“Hidden” Momentum in a Coaxial Cable
Kirk T. McDonald
Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544
(March 28, 2002; updated November 23, 2015)
1
Problem
Calculate the electromagnetic momentum and identify the “hidden” mechanical momentum
in a coaxial cable of length L, inner radius a, outer radius b, when a battery of voltage V is
connected to one end and a load resistor R0 is connected to the other. The current may be
taken as uniformly distributed over the inner conductor, which has resistivity ρ. The outer
conductor has negligible resistivity, and the current flows on it in a thin sheet at radius b.
The battery has negligible internal resistance.
Deduce the charge per unit length on the outer surface of the inner conductor. Then,
suppose the battery can be turned off in such a way that the current in the cable falls to
zero with some time dependence I(t). Calculate the impulse on the charge on the surface of
the inner conductor due to the electric field induced by the transient current.
This problem is based on sec. 17 of [1], and on prob. 7.57, ex. 8.3 and ex. 12.12 of [2].1
2
Solution
2.1
Electromagnetic Fields and Field Momentum
We perform the analysis in the rest frame of the cable + battery + resistor, which we call
the cable frame.
We denote the resistance per unit length along inner conductor as
ρ
R=
.
(1)
πa2
Then, the total resistance of the cable plus load resistor is R0 + RL. To have current I in
the system, the battery must have voltage (relative to V = 0 on the outer conductor)
V = I(R0 + RL).
(2)
The current I causes a magnetic field B that is nonzero only inside the cable. The field is
readily calculated via Ampère’s law to be (in Gaussian units, and in a cylindrical coordinate
system (r, φ, z) with the coaxial cable centered on the z axis),2
⎧
⎪
r
⎪
⎪
2
⎪
⎪
⎨ a
2I
B(z inside cable) = φ̂
c ⎪
⎪
⎪
⎪
⎪
⎩
1
(r < a),
1
r
(a < r < b),
0
(r > b).
(3)
The fields and Poynting vector found in sec. 2.1 below were discussed qualitatively by Heaviside on
pp. 94-95 of [3] and on pp. 254-55 of the text [4], and quantitatively in [5].
2
If R = 0 the current flows on the surface of the inner conductor and B = 0 for r < a.
1
Inside the inner conductor the electric field is E(r < a, z inside cable) = IRẑ, as needed to
drive the current I against the resistivity ρ.3 Since the tangential component of the electric
field is continuous across a boundary, there must be some electric field in the region r > a
as well. Indeed, a charge distribution Q(z) is needed on the surface of the inner conductor
to shape the interior electric field to be purely longitudinal.
An analysis of the electric field can be based on the convention that the electric potential
V (r, z) is equal to zero on the outer conductor, and is also zero on the plane z = 0 (which
is not necessarily inside the wire of length L). That is, we suppose the cable extends from
z = −L − R0 /R (the position of the battery) to z = −R0 /R (the position of the resistor),
so that the electric potential for r ≤ a can be written as
V (r ≤ a, z inside cable) = −IRz.
(4)
Thus, the potential of the inner conductor at the position of the load resistor is IR0, and
the potential at the position of the battery is IR(L + R0 /R), i.e., the battery voltage (2).
The capacitance per unit length between the inner and outer conductors of the coaxial
cable is well known to be
1
.
(5)
C=
2 ln(b/a)
The charge Q(z) per unit length on the inner conductor is therefore
Q(z) = CV (r = a, z) = −
IRz
IRz
=
,
2 ln(b/a)
2 ln(a/b)
(6)
assuming that L b so that Q(z) is essentially constant over length Δz b. Further, the
potential in the region a < r < b is essentially that for a long wire of charge density Q(z),
matched to the condition that V (r = b) = 0, namely
V (a < r < b, z) = −2Q(z) ln(r/b) = −
IRz ln(r/b)
,
ln(a/b)
(7)
which also matches eq. (4) at r = a. The potential (7) can also be obtained by a separationof-variables solution to Laplace’s equation [1].
The electric field is obtained by taking the gradient of eq. (7), and we find4
E = IR
⎧
⎪
⎪
⎪
⎪
⎪
⎨
ẑ
(r < a),
ln(r/b)
z
ẑ + r ln(a/b) r̂
ln(a/b)
⎪
⎪
⎪
⎪
⎪
⎩
0
(a < r < b),
(8)
(r > b).
If we ignore the resistance R of the inner conductor (as will be done in sec. 2.2), a simplified analysis
can be made. The battery can be taken to lie in the plane z = 0 and the resistor in the plane z = L. For the
outer conductor at zero potential, the inner conductor (r ≤ a, 0 ≤ z ≤ L) has V0 = IR0 = Vbattery = Vresistor ,
and the electric field is nonzero only inside the cable, (a < r < b, 0 ≤ z ≤ L), where it has only the radial
component Er = V0 /r ln(b/a) = −V0 /r ln(a/b). The potential in this region is V = V0 ln(r/b)/ ln(a/b). The
inner conductor has charge Q = V0 /2 ln(b/a) per unit length on its surface.
