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Question

Consider two dipoles $({p_1}\hat{i}$ and ${-p_2}\hat{j})$ kept in the $x-y$ plane at $(0,0)$ and $(d,0)$ respectively. Calculate the torque about the COM.

Approach 1

Suppose we select the COM as the origin. Then, we label the 4 particles $1,2,3,4$. Let their position vectors be $r_{1},r_{2}..etc$. Then, when we consider the Force on the $i$th particle due to the $j$th particle= $F_{ij}$.

The Torque about the COM.= $r_{i}\times F_{ij}$. When we sum these torques over all particles, and use the following facts:

  • $F_{ji}=-F_{ij}$
  • $r_{i}-r{j}=r_{ij}$
  • $r_{ij}\times F_{ij}=0$

$$r_{1}\times({F_{21}+F_{31}+F_{41})}+r_{2}\times({F_{12}+F_{32}+F_{42})}$$ $$+r_{3}\times({F_{13}+F_{23}+F_{43})}+r_{4}\times({F_{14}+F_{24}+F_{34})}$$ $$= (r_{1}-r_{2})\times F_{12}+(r_{1}-r_{3})\times F_{13} ....$$ $$=0 + 0 ...0$$

We conclude that the net Torque about the COM is is zero. This is somewhat expected, there are no "external" elements in our system.

(In fact, this argument suggests that the Torque about any point in the x-y plane is zero. We will still end up with the same expression, as the difference of two position vectors is independent of the choice of the origin.)

Approach 2

However, qualitatively, when we use our knowledge of electric fields, we can see that both dipoles seem to experience an anticlockwise torque. They will add up, and thus The net torque is non zero.

This appears to be a contradiction. The latter finding is bizarre in the sense that it seems to suggest that the angular momentum of an isolated system is not constant.

And the only way that argument 1 can be wrong , is if:

$$F_{ij}=-F_{ji}$$

does not hold.

I have seen this happen, but the explanation was that the "3rd law of motion" is actually a statement of the conservation of momentum, and in that particular case, some portion of the momentum (of the particles) was carried away by the associated electromagnetic field.

I don't think that something of this sort is responsible for my particular case, since there is no $B$ field, and therefore no $S$, which appears in the momentum density of the fields.

Apart from this, the two arguments seem to be quite valid individually. So what's going on here? My initial thought was that the fields might be storing angular momentum, but like I said, there is no $B$.

Edit 1:

I think I have a hypothesis: In approach 2, as the dipoles rotate for some time say $dt$ , they start producing a $B$ field (since we now have moving charges).

We now have both $E$ and $B$ fields, so we actually have an angular momentum associated with the field, which perhaps cancels out with the angular momentum produced by the dipoles(=$(T1 +T2)dt$).

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Two dipoles

So yes, the left dipole gets a downward force on its negative charge and an upwards force on its positive charge, giving an anticlockwise torque. But the upward force is bigger as the + charge is nearer the second dipole, so there's also a net upwards force.

The right dipole gets a leftward force on the negative charge and an equal rightward force on the positive charge, giving another anticlockwise torque. But there is also a vertical component of force which is downward in both cases, again as the + on the LH dipole is closer than the -.

So each dipole gets an anticlockwise torque, but the joint system gets a clockwise torque. The maths in part 1 shows they balance. dipoles with forces

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Both dipoles appear to experience an anticlockwise torque, around their own centers of mass. This is not the same as an anticlockwise torque around the origin, or around any other single point.

Each individual dipole rotates anticlockwise, but the two dipoles also move, in a clockwise direction around the origin. The overall angular momentum is conserved.

