# The torque on a dipole in uniform motion

```The torque on a dipole in uniform motion
David J. Griffiths∗
Department of Physics, Reed College, Portland, Oregon 97202
V. Hnizdo
National Institute for Occupational Safety and Health, Morgantown, West Virginia 26505
arXiv:1308.4569v3 [physics.class-ph] 18 Jan 2014
We calculate the torque on an ideal (point) dipole moving with constant velocity through uniform
electric and magnetic fields.
I.
INTRODUCTION
The torque on an electric dipole p, at rest in a uniform
electric field E, is
N = p &times; E.
-q
r-
(1)
q
d
r+
The torque on a magnetic dipole m, at rest in a uniform
magnetic field B, is
N = m &times; B.
But what if the dipole is moving, at a constant velocity
v? It is well known that a moving electric dipole acquires
a magnetic dipole moment1
mv = −(v &times; p),
(3)
so one might guess that the torque on a moving electric
dipole in uniform electric and magnetic fields would be2
N = (p &times; E) + (mv &times; B).
(4)
Similarly, a moving magnetic dipole acquires an electric
dipole moment3
pv =
1
(v &times; m),
c2
(5)
and one might guess that the torque on a moving magnetic dipole would be
N = (m &times; B) + (pv &times; E).
(6)
But these formulas are incorrect; in each case there is a
third term:4
Np = (p &times; E) + (mv &times; B) + v &times; (p &times; B),
Nm = (m &times; B) + (pv &times; E) + v &times; (pv &times; B).
(7)
(8)
In general, then,
N = (p &times; E) + (m &times; B) + v &times; (p &times; B).
Our purpose in this note is to derive that result.5
FIG. 1. Electric dipole.
(2)
(9)
II.
ELECTRIC DIPOLE IN MOTION
An electric dipole consists of positive and negative
charges &plusmn;q separated by a displacement d (Figure 1).
The force on a point charge is
F = q (E + v &times; B) ,
(10)
6
so the torque on a dipole of moment p = qd is
N = (r+ &times; F+ ) + (r− &times; F− )
= (r+ − r− ) &times; [q (E + v &times; B)]
= qd &times; (E + v &times; B)
= (p &times; E) + p &times; (v &times; B).
(11)
Using the vector identity p &times; (v &times; B) = v &times; (p &times; B) +
(p &times; v) &times; B, the previous result becomes
N = (p &times; E) + (mv &times; B) + v &times; (p &times; B).
(12)
That is surely the easiest way to obtain Eq. (7) (it
borrows from an argument by V. Namias7 ). But the
magnetic analog is not so straightforward,8 so let’s repeat the derivation, this time treating the electric dipole
as the point limit of a uniformly polarized object.9 Polarization (P) and magnetization (M) together constitute
an antisymmetric second-rank tensor:


0
cPx cPy cPz
0 −Mz My 
−cPx
P &micro;ν = 
,
(13)
−cPy Mz
0 −Mx 
−cPz −My Mx
0
and the transformation rule gives10
v
v
Pz =Pz0 , Px =γ(Px0 − 2 My0 ), Py =γ(Py0 + 2 Mx0 ), (14)
c
c
Mz =Mz0 , Mx =γ(Mx0 +vPy0 ), My =γ(My0 −vPx0 ). (15)
2
We use primes for the “proper” (rest) system of the dipole
(S 0 ); no prime means the “lab” frame (S), through which
the dipole is moving with velocity v ẑ. Notice that if
M0 = 0, then M = −(v &times; P), confirming Eq. (3), and if
P0 = 0, then P = (1/c2 )(v &times; M), confirming Eq. (5).
Now, suppose we have an electric dipole p0 at rest at
the origin in S 0 :
P0 (x0 , y 0 , z 0 ) = p0 δ(x0 )δ(y 0 )δ(z 0 ).