4
If R = 0 then z → −R0 /R → −∞ and E = IR0 z r̂/r ln(a/b) > 0 is nonzero only for a < r < b.
3
2
The electromagnetic momentum density is
− ar2 r̂
S
E×B
I R
= 2 =
=
− rln(r/b)
r̂ +
ln(a/b)
c
4πc
2πc2 ⎪
⎪
⎪
⎪
⎪
⎩ 0
2
pEM
⎧
⎪
⎪
⎪
⎪
⎪
⎨
(r < a),
z
ẑ
r 2 ln(a/b)
(a < r < b),
(9)
(r > b).
The Poynting vector S quantifies the flow of energy from the battery in the region (a <
r < b, z = −L − R0 /R) to the inner conductor and to the load resistor, where the energy is
dissipated in Joule heating.
The figure on the next page (from [1]) shows lines of electric field and of Poynting flux
in a coaxial cable that has no terminating resistor, but rather is symmetric about the origin
and with power sources at both ends. The example considered here corresponds to, say, the
left third of the figure, plus a terminating resistive plate; the power source is at the left of
the figure.
The total electromagnetic momentum in the cable is
PEM
2.2
I 2 R ẑ
=
pEM dVol =
2πc2 ln(a/b)
I 2 RL(L + 2R0 /R)
=
ẑ.
2c2
b
a
2πr dr
−R0 /R
−L−R0 /R
dz
z
r2
(10)
Analysis of Transient Forces on the System in the Cable Frame
A confirmation of the result (10) for the electromagnetic field momentum PEM can be found
by supposing the current rises from zero to the final value I with time. The changing
magnetic field induces a longitudinal electric field that pushes on the charge on the surface
of the inner conductor. The force on the conduction electrons opposes the current, which
transfers the force to coaxial cable.
3
By Faraday’s law, the induced electric field at r = a is5
Ez,induced(r = a) = −
1d
c dt
b
a
Bφ dr = −
2 dI
ln(b/a),
c2 dt
(12)
noting that Ez,induced(r = b) = 0 since the outer (perfect) conductor can support no tangential
electric field. The additional force on the surface charge is
Fz,induced =
−R0 /R
−L−R0 /R
Q(z)Ez,induced(r = a) dz = −
RL(L + 2R0 /R) dI
,
2c2
dt
(13)
using eq. (6). The momentum kick to the wire as the current increase from zero to I is
therefore
I 2RL(L + 2R0 /R)
(14)
ẑ = −PEM .
ΔPmech = ẑ Fz,induced dt = −
2c2
Thus, the back reaction to the process of emission of the electromagnetic energy into the
coaxial cable results in a very small mechanical momentum of the cable as a whole in the
direction opposite to the energy flow.
This result reinforces the interpretation of eq. (10) as field momentum stored in the
system, that could be converted to back into mechanical momentum when the current drops
to zero. However, this mechanical momentum turns out to be “hidden” at the macroscopic
level, and its existence does not imply that the center of mass/energy of the system as a
whole is moving in the cable frame, as discussed in the next section.
2.3
Mechanical and Total Momentum in the Cable Frame
Suppose the entire system of coaxial cable, battery and load resistor is isolated from the rest
of the Universe and that the we work in the frame in which these items are at rest, which
we have previously called the cable frame. Then, we might expect the total momentum of
the system to be zero.6 While there is internal motion associated with the electrical current,
we (naı̈vely) expect the net momentum of the current to be zero, since the steady current
density J obeys
J dVol = 0.
(15)
This suggests that there is mechanical momentum “hidden” somewhere in the system
such that the total mechanical momentum cancels the electromagnetic momentum (10).7
5
Alternatively, we can use Faraday’s law in differential form,
(∇ × Einduced )φ =
∂Er,induced
∂Ez,induced
1 ∂Bφ
2I˙
−
=−
=− 2
∂z
∂r
c ∂t
c r
(a < r < b).
(11)
There will be no radial component to the induced field, so eq. (11) integrates to the form (12) after enforcing
the condition that Ez,induced (r = b) = 0.
6
The total momentum of an isolated system is zero in the frame in which the microscopic center of
mass/energy is at rest [6]. In this note, we designate the lab frame as this special frame. It is not immediately
clear that the cable frame is the lab frame, and indeed it is not. Yet, the total momentum in the cable frame
will turn out to be zero.
7
The term “hidden” momentum was introduced in [7].
4
Jon Thaler (private communication, Aug. 26, 2007) reminds us that the present example
is very close to that considered by Einstein in 1905 [8] from which he deduced that the
emission of light of energy U lowers the mass of the emitting body according to8
U = Δmc2.
