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arXiv:physics/0612202v1 [physics.class-ph] 21 Dec 2006
Relativistic momentum and kinetic energy, and E=mc2
Ben Yu-Kuang Hu∗
Department of Physics, University of Akron, Akron, OH 44325-4001.
(Dated: February 2, 2008)
Abstract
Based on relativistic velocity addition and the conservation of momentum and energy, I present
derivations of the expressions for the relativistic momentum and kinetic energy, and E=mc2.
1
I. INTRODUCTION
The standard formal way that expressions for the relativistic momentum pand the rela-
tivistic kinetic energy T, and the mass–energy relationship E=mc2are derived in upper-
level undergraduate textbooks1is by first introducing Lorentz transformations and 4-vectors,
and then defining the 4-momentum vector pµ=m dxµ/dτ (µ= 0,1,2,3), where τis the
proper time. It is then postulated, backed up by extensive experimental observations, that
in an isolated system all the components of pµare conserved. The spatial components of
pµreduce to mvin the non-relativistic limit, and hence correspond to the components of
the relativistic momentum. The temporal component reduces in the non-relativistic limit
to mc2+mv2/2, and therefore is identified as the total energy, composed of the rest-mass
and kinetic energies. Logically, there is of course nothing wrong with this approach. Ped-
agogically, however, it is probably helpful to have more intuitive derivations. To this end,
through the years many have been published.2
This paper describes relatively simple and concise derivations of the relativistic forms of
pand T, and E=mc2, based on (i) conservation of momentum and energy in the collisions
of two particles, and (ii) the velocity addition rules. Momentum and energy conservation
should be familiar concepts to students, and the velocity addition rules can be quite simply
derived from the constancy of the speed of light in all inertial reference frames.3In each
derivation, collisions are viewed in the center-of-momentum frame of reference, Scm, in which
both particles have momenta that are equal in magnitude and opposite in direction, and
in the laboratory frame of reference, Slab , in which one of the particles initially is at rest.
Imposition of the conservation laws gives the desired expressions.
To simplify the algebra, velocities in this paper are expressed in units of c, the speed of
light. Hence velocities are dimensionless, and c= 1. To obtain the standard dimensional
expressions, replace all velocities in the expressions given here by v→v/c, and multiply all
masses by c2in order to obtain energy. Also, in this paper primes on variables denote “after
collision.”
2
II. THE DERIVATIONS
First, let us recall the relativistic velocity transformation rules. Let ˜
Sbe an inertial
frame moving with velocity (u, 0) with respect to frame S. If a particle has velocity (vx, vy)
in frame S, the components of its velocity in frame ˜
Sare1,3
˜vx=vx−u
1−vxu,(1a)
˜vy=vy√1−u2
1−vxu.(1b)
A. Relativistic momentum
From dimensional analysis and the vector4nature of momentum, the momentum of a
particle of mass mtravelling with velocity vmust have the form
p=mγvv,(2)
where γvis an unknown function to be determined. Since p=mvfor non-relativistic
velocities, γv→0= 1.
Consider the case where the particles are identical, hence m1=m2=m. Let the motion
of the particles be in the x–yplane and their initial velocities in Scm be ±(v, 0). Assume
that the particles barely graze each other, so that in the collision each particle picks up
a very small y-component of the velocity of magnitude δv in Scm. [See Fig. 1(a).] Their
speeds in Scm do not change because the collision is elastic, and hence their velocities after
the collision are ±(pv2−(δv)2, δv) = ±(vp1−(δv/v)2, δv)≈ ±(v, δv), to first order in
δv. Because δv is assumed to be very small, we ignore all terms of order (δv)2and higher.
Now consider the collision in the laboratory frame of reference Slab that is moving with
velocity (−v, 0) with respect to Scm . [See Fig. 1(b).] The pre-collision velocities of the
particles Slab, using Eqs. (1) on the Scm velocities ±(v, 0), are v1,lab = (w, 0), where
w=2v
1 + v2,(3)
and v2,lab = (0,0). After the collision, transforming the post-collision Scm velocities ±(v, δv)
3
to the Slab frame we obtain, to first order in δv,
v′
1,lab ≈w, δv √1−v2
1 + v2,(4a)
v′
2,lab ≈0,−δv √1−v2
1−v2.(4b)
The y-component of the total momentum before the collision is zero, and hence by con-
servation of momentum, after the collision
p′
1,lab +p′
2,laby=m γ|v′
1,lab|v′
1,lab,y +m γ|v′
2,lab|v′
2,lab,y = 0.(5)
Since v′
lab,y terms are of order δv and we are ignoring terms of order (δv)2, it is sufficient
in Eq. (5) to evaluate |v′
1,lab|and |v′
2,lab|to zeroth order in δv (i.e., ignoring δv altogether).
Substituting |v′
1,lab| ≈ w,|v′
2,lab| ≈ 0, and the y-components of the velocities from Eqs. (4a)
and (4b) into Eq. (5) and using γv′→0= 1 yields
γw
1 + v2−1
1−v2= 0.(6)
This, together with Eq. (3) gives
γw=1 + v2
1−v2= 1−2v
1 + v22!−1/2
=1−w2−1/2.(7)
B. Relativistic kinetic energy
Dimensional analysis and the scalar4property of kinetic energy imply that its form is
T=m G(v),(8)
where mis the mass of the particle, v=|v|is its speed and the function G(v) is to be
determined.