(16)
Assume first that p0 is perpendicular to v, say
p0 = p0 x̂.
so
N = p0 x̂&times;[E+(v &times;B)] = (p0 &times;E)+p0 &times;(v &times;B), (29)
which agrees with Eq. (11), and hence confirms Eq. (7)
for the case p0 perpendicular to v. (For this orientation
the electric dipole moments in S and S 0 are the same—
there is no Lorentz contraction.
R This can be confirmed
by integrating Eq. (18): p ≡ P d3 r = p0 .)
If p0 is parallel to v,
p0 = p0 ẑ,
(17)
(30)
giving
Then
Px = γp0 δ(x0 )δ(y 0 )δ(z 0 ) = γp0 δ(x)δ(y)δ(γ(z − vt))
= p0 δ(x)δ(y)δ(z − vt),
(18)
0
0
0
My = −vγp0 δ(x )δ(y )δ(z )
= −vp0 δ(x)δ(y)δ(z − vt),
(19)
Pz = p0 δ(x0 )δ(y 0 )δ(z 0 ) =
(20)
∂P
.
∂t
(21)
ρ = −p0 δ 0 (x)δ(y)δ(z − vt),
(22)
(24)
(33)
p0
δ(x)δ(y)δ 0 (z − vt)[E + (v &times; B)].
γ
(34)
The torque is
N=
∇&times;M=vp0 [δ(x)δ(y)δ 0 (z−vt) x̂−δ 0 (x)δ(y)δ(z−vt) ẑ] ,
(23)
∂P
= −vp0 δ(x)δ(y)δ 0 (z − vt) x̂,
∂t
vp0
δ(x)δ(y)δ 0 (z − vt) ẑ,
γ
so that
f =−
In our case,12
(32)
and
J=−
J=∇&times;M+
p0
δ(x)δ(y)δ 0 (z − vt)
γ
ρ=−
where11
ρ = −∇ &middot; P,
(31)
with all other components of P and M being zero. In
this case,
and all other components are zero. According to the
Lorentz law, the force density is
f = ρE + J &times; B,
p0
δ(x)δ(y)δ(z − vt),
γ
p0
&times; [E + (v &times; B)].
γ
(35)
Because of Lorentz contraction, p = p0 /γ—as is confirmed by integrating Eq. (31)—and again we recover
Eq. (7).
so
J = −vp0 δ 0 (x)δ(y)δ(z − vt) ẑ = −p0 δ 0 (x)δ(y)δ(z − vt) v
(25)
and
0
f = −p0 δ (x)δ(y)δ(z − vt)[E + (v &times; B)].
(26)
MAGNETIC DIPOLE IN MOTION
Now, suppose we have a magnetic dipole m0 , at rest
at the origin, in S 0 :
M0 (x0 , y 0 , z 0 ) = m0 δ(x0 )δ(y 0 )δ(z 0 ).
The torque is
Z
N = (r &times; f ) d3 r
Z
0
= − p0
δ (x)δ(y)δ(z − vt)r dx dy dz
&times;[E + (v &times; B)],
III.
(36)
Assume first that m0 is perpendicular to v, say
m0 = m0 x̂.
(27)
where r = (x, y, z). The y and z components of the
integral are zero; the x component is13
Z
Z
0
xδ (x)δ(y)δ(z − vt) dx dy dz = xδ 0 (x) dx = −1,
(28)
(37)
From the transformation rule,
vm0
vm0
Py =γ 2 δ(x0 )δ(y 0 )δ(z 0 )= 2 δ(x)δ(y)δ(z−vt), (38)
c
c
Mx =γm0 δ(x0 )δ(y 0 )δ(z 0 )=m0 δ(x)δ(y)δ(z−vt). (39)
This time
ρ = −∇ &middot; P = −
vm0
δ(x)δ 0 (y)δ(z − vt),
c2
(40)
3
∇&times;M=m0 [δ(x)δ(y)δ 0 (z−vt) ŷ−δ(x)δ 0 (y)δ(z−vt) ẑ] ,
(41)
v2
∂P
= − 2 m0 δ(x)δ(y)δ 0 (z − vt) ŷ,
∂t
c
(42)
so
J = m0
1
0
0
δ(x)δ(y)δ (z − vt)ŷ − δ(x)δ (y)δ(z − vt)ẑ ,
γ2
(43)
and
vm0
δ(x)δ 0 (y)δ(z − vt)E
c2
1
+ m0 2 δ(x)δ(y)δ 0 (z − vt)ŷ − δ(x)δ 0 (y)δ(z − vt)ẑ
γ
&times;B.