(16)
Here, the mass of the battery is reduced as the electromagnetic field carries energy away to
the resistive inner conductor and the load resistor. As the latter absorb the energy their
masses increase (ignoring possible thermal transport of the absorbed energy). Hence, the
mass of the system at positive z is increasing with time (in the cable frame = rest frame of the
battery + cable + load resistor), which implies that the cable must moves in the negative-z
direction with respect to the lab frame, such that the center of mass/energy remains fixed
in the lab frame. This argument also suggests that center of mass/energy of the system as a
whole is moving in positive z direction with respect to the cable frame, but as will be shown
in sec. 2.2.2 this is not the case, and the microscopic center of mass/energy is at rest in the
cable frame (and the total momentum is zero in this frame), taking into account a kind of
Zitterbewegung of the center of mass/energy of the currents in the coaxial cable.
2.3.1
Steady-State Momentum in the Cable Frame
We return to the steady-state example, and continue the analysis in the cable frame (in
which the battery + cable + resistor is at rest). A goal is to decide whether or not the
center of total mass/energy is at rest in this frame.
For simplicity, we now suppose that both conductors of the coaxial cable have zero
resistance (with the battery at z = 0 and the resistor at z = L; see footnotes 1 and 2), in
which case the electromagnetic field momentum (10) reduces to
PEM
I 2R0 L
=
ẑ.
c2
(17)
We next consider the system to consist of two subsystems:
1. The EM subsystem, consisting of the macroscopic electromagnetic fields, which are
nonzero only in the volume (a < r < b, 0 < z < L). However, formally the EM
subsystem extends over all space.
2. The matter subsystem, consisting of the “matter” of the battery + cable + resistor,
including the moving (conduction) charges of the electrical current, as well as the microscopic electromagnetic fields of all these items.9,10 Formally, the matter subsystem
extends over all space.
8
This point was also recently made by Timothy Boyer (private communication, Sept. 21, 2007), [9].
The electrical current is electrically neutral on the macroscopic scale.
10
We count any elastic or thermal energy densities, in the system, uelastic or uthermal with effective mass
densities uelastic /c2 and uthermal /c2 , as part of the “matter.”
9
5
These subsystems have mass/energy densities uEM and umatter,11 such that the total energies
of the subsystems are
UEM =
uEM dVol ≡ MEM c2,
Umatter =
umatter dVol ≡ Mmatterc2 ,
(18)
and the centers of mass/energy of the subsystems are
xce,EM =
uEM x dVol
,
UEM
xce,matter =
umatter x dVol
.
Umatter
(19)
Velocity of the Macroscopic Electromagnetic Center of Energy
The macroscopic electromagnetic field energy density of the system is constant over time,
so the velocity of the center of energy of the macroscopic electromagnetic field in the cable
frame is zero,
vce,EM =
dxce,EM
= 0,
dt
xce,EM = const =
L
.
2
(20)
Velocity of the Macroscopic Center of Mass/Energy of the Matter Subsystem
In contrast, the macroscopic center of mass/energy of the matter subsystem is moving
because the electromagnetic field transfers energy from the battery to the resistor at rate
dUresistor/dt = I 2R0 = −dUbattery/dt, while the total energy Umatter is constant.12 Hence,13
xbattery
dx̄ce,matter
=
v̄ce,matter =
dt
dUbattery
dt
+ xresistor dUresistor
dt
Umatter
=
I 2 R0L
I 2R0 L
ẑ =
ẑ, (21)
Umatter
Mmatter c2
recalling that zbattery = 0 and zresistor = L. Thus,
Mmatterv̄ce,matter =
I 2 R0L
ẑ = PEM
c2
(macroscopic).
(22)
That is, the momentum associated with v̄ce,matter is not mechanical but electromagnetic, as
energy in the battery is converted to field energy, which is transported by the field to the
resistor, where it is converted back into mechanical energy.
One might say that there is something “hidden” about momentum in this example.
Total Momentum of the Matter Subsystem
The only matter in the cable frame that has motion, on time scales larger than the
interatomic distance divided by the speed of sound, is the charge carriers of the current I.14
The mass density ρmatter contributes ρmatterc2 / 1 − v2 /c2 to the energy density umatter, where v is the
velocity of the mass element.
12
At the microscopic level, there is an additional contribution to the center-of-mass/energy velocity of the
matter, associated with the microscopic Zitterbewegung motion of the center-of-mass/energy of the charges
in the conductors, eq. (27). See [10] for further discussion. This contribution brings the microscopic velocity
of the center of mass/energy to zero, but its effect on the macroscopic position of the center of mass/energy
is averaged to zero, as illustrated in the figure on p. 9.
13
Quantities whose macroscopic value differs from its microscopic value are denoted with a bar.
14
The notion that electrical current consists of moving charge carriers was not part of Maxwell’s original
electrodynamics [11], and can be considered an aspect of microscopic electrodynamics.
