-\frac{1}{(\rho^2+z^2)^{3/2}}\right)
\end{cases}$$Now for the components of the force we have, $$\begin{align} & F_{\rho}=(\vec p \cdot \vec{\nabla})E_{\rho}=p\frac{\partial}{\partial z}\left[kp\frac{3\rho z}{(\rho^2+z^2)^{5/2}}\right]\nonumber \\ & F_{z}=(\vec p \cdot \vec{\nabla})E_{z}=p\frac{\partial}{\partial z}\left[kp\left(\frac{3z^2}{(\rho^2+z^2)^{5/2}}-\frac{1}{(\rho^2+z^2)^{3/2}}\right) \right].\nonumber\end{align}$$ If one takes the derivatives and sets $z=0$, the axial component vanishes but the radial component does not.

 

[Orodruin]

Now for the components of the force we have, $$\begin{align} & F_{\rho}=(\vec p \cdot \vec{\nabla})E_{\rho}=p\frac{\partial}{\partial z}\left[kp\frac{3\rho z}{(\rho^2+z^2)^{5/2}}\right]\nonumber \\ & F_{z}=(\vec p \cdot \vec{\nabla})E_{z}=p\frac{\partial}{\partial z}\left[kp\left(\frac{3z^2}{(\rho^2+z^2)^{5/2}}-\frac{1}{(\rho^2+z^2)^{3/2}}\right) \right]\nonumber\end{align}.$$ If one takes the derivatives and sets $z=0$, the axial component vanishes but the radial component does not.

Good catch, but a caveat:
The above (taking the directional derivatives of the components) only works in this case because the chosen vector basis is constant in the z-direction. Had the dipole instead been oriented in the $\phi$ direction, you would need to use the covariant derivative here as $\hat\rho$ depends on $\phi$.

 

[kuruman]

Good catch, but a caveat:
The above (taking the directional derivatives of the components) only works in this case because the chosen vector basis is constant in the z-direction. Had the dipole instead been oriented in the $\phi$ direction, you would need to use the covariant derivative here as $\hat\rho$ depends on $\phi$.

That is why I prefer to find an expression for the scalar potential and take the derivatives of that.

 

[Orodruin]

That is why I prefer to find an expression for the scalar potential and take the derivatives of that.

Indeed. I think it is also worth pointing out why the two expressions are the same because if one just considers them at face value, they do not match, i.e., if $\vec E$ would be a general vector field and $\vec p$ a constant vector then $(\vec p \cdot \nabla) \vec E \neq \nabla (\vec p \cdot \vec E)$. However, $\vec E$ is not just any field, but as the electric field is governed by Maxwell's equations. In particular, $\nabla \times \vec E = 0$ for a static field and so
$$
0 = \vec p \times (\nabla \times \vec E) = \nabla(\vec p \cdot \vec E)-(\vec p \cdot \nabla) \vec E
$$
and so $\nabla(\vec p \cdot \vec E) = (\vec p \cdot \nabla) \vec E$.

Edit: Alternatively, this can be argued in a different way as $\nabla \times\vec E$ implies that $\vec E$ has a potential $\phi$ such that $\vec E = - \nabla\phi$. This implies that
$$
\nabla(\vec p \cdot \vec E) = -\nabla(\vec p \cdot \nabla \phi) = -(\vec p \cdot \nabla)\nabla\phi = (\vec p\cdot\nabla)\vec E.
$$
Of course, the origin of both arguments is the same, the curl of $\vec E$ being equal to zero.

 

[kuruman]

One can even be more general than that and go all the way back to the vector calculus identity $$\mathbf{\nabla}(\mathbf{F}\cdot\mathbf{G})=(\mathbf{F}\cdot \mathbf{\nabla} )\mathbf{G}+(\mathbf{G}\cdot \mathbf{\nabla} )\mathbf{F}+\mathbf{F}\times( \mathbf{\nabla}\times\mathbf{G})+\mathbf{G}\times( \mathbf{\nabla}\times\mathbf{F}) $$ then set one of the vectors equal to the constant dipole moment and the other equal to the conservative electric field and see what's left standing.

 

[robphy]

A follow-up comment to the one made by @Orodruin in #6

See, the force acting on a dipole is $$\vec F = (\vec p \cdot \vec \nabla ) \vec E$$

So the force is parallel to the electric field.

This is an example where the force $\vec F$ is linearly-dependent on $\vec E$
(e.g., you can double $\vec E$ or superpose two sources (say, $\vec E_1$ and $\vec E_2$)),
but the force $\vec F$ is not in general parallel to $\vec E$.

Note that when the gradient of $\vec E$ is zero (e.g. a nonzero uniform electric field),
the dipole feels zero force.

 

[PhDeezNutz]

OP I apologize for my half brained suggestions earlier. Allow me to redeem myself.

Apparently $\vec{F}_2 = \left(\vec{p}_2 \cdot \nabla \right) \vec{E}_1$ in spherical basis is

Where $\vec{A} = \vec{p}_2$ and $\vec{B} = \vec{E}_1$

Now to express $\vec{p}_2 = p_2 \hat{z}$ and $\vec{E}_1$ in spherical basis

$\vec{p}_2 = p_2 \cos \theta \hat{r} - p_2 \sin \theta \hat{\theta}$

A common expression found in say Griffiths is the spherical representation of the field for a dipole located at the origin and oriented in the positive $z$ direction$\vec{E}_1 = \frac{p_1}{4 \pi \epsilon_0} \left( \frac{2 \cos \theta}{r^3} \hat{r} + \frac{2 \sin \theta}{r^3} \hat{\theta} \right)$

When you carry this out you get the general expression

$\vec{F}_2 = \frac{p_1}{4 \pi \epsilon_0}\left[\left( - \frac{6 p_2 \cos^2 \theta}{r^4} + \frac{2 p_2 \sin^2 \theta}{r^4} + \frac{p_2 \sin^2 \theta}{r^4} \right) \hat{r} + \left( -\frac{3 p_2 \cos^2 \theta}{r^4} - \frac{3 p_2 \cos^2 \theta}{r^4} - \frac{2 p_2 \sin \theta \cos \theta }{r^4} \right) \hat {\theta}\right]$

Plugging in $\theta = \frac{\pi}{2}$

$\vec{F}_2 = \frac{3 p_1 p_2}{4 \pi \epsilon_0} \frac{1}{r^4} \hat{r}$

Hopefully I did that right. I guess the expressions for the "material derivative" in basis other than cartesian are not so simple.

Edit: messed up on a minus sign somewhere, don't know where though.