(44)
f =−
IV.
HIDDEN MOMENTUM
In the presence of an electric field, a magnetic dipole
(a loop of electric current) carries “hidden momentum”14
ph = (1/c2 )(m &times; E) and hence also hidden momentum
1
Lh = 2 r &times; (m &times; E).
(51)
c
It is “hidden” in the sense that it is not associated with
overt motion of the object as a whole; it occurs in systems
with internally moving parts, such as current loops.15
The total angular momentum is the sum of “overt” and
hidden components:
L = Lo + Lh ,
(52)
The torque is
vm0
m0
(ŷ &times; E) − 2 ẑ &times; (ŷ &times; B) + m0 ŷ &times; (ẑ &times; B)
c2
γ
= (pv &times; E) + (m0 &times; B) + v &times; (pv &times; B),
(45)
N=
and the torque is its rate of change:16
dLo
dLh
+
dt
dt
1
= No + 2 v &times; (m &times; E).
c
N=
in agreement with Eq. (8), for m0 perpendicular to v. (In
this orientation, the magnetic dipole moments in S and
S 0 areR the same, as we confirm by integrating Eq. (39):
m ≡ M d3 r = m0 .)
If m0 is parallel to v,
m0 = m0 ẑ,
(46)
(53)
Thus, combining Eqs. (9) and (53), the “overt” torque
on a moving dipole is
giving
Mz = m0 δ(x0 )δ(y 0 )δ(z 0 ) =
m0
δ(x)δ(y)δ(z − vt), (47)
γ
with all other components of P and M being zero. In
this case ρ and ∂P/∂t vanish, leaving
J=
m0
[δ(x)δ 0 (y)δ(z − vt) x̂ − δ 0 (x)δ(y)δ(z − vt) ŷ] ,
γ
(48)
so
No = (p&times;E)+(m&times;B)+v&times;(p&times;B)−
1
v&times;(m&times;E). (54)
c2
By contrast, a (stationary) electric dipole has no
internally moving parts, and it harbors no hidden
momentum.17 However, when it is in motion it picks
up a magnetic dipole moment mv = −(v) &times; p [Eq. (3)]
that does carry hidden momentum,18 so Eq. (54) (with
m = mv ) applies to this case as well.19
m0
[δ(x)δ 0 (y)δ(z − vt) x̂ − δ 0 (x)δ(y)δ(z − vt) ŷ] &times; B,
γ
(49)
and the torque is
f=
1
N = (m0 &times; B),
γ
ACKNOWLEDGMENT
(50)
which is again consistent with Eq. (8) because in this
orientation pv = 0 and, integrating Eq. (47), m = m0 /γ.
∗
1
See, for instance, V. Hnizdo, “Magnetic dipole moment of
a moving electric dipole,” Am. J. Phys. 80, 645-647 (2012).
V.H. co-authored this note in his private capacity; no
official support or endorsement by the Centers for Disease
Control and Prevention is intended or should be inferred.
2
In this paper, except for Eqs. (1) and (2), p and m are the
dipole moments of the moving dipoles—not the “proper”
moments in the dipoles’ rest frame.
4
3
4
5
6
7
8
9
10
See, for example, K. W. H. Panofsky and M. Phillips, Classical Electricity and Magnetism, 2nd ed. (Dover, New York,
2005), Sec. 18-4.