Consider an elastic head-on collision between two particles, of mass mand M≫m, with
speeds in Scm of vand V, respectively. In Scm , the particles simply reverse directions, and
the motion is one-dimensional. [See Fig. 2(a).] Assume that the mass Mis so large that
in frame Scm its speed V≪1, and hence we can use the non-relativistic expressions for the
momentum and kinetic energy of mass M. The magnitudes of the momenta of mand M
are equal in Scm, implying
mγvv=MV. (9)
4
The Scm frame pre- and post-collision velocities of mass mare vcm =vand v′
cm =−v
respectively, and of mass Mare Vcm =Vand V′
cm =−V, respectively. Transforming these
to the Slab frame, which is moving at velocity −Vwith respect to Scm [see Fig. 2(b)], gives
vlab = (v+V)/(1 + vV ), v′
lab = (−v+V)/(1 −vV ), Vlab = 0 and V′
lab = 2V/(1 + V2). By
conservation of kinetic energy in an elastic collision in the Slab frame and Eq. (8),
m G(|vlab|) = m G(|v′
lab|) + M
2|V′
lab|2.(10)
Expanding |vlab|,|v′
lab|and |V′
lab|to first order in V,
|vlab| ≈ (v+V)(1 −vV )≈v+V(1 −v2),(11a)
|v′
lab| ≈ (v−V)(1 + vV )≈v−V(1 −v2),(11b)
|V′
lab| ≈ 2V(1 −V2)≈2V, (11c)
and substituting these into the Taylor expansions of the Gterms about vin Eq. (10) gives,
to first order in V,5
mG(v) + dG(u)
du v
V(1 −v2)=mG(v)−dG(u)
du v
V(1 −v2)+ 2MV 2.(12)
Substituting 2MV 2= 2mγvvV [from Eq. (9)] into Eq. (12) leads to
dG
du v
=γvv
1−v2=v
(1 −v)3/2,(13)
which upon integration yields
G(v)−G(0) = 1
(1 −u2)1/2u=v
u=0
=γv−1.(14)
Since the kinetic energy vanishes when vis zero, G(0) = 0, and hence Eqs. (8) and (14)
imply that (reintroducing c)T=mγv−1c2.
C. E=mc2
Consider the initial situation as in Sec. II B, except that the speed Vof mass Mcan be
relativistic, and after collision the two particles merge into one composite particle. In Scm,
MγVV=mγvv, and after the collision the composite particle is stationary. In Slab which is
moving with velocity −Vwith respect to Scm , before the collision particle Mis stationary
5
and particle mmoves with velocity vlab = (v+V)/(1 + vV ), and after the collision the
composite particle moves with velocity V.
The total momentum in Slab before the collision is Plab =mγ|vlab|vlab =mγvγV(u+
V). If the mass of the composite particle does not change, then the momentum of the
composite particle after the collision in Slab would be (M+m)γVV6=Plab in general,
violating conservation of momentum. Therefore, the mass of the composite particle must
change by ∆msuch that momentum is conserved in Slab;i.e.,
mγvγV(v+V) = (M+m+ ∆m)γVV. (15)
Substituting mγvv=MγVVon the left hand side and cancelling γVVon both sides gives
∆m=m(γv−1) + M(γV−1).(16)
From Sec. II B, the right hand side of Eq. (16) is equal to −∆T, the total change in kinetic
energy in Scm (since particles mand Mstart with speeds vand V, respectively, and both
are stationary at the end). By conservation of total energy, ∆E+ ∆T= 0, where ∆Eis the
energy associated with the change in mass. Hence, ∆E=−∆T= ∆mor (reintroducing
c, and making the plausible assumption that a zero mass object with zero velocity has zero
energy) E=mc2. Finally, combining the results of Sections II B and II C gives the total
energy of a particle of mass mmoving with speed v,E+T=Etotal =mγvc2.
III. CONCLUDING REMARKS
It should be noted that these derivations do not guarantee that the momentum and total
energy are conserved in all inertial reference frames or in all collisions. They only show the
forms that the momentum, kinetic energy and energy–mass relation must have, given mo-
mentum and energy conservation. Once these expressions are known, when the 4-momentum
is introduced its components will be recognized as the total energy and momentum. The
covariance of the momentum 4-vector can then be used to demonstrate momentum and to-
tal energy conservation in all inertial frames. Conservation of momentum can be shown to
be a consequence of conservation of energy,6and, as befitting an experimental science, the
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conservation of energy ultimately depends on experimental observations.
∗On sabbatical leave at Department of Physics, University of Maryland, College Park, MD 20742-
4111.; Electronic address: yhu@umd.edu
1See e.g., David J. Griffiths, Introduction to Electrodynamics, 3rd ed. (Prentice-Hall, Upper Sad-
dle River, NJ, 1999); John R. Taylor, Classical Mechanics (University Science Books, Sausalito,
CA, 2005).
2See, e.g., G. L. Lewis and R. C. Tolman, “The principle of relativity, and non-Newtonian me-
chanics,” Philos. Mag. 18, 510–523 (1909); R. Penrose and W. Rindler, “Energy conservation
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3N. David Mermin, “Relativistic addition of velocities directly from the constancy of the ve-
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4Here, the terms “vector” and “scalar” are used in the non-relativistic (i.e., not the 4-vector)
7
Figures
FIG. 1: Grazing collision between two particles of equal mass, in (a) center-of-momentum and
(b) laboratory frames of reference. Dashed and solid lines indicate before and after the collision,
respectively.
FIG. 2: Head-on collision between particles of mass mand M≫m, in (a) center-of-momentum and
(b) laboratory frames of reference. Dashed and solid lines indicate before and after the collision,
respectively.
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