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Hidden Momentum

A Thesis

Presented to

The Division of Mathematics and Natural Sciences

Reed College

In Partial Fulfillment

of the Requirements for the Degree

Bachelor of Arts

Robin Bjorkquist

May 2009

Approved for the Division(Physics)

David Griffiths

Acknowledgments

I am grateful to the following people:

David Griffiths. Your kindness and wisdom have carried me forward. I am de-lighted to have had the opportunity to both begin and end my college education asyour student (not to mention all the parts in the middle).

My teachers and mentors: Amy Feller, who knew that I would be a physicist beforeI did. Ron Zaraza, for believing in me (and telling me so). Barbara Culpepper, whosent me to Reed at the beginning and always welcomed me back. Johnny Powell,the saving grace of my sophomore year. Darrell Schroeter, for electrodynamics andconversations. Joel Franklin, for his enthusiasm. Mary James, who is always willingto listen and offer solid advice.

The women of WARP (whether or not you came to meetings), especially SophieBerkman, Rosie Cottingham, Laurel Stephenson-Haskins, Renee Dosick, May-Ling Li,Mary Solbrig, Moriah Tobin, Sarah MacQueen and Ellen McManis. Your companyhas been invaluable.

Whitney Mims, for four years of companionship; I couldn’t ask for a better friend.Mary Solbrig, who painted my office and came downstairs to visit me (among otherthings). Ellen McManis, for making my birthday happy.

Friends in faraway places, who sent letters, chocolate, and love through the mail:Suzanne Snell, Janelle Prouty, Sarah Mitteldorf and Amado Glick.

My family, for everything.The reactor folk, past and present. It is an honor and a joy to work among you.

The porpoise is for you.

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 1: Motivation and Beginnings . . . . . . . . . . . . . . . . . . 31.1 Historical Overview: Part One . . . . . . . . . . . . . . . . . . . . . . 31.2 Electromagnetic Field Momentum . . . . . . . . . . . . . . . . . . . 31.3 Example Configurations . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Simple Localized Charge Distributions . . . . . . . . . . . . . 81.3.2 Simple Localized Current Distributions . . . . . . . . . . . . . 91.3.3 Point Charge and Toroid . . . . . . . . . . . . . . . . . . . . . 101.3.4 Point Charge and Magnetic Dipole . . . . . . . . . . . . . . . 111.3.5 Electric Dipole in Uniform Magnetic Field . . . . . . . . . . . 111.3.6 Magnetic Dipole in Uniform Electric Field . . . . . . . . . . . 121.3.7 Nested Uniformly Polarized and Uniformly Magnetized Spheres 12

1.4 Interlude: Return to the Shockley/James Paradox . . . . . . . . . . . 131.5 Introduction to the Stress-Energy Tensor . . . . . . . . . . . . . . . . 131.6 Center of Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . 161.7 Historical Overview: Part Two . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 2: Locating the Hidden Momentum: Various Current Models 192.1 General Stress Tensor Argument . . . . . . . . . . . . . . . . . . . . . 192.2 Incoherent Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2.1 Example: Rectangular Loop . . . . . . . . . . . . . . . . . . . 212.2.2 General Derivation . . . . . . . . . . . . . . . . . . . . . . . . 23

2.3 Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.4 Incompressible Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.4.1 Example: Rectangular Loop . . . . . . . . . . . . . . . . . . . 242.4.2 Why There is Momentum Associated with Pressure . . . . . . 252.4.3 Return to the Rectangular Loop . . . . . . . . . . . . . . . . . 272.4.4 General Derivation . . . . . . . . . . . . . . . . . . . . . . . . 30

Chapter 3: Expanding the Definition: What is Hidden Momentum? 333.1 Gravitational Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 343.2 Moving Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2.1 Example: Moving Capacitor . . . . . . . . . . . . . . . . . . 353.2.2 Tentative Definition of Hidden Momentum . . . . . . . . . . . 37

3.2.3 Two Electromagnetic Subsystems . . . . . . . . . . . . . . . . 383.2.4 Two Mechanical Subystems . . . . . . . . . . . . . . . . . . . 38

Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Abstract

According to the laws of classical electrodynamics, electromagnetic fields carry mo-mentum, even when the sources themselves are stationary. But a general theorem ofrelativistic field theory says that if the center of energy of a localized system is at rest,then the total momentum is zero. Evidently, there must be some other momentumin such cases—momentum that, because it is in some sense “unexpected,” has cometo be called “hidden momentum.” In this thesis, I explore some examples of hiddenmomentum, with a view to formulating a general theory of the phenomenon.

Introduction

My thesis is about momentum: electromagnetic and mechanical momentum; momen-tum in moving systems and in stationary systems; relativistic and non-relativisticmomentum; the momentum associated with pressure; the way that momentum fitsinto the stress-energy tensor; the relationship between momentum and energy; and,of course, hidden momentum.

What is hidden momentum? The story begins with momentum of a different sort:electromagnetic momentum. Charges produce electric fields and currents producemagnetic fields. When combined, electric and magnetic fields can carry momentum.This is certainly true for light (which is an electromagnetic wave), but it is also trueeven when the fields in question are static, with no propagation in sight. It is possibleto build a system out of stationary charges and steady currents which has non-zeroelectromagnetic momentum.

But momentum is associated with motion; it seems uncanny to have a staticsystem with net momentum. Indeed, that sort of thing is not allowed. There is ageneral theorem (I call it the Center of Energy Theorem) which says that if the centerof energy of a localized system is at rest, then the total momentum of the systemmust be zero.

This means that a system of stationary charges and steady currents which hasnon-zero electromagnetic momentum must also have some other momentum, so thatthe total momentum is zero. This other momentum is the “hidden momentum.”

In the first chapter, I calculate the electromagnetic momentum of some examplesystems and prove the Center of Energy Theorem. In Chapter 2, I explain how hiddenmomentum works for two different current models. Finally, in Chapter 3, I discussthe general characteristics of hidden momentum and consider systems other than thestatic electromagnetic systems which began the story.

Chapter 1

Motivation and Beginnings

1.1 Historical Overview: Part One

In a 1967 paper [1], W. Shockley and R. P. James pointed out the following paradox:if you have a magnet and a point charge, and you proceed to turn off the magnet,1

the changing magnetic field at the location of the charge induces an electric field, andthe charge accelerates. However, there is no corresponding force on the magnet. Wehave here a situation in which everything starts out at rest (momentum = 0) andthen one part of the system starts moving (momentum 6= 0). This is decidedly odd,and Shockley and James concluded that there must be more to the story. In order toconserve momentum, there must be a force on the magnet. They called it the hiddenmomentum force, and attributed it to . . . well, to the hidden momentum.

But I am getting ahead of myself. Before we can talk about the mysterious hiddenmomentum force, we need some background:

1.2 Electromagnetic Field Momentum

The first thing to understand is that electromagnetic fields carry momentum. Themomentum density of an electromagnetic field is

gem = ε0E×B. (1.1)

To get the total field momentum, integrate the momentum density over all space (or,at the very least, over all the places where the momentum density is non-zero):

Pem = ε0

∫E×B dτ. (1.2)

Equation (1.2) is always true, but in certain special circumstances, we can rewriteit in other useful forms. First, suppose that we have a localized, static system. By

1Shockley and James made their magnet out of two counter-rotating disks with oppositely chargedrims, and allowed the slight friction between the disks to bring them to a stop; keep in mind that thisis a thought experiment. They also had two charges, not just one (so that the angular momentumwould be zero), and attached the charges to the magnet with rigid rods. We don’t really need all ofthat to see the paradox, however, so I am leaving it out.

4 Chapter 1. Motivation and Beginnings

localized, I mean that the sources of the electric and magnetic fields—the charges andcurrents—are confined to a particular region of space. They do not extend to infinity,so the electric and magnetic fields vanish if you go far enough away.2 Static meansthat the fields are not changing in time, so Maxwell’s equations are

∇ · E =ρ

ε0∇× E = 0

∇ ·B = 0 ∇×B = µ0J. (1.3)

The requirement that the system be static is equivalent to the requirement that thecurrent be steady, so that charge is not building up anywhere, and the electric andmagnetic fields do not change. Steady currents are divergenceless: ∇ · J = 0.

Assuming static fields, we may rewrite the electric field in Equation (1.2) in termsof the potential, E = −∇V , and integrate by parts to get

Pem = ε0

∫[V (∇×B)−∇× (VB)] dτ. (1.4)

The second term is a boundary term; it vanishes for a localized system. For staticsystems, ∇×B = µ0J, so

Pem = ε0µ0

∫V J dτ =

1

c2

∫V J dτ. (1.5)

Equation (1.5) is true for any localized system of stationary charges and steadycurrents. If we go one step further, we will reach another expression, which we willhave occasion to use later on. The “one step further” is to consider a system inwhich the electric field is uniform throughout the region where the current flows.3

Constant E implies linear V , so we may replace V with a first-order Taylor expansion:V = V0 + r · ∇0V = V0 − r · E, where V0 is the potential at the origin and E is theuniform electric field. Then Equation (1.5) becomes

Pem =1

c2V0

∫J dτ − 1

c2

∫(r · E)J dτ. (1.6)

Now, for a localized, divergenceless J, the following identities are true:∫J dτ = 0 (1.7)

and ∫(r · c)J dτ = −1

2c×∫

(r× J) dτ (1.8)

2Localization will allow us to throw away boundary terms on integrals such as Equation (1.2).Each of the fields falls off at least as fast as 1/r2 (once you get far enough away from the sources).Thus, surface integrals involving the product of E and B will vanish for a sufficiently large integrationvolume (such as all of space).

3Of course, the electric field cannot be uniform everywhere (unless it is uniformly zero, which isnot a particularly interesting case), because it is produced by a localized charge distribution.