11
6
The argument given at the beginning of sec. 2.2 (that J
dVol = 0 implies Pmatter = 0),
requires that all charge carriers have the same ratio of m/e 1 − v 2/c2 , where m is the rest
mass of the charge.
However, the charge carriers change their energy and velocity as they pass through the
resistor and the battery. In this example, the inner conductor is at the higher potential,
so the electric field inside the resistor points radially outward. The delicate argument is
that the charges e gain energy eV = eIR0 as they move outwards through the resistor, even
though they undergo collisions every few Angstroms along the way. As as result, there is a
difference in the relativistic masses of the carriers in the inner and outer conductors,15,16
(mo γ o − miγ i )c2 = ΔU = eIR0.
(23)
Labeling the carrier number density per unit length in the inner and outer conductors as ni
and no , we have
(24)
I = eni vi = eno vo ,
and the momentum of the carriers (and hence the momentum of the matter in the cable
frame) is
Pmatter = Pcharges = (ni Lmi γ i vi − no Lmo γ o vo ) ẑ = −
LIΔU
I 2R0 L
ẑ = −PEM . (25)
ẑ
=
−
c2 e
c2
The Total Momentum is Zero in the Cable Frame
As remarked above, the only momentum of matter in the cable frame is that associated
with the conduction charges. Thus, from eq. (25),
Ptotal = Pmatter + PEM = Pcharges + PEM = 0,
(26)
even though the macroscopic center of mass/energy of the system is moving, eq (30).
The Velocity of the Center-of-Energy is Zero in the Cable Frame
Furthermore, there is motion (called microscopic here) of the center of mass/energy of
the (conduction) charges, associated with the net mechanical momentum (in the negative-z
direction) of the charges,
Mchargesvce,charges = (ni Lmγ i vi − no Lmγ o vo ) ẑ = Pcharges = −PEM = −Mmatterv̄ce,matter, (27)
which is equal and opposite to the macroscopic term Mmatterv̄ce,matter found in eq. (22).
Hence, the microscopic velocity of the center of mass/energy is zero in the cable frame,17
Mvce = Mmattervce,matter + MEM vce,EM
= Mchargesvce,charges + Mmatterv̄ce,matter + MEM vce,EM = 0.
15
(microscopic).(28)
A version of this argument was first given on p. 215 of [12]. See also [10].
Inside conductors and batteries, the conduction electrons interact with the “lattice,” such that their
average (drift) velocity is essentially constant throughout the circuit. However, the (quantum) interaction
of the electrons with the lattice is such that their motion should be described by an effective mass (Peierls
(1929) [13]) that can vary with position, as in eq. (23).
17
If the electrical current ceases after having flowed for some time, the center of energy of the system moves
during the transient reduction of the current to the location associated with the final lower mass/energy of
the battery and higher mass/energy of the resistor.
16
7
Note also that the velocity of the microscopic center of mass/energy of the matter subsystem is zero,
Mmattervce,matter = Mchargesvce,charges + Mmatter v̄ce,matter = 0
(microscopic).
(29)
Position of the Microscopic Center of Mass/Energy of the Entire System
The nonzero value of vce,charges is observable only at the microscopic level. Associated
with this nonzero velocity is a microscopic motion of the center xce,charges of mass/energy
of the moving charges in that when charges enter/leave the inner or outer conductor the
microscopic center of mass/energy of the charges is instantaneously shifted by amount d/nL
where d is the spacing between charges.18 The center of mass/energy zce,chargesof the charges
executes a Zitterbewegung such that the microscopic velocity of the center of mass/energy
is nonzero, eq. (27), but on average the (macroscopic) center of mass/energy z̄ce,charges of the
charges is at rest (at z = L/2).
Likewise, in a microscopic description the electrical current I (charge passing some point
per unit time) is not constant if the unit time is very small. In a macroscopic view in which
quantities are averaged over times longer than the time for a charge to move one gap length
(≈ one atomic radius), the macroscopic current I is constant and the macroscopic velocity
of the center of mass/energy of the charges is zero.
The macroscopic velocity v̄ce,charges is zero, and the macroscopic velocity v̄ce of the entire
system, with respect to the cable frame, is not zero (recalling eqs. (20) and (22)),
M v̄ce = Mmatterv̄ce,matter + MEM vce,EM = Mmatterv̄ce,matter
I 2R0 L
= PEM =
ẑ
(macroscopic).
c2
18
(30)
In sec. 3.3 we note that the distributions of positive and negative charges inside resistive conductors are
better described as continuous, rather than discrete. In this view, the center of mass/energy of the moving
charges is just x̄ce,charges.
8
Although the coax/battery/resistor are not moving in the cable frame, the macroscopic center
of mass/energy v̄ce of the entire system has a tiny velocity, calculable but not practically
observable as a motion of matter in the system. Indeed,
v̄ce =
UEM vce,EM + Umatter v̄ce,matter
I 2R0 L
≈
ẑ = v̄ce,matter ,
UEM + Umatter
Umatter
(31)
noting that UEM Umatter.
