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so, here are two dipoles distance $r$ apart i have calculated following:

(i)$\vec {F_{p_{1}}} $ (force on $p_{1}$ due to $p_{2}$)$=0$

(ii)$\vec {F_{p_{2}}}$(force on $p_{2}$ due to $p_{1}= -\dfrac{3p_{1}p_{2}}{4\pi \epsilon_{0}r^4}\neq 0$

Why two forces are unequal? Is it a violation of newton's 3rd law?

I also found torque on one due to another it is also coming unequal. I want to know, why it is the case for above dipoles configuration?


part (i)

The electric field at point $p_{1}$ due to $p_{2}$ will be

$\mathbf {\vec E}=-\dfrac{p_{2}}{2\pi \epsilon_{0} r^3 }\mathbf{\hat r} $

so, force on $p_{1}$ will be $F_{p_{1}}=(\vec p_{1} \cdot \nabla)\vec E\implies -p_{1}\dfrac{d \vec E }{r \ d\theta}=0$

$\mathbf {\hat {r}}$ is unit vector along left to right


$\mathbf {\hat{\theta}}$ is unit vector from up to down

for (ii)part

electric field at point $p_{2}$ due to $p_{1}$ will be

$\mathbf {\vec E'}=\dfrac{p_{1}}{4\pi \epsilon_{0} r^3 }\mathbf{\hat \theta} $

so, force on $p_{2}$ will be

$F_{p_{2}}=(\vec p_{2} \cdot \nabla)\vec E'\implies p_{2}\dfrac{d \vec E'}{dr}=-\dfrac{3p_{1}p_{2}}{4\pi \epsilon_{0} r^4}\mathbf{\hat \theta} $

What is the physical explanation of phenomena?

  • 3
    $\begingroup$ It looks like something is wrong with the forces you calculate. Also, just because the forces are equal and opposite does not mean the torques will be. This is because the vector connecting the two dipoles is parallel to one dipole and perpendicular to the other. $\endgroup$ – Aaron Stevens Aug 31 '18 at 20:06
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    $\begingroup$ why do you have $d\vec E/d\theta=0$? $\endgroup$ – JEB Aug 31 '18 at 20:43
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    $\begingroup$ @veereshpandey It might be easier to do this problem in Cartesian coordinates. This is because performing the dot product $\vec p \cdot \nabla$ is not as straightforward in spherical coordinates. Also, this is not a "check my work" site. $\endgroup$ – Aaron Stevens Aug 31 '18 at 20:45
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    $\begingroup$ @veereshpandey But it is a function of $\theta$. $\endgroup$ – JEB Aug 31 '18 at 20:46
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    $\begingroup$ @veereshpandey That statement is only true in Cartesian coordinates since the unit vectors do not depend on position. In spherical coordinates the unit vectors do depend on position, so you need to be careful when doing things like integrals and derivatives with unit vectors in curvilinear coordinates. $\endgroup$ – Aaron Stevens Aug 31 '18 at 20:49

The comments by others have provided the answers to both of the key issues in the question. Aaron pointed out the conceptual error regarding the torques: they do not necessarily cancel, in general. JEB, with an additional comment from Aaron, pointed out the mathematical error regarding the way you are calculating the derivative of the electric field. My answer here has a broader aim.

Since you mentioned Newton's third law, and asked for some physical explanation, I'm providing an alternative derivation which may enable you to check your own work. It is also, in my opinion, practically useful, closer to the underlying physical principles, and less error-prone.

Here's the essential physics. Assume that the forces $\vec{F}_1$, $\vec{F}_2$, and torques $\vec{T}_1$, $\vec{T}_2$, are derived from a potential energy function $V$ of the positions and orientations of the two objects. If $V$ is invariant to translations of the origin of coordinates, then $$ \vec{F}_1 + \vec{F}_2 = \vec{0}. $$ If $V$ is invariant to rotations about the origin of coordinates, then $$ \vec{T}_1 + \vec{T}_2 + \vec{r}_1\times\vec{F}_1 + \vec{r}_2\times\vec{F}_2 = \vec{T}_1 + \vec{T}_2 + (\vec{r}_1-\vec{r}_2)\times\vec{F}_1 = \vec{0}. $$ Both of these symmetries, of course, are related to conservation laws, for linear and angular momentum respectively, through Noether's theorem applied to the Lagrangian. Given the above assumption, the expressions for forces and torques must satisfy these equations, otherwise they are wrong; Newton's third law doesn't come into question. I won't give a formal proof of the above equations, but the derivation below contains the essentials.

