A pedestrian approach to Einstein's formula $E=mc^2$ with an application to photon dynamics

Abstract

There are several ways to derive Einstein's celebrated formula for the energy of a massive particle at rest, E=mc2E=mc^2. Noether's theorem applied to the relativistic Lagrange function provides an unambiguous and straightforward access to energy and momentum conservation laws but those tools were not available at the beginning of the twentieth century and are not at hand for newcomers even nowadays. In a pedestrian approach, we start from relativistic kinematics and analyze elastic and inelastic scattering processes in different reference frames to derive the relativistic energy-mass relation. We extend the analysis to Compton scattering between a massive particle and a photon, and a massive particle emitting two photons. Using the Doppler formula, it follows that E=ℏωE=\hbar \omega for photons at angular frequency ω\omega where ℏ\hbar is the reduced Planck constant. We relate our work to other derivations of Einstein's formula in the literature.

Full text

A pedestrian approach to Einstein’s formula E=mc2with an application to photon
dynamics
A. V. Nenashev,1, 2, 3 S. D. Baranovskii,1, 4 and F. Gebhard1
1Department of Physics and Material Sciences Center, Philipps-University, D-35032 Marburg, Germany
2Institute of Semiconductor Physics, 630090 Novosibirsk, Russia
3Novosibirsk State University, 630090 Novosibirsk, Russia
4Department f¨ur Chemie, Universit¨at zu K¨oln, Luxemburger Strasse 116, 50939 K¨oln, Germany
(Dated: August 8, 2023)
There are several ways to derive Einstein’s celebrated formula for the energy of a massive particle
at rest, E=mc2. Noether’s theorem applied to the relativistic Lagrange function provides an
unambiguous and straightforward access to energy and momentum conservation laws but those tools
were not available at the beginning of the twentieth century and are not at hand for newcomers
even nowadays. In a pedestrian approach, we start from relativistic kinematics and analyze elastic
and inelastic scattering processes in different reference frames to derive the relativistic energy-mass
relation. We extend the analysis to Compton scattering between a massive particle and a photon,
and a massive particle emitting two photons. Using the Doppler formula, it follows that E=ℏω
for photons at angular frequency ωwhere ℏis the reduced Planck constant. We relate our work to
other derivations of Einstein’s formula in the literature.
I. INTRODUCTION
The energy-mass relation for a particle of mass mat
rest (c: speed of light),
E0=mc2(1)
is one of the most popular formulas in physics. It is the
basis for our understanding of the energy production by
fusion in stars and by fission in nuclear power plants. For
this reason, it is desirable to make it accessible to begin-
ners, not only at university level but preferably already
at high-school level. Consequently, physicists seek to pro-
vide an elementary derivation of the famous formula (1),
as are the actual titles of Einstein’s paper in 1935 [1] and
of Rohrlich’s paper in 1990 [2].
The notion of “elementary derivation” implies at least
three points.
1. all physical concepts are stated clearly;
2. notions extrinsic to mechanics, e.g., those from
electrodynamics or even quantum mechanics, are
avoided;
3. sophisticated mathematics are kept to a minimum.
Naturally, all derivations must be correct. This is
not always guaranteed, see the comment by Ruby and
Reynolds [3] who pointed out that the non-relativistic
Doppler formula used by Rohrlich is insufficient to derive
the relativistic relation (1). Moreover, the derivations
should be self-contained and not use short-cuts that are
justified only a-posteriori.
The second postulation appears impossible to meet be-
cause Einstein’s formula invokes the speed of light c. In
the preface to the celebrated textbook The Classical The-
ory of Fields by Landau and Lifshitz [4] where relativ-
ity and electromagnetism are treated in one volume, the
authors explicitly state that ‘A complete, logically con-
nected theory of the electromagnetic field includes the
special theory of relativity, so the latter has been taken
as the basis of the presentation.’ Likewise, Einstein’s
original considerations [5] in 1905 are not based on clas-
sical mechanics alone. Einstein addresses a process of
emitting electromagnetic waves by a massive object. He
concludes from the energy balance that when the object
loses some amount ∆Eof its energy, it simultaneously
loses the amount ∆m= ∆E/c2of its mass,
∆E0= ∆mc2.(2)
In his arguments, Einstein relies on Maxwell’s electrody-
namics, more precisely, to the results from the electro-
dynamic part of his famous work ‘Zur Elektrodynamik
bewegter K¨orper’ [6]. Also, the relativistic formula for
kinetic energy is obtained by Einstein from electrody-
namic arguments, namely, from equations of motion of a
charged particle in an electric field [6]. This little histor-
ical excursion demonstrates how closely the relativistic
theory is tied to the theory of electromagnetism.
Already in 1906, Planck recognized that the relativistic
dynamics fits into the framework of the principle of least
action [7]; according to Pais [8] this was most probably
the first work on Einstein’s special relativity not writ-
ten by Einstein himself. Then, in 1907, Minkowski gave
a talk in which he introduced four-vectors in spacetime
and showed that the kinetic energy of a massive body
is related to the temporal component of body’s four-
velocity [9]. Thus, very soon after the invention of special
relativity there appeared at least three ‘points of support’
that permit to derive relativistic dynamics (energy, mo-
mentum, etc.) from relativistic kinematics (time dilation,
length contraction, Lorentz transformations, etc.). These
‘points of support’ remain electrodynamics, the principle
of least action, Noether’s theorem [10], and vectors in
Minkowski spacetime.
arXiv:2308.02612v1 [physics.class-ph] 4 Aug 2023