Relation between the Cable and Lab Frames
Although the microscopic velocity vce of the center of mass/energy is zero in the cable
frame, eq. (28), the microscopic position zce of the center of mass/energy changes with time,
as shown in the figure on the previous page, and the macroscopic average position z̄ce varies
linearly with time. The time derivative v̄ce = dz̄ce /dt is nonzero constant in the cable frame.
In the lab frame, the microscopic velocity of the center of mass/energy of the entire
system is zero (by definition of the lab frame), so the macroscopic average velocity of the
center of mass/energy must be zero in this frame. Hence, we deduce that the cable frame
= −v̄ce with respect to the lab frame.
has velocity vcable
Further discussion of quantities in the lab frame is given in sec. 2.5.
2.4
“Hidden” Momentum in the Cable Frame
The (equal and opposite) momenta PEM and Pcharges are tiny effects of order 1/c2 , and so
not readily noticed. This has led to the term “hidden” momentum, whose meaning has been
ambiguous/controversial in the literature [6, 7, 12, 14, 15, 17, 16, 18, 19, 20, 21, 22, 2, 23,
24, 25].
2.4.1
A Definition of “Hidden” Momentum
We now write the momentum P of each of the subsystems (or of the whole system) as
U
P = 2 vce + Phidden −
c
boundary
(x − xcm ) (p − ρvb) · dArea,
(32)
which defines notion of “hidden” momentum,19 where vce = dxce /dt is the center-of-mass/energy
velocity of the subsystem, p is its momentum density, ρ is its macroscopic mass density, and
vb is the velocity (field) of the boundary. In the present example the boundaries of the
subsystems are at infinity,20 and21
Phidden = P −
U
vce
c2
(boundary at infinity).
19
(33)
This definition is due to Daniel Vanzella, as elaborated upon in [26].
See [27] for application of definition (32) in an all-mechanical example with mass flow between two
subsystems.
21
The form (33) was given as a general definition of “hidden” momentum in eq. (78) of [20].
20
9
As discussed in sec. 3.2 of [26], for an electromagnetic subsystem the quantities vce, xce
and p should be macroscopic averages rather than the microscopic values.22 It is ambiguous whether for a matter subsystem these quantities should be taken as macroscopic or
microscopic. Indeed, the present problem serves to illustrates this ambiguity.
“Hidden” Electromagnetic Momentum
Using the definition (33), the “hidden” momentum of the electromagnetic field in the
cable frame is
Phidden,EM = PEM −
UEM
I 2 R0L
=
P
=
v̄
ẑ,
ce,EM
EM
c2
c2
(34)
in view of eq. (20), noting that the macroscopic and microscopic velocities of the center
of electromagnetic energy are both zero in the cable frame. That is, all electromagnetic
momentum in this example, where the electric and magnetic fields are static and there is no
electromagnetic wave propagation, is considered to be “hidden” according to definition (33).
In practice, the momentum (34) is unobservably small, being of order 1/c2 .
“Hidden” Mechanical Momentum
For the matter subsystem we must decide whether
Phidden,matter = Pmatter − Mmatterv̄ce,matter
or
Pmatter − Mmattervce,matter.
(35)
To evaluate the “hidden” mechanical momentum according to eq. (35), we must chose
between use of the microscopic center-of-mass velocity, vce,matter of eq. (29), and the macroscopic velocity v̄ce,matter of eq. (22),
?
Phidden,matter = Pmatter − Mmattervce,matter. = Pcharges − 0 = −PEM ,
(36)
or
?
Phidden,matter = Pmatter − Mmatterv̄ce,matter = Pcharges − PEM = −2PEM ,
(37)
There is a general expectation (see, for example, sec. 4.1 of [26]) that in quasistatic
systems the “hidden” electromagnetic and mechanical momenta be equal and opposite, which
favors eq. (36). On the other hand, it appears preferable to use the macroscopic quantity
v̄ce,EM, rather than the microscopic quantity vce,EM (in those cases where these two quantities
differ), when computing “hidden” electromagnetic momentum.23
As will be reinforced by secs. 2.5 and 3, it seems most consistent to accept eq. (36),
Phidden,matter = Pmatter − Mmattervce,matter. = Pmatter = −PEM ,
(38)
Then, the total “hidden” momentum of the system (in the cable frame) is zero,
Phidden,total = PEM + Phidden,matter = 0 = Ptotal = Ptotal − Mvce,
(39)
again using the microscopic center-of-mass/energy velocity in the general form (33).
22
In the present problem there is no difference between the macroscopic and microscopic values of the
electromagnetic field of the EM subsystem, except near the inner surface of the coaxial cable.
23
See sec. 3.2 of [26].
10
2.5
Analysis in the Lab Frame
We define the lab frame to be such that the microscopic (not macroscopic) center of mass/energy
is at rest.