For linear objects, whose orientations are specified by unit vectors $\hat{e}_1$ and $\hat{e}_2$, $V$ can very often be expressed as a function of four quantities, all of which are translationally and rotationally invariant: $$r=|\vec{r}_1-\vec{r}_2|,\quad c=\hat{e}_1\cdot\hat{e}_2, \quad C_1=\hat{e}_1\cdot(\vec{r}_1-\vec{r}_2), \quad C_2=\hat{e}_2\cdot(\vec{r}_1-\vec{r}_2). $$ For the case of dipoles $p_1\hat{e}_1$, $p_2\hat{e}_2$: $$ V = \frac{A}{r^3} \left[ c - 3 \frac{C_1C_2}{r^2} \right] \quad\text{where}\quad A = \frac{p_1p_2}{4\pi\epsilon_0} $$ Deriving the forces and torques is quite straightforward: it is only necessary to differentiate $V$ with respect to these four quantities, and use the chain rule. We only require the gradients (with respect to position and orientation) of these four functions. Here they are: $$ \begin{array}{c|cccc} \varphi & \vec{\nabla}_{r_1}\varphi & \vec{\nabla}_{r_2}\varphi & \vec{\nabla}_{e_1}\varphi & \vec{\nabla}_{e_2}\varphi \\ \hline r & (\vec{r}_1-\vec{r}_2)/r & -(\vec{r}_1-\vec{r}_2)/r & 0 & 0 \\ c & 0 & 0 & \hat{e}_2 & \hat{e}_1 \\ C_1 & \hat{e}_1 & -\hat{e}_1 & \vec{r}_1-\vec{r}_2 & 0 \\ C_2 & \hat{e}_2 & -\hat{e}_2 & 0 & \vec{r}_1-\vec{r}_2 \end{array} $$ The forces and torques may then be expressed $$ \vec{F}_1 = -\vec{\nabla}_{r_1} V, \quad \vec{F}_2 = -\vec{\nabla}_{r_2} V, \quad \vec{T}_1 = -\hat{e}_1\times\vec{\nabla}_{e_1} V, \quad \vec{T}_2 = -\hat{e}_2\times\vec{\nabla}_{e_2} V $$ The last two equations follow from considering the derivative $-\partial V/\partial\psi$ resulting from rotating the object by an angle $\psi$ about an arbitrary unit axis vector $\hat{n}$, which by definition gives the $\hat{n}$ component of the torque. In considering $\vec{\nabla}_{e_i}$, there is no need to worry about the constraint that $\hat{e}_i$ is a unit vector, because the cross-product $\hat{e}_i\times$ eliminates any unphysical component of the gradient along $\hat{e}_i$.

For the dipole-dipole case we get, in very few lines, \begin{align*} \vec{F}_1 =-\vec{F}_2 &= \frac{3A}{r^5} \left[ \left(c-5\frac{C_1C_2}{r^2}\right) (\vec{r}_1-\vec{r}_2) + C_2\hat{e}_1 + C_1 \hat{e}_2 \right] \\ \vec{T}_1 &= -\frac{A}{r^3} \hat{e}_1\times\left[\hat{e}_2-3\frac{C_2}{r^2}(\vec{r}_1-\vec{r}_2)\right] \\ \vec{T}_2 &= -\frac{A}{r^3} \hat{e}_2\times\left[\hat{e}_1-3\frac{C_1}{r^2}(\vec{r}_1-\vec{r}_2)\right] \end{align*} These satisfy the equations at the top of this answer, and indeed this is guaranteed from the way they were derived.

For your particular geometry, setting $\vec{r}_1 = \vec{0}$, $\vec{r}_2=r\hat{x}$, $\hat{e}_1=\hat{y}$, $\hat{e}_2=\hat{x}$, $c=0$, $C_1=0$, $C_2=-r$, $$ \vec{F}_1 = -\frac{3A}{r^4}\hat{y},\quad \vec{F}_2 = \frac{3A}{r^4}\hat{y},\quad \vec{T}_1 = -\frac{2A}{r^3}\hat{z},\quad \vec{T}_2 = -\frac{A}{r^3}\hat{z} $$ So indeed, the torques are not equal and opposite, but they do satisfy rotational invariance when combined with $(\vec{r}_1-\vec{r}_2)\times\vec{F}_1$. The force $\vec{F}_1$ is not zero. The formula for $\vec{F}_2$ agrees with your result ($\hat{y}$ points upward, your $\hat{\theta}$ points downward).

  • $\begingroup$ @ LoneyProf: thanks a lot . you mathematically proved it that torques aren't equal but they do owe what is more necessary the " ROTATIONAL INVARIANCE" $\endgroup$ – Faraday Pathak Sep 1 '18 at 14:09
  • $\begingroup$ I didn't actually give a full proof in the general case, but I hope that I showed the physical idea. Each of the three terms in the rotational invariance equation shows how $V$, or rather $-V$, changes when you rotate the directions of the three crucial vectors $\hat{e}_1$, $\hat{e}_2$, and $\vec{r}_1-\vec{r}_2$, without changing their lengths. You need to rotate them all, not just the orientations $\hat{e}_1$, $\hat{e}_2$, in order that $V$ does not change, and this corresponds to the overall rotational symmetry. $\endgroup$ – user197851 Sep 1 '18 at 14:20

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