1.2. Electromagnetic Field Momentum 5

for any constant vector c (in this case, we will use c = E).4 With these identities,the electromagnetic momentum becomes

Pem =1

c2E×m, (1.14)

where m = 12

∫(r× J) dτ is the magnetic dipole moment of the current distribution.

Backing up, we see that we can proceed a different way: Instead of starting fromEquation (1.2) and writing E in terms of V , we can write B in terms of the vectorpotential A:

Pem = ε0

∫E× (∇×A) dτ. (1.15)

The product rules

∇(E ·A) = E× (∇×A) + A× (∇× E) + (E · ∇)A + (A · ∇)E (1.16)

4 Proof of Equation (1.7): To prove this, I will prove something slightly more general: If twolocalized vector fields have the same divergence (∇ ·A = ∇ · B), then

∫A dτ =

∫B dτ , with the

integrals taken over the volume of the fields. To see this, look at just one component:∫Ai dτ =

∫A · (∇xi) dτ =

∫∇ · (xiA) dτ −

∫xi(∇ ·A) dτ, (1.9)

where we have used a product rule to split the integral into two parts. The first term becomes asurface integral and vanishes (that’s where the localization of the fields comes in), leaving us with∫

Ai dτ = −∫xi(∇ ·A) dτ = −

∫xi(∇ ·B) dτ =

∫Bi dτ. (1.10)

Then, because ∇ · J = ∇ · 0, it must be true that∫

J dτ =∫

0 dτ = 0.

Proof of Equation (1.8): First, use product rules to write

∇ · (rirjJ) = ri[∇ · (rjJ)] + (rjJ) · (∇ri)= ri [rj(∇ · J) + J · (∇rj)] + rjJ · (∇ri)= riJj + rjJi (1.11)

Because J is localized,∫∇ · (rirjJ) =

∮rirjJ · da = 0. Thus,∫riJj dτ = −

∫rjJi dτ∫

riJ dτ = −∫

rJi dτ

c ·∫riJ dτ = −c ·

∫rJi dτ∫

r(c · J) dτ = −∫

J(c · r) dτ. (1.12)

We can use this relation to write∫(r · c)J dτ =

12

[∫J(c · r) dτ −

∫r(c · J) dτ

]= −1

2c×

∫(r× J) dτ, (1.13)

which is the desired result.


and

∇× (E×A) = (A · ∇)E− (E · ∇)A + E(∇ ·A)−A(∇ · E), (1.17)

along with Maxwell’s equations for electrostatics (Equation 1.3) and the fact that∇ ·A = 0 in the Coulomb gauge, give

E× (∇×A) = ∇(E ·A)− (E · ∇)A− (A · ∇)E

= ∇(E ·A) +∇× (E×A)− 2(A · ∇)E + A(∇ · E)

= ∇(E ·A) +∇× (E×A)− 2(A · ∇)E +ρ

ε0A. (1.18)

The first three terms contribute nothing to the integral in Equation (1.15),5 so weare left with

Pem =

∫ρA dτ. (1.22)

Equation (1.22), like Equation (1.5), is true for any localized system of stationarycharges and steady currents. We can associate the momentum with currents in anexternal electric field (Equation 1.5), or we can associate it with charges in an externalmagnetic field (Equation 1.22). Of course, we can always return to Equation (1.2), inwhich we associate the momentum with the electric and magnetic fields. All of theseexpressions are equivalent, and we can use whichever is most convenient.

Just as we did with Equation (1.5), we can go one step further from Equation(1.22) by assuming that the magnetic field is uniform over the entire charge distribu-tion. For uniform B, the vector potential is given by

A = A0 −1

2(r×B), (1.23)

5This is a consequence of the localization of the system.First term: The integral of a total derivative becomes a surface integral; E and A both go to zeroat infinity, so their product drops off quickly enough for the surface integral to vanish:∫

∇(E ·A) dτ =∮

(E ·A) da = 0. (1.19)

Second term: Similarly, ∫∇× (E×A) dτ = −

∮(E×A)× da = 0 (1.20)

(See Griffiths [2] Problem 1.60 for the integral identities).

Third term: For this one, write it out in index notation (summation over repeated indices isimplied), and integrate by parts; the boundary term vanishes, as we have come to expect:∫

(A · ∇)Ei dτ =∫Aj∂Ei∂xj

dτ = −∫∂Aj∂xj

Ei dτ = −∫

(∇ ·A)Ei dτ, (1.21)

and ∇ ·A = 0.

1.2. Electromagnetic Field Momentum 7

where A0 is the vector potential at the origin.6 Thus, Equation (1.22) becomes

Pem = A0

∫ρ dτ − 1

2

(∫ρr dτ

)×B = QA0 −

1

2p×B, (1.26)

where Q =∫ρ dτ is the total charge and p =

∫ρr dτ is the electric dipole moment

of the charge distribution.To sum it all up, we have five different ways of expressing the momentum in an

electromagnetic field:

Pem = ε0

∫E×B dτ (1.27)

Pem =1

c2

∫V J dτ (localized, static) (1.28)

Pem =1

c2E×m (uniform E, localized, static) (1.29)

Pem =

∫ρA dτ (localized, static) (1.30)

Pem = QA0 +1

2B× p (uniform B, localized, static). (1.31)

Equation (1.2) is always true. The other four equations require that the system belocalized (the charges and currents are contained within a finite region of space) andstatic (the fields don’t change in time). Additionally Equation (1.29) requires that

6This is Problem 5.24 in Griffiths [2]. The easiest way to see that it is true is to check that∇×A = B and ∇ ·A = 0:

∇×(A0 −

12

(r×B))

= −12

[(B · ∇)r−B(∇ · r)] = −12

[B− 3B] = B; (1.24)

∇ ·(A0 −

12

(r×B))

= −12

[B · (∇× r)] = 0. (1.25)

However, I discovered recently that there is a problem with Equation (1.23). The trouble is thata vector field is not uniquely determined by its divergence and curl. Just because A0 − 1

2 (r × B)has the right derivatives does not necessarily mean that it is the correct vector potential. We canadd ∇λ to A for any scalar λ which satisfies ∇2λ = 0 without altering the divergence or curl of A.Additionally, we require A→ 0 at infinity (so that the integrals in footnote 5 vanish appropriately);this boundary condition, along with ∇ × A = B and ∇ · A = 0, uniquely determines A—andtherefore determines the ∇λ which we must add to Equation (1.23) in order to get the correct A.Sometimes, Equation (1.23) will not require modification, but sometimes it will. Equation (1.23) istrue for the inside of a uniformly magnetized sphere. It is also true for the field inside an infinitesolenoid. However, it is not true for the field of a magnetic dipole (even though a dipole does notproduce a uniform magnetic field, the result should hold within sufficiently small regions of space—itdoes not). I discovered that Equation (1.23) is invalid (and also Equation (1.31), which follows fromEquation (1.23)) while calculating the electromagnetic momentum of a pair of dipoles (one electricand one magnetic); I found that Equation (1.29) (evaluated at the location of the magnetic dipole)and Equation (1.31) (evaluated at the location of the electric dipole) give different results. In thisthesis, I use Equation (1.31) in Sections 1.3.5 and 1.3.7; in both cases, the current distribution is auniformly magnetized sphere, so my results are correct. However, Equation (1.31) should only beused with extreme caution.


the electric field be uniform over the volume of the current distribution and Equation(1.31) requires that the magnetic field be uniform over the volume of the chargedistribution. Within these constraints, we are free to use whichever expression bestsuits our needs.

1.3 Example Configurations

In this section, we look at some examples of localized, static systems which carry non-zero electromagnetic momentum. These systems will turn out to be of great interest,for they are the ones in which we will find the mysterious hidden momentum. Indeed,Shockley and James’s magnet-and-charge is one such system.

For a system to have electromagnetic momentum, it must have a charge distri-bution (the source of the electric field) and a current distribution (the source of themagnetic field). Equations (1.27–1.31) work for any localized, static system, but thecalculations are easier for simple charge and current distributions. My goal in thissection is not to discuss electromagnetic momentum generally (I’ve already done that,in Section 1.2), but rather to illustrate the concept with a few very simple examples.

1.3.1 Simple Localized Charge Distributions

In my examples, I will use three charge distributions: a point charge, an electric dipole,and a uniformly polarized sphere. To calculate the electromagnetic momentum, wewill need the electric field E, the potential V , and/or the electric dipole moment p,depending on which of Equations (1.27–1.31) we use.

Point Charge The electric field of a point charge q at the origin is

E =1

4πε0

q

r2r (1.32)

and the potential is

V =1

4πε0

q

r. (1.33)

Electric Dipole The field of a pure electric dipole with moment p is7

E =1

4πε0r3[3(p · r) r− p]− 1

3ε0p δ3(r). (1.34)

The potential is

V =1

4πε0

p · rr2

. (1.35)

7See Griffiths [2] Section 3.4.4. In particular, Problem 3.42 discusses the delta function term. Thefirst term applies outside a sphere of vanishingly small radius, and the second term applies insidethat sphere. This becomes important if you use Equation (1.27) to calculate the field momentum,because you must integrate over all space, including the location of the dipole (where the deltafunction holds sway).

1.3. Example Configurations 9

Uniformly Polarized Sphere A uniformly polarized sphere8 with polarization Pand radius R has a uniform electric field inside the sphere and a dipole field outside:

E =

− 1

4πε0R3p inside the sphere

1

4πε0

1

r3[3(p · r) r− p] outside the sphere.

(1.36)

The potential is

V =

p · r

4πε0R3inside the sphere

1

4πε0

p · rr2

outside the sphere.