Denoting quantities in the lab frame with the superscript , we have that
zce
= const ≡ z0 .
(40)
The coordinate transformation between the cable frame and the rest frame is
x = x,
z (t) = z(t) − zce(t) + z0,
y = y,
(41)
such that the velocity transformation is
vx = vx,
vz = vz −
vy = vy ,
dzce
= vz
dt
(42)
where the relation for vz holds for all times except when charges enter/leave the cable (a set
of measure zero). Hence, all expressions involving velocities and/or momenta in the cable
frame hold in the lab frame as well, with the replacements v → v and P → P .
This reinforces the choice of eq. (36) over (37), in which case we have that
Ptotal = Pmatter + PEM = 0,
Phidden,matter = −PEM = −Phidden,EM,
Phidden,total = Phidden,matter + Phidden,EM = 0,
(43)
(44)
(45)
in the lab frame.
3
Comments
3.1
Origin of the Term “Hidden” Momentum
The name “hidden” momentum originated in examples where a permanent (Ampèrian) magnet resides in a static electric field such that electromagnetic momentum,
PEM =
S
dVol =
c2
E×B
dVol,
4πc
(46)
is nonzero [6, 7, 12, 14, 15, 17, 16, 18, 19, 20, 21, 22, 2, 23, 24, 25]. In such examples there is
no flow of energy between a source and a sink, and the center of mass/energy of the whole
system, as well as those of the electromagnetic fields and the matter, are at rest in the lab
frame. The present example extends the set of examples of “hidden” momentum to include a
case in which there is a net flow of energy in the (lab) frame where the center of mass/energy
is at rest. When this energy flow is associated with an electromagnetic field there can be
“hidden” momentum.24
24
for a case in which the energy flow is associated with mass transport see [28], and for an example with
mass transfer via a sound wave see [29].
11
3.2
An Alternative Expression for “Hidden” Momentum
The “hidden” momentum (of a subsystem, according to definition (32)) can also be written
as [26]
Phidden = −
f0
(x − xce) dVol,
c
(47)
where
fμ =
∂T μν
∂xν
(48)
is the 4-force density exerted on the subsystem by on all other subsystems, and T μν is the
stress-energy-momentum tensor of the subsystem.25
The (Lorentz) 4-force density of the electromagnetic field acting on the charged matter
has time component E · J/c, which is nonzero only inside the battery and the resistor in the
present example. There, the electric field is E = IR0 r̂/r ln(b/a), where r is the radial vector
in a cylindrical coordinate system, and the current density is J = ±I r̂/2πrArea, with the
− sign inside the battery and the + sign inside the resistor. Thus, eq. (47) leads to
Phidden,matter = −
I 2R0 L
ẑ = −PEM .
c2
(49)
For the electromagnetic subsystem, the top row of the stress-energy-momentum tensor has
0μ
0
the form TEM
= (uEM , c pEM ) = (uEM , S/c), and S is the Poynting vector. Thus, fEM
=
∂uEM/∂ct + ∇ · S/c = ∂uEM/∂ct − ∂uEM/∂ct − J · E/c = −J · E/c, and
Phidden,EM =
J·E I 2R0 L
EM
x
−
x
dVol
=
ẑ = PEM .
ce
c2
c2
(50)
This further validates the use of the microscopic velocity of the center-of-mass/energy of
the matter subsystem in eq. (38).
3.3
On the Microscopic Nature of Electrical Currents
The notion of electrical currents as consisting of moving electric charges is a key feature of
microscopic electrodynamics. In, say, copper wires, the number of charge carriers is two per
atom, so the characteristic spacing between charge carriers is an atomic radius. As such, a
quantum description of the charge carriers is more appropriate than a classical approximation
of them as moving point charges.
The quantum wave function of each charge carrier has characteristic size of an atom, so
the effective density of the charge carriers is continuous, rather than discrete as for a collection
of point charges. This quantum feature is consistent with the Maxwellian view that charge
and current densities associated with conductors are continuous in classical electrodynamics.
For an isolated, closed system with total stress-energy-momentum tensor T μν , the 4-divergence of the
μ
latter is zero, ∂T μν /∂xν = 0. If the system contains two subsystems A and B then fA
= ∂TAμν /∂xν =
μν
μ
μ
ν
−∂TB /∂x = −fB , where fA is the 4-force density exerted by subsystem B on A.
25
12
Furthermore, the interior of an electrical conductor is electrically neutral to a very good
approximation,26 with the implication the there exist continuous densities of both positive
and negative charges that overlap one another. The interactions between the constituent
charges that would exist in the approximation of point particles do not exist inside electrically
neutral conductors.27 Such interactions were tacitly neglected in sec. 2, where in eqs. (23)(27) it was convenient to consider the moving charges has having a discrete rather than
continuous density.
