How to manage dipole interactions?

Griffiths offers this problem: two point dipoles p1 and p2 are given.They are r distance away and they are perpendicular.We are asked to find torue of p1 (about p1's centre) due to p2 and vice versa.
Well, the situation is that I knoe it is =p cross E
even I know the formula for a point dipole pointing in the z direction:
E_dip=(1/(4*pi*epsilon))(p/r^3)[2cos(theta) (r^)+sin (theta) (theta^)
where (r^) and (theta^) are the unit vectors in a polar system coincident with the xyz system.
I can surely find the result for 1 part.Assuming p1 points in the z direction,it can be evaluated.What could be done in the second case?

Hi kolahal_b,

Your formula for E due to a dipole is correct, but can be put into a coordinate free form. I know Griffiths does this somewhere in an exercise, but the answer is (1/(4pi*epsilon*r^5))(3(p·r)r - (r·r)p), where p is the dipole vector, r is the relative position vector (NOT the unit vector r^) and · is the vector dot product. You can easily check that if p = |p|z^, this formula reduces to the one you quoted.

Once the formula is written in a coordinate independent form, you can use it with any coordinate system you wish, including cartesian, with the origin at either dipole. This is the most useful form of the formula.

When you calculate the answer, there should be a small surprise, which is not really a surprise if you think about it. (Hint: Newton's third law applies to torque as well as to force.)

Hope this helps!

--Stuart Anderson

Oh!It's funny! Just after posting here I got the idea last night.
However,I think both of us wrote the same formula for the field.What I did was to extract a factor of |r|^2 from the formula you wrote and the end result is the same.I will get (r^) unit vectors for the 1st term.For the 2nd term (r.r)=(r^2)

Well,what I got was:
torque on p2: N2=-(1/4*pi*epsilon)(1/r^3)(p2 cross p1)
torque on p1= N1=(1/4*pi*epsilon)(1/r^3)[2(p1 cross p2)]

The two answers differ.griffiths says that if it bothers me then to refer to another problem at the end of the chapter.HaHa!Ok,I did not solve that one so far.Can you say the reason why do they differ?

I see that we were thinking alike! There are two levels of understanding why the torques are not equal. The first level is simply to note that the E field along the axis of a dipole is twice the strength of the E field at the equator, for equal distances. This of course will make one torque twice the size of the other.

But there is a deeper level of explanation also. Since angular momentum is conserved, the net torque on the system must be zero. Usually, you expect that the torque version of Newton's third law will take care of this, so that the torques will be equal and opposite, so that the change in angular momentum for the whole system will be zero. However, here that is not true, which is puzzling. In fact, not only are the torques unequal, they both turn in the same direction! (Notice that the - sign in your first formula cancels the sign from reversing the order of the vector cross product.)

You must also think about the forces these dipoles exert on each other. These forces are of course equal and opposite by Newton's third law, but (this is the unusual part) these forces are not directed along the line joining the dipoles. The force on a dipole is (p·grad)(E) which lets you calculate the forces on the dipoles easily, and you will find that they are not parallel to the vector r. Therefore, the forces exerted by the dipoles on each other will produce a NONZERO torque. Since the NET torque must be zero for an isolated system, the torques directly exerted on each dipole must cancel out this torqe, which means that they must NOT cancel each other. This is why they are not equal and opposite.

Assume that P1 is at the origin and oriented towards the +z axis, and that P2 is on the +z axis and oriented toward the +x axis, then the force on P2 comes out to be (3|P1||P2|x^)/(4pi*|r|^4). The force on P1 is of course equal and opposite, which makes the torque = (3|P1||P2|y^)/(4pi*r^3). This is exactly equal and opposite to the sum of the torques you calculated on the dipoles directly. Therefore, the net torque on the system is indeed zero, as required by conservation of angular momentum.

I think that makes the answer to the puzzle clear.

--Stuart Anderson

I believe each separate dipole constitutes a system here.Well,then,what I understood from your post,we must have for each single dipole:
N=(p cross E)+(r cross F)=0 where F=(p.grad)E
(please note that I have used p1 pointing in the z dirn. p2 is pointing in y dirn.)
N1=p cross E=-(1/(4*pi epsilonr^3))(p1*p2)(x^) [using my previous formula]
and r cross F=(1/(4*pi epsilon*r^3))(3p1*p2)(x^)
So as you can see,the two torques are opposite but not equal.
So,they are not cancelling!

(by the way,I am having a tough time with this formula:
N=(p cross E)+(r cross F)=0 where F=(p.grad)E
Griffiths develops the 1st term about a dipole's own centre and develops the 2nd term about any other point,and simply summed the two.I tried to prove it about a general point.But that eventually assumes that for 1st term electric field at + and - charge positions are the same,i.e.we are dealing with uniform E and for the 2nd term,the Electric field is position dependent!
can you help at this?

The concept of "system" in physics is very flexible. A system is ANY collection of things you desire to consider together. Therefore, you certainly can choose to consider each dipole separately as a system, or you can choose to consider both together as a system.