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  • $\begingroup$ If I read that right, you mean in the above approaches we are missing the translation caused right? $\endgroup$ Apr 2 at 14:45
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    $\begingroup$ @RishiNandhaVanchi Approach 1 in the OP is just correct (in electrostatics). Approach 2 misses the torque involved in translating the dipoles, yes. $\endgroup$
    – Chris
    Apr 2 at 22:43
  • $\begingroup$ but what about angular momentum being exchanged with the field? $\endgroup$
    – satan 29
    Apr 3 at 10:00
  • $\begingroup$ @satan29 In electrostatics (and magnetostatics), the field does not carry any momentum, angular or otherwise. $\endgroup$
    – Chris
    Apr 3 at 11:46
  • $\begingroup$ Ahh ok I think I know what you mean. At t=0, this is an electrostatic situation. but subsequently , the dipoles will start moving, and presumably then we would have interactions with the field? $\endgroup$
    – satan 29
    Apr 3 at 13:17
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I would respectfully disagree with answers and comments saying that the explanation is generation of magnetic field and hence angular momentum of electromagnetic field. Magnetic field could give some contribution, but it is only partially relevant.

Let me start with $\textbf{approach 1}$. There is nothing wrong with it, but it implicitly assumes that electric field propagates instantaneously (namely, in the part where you say $F_{ij} = -F_{ji}$. If it was not instantaneous, you would need to take retardation into account). In other words, we could say that the first approach deals with the limit $c\rightarrow\infty$. However, in this limit magnetic field becomes irrelevant (as you may know, electric and magnetic field are unified in special relativity. At small speeds effects of the magnetic field are basically of order $\dfrac{v}{c}$ compared to effects of electric field. In the limit of $c\rightarrow\infty$ the theory of electric field alone is self-consistent, and you don't need to add magnetic field to it). But it would seem that the paradox persists even in this limit - if we believe that approach 2 is complete.

But $\textbf{approach 2}$ is incomplete. Besides rotation of dipoles there would also be translational motion and you need to take the corresponding forces into account when calculating the net torque. More precisely, there is force $\vec{F_1}$ acting on dipole 1 and $\vec{F_2}$ acting on dipole 2. When you calculate the net torque, these forces contribute as well. By Newton's third law (valid in the usual sense in the limit $c\rightarrow\infty$), $\vec{F_1} + \vec{F_2} = 0$, so you could evaluate this torque with respect to any point and it won't matter. If we pick dipole 1 as the reference point only $\vec{F_2}$ would contribute to this torque. $\vec{F_2}$ is calculated from the gradient electric field created by the first dipole $\vec{E_1}$ as follows

$$\vec{F_2} = (\vec{p_2}\cdot\vec{\nabla}) \vec{E_1} = - p_2 \nabla_y \vec{E_1}$$

and we only need the $y$ component, which is (from general expression of electric field of a dipole)

$$F_{2y} = - \frac{3p_1 p_2}{d^4}.$$

So, the torque from pair of forces $\vec{F_1}$ and $\vec{F_2}$ with respect to any point is

$$\vec{T_F} = - \frac{3p_1p_2}{d^3}\hat{k}$$

which is clockwise. Anticlockwise torques, which you mentioned, are $\vec{T_1} = \vec{p_1}\times\vec{E_2} = \dfrac{p_1p_2}{d^3}\hat{k}$ and $\vec{T_2} = \vec{p_2}\times\vec{E_1} = \dfrac{2p_1p_2}{d^3}\hat{k}$. Yay! It adds up to zero nicely:

$$\vec{T_1} + \vec{T_2} + \vec{T_F} = 0.$$

Finally, let me come back a bit to the issue with magnetic field. As I said in the beginning, it is partially relevant - but only if we stop working it the limit $c\rightarrow\infty$. Then electric field does not propagate instantaneously and approach 1 requires a modification to account for retardation. Approach 2 requires a similar modification though (they are not so different after all - they only differ in how you chose to group the forces). This modification would lead to apparent non-conservation of angular momentum and it would be resolved by introducing angular momentum of the e/m field.

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  • $\begingroup$ are you sure we can use $\tau=p \times E$ here? The field varies non-uniformly. $\endgroup$
    – satan 29
    Apr 2 at 5:00
  • $\begingroup$ This formula is valid even for non-uniform field as long as the field does not change much on the length scale of the dipole (and dipoles are often considered to be pointlike, so this is usually a very good approximation). $\endgroup$
    – Viking
    Apr 2 at 14:16

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