References
[1] A. Sommerfeld, Electrodynamics (Academic Press, New York, 1952).
[2] D.J. Griffiths, Introduction to Electrodynamics, 3rd ed. (Prentice Hall, Upper Saddle
River, New Jersey, 1999),
http://physics.princeton.edu/~mcdonald/examples/EM/griffiths_8.3_12.12.pdf
[3] O. Heaviside, Electrical Papers, vol. 2 (Macmillan, 1894),
http://physics.princeton.edu/~mcdonald/examples/EM/heaviside_electrical_papers_2.pdf
[4] W.S. Franklin and B. MacNutt, Advanded Theory of Electricity and Magnetism
(Macmillan (Franklin & Charles), 1914),
http://physics.princeton.edu/~mcdonald/examples/EM/franklin_macnutt_adv_electricity.pdf
[5] A. Marcus, The Electric Field Associated with a Steady Current in Long Cylindrical
Conductor, Am. J. Phys. 9, 225 (1941),
http://physics.princeton.edu/~mcdonald/examples/EM/marcus_ajp_9_225_41.pdf
[6] S. Coleman and J.H. Van Vleck, Origin of “Hidden Momentum” Forces on Magnets,
Phys. Rev. 171, 1370 (1968),
http://physics.princeton.edu/~mcdonald/examples/EM/coleman_pr_171_1370_68.pdf
[7] W. Shockley and R.P. James, “Try Simplest Cases” Discovery of “Hidden Momentum”
Forces on “Magnetic Currents”, Phys. Rev. Lett. 18, 876 (1967),
http://physics.princeton.edu/~mcdonald/examples/EM/shockley_prl_18_876_67.pdf
W. Shockley, “Hidden Linear Momentum” Related to the α·E Term for a Dirac-Electron
Wave Packet in an Electric Field, Phys. Rev. Lett. 20, 343 (1968),
http://physics.princeton.edu/~mcdonald/examples/EM/shockley_prl_20_343_68.pdf
26
Resistive, current-carrying electrical conductors are not precisely neutral, but have a net negative charge
density of order 1/c2 . See, for example, [30].
27
Hence, a recent claim [31] is bogus that past analyses of “hidden” momentum in electromagnetic were
wrong because they neglected the interactions among
the point charge carriers of the current. In particular,
the electromagnetic momentum PEM,charges = i qi Ai /c associated with charges qi inside the conductor is
zero as the conductor is
electrically neutral and better described as have continuous rather than discrete
charges: PEM,charges = ρA/c = 0 since the total charge density is zero inside a resistive conductor.
Of course, in examples such as an electron storage ring or a vacuum tube, where only electrons comprise the
current (and their spacing is typically large compared to an atom), such the electron-electron (space-charge)
interactions should not be neglected.
13
[8] A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?, Ann.
Phys. 18, 639 (1905), http://physics.princeton.edu/~mcdonald/examples/EM/einstein_ap_18_639_05.pdf
Translation: Does the Inertia of a Body Depend upon its Energy-Content?,
http://physics.princeton.edu/~mcdonald/examples/EM/einstein_ap_18_639_05_english.pdf
[9] T.H. Boyer, Concerning hidden momentum, Am. J. Phys. 76, 190 (2008),
http://physics.princeton.edu/~mcdonald/examples/EM/boyer_ajp_76_190_08.pdf
[10] K.T. McDonald, “Hidden” Momentum in a Current Loop, (June 30, 2012),
http://physics.princeton.edu/~mcdonald/examples/penfield.pdf
[11] J.C. Maxwell, A Dynamical Theory of the Electromagnetic Field, Phil. Trans. Roy. Soc.
London 155, 459 (1865),
http://puhep1.princeton.edu/~mcdonald/examples/EM/maxwell_ptrsl_155_459_65.pdf
[12] P. Penfield, Jr. and H.A. Haus, Electrodynamics of Moving Media, p. 215 (MIT, 1967),
http://physics.princeton.edu/~mcdonald/examples/EM/penfield_haus_chap7.pdf
[13] R. Peierls, Zur Theorie der galvanomagnetischen Effekte, Z. Phys. 53, 255 (1929), http:
//physics.princeton.edu/~mcdonald/examples/QM/peierls_zp_53_255_29.pdf
[14] O. Costa de Beauregard, A New Law in Electrodynamics, Phys. Lett. 24A, 177 (1967),
http://physics.princeton.edu/~mcdonald/examples/EM/costa_de_beauregard_pl_24a_177_67.pdf
Physical Reality of Electromagnetic Potentials?, Phys. Lett. 25A, 375 (1967),
http://physics.princeton.edu/~mcdonald/examples/EM/costa_de_beauregard_pl_25a_375_67.pdf
“Hidden Momentum” in Magnets and Interaction Energy, Phys. Lett. 28A, 365 (1968),
http://physics.princeton.edu/~mcdonald/examples/EM/costa_de_beauregard_pl_28a_365_68.pdf
[15] H.A. Haus and P. Penfield Jr, Forces on a Current Loop, Phys. Lett. 26A, 412 (1968),
http://physics.princeton.edu/~mcdonald/examples/EM/penfield_pl_26a_412_68.pdf
[16] W.H. Furry, Examples of Momentum Distributions in the Electromagnetic Field and in
Matter, Am. J. Phys. 37, 621 (1969),
http://physics.princeton.edu/~mcdonald/examples/EM/furry_ajp_37_621_69.pdf
[17] M.G. Calkin, Linear Momentum of the Source of a Static Electromagnetic Field, Am.