(1.37)

The dipole moment is

p =4

3πR3P. (1.38)

1.3.2 Simple Localized Current Distributions

I will use three current distributions: a toroidal coil, a magnetic dipole, and a uni-formly magnetized sphere. To calculate Pem, we will need to know the magnetic fieldB, the vector potential A and/or the magnetic dipole moment m.

Toroid The toroidal coil is a nice current distribution because the magnetic fieldis entirely contained within the volume of the toroid. (An infinite solenoid has thesame property, but is not a localized current distribution.) The magnetic field insidea toroidal coil in the x-y plane centered at the origin is

B =µ0NI

2πsφ, (1.39)

where N is the number of windings and I is the current. The magnetic field of thetoroid is simple and easy to derive, but the vector potential is not ; it will be best ifwe stick with B and avoid A for this particular distribution.

Magnetic dipole The field of a pure magnetic dipole with moment m is9

B =µ0

4π

1

r3[3(m · r) r−m] +

2µ0

3m δ3(r). (1.40)

8See Griffiths [2] Example 4.2. When I say “uniformly polarized sphere,” what I really mean isa spherical shell carrying charge distribution σ(θ) = P cos θ, with θ measured from the direction ofthe dipole moment p. Such a distribution has the same electric field as a uniformly polarized sphere,but allows me to steer clear of a discussion of electromagnetic momentum in matter. The reason Istill call it a “uniformly polarized sphere” (even though that’s not really what I mean) is because itis easier to say than “a spherical shell carrying charge density proportional to cos θ,” and because itpresents a useful mental image.

9The magnetic dipole field, like the electric dipole field, has a delta function at its center (seeGriffiths [2] Problem 5.59).


The vector potential is

A =µ0

4π

m× r

r2. (1.41)

Uniformly magnetized sphere The field of a uniformly magnetized sphere10 withmagnetization M and radius R is

B =

µ0

2πR3m inside the sphere

µ0

4π

1

r3[3(m · r) r−m] outside the sphere,

(1.42)

similar to the electric field of a uniformly polarized sphere. The vector potential is

A =

µ0

4πR3(m× r) inside the sphere

µ0

4π

(m× r)

r2outside the sphere.

(1.43)

The dipole moment is

m =4

3πR3M. (1.44)

Now I will find the electromagnetic momentum for various combinations of thesesimple charge and current distributions.

1.3.3 Point Charge and Toroid

R

x

y

z

q

Figure 1.1: Toroidal coil and point charge.

Place the point charge at the origin, and center the toroidal coil around it, asshown in Figure 1.1.11 Suppose that the toroid is thin so that the magnetic and

10See Griffiths [2] Example 6.1. Just as in the case of the uniformly polarized sphere, I actuallymean a spherical shell carrying the current distribution which produces the same magnetic field as auniformly magnetized sphere. This happens to be the same as the current distribution of a spinning,uniformly charged spherical shell (just replace M with σRω).

11Vaidman [3] uses this example in his discussion of hidden momentum.

1.3. Example Configurations 11

electric fields have (approximately) the same magnitudes everywhere inside the coil.From Equations (1.32) and (1.39), the electric field inside the toroidal coil is

E =q

4πε0R2s (1.45)

and the magnetic field is

B =µ0NI

2πRφ. (1.46)

Because the magnetic and electric fields both have constant magnitude and E×Bis always in the z direction, it is easy to use Equation (1.27) to calculate the fieldmomentum:

Pem = ε0q

4πε0R2

µ0NI

2πR(2πRA) z =

µ0qNIA

4πR2z, (1.47)

where A is the cross-sectional area of the toroidal coil. A thick toroidal coil wouldalso have net electromagnetic momentum in the z direction, but it would be moredifficult to calculate.

1.3.4 Point Charge and Magnetic Dipole

xy

zq

m r

Figure 1.2: Magnetic dipole and point charge.

Suppose that we have a magnetic dipole m = mz at the origin and a point chargeq located at position r, as shown in Figure 1.2.12 Because there is only one charge inthis system, Equation (1.30) becomes

Pem = qA =µ0qm

4πr2sin θ φ, (1.48)

where we have used Equation (1.41) for the vector potential at the location of thepoint charge. The electromagnetic momentum is non-zero if the point charge is offthe z-axis.

1.3.5 Electric Dipole in Uniform Magnetic Field

From Equation (1.31), we know that the electromagnetic momentum of an electricdipole (with zero net charge) located in a uniform magnetic field is simply13

Pem =1

2B× p. (1.49)

12This situation is analogous to Shockley and James’s [1] magnet and point charge configuration.13See footnote 6.


This is non-zero if the electric dipole moment p and the magnetic field B are orientedin different directions. In order to use Equation (1.31), the source of the uniform mag-netic field must be localized;14 luckily, we already know that a uniformly magnetizedsphere produces just such a field (Equation 1.42).

1.3.6 Magnetic Dipole in Uniform Electric Field

The electromagnetic momentum of a magnetic dipole located in a uniform electricfield is (Equation 1.29)

Pem =1

c2E×m, (1.50)

which is non-zero if the electric field E and magnetic dipole moment m point indifferent directions. The standard source of a uniform electric field is an infiniteparallel-plate capacitor; however, as in the magnetic case, we can also produce thedesired electric field with a localized charge distribution: the uniformly polarizedsphere.

1.3.7 Nested Uniformly Polarized and UniformlyMagnetized Spheres

Now consider a system composed of concentric spheres: one with uniform polarizationand one with uniform magnetization.15 Let R be the radius of the outer sphere. Ifthe uniformly magnetized sphere is inside the uniformly polarized sphere, then theelectric field is uniform throughout the entire current distribution and we can useEquation (1.29):

Pem =1

c2E×m =

1

c2

(− 1

4πε0R3p

)×m =

µ0

4πR3(m× p). (1.51)

If, on the other hand, the uniformly polarized sphere is inside the uniformly magne-tized sphere, then the magnetic field is uniform throughout the charge distribution,and Equation (1.31) is valid:16

Pem =1

2B× p =

1

2

( µ0

2πR3m)× p =

µ0

4πR3(m× p), (1.52)

The momentum is the same either way, implying that even if the two spheres havethe same radius, the electromagnetic momentum is

Pem =µ0

4πR3(m× p) =

4

9πµ0R

3(M×P). (1.53)

14Actually, it might be sufficient for just one of the fields (electric or magnetic) to be localized. Iam playing it safe by only using localized charge and current distributions. The field momentum ofan electric dipole in a uniform magnetic field is worked out in detail in Reference [4] for both cases:with the magnetic field produced by a spinning charged spherical shell and with the magnetic fieldproduced by an infinite solenoid. The result is the same either way.

15This is essentially the example Romer [5] uses when discussing electromagnetic field momentum:a shell of radius a with charge density proportional to cos θ and a shell of radius b with current densityproportional to sin θ′. He does the case b > a.

16See footnote 6.

1.4. Interlude: Return to the Shockley/James Paradox 13

1.4 Interlude: Return to the Shockley/James

Paradox

Recall the paradox with which we began: If you have a magnetic dipole and a pointcharge,17 and you turn off the magnetic field, the induced electric field delivers animpulse to the charge. Before turning off the magnetic field, the dipole and chargeare both at rest, so the total momentum of the system is zero. After turning off thefield, the charge is moving and the dipole is not moving, so the total momentum isnon-zero. This is a paradox because momentum is supposed to be conserved.

But wait! The original system did have momentum—electromagnetic momentum.We calculated it in Section 1.3.4. Indeed, the electromagnetic momentum originallystored in the fields is precisely the same as the impulse delivered to the charge whenthe magnetic field is turned off.18 Conservation of momentum is saved!

However, that’s not the end of the story.19 Not only does the system have toconserve momentum, but it also must have the correct amount of momentum.20 Atthe beginning, I told you that the total momentum of a static system is zero; youprobably believed me, because it is eminently reasonable. It is more than simplyreasonable; it is required. There is a general theorem in relativity which says that alocalized system with stationary center of energy has zero total momentum.21 I willprove the Center of Energy Theorem in Section 1.6, but I will need some specializedlanguage. Thus, I will now cover some more background material:

1.5 Introduction to the Stress-Energy Tensor

In special relativity, a four-vector is an object (with four components) which trans-forms like the spacetime coordinates (ct, x, y, z) when you switch between reference

17Or, equivalently, any of the example systems from Section 1.3.18Shockley and James [1] were well aware of this fact; they show that it is true for their system

of counter-rotating disks and attached charges. In Reference [4], we do the same for an electricdipole at the center of a spinning, uniformly charged sphere and an electric dipole inside an infinitesolenoid. Interestingly, if you turn off the electric field instead of the magnetic field, there is animpulse to the current-carrying body as well as an impulse to the charge-carrying body; the totalimpulse is still equal to the original field momentum.

19You knew it wouldn’t be, because I have persisted in writing this thesis, and I have not yetexplained the hidden momentum.

20Coleman and Van Vleck [6] make this point nicely; they emphasize that the two requirementsare distinct in a relativistic theory.

21The center of energy is the relativistic analog to the center of mass. The energy distribution forour stationary magnetic-dipole-and-charge system doesn’t change in time; therefore, the location ofthe center of energy also does not change. The Center of Energy Theorem certainly holds for sucha system, but it also holds for systems that have moving parts, provided that the center of energyis at rest.


frames. For a Lorentz transformation in the x direction,ct′

x′

y′

z′

=

γ −γβ 0 0−γβ γ 0 0

0 0 1 00 0 0 1

ctxyz

, (1.54)

where γ ≡ (1 − β2)−1/2, β ≡ v/c and v is the velocity (in the x direction) of frameS ′ with respect to the original frame S. We can write this transformation in a morecompact form:22

(xµ)′ = Λµνx

ν . (1.56)

Here, xµ can be the spacetime coordinates, but it can just as well be any other four-vector, because all four-vectors transform like this. The matrix Λ (with elementsΛµν ) encodes the Lorentz transformation. For a transformation in the x direction, the

Lorentz transformation matrix is what appears in Equation (1.54); if the transforma-tion is in some other direction, Λ is more complicated, but the transformation worksthe same way.