In a recent article [D. J. Griffiths and V. Hnizdo,
“Mansuripur’s paradox,” Am. J. Phys. 81, 570-574 (2013)]
we stated that there is no “third term” in the torque formula. This was misleading: there is a third term (in general), but it was zero (in that context) because B = 0 at
the location of the dipole.
If the fields are nonuniform, the torque (9) is supplemented
by r &times; [(p &middot; ∇)E + ∇(m &middot; B) + p &times; (v &middot; ∇)B], where r is
the dipole’s displacement from the point with respect to
which torques are calculated. The term inside the brackets
comprises the force on a moving dipole [G. E. Vekstein,
“On the electromagnetic force on a moving dipole,” Eur.
J. Phys. 18, 113–117 (1997)]. This additional term vanishes
when torques are calculated with respect to the position
of the dipole.
We shall calculate all torques and angular momenta with
respect to the origin.
V. Namias, “Electrodynamics of moving dipoles: The case
of the missing torque,” Am. J. Phys. 57, 171-177 (1989).
Of course, you could model m as separated monopoles (this
is what Namias did in Ref. 7), but it turns out the torque is
different for “Gilbert” dipoles and “Ampère” dipoles (tiny
current loops). See Ref. 4.
This approach is based on an argument by M. Mansuripur,
“Trouble with the Lorentz Law of Force: Incompatibility with special relativity and momentum conservation,”
Phys. Rev. Lett. 108, 193901-1-4 (2012). A simpler version
starts by representing stationary electric and magnetic by
their charge and current densities:
ρ(r, t) = −(p0 &middot; ∇)δ 3 (r), J(r, t) = 0;
ρ(r, t) = 0, J(r, t) = −(m0 &times; ∇)δ 3 (r),
and Lorentz transforming the 4-vector (cρ, J) to get ρ and
J for moving dipoles [Eqs. (22), (25), (40), and (43)].
See, for example, D. J. Griffiths, Introduction to Electrodynamics, 4th ed. (Pearson, Boston, 2013), Eq. (12.118).
11
12
13
14
15
16
17
18
19
Ref. 10, Section 7.3.5.
When attached to a delta function, the prime denotes differentiation with respect to its argument.
Here we integrate by parts.
W. Shockley and R. P. James, “ ‘Try simplest cases’ discovery of ‘hidden momentum’ forces on ‘magnetic currents’,”
Phys. Rev. Lett. 18, 876-879 (1967); W. H. Furry, “Examples of momentum distributions in the electromagnetic
field and in matter,” Am. J. Phys. 37, 621-636 (1969);
L. Vaidman, “Torque and force on a magnetic dipole,”
Am. J. Phys. 58, 978-983 (1990).
Hidden momentum is mechanical in nature, even though
it arises (typically) in electromagnetic contexts. For details
see Ref. 4, or Ref. 10 (Example 12.13).
If the electric field is a function of time, there will be an
additional term of the form (1/c2 )r &times; [m &times; (dE/dt)], but
it vanishes if the torque is calculated with respect to the
dipole’s location. There is no term involving dm/dt, since
the dipole moments are assumed to be constant.
D. J. Griffiths,“Dipoles at rest,” Am. J. Phys. 60, 979-987
(1992).
Note that a moving electric dipole carries a component of
circulating current; see Hnizdo, Ref. 1.
A. L. Kholmetskii, O. V. Missevitch, and T. Yarman,
“Torque on a Moving Electric/Magnetic Dipole,”
Prog. Electromagn. Res. B 45, 83-99 (2012). These
authors claim that the last two terms are new discoveries,
but both these terms were obtained already by Namias
(Ref. 7, Eqs. (1) and (3); the 3rd term is implicit in the
former equation). Incidentally, the same result [Eq. (54)]
holds for p and m constructed from magnetic monopoles
and their currents, though (curiously) the roles of the
last two terms are reversed: the fourth appears already
in the analog to Eq. (9), while the third is due to hidden
momentum.
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