J. Phys. 39, 513-516 (1971),
http://physics.princeton.edu/~mcdonald/examples/EM/calkin_ajp_39_513_71.pdf
[18] L. Vaidman, Torque and force on a magnetic dipole, Am. J. Phys. 58, 978 (1990),
http://physics.princeton.edu/~mcdonald/examples/EM/vaidman_ajp_58_978_90.pdf
[19] D.J. Griffiths, Dipoles at rest, Am. J. Phys. 60, 979 (1992),
http://physics.princeton.edu/~mcdonald/examples/EM/griffiths_ajp_60_979_92.pdf
[20] V. Hnizdo, Hidden momentum and the electromagnetic mass of a charge and current
carrying body, Am. J. Phys. 65, 55 (1997),
http://physics.princeton.edu/~mcdonald/examples/EM/hnizdo_ajp_65_55_97.pdf
14
[21] V. Hnizdo, Hidden momentum of a relativistic fluid in an external field, Am. J. Phys.
65, 92 (1997), http://physics.princeton.edu/~mcdonald/examples/EM/hnizdo_ajp_65_92_97.pdf
Hidden mechanical momentum and the field momentum in stationary electromagnetic
and gravitational systems, Am. J. Phys. 65, 515 (1997),
http://physics.princeton.edu/~mcdonald/examples/EM/hnizdo_ajp_65_515_97.pdf
Covariance of the total energy-momentum four-vector of a charge and current carrying
macroscopic body, Am. J. Phys. 66, 414 (1998),
http://physics.princeton.edu/~mcdonald/examples/EM/hnizdo_ajp_66_414_98.pdf
On linear momentum in quasistatic electromagnetic systems (July 6, 2004),
http://arxiv.org/abs/physics/0407027
[22] E. Comay, Exposing “hidden momentum”, Am. J. Phys. 64, 1028 (1996),
http://physics.princeton.edu/~mcdonald/examples/EM/comay_ajp_64_1028_96.pdf
Lorentz transformation of a system carrying Hidden Momentum, Am. J. Phys. 68, 1007
(2000), http://physics.princeton.edu/~mcdonald/examples/EM/comay_ajp_68_1007_00.pdf
[23] T.H. Boyer, Connecting linear momentum and energy for electromagnetic systems, Am.
J. Phys. 74, 742 (2006), http://physics.princeton.edu/~mcdonald/examples/EM/boyer_ajp_74_742_06.pdf
Interaction of a point charge and a magnet: Comments on hidden mechanical momentum due to hidden nonelectromagnetic forces, http://arxiv.org/abs/0708.3367
[24] D. Babson et al., Hidden momentum, field momentum, and electromagnetic impulse,
Am. J. Phys. 77, 826 (2009),
http://physics.princeton.edu/~mcdonald/examples/EM/babson_ajp_77_826_09.pdf
[25] K.T. McDonald, “Hidden” Momentum of a Steady Current Distribution in a System at
“Rest,” (Apr. 21, 2009), http://physics.princeton.edu/~mcdonald/examples/current.pdf
[26] K.T. McDonald, On the Definition of “Hidden” Momentum (July 9, 2012),
http://physics.princeton.edu/~mcdonald/examples/hiddendef.pdf
[27] K.T. McDonald, “Hidden” Momentum in an Oscillating Tube of Water, (June 24, 2012),
http://physics.princeton.edu/~mcdonald/examples/utube.pdf
[28] K.T. McDonald, “Hidden” Momentum in Link of a Moving Chain, (June 28, 2012),
http://physics.princeton.edu/~mcdonald/examples/link.pdf
[29] K.T. McDonald, “Hidden” Momentum in a Sound Wave, (Oct. 31, 2007),
http://physics.princeton.edu/~mcdonald/examples/hidden_sound.pdf
[30] K.T. McDonald, Charge Density in a Current-Carrying Wire, (Dec. 10, 2010),
http://physics.princeton.edu/~mcdonald/examples/wire.pdf
[31] T.H. Boyer, Interaction of a magnet and a point charge: Unrecognized internal electromagnetic momentum, Am. J. Phys. 83, 433 (2015),
http://physics.princeton.edu/~mcdonald/examples/EM/boyer_ajp_83_433_15.pdf
15