Four-vectors are a sub-category of a larger class of objects called tensors. A four-vector has one index, and transforms according to the Lorentz transformation, byway of the elements of the matrix Λ. Tensors can have any number of indices, andtransform with the corresponding number of Λ matrices. The number of indices iscalled the rank of the tensor. Thus, a second-rank tensor transforms like this:23

(T µν)′ = ΛµλΛν

σTλσ. (1.57)

The tensor of most interest to us is the stress-energy tensor. This is a symmetricsecond-rank tensor which contains information about the energy, momentum, andforces within a system. The energy density is

u = T 00, (1.58)

the i-th component of the momentum density is

gi =T 0i

c, (1.59)

and the other components of the tensor (T ij) contain the stresses, or internal forces,within the system.24 If we want to know the total energy or momentum in our system,

22Using the Einstein summation convention, summation over repeated indices is implied. Writtenout, this expression is

(xµ)′ = Λµ0x0 + Λµ1x

1 + Λµ2x2 + Λµ3x

3. (1.55)

Greek indices run from 0 to 3, and are used for the components of four-vectors; if we are talkingabout the spacetime coordinates, then x0 = ct, x1 = x, x2 = y and x3 = z. Latin indices run from1 to 3, and are used for the spatial portions of four-vectors.

23This expression would have sixteen terms if we wrote out the summation explicitly; we quicklybegin to appreciate the index notation.

24I will not be dealing extensively with the stress part of the stress-energy tensor. However,Feynman [7] has a lovely discussion of the three-dimensionsal stress-tensor in section 31-6. He goeson to discuss the four-dimensional stress-energy tensor in section 31-8; he particularly emphasizesthe relation between the different components of the tensor (momentum as the flow of energy andstresses as the flow of momentum).

1.5. Introduction to the Stress-Energy Tensor 15

we integrate the appropriate component of the stress-energy tensor over all space:

U =

∫T 00 dτ and Pi =

∫T 0i

cdτ. (1.60)

For an isolated system, the (four-dimensional) divergence of the stress-energytensor is zero:25

∂µTµν = 0. (1.61)

This is a statement of the local conservation of energy and momentum. To seewhy, first consider another familiar conservation law: the conservation of charge isexpressed by the continuity equation

∂ρ

∂t+∇ · J = 0, (1.62)

where ρ is the charge density and J is the current density. The continuity equationsays that if the amount of charge at a particular location in space is changing, theremust be a corresponding flow of charge to or away from that place; charge cannotbe created or destroyed, and it also can’t move discontinuously from one location toanother.

Now, Equation (1.61) is really four equations—one for each row of the stress-energy tensor. For the ν-th row,

∂µTµν = ∂0T

0ν + ∂iTiν = 0

1

c

∂

∂t(T 0ν) + ∂iT

iν = 0

∂

∂t(T 0ν) + ∂i(cT

iν) = 0. (1.63)

Substituting in the energy and momentum densities for the appropriate tensor ele-ments, this becomes

∂u

∂t+∇ · (c2g) = 0 and

∂gj∂t

+ ∂iTij = 0. (1.64)

These equations have the same form as Equation (1.62), the continuity equation forcharge:

∂

∂t(density) +∇ · (flow vector) = 0.

Thus, we see that Equations (1.64) are continuity equations for energy and momen-tum, and therefore express the local conservation of those quantities. The momentumdensity g (with a couple factors of c), describes the flow of energy, and the stress com-ponents T ij describe the flow of momentum.

To get global conservation laws, we integrate Equation (1.63) over all space:

d

dt

∫T 0ν dτ +

∫∂i(cT

iν) dτ = 0. (1.65)

25∂µ means ∂/∂xµ.


The second term is the volume integral of a divergence, which becomes a surfaceintegral. For a localized system, all elements of the stress tensor vanish at infinity, sowe have

d

dt

∫T 0ν = 0, (1.66)

ordU

dt= 0 and

dP

dt= 0, (1.67)

the global conservation of energy and momentum.All of that was for an isolated system, which (by definition) cannot exchange

energy or momentum with the outside world. The stress-energy tensor of a non-isolated system is not divergenceless, because the energy and momentum need not beconstant. The generalization of Equation (1.61) is

∂µTµν = f ν , (1.68)

where f ν is the external force density acting on the system.26 An isolated system, ofcourse, has f ν = 0, returning Equation (1.61).

1.6 Center of Energy Theorem

We have seen that momentum is related to the flow of energy (Equation 1.64). Thatbrings us to the Center of Energy Theorem, which I mentioned back in Section 1.4: Ifthe center of energy of a localized, isolated system is at rest, then the total momentumof the system must be zero. If there were a net momentum, then there would be acorresponding flow of energy, and the center of energy would move.27

The center of energy of the system is defined as

R ≡ 1

U

∫ur dτ, (1.69)

where u = T 00 is the energy density and U is the total energy. To find the velocityof the center of energy, take the time derivative of Equation (1.69):

dRi

dt=

d

dt

(1

U

∫uxi dτ

). (1.70)

The total energy is constant (because the system is isolated), so

dRi

dt=

1

U

∫∂(uxi)

∂tdτ =

1

U

∫∂u

∂txi dτ. (1.71)

From Equation (1.64), we know

∂u

∂t= −c2∇ · g ; (1.72)

26Note that while the force F = dpµ/dt is not a four-vector, the force density is a four-vector.27Coleman and Van Vleck [6] provide a lovely proof of this theorem. I am following their lead.

1.7. Historical Overview: Part Two 17

using this continuity relation followed by a vector identity,28 we have

dRi

dt= −c

2

U

∫(∇ · g)xi dτ = −c

2

U

(∫∇ · (xig) dτ −

∫gi dτ

)=c2

UPi. (1.73)

The first integral vanishes due to the localization of the system. Thus, we have arrivedat the expression

dR

dt=c2

UP, (1.74)

which explicitly relates the motion of the center of energy to the the total momentumof the system. If the center of energy doesn’t move, the total momentum must bezero.

1.7 Historical Overview: Part Two

Where does this leave us with regard to the Shockley/James paradox? We now know,beyond any doubt, that the total momentum of the system must be zero—both beforeand after we turn off the magnetic field.

Before turning off the field There is momentum in the electromagnetic fields,but the total momentum must be zero. Therefore, there must be some other, non-electromagnetic momentum in the system. Shockley and James [1] called it hiddenmomentum, and said that it must be located in the current loop of the magneticdipole. They knew that the missing momentum must be hiding in the current becausethey thought about the energy flow pattern of the system. The flow of electromagneticenergy is described by the Poynting vector,29

S =1

µ0

E×B. (1.75)

For Shockley and James’s system of a current loop (magnetic dipole) and charge,the Poynting vector indicates that the electromagnetic field carries energy away fromone side of the current loop and returns it to the other side. The energy density isconstant everywhere, so there must be a return flow of energy inside the current loop.Energy flow and momentum go hand in hand, so if there is an energy flow within thecurrent loop, there is also hidden momentum.

While turning off the field The changing magnetic field induces an electric field,which exerts a force on the charge. In order for momentum to be conserved, theremust be an equal and opposite force on the current loop. Shockley and James calledit the hidden momentum force. The hidden momentum force is not an external forceon the current loop; rather, it is the result of the hidden momentum in the current

28By a product rule, ∇ · (xig) = xi(∇ · g) + g · (∇xi) = xi(∇ · g) + gi.29The magnitude of the Poynting vector is the energy per area per time passing through a surface

normal to its direction. Notice that S = c2gem, just as we expect from Equation (1.64).


loop turning into non-hidden momentum. The hidden momentum only exists inthe presence of electric and magnetic fields; if we turn off either field, the hiddenmomentum cannot remain hidden, so it delivers an impulse to the current loop.

After turning off the field Without a magnetic field, there is no electromagneticmomentum and no hidden momentum. Due to impulses delivered while the magneticfield was being turned off, the charge and dipole travel in opposite directions. Thetotal momentum is still zero.

However, we have not yet identified the nature of the hidden momentum. Shockleyand James suggested (very briefly) a couple possible mechanisms for the energy flowinside the current loop, but did not give a detailed explanation. Later, other peoplepicked up the story and carried the topic forward. For the past forty years, variouspeople have offered interpretations and examples. In the next chapter, I will explainthe hidden momentum, drawing on these interpretations and examples.

Chapter 2

Locating the Hidden Momentum:Various Current Models

We have seen that a localized system with stationary center of energy must have zerototal momentum. Thus, if a static electromagnetic system (composed of stationarycharges and steady currents) has non-zero field momentum, it must also have someother, equal and opposite momentum. Shockley and James [1] coined the term hiddenmomentum to describe this mysterious other momentum, perhaps because at firstglance it is not entirely clear where in the system it is located.

By looking carefully at the flow of energy in the electromagnetic field, we canconclude (as Shockley and James did) that the hidden momentum is located in thecurrents. The hidden momentum is mechanical momentum associated with the move-ment of the charges which make up the current. However, the currents are steady, soall the moving charges follow closed paths. It is counter-intuitive (to say the least)that such a system would have non-zero mechanical momentum.

2.1 General Stress Tensor Argument

Using the properties of the stress tensor, we can derive a general form for the me-chanical momentum of a current acted on by an external force.1 The only thing wemust assume is that the charge carriers are described by a stress tensor.2 Recall fromChapter 1 (Equation 1.68) that the force per unit volume acting on the material is

f ν = ∂µTµν . (2.1)

1This argument follows Hnizdo [8].2This is not a very stringent requirement.