In the case of both dipoles together, there is no outside force or torque applied to the system. The only force or torque present is what comes from each dipoles own field acting upon the other, which means that this force and torque are internal to the system. Therefore, in this case, the NET force and torque on the system must be zero.

In the case of a single dipole, there IS outside force and torque, because the other dipole is now not part of the system. Therefore, the same force and torque which were considered to be internal for the two dipole system are now considered to be external for the one dipole system. Since the system has external force and torque, the NET force and torque are nonzero.

This is as it should be. If each individual particle in the universe, considered as a separate system, had zero force and torque, then nothing would ever move! If systems of several particles had a net nonzero force from their internal interactions, then systems would start to move with no outside force applied to them, which would violate conservation of momentum. Therefore, the only way things can behave sensibly is if internal forces and torques in each system must sum up to zero, but external forces and torques must often be nonzero. This is exactly what we observe in the present case.

Yes, this is what happens when the theory is approached too abstractly. Sometimes the explanations in terms of calculus appear to be magic, or appear to involve inconsistent assumptions, as in this case.

Instead, let us look at the situation more concretely. A dipole physically consists of two equal and opposite electric charges separated by a small but constant distance. We will make ONE calculus based assumption: the dipole is so small that only changes in the field to first order will be detectable. In other words, we assume that in the region of the dipole, the E field appears to be changing linearly with position. For small enough dipoles, this will be true to as good an approximation as we wish.

Now we can compute the forces on each charge directly. Let the center of the dipole be at position r, and the two charges +q and -q at positions r+dr/2 and r-dr/2, so that the dipole is P = qdr. Let the field at the center be E. Then since the E field is changing linearly, the field at the location of the two charges will be E+dE/2 and E-dE/2. Consider the problem now as a superposition of two fields. The first field is E at both r+dr/2 and r-dr/2, and the second field is +dE/2 at r+dr/2 and -dE/2 at r-dr/2. These fields clearly sum up to the actual field.

What forces to these fields exert on the dipole? The first field exerts forces qE and -qE, which are equal and opposite. Therefore, this field exerts no net force on the dipole. However, equal and opposite forces at DIFFERENT locations CAN and do exert a nonzero torque. The torque computed around the center of the dipole will be (dr/2)X(qE) + (-dr/2)X(-qE) = (qdr)XE = pXE. So the first field exerts a torque but no net force.

What does the second field do? This field exerts forces q(dE/2) at r+r/2 and -q(-dE/2) at r-r/2. These forces are both the same, qdE/2, so they add up to a net force qdE. This is the same as q(dx*dE/dx + dy*dE/dy + dz*dE/dz) = q(dx*d/dx + dy*d/dy + dz*d/dz)E = q(dr·grad)E = (P.grad)E. On the other hand, the torque they create around the center of the dipole is (dr/2)X(qdE/2) + (-dr/2)X(qdE/2) = 0. Therefore, the second field exerts a force but no net torque.

So you can see that no conflicting assumptions were made; really the field was assumed to be smooth enough that at small scales it changed linearly with position. What is happening is that the average value of the field contributes to the torque but not the force, and the gradient of the field contributes to the force but not the torque. Therefore in the final formula, it APPEARS that conflicting assumptions were made, but really thisis only because different aspects of the field control the force and the torque.

Hope this helps!

--Stuart Anderson

I should say that I misinterpreted first time the following portion of your post:

OK,it is most clear.However,I am stuck at the edge.Fro the following example given by you:

I am not getting the sum zero.The two rXF torques adds to 6(...) while the two pXE torques 3(...).It is the same whether I use your co-ordinates or I use my co-ordinates.for your co-ordinates,torque on p2 due to p1 points in the (y^) dirn.(which points into page).Also torque on p1 due to p2 points into
(-z^)X(-x^) i.e. in x^ dirn.So,both add to give 6(...).Please help.

The development of the formula was really fantastic.

There is a mistake in the last but one line.(-z^)X(-x^)~(y^).What I got was
6(...)(y^)+3(...)(-y^).

I think you have made an error in adding the torques. You probably recall a theorem of mechanics, which states that when the sum of forces is zero, the net torque is independent of the location of the axis. This means that you can place the axis on top of one charge, so that you only get ONE copy of the torque by rXF. Alternatively, you can place the axis at the other charge and get the other torque (-r)X(-F), or place the axis at the midpoint and get (r/2)XF + (-r/2)X(-F). But in every case, the total will be a single rXF.

If you were adding rXF + (-r)X(-F), you were using two different axes at once to compute the torque, which is an inconsistent calculation. That is why the rXF torque did not cancel the PXE torque for you.

I hope all is clear now. Oh... I'm glad you liked my derivation of the formula!

--Stuart Anderson

OK,just before connecting to internet,I was wishing you not to reply.It's a childish thing and I made it complicated.Really,I forgot that we should use a single origin with respect to which the torques may be calculated.I really need to shameful.
Thank you very much for your help and the fragrance of the dynamicity of our life which I got here:

It is OK. We all make mistakes. A while ago I made exactly the same mistake you just did, so I know that feeling.

--Stuart Anderson