20 Chapter 2. Locating the Hidden Momentum: Various Current Models

In this case, the external force is electric, so f ν = (J·E/c, ρE). The zeroth componentof Equation (2.1) is

∂µTµ0 = f 0

∂0T00 + ∂iT

i0 = f 0

1

c

∂u

∂t+ c∇ · g =

1

cJ · E. (2.2)

For a static system, ∂u/∂t = 0, so

∇ · g =1

c2J · E. (2.3)

Writing E in terms of the potential V and using a vector identity, this becomes

∇ · g = − 1

c2J · (∇V ) = − 1

c2[∇ · (V J)− V (∇ · J)] . (2.4)

The current is steady (∇ · J = 0), so we are left with

∇ · g = ∇ ·(− 1

c2V J

). (2.5)

Now, if two localized vector fields A and B have the same divergence, then∫

A dτ =∫B dτ .3 That means the total momentum carried by a localized, steady current in

the presence of a static electric field is

Pcurrent =

∫g dτ = − 1

c2

∫V J dτ. (2.6)

By comparison to Equation (1.28), we see that the momentum in the current isprecisely the opposite of the electromagnetic field momentum. Thus, the hiddenmomentum is located in the current, just as we suspected.

We got this result without saying anything specific about the current. We neversaid what the stress tensor for the current is, just that it exists. This general argumentconfirms our guess that the hidden momentum is mechanical momentum located inthe currents, but it does not illuminate the nature of the hidden momentum. Ifwe want to understand the mechanism of the hidden momentum, we must choose aparticular model for the current.

Vaidman [3] presents three ways of modeling a current: a gas of non-interactingcharged particles moving through a neutral tube, charges moving through a con-ducting tube, and an incompressible charged fluid moving through a neutral tube.Following his lead, I will consider each of these models in turn.

2.2. Incoherent Gas 21

I

r

vl

l

E w

v

Figure 2.1: Rectangular current loop in a uniform electric field. The Nr particles inthe top segment move to the right with speed vr and the Nl particles in the bottomsegment move to the left with speed vl.

2.2 Incoherent Gas

2.2.1 Example: Rectangular Loop

To get a handle on what’s going on, it helps to consider the simple example of arectangular current loop in a uniform electric field.4 Suppose the current is made offreely moving positively charged particles which do not interact with each other. Asthe charges move up the left side of the loop, they gain energy from the electric field,speed up and spread out; as they move down the right side of the loop, they slowdown and bunch together.5 Whatever it is that makes them turn the corners of theloop does not affect their energy. Thus, the top charges move faster than the bottomcharges, but the density of particles is less at the top than the bottom. The currentis the same everywhere:

I = λv =qN

lv, (2.7)

where N is the total number of particles in the segment, and q is the charge of oneparticle. The right-going current in the top segment is the same as the left-goingcurrent in the bottom segment, so

I =qNr

lvr =

qNl

lvl =⇒ Nrvr = Nlvl =

Il

q. (2.8)

We know from the general argument in the previous section that the hidden mo-mentum is located in the current, so we should find the total mechanical momentumof all the moving charges. Classically, the momentum of a particle with mass m and

3I proved this identity in footnote 4 on page 5.4See Griffiths [2] Example 12.12.5I use “up” and “down” in the context of the diagram on the page; I don’t actually mean that

the loop is positioned vertically in space. Note that gravity has nothing to do with any of this.


speed v is P = mv. Thus, the total mechanical momentum of the charges in thecurrent loop is

Pclassical = mNrvr −mNlvl = 0 (2.9)

(the sides of the loop cancel). However, the relativistic momentum is P = γmv.The factor of γ effectively makes higher velocities count more than lower velocities,resulting in a net momentum to the right:

Prelativistic = γrmNrvr − γlmNlvl =mIl

q(γr − γl). (2.10)

Relativistic energy is γmc2, and the energy gained by a charge as it goes up the leftside of the loop is qEw, so

(γr − γl)mc2 = qEw (2.11)

and that means

Phidden =mIl

q

qEw

mc2=IlEw

c2. (2.12)

Or, as a vector,

Phidden =1

c2(m× E), (2.13)

where m = Ia = Ilw is the magnetic dipole moment of the loop.This is a situation where we have a localized, steady current in a uniform electric

field, so the electromagnetic field momentum is given by Equation (1.29),

Pem =1

c2E×m. (2.14)

The hidden momentum and field momentum are equal and opposite, so the totalmomentum is zero, as it should be.

All of that was for a current of positive charges moving clockwise around theloop. We can run the same argument for a current of negative charges movingcounterclockwise—the only difference is that now the right-going current (which stillhas greater momentum than the left-going current) is located in the bottom segmentof the loop rather than the top segment. In order to have a neutral current, we needboth positive and negative charges, running in opposite directions.6

6If you have a sharp eye, you may have noticed that for this current model, using two counter-propagating oppositely-charged gases does not quite ensure the neutrality of the current distribution.The speeding up and slowing down of the charges as they move around the loop results in differentcharge densities at different places. There is a net positive charge on the bottom segment and anegative charge on the top segment. Do we need to worry about the electric field produced by thesecharges, and its effect on the hidden momentum? Hnizdo [8] assumes we can ignore it; I am notentirely convinced. Perhaps we can take care of the problem by plastering some extra static chargeonto the tube in just such a way as to make the charge density zero. This becomes problematic if wecontemplate changing the current in any way (for instance, if we want to turn down the magneticfield in order to cause a hidden momentum force on the current loop), because our plastered-oncharge is only right for the original current. I chose to put this issue aside, but it has remained atthe back of my mind. My hunch is that it is not important, but I lack compelling proof.

2.3. Conductor 23

2.2.2 General Derivation

Everything works out beautifully for the rectangular current loop in a uniform electricfield, but what about other charge and current distributions?

The relativistic mechanical momentum density is g = γmnv, where we allow theparticle density n and velocity v to vary in space (but not time—we still consideronly static systems). The current density is J = qnv, so

g = γm

qJ. (2.15)

As with the rectangular loop, we relate γ to the particle energy: the total energy ofa particle (rest energy + kinetic energy + electric potential energy) is constant, andis equal to

E = γmc2 + qV =⇒ γm =1

c2(E − qV ). (2.16)

Thus, the momentum density is

g =1

c2

(Eq− V

)J, (2.17)

and the total hidden momentum is

Phid =

∫g dτ =

1

c2

(Eq

∫J dτ −

∫V J dτ

)= − 1

c2

∫V J dτ, (2.18)

where we have used∫

J dτ = 0 for a steady current (Equation 1.7). This is equal andopposite to the electromagnetic momentum (Equation 1.28), just as it should be.

2.3 Conductor

All right, you say, That’s fine for a charged gas moving through a tube. But that’s aridiculous model for current. What about electrons moving through a perfect conduc-tor? Where’s the hidden momentum there?

The key to hidden momentum is that the current is located in an electric field sothat the moving charges are subject to an external force. There cannot be an electricfield inside a perfect conductor, so there is no hidden momentum. That’s all right,because there’s no electromagnetic momentum, either:

Pem =1

c2

∫V J dτ, (2.19)

but V is constant in a conductor, so

Pem =V

c2

∫J dτ = 0. (2.20)

Evidently, the charge induced on the surface of the conductor is just right to kill offthe electromagnetic momentum.


2.4 Incompressible Fluid

Suppose that our current is not a gas of charged particles, but rather a charged,incompressible fluid.7 The mechanism for the hidden momentum is no longer thesame, as we can see by considering again the rectangular current loop:

2.4.1 Example: Rectangular Loop

l

wEI

higher pressure

lower pressure

Figure 2.2: Rectangular current loop in a uniform electric field. Charged fluid movesthrough the loop with constant speed, but the pressure in the top segment (pr) ishigher than the pressure in the bottom segment (pl).

The fluid is incompressible, so (assuming a constant cross-section) there can beno variation in the velocity or density of charge carriers. Thus, it cannot be thecase that the charges in the top segment move faster than the charges in the bottomsegment. For this current model, there is no variation in γ, so some other effect mustbe responsible for the hidden momentum.

The electric field causes the pressure to be higher in the top segment than in thebottom segment—and here we find our answer. A moving fluid under high pressurehas more momentum than the same fluid under lower pressure, even if the two fluidshave the same mass and velocity.8

7I find this model more satisfying than the incoherent gas model for a few reasons: (1) Theincompressible fluid model is analogous to the case of a current created by spinning a chargedobject, such as a uniformly charged sphere. (2) There is no problem with charge densities (seefootnote 6). (3) We do not ignore inter-particle interactions, the way we did for the other currentmodel.

8This is a relativistic effect, just like the hidden momentum mechanism for the free charge model;that’s why it doesn’t show up if we use the classical momentum formula P = mv.

2.4. Incompressible Fluid 25

2.4.2 Why There is Momentum Associated with Pressure

Consider this simple system (Figure 2.3): a box contains two masses m each moving atspeed u.9 They bounce back and forth off the walls, hitting the sides simultaneouslyand meeting periodically at the center of the box.10 The total momentum of thesystem is obviously zero (P = 0).

u

0

SA

S’v m mu

L

Figure 2.3: Box containing two bouncing masses.

Each time a mass hits the side of the box, it transfers momentum ∆P = 2γumuto the wall. The wall is struck once every time interval ∆t = L0/u. Thus, theaverage force on the wall is F = ∆P/∆t = 2γumu

2/L0, and the average pressure isp = 2γumu

2/L0A, where A is the area of the wall.Now consider the same system from the perspective of frame S ′, which travels to

the left at speed v. The two masses still meet periodically in the center of the box,but they no longer travel at the same speed. The left-going mass has velocity

ul =v − u1− uv

c2

(2.21)

and the right-going mass has velocity

ur =v + u

1 + uvc2

. (2.22)

9The argument in this section is due to Jagannathan [9]. Comay [10] runs a very similar argument.I am using Jagannathan’s two-mass setup because it has a lovely simplicity, and makes the basicmechanism of the pressure-momentum clear.

10We are actually imagining a box containing many particles bouncing around in all directions(they are the particles of our fluid—a gas, in this case). To see what’s going on, it suffices to considerjust two particles. We will take time averages for the two-particle system, which is equivalent toaveraging over all the particles in a many-particle system. Also, the two particles travel in just onedirection; in general, the velocities could be in any direction, but we can run the same argument usingjust the x-components (this will be the direction of our Lorentz transformation); the momentum inthe other directions averages to zero.


The total momentum, then, is

P ′ = γlmul + γrmur

= m

v − u(1− uv

c2

)√1−

(u−vc−uv

c

)2+

v + u(1 + uv

c2

)√1−

(u+vc+ uv

c

)2

= m

v − u√(1− uv

c2

)2 −(u−vc

)2+

v + u√(1 + uv

c2

)2 −(u+vc

)2

= m

v − u√1 +

(uvc2

)2 −(uc

)2 −(vc

)2+

v + u√1 +

(uvc2

)2 −(uc

)2 −(vc

)2

=

2mv√1−

(uc

)2√

1−(vc

)2= 2γuγvmv. (2.23)

In this frame, the volume of the box is V ′ = AL = AL0/γv, so the momentumdensity of the system is

g =γ2v2γumv

AL0

= γ2vρv, (2.24)

with ρ = 2γum/AL0, the energy density (over c2) in frame S. This does not dependon pressure. However, that’s not the end of the story. In frame S ′, the two masses donot reach the ends of the box simultaneously—the one on the left changes directionfirst.11 For a short time, both masses move to the right, thereby increasing theaverage momentum of the system! This means that there is a small correction tothe momentum of Equation (2.24). We will find that correction term.

Suppose that the spacetime origin is at the center of the box, when both massesare there. Consider the following events:

Event 1: mass hits left wall

x1 = −L0

2x′1 = γv(vt1 + x1) = γv

L0

2

(vu− 1)

(2.25)

t1 =L0

2ut′1 = γv

(t1 +

v

c2x1

)= γv

L0

2u

(1− uv

c2

)(2.26)

Event 2: mass hits right wall

x2 =L0

2x′2 = γv(vt2 + x2) = γv

L0

2

(vu

+ 1)

(2.27)

t2 =L0

2ut′2 = γv

(t2 +

v

c2x2

)= γv

L0

2u

(1 +

uv

c2

)(2.28)

11Events that are simultaneous in one reference frame are not necessarily simultaneous in a dif-ferent frame.


Event 3: two masses meet in the center of the box

x3 = 0 x′3 = γv(vt3 + x3) = γvL0

uv (2.29)

t3 =L0

ut′3 = γv

(t3 +

v

c2x3

)= γv

L0

u(2.30)

One period of the motion (in S ′) is T = t′3 = γvL0/u. The short time betweencollisions is t′2 − t′1 = γvL0v/c

2. The momentum during the times when both massesmove to the right is

2γrmur = 2γuγvm(v + u), (2.31)

which is greater than the momentum in Equation (2.23) by 2γuγvmu. Thus, the extramomentum, averaged over one cycle, is

Pextra = 2γuγvmu

(t′2 − t′1T

)= 2γumu

2γvv

c2. (2.32)

The average additional momentum density is

Pextra

AL0/γv=

2γumu2

L0Aγ2v

v

c2= pγ2

v

v

c2, (2.33)

which is proportional to the pressure. Here, then, is the relationship between pressureand momentum. In a moving reference frame, the non-simultaneity of collisions re-sults in a pressure-related correction to the momentum of the fluid, so that the totalmomentum density for the fluid is

gfluid = γ2vρv + pγ2

v

v

c2; (2.34)

the first term is related to mass and the second is related to pressure.12

2.4.3 Return to the Rectangular Loop

We have seen that the (relativistic) momentum of a fluid has a term associated withpressure. In Section 2.4.2, we looked at a kinetic type of pressure. However, wewere originally (Section 2.4.1) talking about an incompressible fluid running througha rectangular loop—it would be more appropriate to use a model of pressure that iselastic rather than kinetic, because a gas is not incompressible. I don’t know how tomake a simple argument justifying the existence of momentum due to pressure for anelastic system. What we can do is take a look at the stress tensor for an ideal fluid,which is well known:13

T µν = ρηµην + p

(ηµην

c2+ gµν

), (2.35)

12Note that the pressure-related term is of order 1/c2, much smaller than the mass-related term.13Hnizdo [11] defines ideal fluids nicely: they have negligible viscosity and thermal conductivity,

and cannot support any shear stresses. The ideal fluid stress tensor is presented in section 67 of [12]and section 10.2 of [13].


where ηµ = γ(c,v) is the four-velocity of the fluid, ρ is the rest-frame mass densityand gµν is the metric tensor, with diagonal entries (−1, 1, 1, 1). Recalling that themomentum density is contained within the stress tensor (gi = T 0i/c), we see that

g =1

c

(ργ2cv + p

γ2cv

c2

)= ργ2v + pγ2 v

c2, (2.36)

precisely the same as Equation (2.34).Now that we know how to relate momentum to pressure (we can use the momen-

tum density from the fluid stress tensor, which we checked for a very simple example),we can return to the rectangular loop and show that this effect results in the correctamount of hidden momentum to balance the electromagnetic momentum.

Now, we don’t know what the pressure in the loop is. What really matters is howmuch the pressure changes between the top of the loop and the bottom. The massdensity and speed of the fluid are the same throughout the rectangular loop, so thefirst term of the momentum (which does not depend on pressure) integrates to zero.The second term (which does depend on pressure) results in a greater right-goingmomentum in the top segment of the loop and a smaller left-going momentum in thebottom segment. The sides cancel, just as for the incoherent gas model. Thus, thetotal mechanical momentum of the fluid in the loop is

Phidden = (pr − pl) γ2 v

c2Al, (2.37)

where A is the cross-sectional area of the tube, and l is the length of the top andbottom segments. In this scenario, (pr−pl) plays the same role as the (γr−γl) in theother current model. As we did then, we will figure out what the hidden momentumis by relating the changing quantity (in this case, the pressure) to the external electricfield.

The electric field exerts a force on the fluid in the side segments; it is this forcewhich causes the pressure to be higher at the top than the bottom. The total forceon one of the side segments is

F = QE = ρAwE, (2.38)

where ρ is now the charge density, w is the height of the loop, and A is, as before, thecross-sectional area of the tube. Now, this force must be balanced by the differentialpressure on the two ends of the loop. One would expect

F = ρAwE = (pr − pl)A, (2.39)

so that the total force on the segment is zero, but that isn’t quite right . . . the actualrelation is

F = ρAwE = γ2(pr − pl)A. (2.40)

How can this be? Doesn’t it mean that the total force on the fluid (due to thepressure and the electric field) in the left segment is not zero? Yes! This is exactly asit should be: although that section of fluid is not accelerating, its (hidden) momentum


is increasing, because the pressure on the fluid increases as it moves up the left side.This means that there is a force on the fluid, because F = dP/dt.14

Where does the factor of γ2 in Equation (2.40) come from? We can answer thatby thinking about a section of fluid moving with speed v through a long straight tube.The tube is oriented in the direction of a uniform electric field (just like the left sideof our rectangular current loop). Mark off a length x of tube in the lab frame S. The

v

S S’v

E

A

x

Figure 2.4: Section of long tube containing charged incompressible fluid moving tothe right with speed v.

electric field causes the pressure at the right end of this section p2 to be greater thanthe pressure at the left end p1. The difference (p2 − p1) ought to be related to theelectric field E. In the fluid’s rest frame S ′, the total force on the chunk of fluid15 iszero:

(p′2 − p′1)A′ = ρ′A′x′E ′ =⇒ p′2 − p′1 = ρ′x′E ′. (2.41)

Pressure is invariant under Lorentz transformations,16 so p′2 = p2 and p′1 = p1. Theelectric field is oriented in the direction of the transformation, so it is also unchanged:E ′ = E.17 That leaves us with two quantities that are different in the lab frame andthe fluid rest frame: the length of the segment and the charge density of the fluid.To move the equation back to the lab frame (which is most natural for us), we needto know how the charge density and the length of the segment transform. Because

14In Vaidman’s discussion of the incompressible fluid model [3], he does not include the factor ofγ2 in Equation (2.40). In order to get the correct hidden momentum result, he works only to firstorder in 1/c (so that γ2 = 1). Making this sort of approximation is suspicious, because the hiddenmomentum itself is a relativistic phenomenon. Hnizdo’s treatment of the same model in [8] and [11]correctly accounts for the γ2.

15Anyone who objects to calling a volume of fluid a “chunk” is welcome to use “wee bit” instead.16Møller [12] asserts this in section 67. Parallel to the transformation, F ′ = F and A′ = A.

Perpendicular to the transformation, F ′ = F/γ and A′ = A/γ. Thus, the pressure in all directionsis the same in both frames (p′ = F ′/A′ = F/A = p).

17Components of electric and magnetic fields parallel to the direction of motion are unchanged byLorentz transformation; see, for example, Griffiths [2] section 12.3.2.


of length contraction, x′ = x/γ. The charged fluid is at rest in frame S ′, so thecharge density transforms as ρ′ = ρ/γ (also because of length contraction). Thesetwo contractions together, one of the tube and the other of the fluid, conspire to giveus a factor of γ2:

p2 − p1 =1

γ2ρxE. (2.42)

In the same manner, the pressure difference between the top and bottom of therectangular loop is

pr − pl =1

γ2ρwE. (2.43)

Thus, from Equation (2.37), the total hidden momentum in the loop is

Phidden =

(1

γ2ρwE

)γ2 v

c2Al =

1

c2(ρAv) lwE. (2.44)

Now, ρA is the charge per length of the tube, so ρAv = I, the current. Thus, we have

Phidden =1

c2(m× E), (2.45)

just as for the other current model (Equation 2.13).

2.4.4 General Derivation

Building off the simple example of a rectangular current loop in a uniform electricfield, we can derive a general expression for the hidden momentum of any localizedsystem with the current modeled as an incompressible charged fluid.

Assume that the current is due to a charged fluid running through constant cross-section tubes. It is an incompressible fluid, so the charge density and speed are thesame throughout. Recall that the momentum density (from the stress tensor) is

g = ργ2v + pγ2 v

c2. (2.46)

We will integrate this around a closed tube of current. The first term integratesto zero (because ρ and γ are constant), so we need only concern ourselves with thesecond term (the one associated with pressure).18

Now, we will relate the pressure to the electric field, much as we did for therectangular loop. This time, however, we won’t require that the electric field beuniform or that our current tube be oriented parallel to the electric field. In the restframe of the fluid, the total force on each infinitesimal chunk of fluid must be zero.Thus, with f equal to the external force on the fluid—that is to say, the force due tothe electromagnetic field and not the internal pressures of the fluid—

f ′ = ∇′p. (2.47)

18Incidentally, it was the first term which gave us the hidden momentum for the other currentmodel, because of the variations in γ throughout the current distribution. There was no pressure inthat model, so the second term vanished.


As before, the primed frame is the rest frame of the chunk of fluid and the unprimedframe is the lab frame. The components of f and ∇p which are perpendicular to thefluid flow are the same in both frames. The components in the direction of the fluidflow are a different matter. Let’s call that the x-direction. We have already said thatthe force is zero in the fluid rest frame:

f ′x =∂p

∂x′. (2.48)

Now, we can transform this equation to the lab frame, where the net force is not zero.The force density is fx = ρEx, so it transforms the same way as the charge density(because Ex = E ′x): fx = γf ′x. For the transformation of ∂p/∂x, use the chain rule:

∂p

∂x′=∂p

∂x

∂x

∂x′+∂p

∂t

∂t

∂x′. (2.49)

In the lab frame, pressures don’t change in time, because the current is steady andthe fields are static, so ∂p/∂t = 0. Therefore,

∂p

∂x′=∂p

∂x

∂x

∂x′= γ

∂p

∂x. (2.50)

Now we are prepared to transform Equation (2.48) into the lab frame. We get

fx = γ2 ∂p

∂x, (2.51)

which contains the now-familiar factor of γ2 (as in Equation 2.40).The electric potential at some point along the tube is

V = −∫

E · dl (2.52)

(we can choose to make V = 0 at the place where we start the integral—it is only thedifference in potential that matters). The integrand E · dl picks out the componentof the electric field in the direction of the fluid flow. Because f = ρE, Equation (2.52)becomes

V = −∫

1

ρ

(γ2∂p

∂l

)dl. (2.53)

The charge density and fluid speed are constant along the tube, so ρ and γ2 can bepulled out of the integral, leaving us with

p = − ρ

γ2V + p0, (2.54)

where p0 is the reference pressure (at the place where V = 0). Only differences inpressure matter, so we can ignore the second term (it will disappear when we plugthe pressure into Equation (2.46) and integrate). Let’s do that now:

Phidden =γ2

c2

∫pv dτ =

γ2

c2

∫ (− ρ

γ2V

)v dτ = − 1

c2

∫V J dτ, (2.55)


which is precisely the result we were looking for; the hidden momentum exactlycancels the electromagnetic momentum in Equation (1.28).

I explained the hidden momentum mechanism for two different current models: theincoherent gas model (in which charged particles move freely through a tube, withoutinteracting with each other) and the incompressible fluid model (in which chargedliquid flows through a tube). We have seen that a steady current in an external electricfield can have net mechanical momentum, and that this momentum exactly balancesthe electromagnetic momentum. The hidden mechanical momentum exists becausethe electric field causes variations in particle velocity (gas model) or pressure (fluidmodel), resulting in one side of a current loop having more (relativistic) momentumthan the opposite side, so that the total mechanical momentum in the loop is non-zero.

Chapter 3

Expanding the Definition:What is Hidden Momentum?

In Chapter 1, I explained why hidden momentum must exist: A system with sta-tionary center of energy must have zero total momentum. Static electromagneticsystems (composed of stationary charges and steady currents) fall into this category,but they can also have non-zero electromagnetic momentum, as we saw in Section1.3. Therefore, in such cases, the system must have some other, non-electromagneticmomentum. Shockley and James [1] called it “hidden momentum,” and the namestuck.

In Chapter 2, I explained the nature of the hidden momentum: It is mechanicalmomentum located in the currents. The precise mechanism by which a steady currentcan have non-zero mechanical momentum depends on how we model the current. Iexplained the mechanism for two models: (1) current as non-interacting gas particles,in which variations in particle speed result in the hidden momentum and (2) current asincompressible fluid, in which variations in pressure result in the hidden momentum.In both cases, the mechanical momentum in the current is equal and opposite to theelectromagnetic momentum, as required by the Center of Energy Theorem.

With this understanding, we are in a position to say some general things abouthidden momentum. As a name, “hidden momentum” has a certain charm to it . . . butit is also somewhat misleading, because it gives the impression that the mechanicalmomentum is somehow more mysterious or harder to understand than the electromag-netic momentum. There is a symmetry here which we obscure by calling one type ofmomentum the “hidden” momentum. The systems we have looked at have two typesof momentum—mechanical and electromagnetic—which are separately non-zero buttogether must be zero. Both types of momentum are perfectly ordinary, and once wesee what’s going on, neither is more hidden than the other. Nevertheless, there arereasons why “hidden momentum” is a fitting name for the mechanical momentum:

• It is relativistic.

• It is small.1

1This goes along with it being relativistic. Comay [10] points out that the hidden momentum isalways of order 1/c2 (so is the electromagnetic momentum, for that matter).

34 Chapter 3. Expanding the Definition: What is Hidden Momentum?

• It is unexpected. Here we have non-zero mechanical momentum in a systemwhich is not moving. In this sense, there is mechanical momentum hidden in astationary system.

But we’ve only looked at a narrow class of examples, and our sense of what hiddenmomentum is is firmly tied to those examples. Can there be hidden momentum insituations other than the static electromagnetic systems we have discussed? If so,how will we recognize it? It depends on how we define hidden momentum. That isthe question I turn to in this final chapter: What is hidden momentum?

3.1 Gravitational Systems

In Chapter 2, we saw that a current loop can have net mechanical momentum whenan external force acts on the moving charges. However, there is no reason why theforce has to be electromagnetic—a gravitational force can cause variations in particlespeed or fluid pressure just as well as an electric force.

We can build systems out of charges and charge currents which have hidden me-chanical momentum. Analogously, we can build systems out of masses and masscurrents, which also have net mechanical momentum—surely it makes sense to call ithidden momentum as well, because it occurs for the same reason as the hidden mo-mentum in the electromagnetic systems. Hidden momentum first attracted attentionin the context of electrodynamics, but the hidden momentum itself has nothing todo with electrodynamics.

Interestingly (and perhaps unexpectedly), this means that static gravitationalfields can carry momentum, just as electromagnetic fields have momentum. We beganthis whole business with the knowledge that electromagnetic fields carry momentum;from that, we knew that hidden momentum must exist (so that the total momentumwould be zero). Now, we can run the same argument the other way: We know thatthere can be net mechanical momentum in a gravitational system, because we havealready worked out the mechanisms. The total momentum must be zero, so there hasto be some other, non-mechanical momentum in the system. The only possibility isthat it resides in the gravitational field.2

3.2 Moving Systems

Thus far, we have only considered systems at rest, for which the total momentum iszero. We found that when a system contains both matter and fields (be they electro-magnetic or gravitational) the momenta of those two subsystems can be separatelynon-zero—this is the essence of the hidden momentum phenomenon. In these sta-tionary systems, the hidden momentum is easy to spot: it is the total mechanicalmomentum.

2Hnizdo [11] discusses hidden momentum in gravitational systems, and the existence of a “gravi-netic” field, the gravitational analog to the magnetic field.

3.2. Moving Systems 35

Now, in a moving system the total momentum is (of course) not zero. If a systemis isolated (that is, the stress-energy tensor is divergenceless) then the total energyand momentum of the system fit together into a four-vector: P µ = (U/c,P). Thus,if we know the total energy U of the system in its rest frame (in which the center ofenergy is at rest, and therefore the momentum P = 0), we can find the total energyand momentum in a moving frame using the Lorentz transformation. If the centerof energy of the system is at rest in frame S and moves with velocity v in frame S ′,then

U ′ = γU + γβPc = γU (3.1)

and

P ′ = γβ

(U

c

)+ γP = γ

v

c2U =

v

c2U ′, (3.2)

where P is the magnitude of the momentum (it is in the direction of v). Note thatthis is exactly what the Center of Energy Theorem (Equation 1.74) says.

So, the question is: how do the momenta of the separate subsystems compare tothe total momentum? And what, if anything, is the hidden momentum?

3.2.1 Example: Moving Capacitor

dy

z

E

v

S

S’

x

σ −σ

Figure 3.1: Parallel plate capacitor, with charge density σ and plate separation d.The plates are kept apart by a gas under pressure.

Consider a parallel plate capacitor with a gas under pressure filling the spacebetween the plates (see Figure 3.1).3 The two plates have charge densities of σ and−σ, respectively, and are separated by a distance d. The electric field between theplates is E = σ/ε0. The (attractive) electric force per unit area on one plate due tothe other plate is σE/2. This force must be balanced by the pressure of the fill-gas,so the pressure is

p = σE

2=

1

2ε0E

2. (3.3)

3To the best of my knowledge, Comay [10] was the first person to extend the ideas of hiddenmomentum to moving systems. The moving-capacitor example is his, and he works it out in moredetail than I will do here. He presents the full stress-energy tensor for each of the subsystems (field,plates and gas) and transforms each of them to a moving frame. I will pull out just a few parts, inorder to highlight what is going on.


The total system contains the charged plates as well as the electric field and gasin the space between the plates; we take the volume of the capacitor as the volumeof the system.4 While the stress-energy tensor for the entire system is divergenceless,the stress-energy tensors for the individual subsystems need not be.

First look at the system in frame S, the rest frame of the capacitor: The energydensity of the electric field is

uem =ε02E2, (3.4)

and the energy density of the gas is

ugas = ρc2, (3.5)

so the total energy contained within our volume of interest is

U =(ε0

2E2 + ρc2

)Ad = (p+ ρc2)Ad, (3.6)

where p is the pressure of the gas and A is the area of each plate. The total momentumis zero (as it must be, by the Center of Energy Theorem). We can use Equations(3.1) and (3.2) to find the energy and momentum of the system in frame S ′, in whichthe capacitor moves to the right (in the direction of the electric field) at speed v:

U ′ = γ(p+ ρc2)Ad (3.7)

andP ′ = γ

v

c2(p+ ρc2)Ad. (3.8)

You might wonder what this has to do with hidden momentum. At first glance,this system is quite different from the hidden momentum systems we looked at before;the capacitor at rest has no currents and no magnetic field. There is neither elec-tromagnetic momentum nor net mechanical momentum. However, in frame S ′ thesystem contains a moving fluid under pressure, which is familiar from our discussionof hidden momentum mechanisms in Chapter 2.

Let’s look at the energy and momentum of the electric field and gas separately.Naively, we might expect the energy/momentum of the field and the energy/momentumof the gas to transform independently as four-vectors:

Naive way of dividing the total energy and momentum

U ′em = γpAd U ′gas = γρc2Ad U ′total = γ(p+ ρc2)Ad

P ′em = γv

c2pAd P ′gas = γρvAd P ′total = γ

v

c2(p+ ρc2)Ad. (3.9)

That’s not what happens, as we can see by calculating U ′em, P ′em, U ′gas and P ′gas. Theelectric field points in the direction of the transformation, so it is the same in bothframes: E ′ = E. This means that the energy density in the moving frame is the same

4We are really looking at one section of an infinite parallel plate capacitor, so the electric field isentirely contained between the plates.

3.2. Moving Systems 37

as the energy density in the rest frame. However, the distance d is contracted in themoving frame (d′ = d/γ), so the total energy in the field is

U ′em = u′A′d′ = uAd

γ=

1

γpAd. (3.10)

As for the electromagnetic momentum, there’s still no magnetic field in frame S ′, soP ′em = 0. To find the correct energy and momentum of the moving gas, we can turnto the fluid stress tensor, Equation (2.35):

u′gas = γ2ρc2 + γ2β2p (3.11)

g′gas = γ2ρv + γ2 v

c2p. (3.12)

The volume in frame S ′ is Ad/γ, so

U ′gas = (γρc2 + γβ2p)Ad (3.13)

P ′gas = (γρv + γv

c2p)Ad. (3.14)

Thus, the corrected version of Equation (3.9) is

Correct way of dividing the total energy and momentum

U ′em =1

γpAd U ′gas = γ(β2p+ ρc2)Ad U ′total = γ(p+ ρc2)Ad

P ′em = 0 P ′gas = γv

c2(p+ ρc2)Ad P ′total = γ

v

c2(p+ ρc2)Ad. (3.15)

The electric field has less than its “share” of the total momentum and energy, and thegas has the extra, so that the total is what the total needs to be. The “extra” termsin the mechanical momentum of the gas are precisely those due pressure, which, aswe have already seen, are responsible for the hidden momentum in the incompressiblefluid model of a current. In this moving system, it is reasonable to identify this extramechanical momentum as hidden momentum. That is, the difference between whatthe mechanical momentum actually is and what it would be if it transformed like afour-vector (which it would if not for the presence of the electric field) is

P ′hidden = γv

c2pAd. (3.16)

In a similar way, we can identify the hidden energy as

U ′hidden = γβ2pAd. (3.17)

3.2.2 Tentative Definition of Hidden Momentum

The Center of Energy Theorem (Equation 1.74) tells us how much momentum isassociated with the movement of the center of energy of a system:

P =U

c2

dR

dt, (3.18)


where dR/dt is the velocity of the center of energy. An isolated system always hasprecisely this amount of momentum. As we saw in Equation (3.2), this is the sameas the statement that the energy and momentum of a system fit together into a four-vector. However, a non-isolated system (such as one subsystem of a larger system),need not obey the Center of Energy Theorem. It might have slightly more or slightlyless momentum than the “expected” amount from Equation (3.18). In Section 3.2.1,I identified the hidden momentum as the difference between the actual momentumand the expected four-vector momentum. This brings me to a possible definition forhidden momentum which applies to systems in motion as well as systems at rest:

Hidden momentum is mechanical momentum which is not associated withmovement of the center of energy of a system.

Does this match our original notion of hidden momentum? Yes. For a stationarysystem, the center of energy is not moving, and the hidden momentum is simply thetotal mechanical momentum.

3.2.3 Two Electromagnetic Subsystems

After discussing the moving parallel plate capacitor of Section 3.2.1, Comay [10] goeson to describe a similar situation: this time, instead of a gas exerting the pressurenecessary to hold the plates of the capacitor apart, there are standing electromagneticwaves. This is a situation in which there is no mechanical momentum at all, onlyelectromagnetic momentum. The electric field of the capacitor is the same as before,and transforms in the same manner; the role that was filled by the gas is now taken onby the standing waves. Does this mean that the electromagnetic waves have hiddenmomentum when the system is moving? There are two possible answers: (1) No,an electromagnetic wave can’t have hidden momentum because hidden momentum isalways mechanical. (2) Yes, the electromagnetic waves fill the same role as the gas, soif the gas had hidden momentum then so do the standing waves. If we take this view,we will have to give up on the idea that hidden momentum is always mechanical.

3.2.4 Two Mechanical Subystems

We can also create a system which has two mechanical subsystems, each with mo-mentum and energy that do not transform as a four-vector. For example, suppose wehave a box filled with a gas under pressure (instead of a capacitor filled with gas);in a moving frame, the gas has a little bit of “extra” momentum, as we saw in Sec-tion 3.2.1; hence, the box must have a little bit less momentum than the “expected”amount. The side walls of the box are under tension; evidently, this acts like negativepressure and slightly reduces the momentum of the body.

Conclusion

When I began this project, I had several goals in mind. First, I aimed to understandhidden momentum: why it must exist, where it is located, and how it works. I wantedto gather all known examples, and write a comprehensive summary of the work thathas been done on the topic in the past forty years, so that everything would be easilyaccessible in one place, written with consistent notation and a systematic approach.Further, I wanted to characterize hidden momentum and provide a clear definition,so that there would no longer be confusion about the subject. What exactly is it thatmakes hidden momentum what it is? What counts as hidden momentum and whatdoesn’t?

While I certainly understand much more about hidden momentum than I did whenI began, I remain uncertain about the definition. It turns out that hidden momentumis not always easy to pin down, particularly in moving systems.

The unifying idea which I take away from the thesis is this: The momentumof an isolated system (for which the stress-energy tensor is divergenceless) alwayscorresponds to motion of the center of energy of the system. However, the samecannot be said for a subsystem of a larger system; sometimes one subsystem will havemore momentum than its “expected” momentum, and the other will have less. It isthe difference between the actual momentum and the expected momentum (expectedbecause it is related to the movement of the center of energy of the system) which isin some cases called hidden momentum.

References

[1] W. Shockley and R. P. James, “ ‘Try simplest cases’ discovery of ‘hidden mo-mentum’ forces on ‘magnetic currents,’ ” Phys. Rev. Lett. 18, 876 (1967).

[2] D. J. Griffiths, Introduction to Electrodynamics (Prentice Hall, 1999), 3rd ed.

[3] L. Vaidman, “Torque and force on a magnetic dipole,” Am. J. Phys. 58, 978(1990).

[4] D. Babson, S. P. Reynolds, R. Bjorkquist, and D. J. Griffiths, “Hidden mo-mentum, field momentum, and electromagnetic impulse,” Am. J. Phys., to bepublished.

[5] R. H. Romer, “Question #26. Electromagnetic field momentum,” Am. J. Phys.63, 777 (1995).

[6] S. Coleman and J. H. Van Vleck, “Origin of ‘hidden momentum forces’ on mag-nets,” Phys. Rev. 171, 1370 (1968).

[7] R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics(Addison-Wesley, 2006), definitive ed.

[8] V. Hnizdo, “Hidden momentum of a relativistic fluid carrying current in anexternal electric field,” Am. J. Phys. 65, 92 (1997).

[9] K. Jagannathan, “Momentum due to pressure: A simple model,” Am. J. Phys.77, 432 (2009).

[10] E. Comay, “Exposing ‘hidden momentum,’ ” Am. J. Phys. 64, 1028 (1996).

[11] V. Hnizdo, “Hidden mechanical momentum and the field momentum in station-ary electromagnetic and gravitational systems,” Am. J. Phys. 65, 515 (1997).

[12] C. Møller, The Theory of Relativity (Oxford University Press, 1952).

[13] R. Adler, M. Bazin, and M. Schiffer, Introduction to General Relativity (McGraw-Hill, 1975), 2nd ed.

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