2
Albeit of fundamental importance, the concepts intro-
duced by Planck and Minkowski are much more involved
than relativistic kinematics alone, and could be seen as
an obstacle for beginners who just want to understand
relativistic dynamics. One line of argument that avoids
electrodynamics, the principle of least action, Noether’s
theorem, and the notion of four-vectors, is based on the
analysis of particle collisions. Already in 1909, Lewis and
Tolman [11] used a collision argument to prove the rela-
tivistic expression for the momentum. Their proof is gen-
erally adopted in various textbooks, such as Spacetime
Physics by Taylor and Wheeler [12], The Feynman Lec-
tures on Physics [13], and, in a somewhat restricted ver-
sion, in the Berkeley Physics Course [14]. Interestingly,
even when they rely on particle collisions, the authors of
textbooks prefer to introduce relativistic formulas for mo-
mentum and energy ad hoc, and use thought experiments
with collisions only as supporting arguments. For exam-
ple, in Spacetime Physics [12] the authors postulate that
momentum and energy are just parts of the four-vector of
relativistic momentum, and write down the correspond-
ing expressions for them. Only at the end of the chapter,
as an exercise, they provide a derivation of the relativistic
momentum through a collision experiment. This reflects
a ‘Babylonian’ approach to physics [15, p. 47] rather than
the ‘Greek’ method in mathematics’ [16].
In this work we shall employ relativistic kinematics
and analyze two-particle scattering processes to derive
the energy-momentum relation that leads to Einstein’s
formula (1). Since we shall analyze scattering processes
with very simple geometries, we require first and second-
order Taylor expansion and the solution of simple first-
order differential equations as mathematical tools to turn
the Babylonian approach to an Euclidean one.
Our paper is organized as follows. In Sec. II we review
the properties of point particles in classical mechanics
and collect the Lorentz transformation and relativistic
Doppler formulas. In Sec. III, to set a point of reference,
we briefly derive the dynamics of a single particle us-
ing the principle of least action, and identify momentum
and energy using Noether’s theorem [4, 10]. In Sec. IV
we provide the pedestrian derivation of those formulas
using only basic concepts of classical mechanics as out-
lined in Sec. II applied to two-particle scattering. Using
the relativistic Doppler effect, we show in Sec. V from
particle-photon (Compton) scattering that a photon with
angular frequency ωhas the energy E=ℏω, where ℏis
the reduced Planck constant [17]. The same result can
be obtained from particle-antiparticle annihilation. Fur-
thermore, in Sec. VI, we briefly review other approaches
to derive Einstein’s formula. Short conclusions, Sec. VII,
close our presentations. Mathematical derivations are
deferred to three appendices.
II. KINEMATICS: POINT PARTICLE AND
LORENTZ TRANSFORMATION
Momentum and energy are concepts of particle dy-
namics. Before we address the equations of motions of
a single particle in Sect. III and derive Einstein’s for-
mula in Sect. IV, we first recall the basic concept of a
point particle in classical mechanics. Next, we collect
the Lorentz transformation formulas for the transforma-
tion of coordinates and velocities between two inertial
systems. Since we address photon dynamics in Sect. V,
we also collect the formulas for the relativistic Doppler
effect in the present section.
A. Point particle in classical mechanics
The first axiom of mechanics defines the setting of
space-time: space-time is four-dimensional, i.e., an event
is given by a point Pwith four coordinates (t, x, y, z) in
some (inertial) reference frame.
The second axiom in classical mechanics states that
a particle is at some spatial point r1at time t1and
arrives at some other spatial point r2at time t2> t1
whereby the world-line that contains all intermediate
points P(t) = (t1≤t≤t2,r(t)) is continuous and (at
least) twice differentiable with respect to the time t.
Kinematics describes the functional dependence of r(t)
on the time t.
B. Lorentz transformation
Apparently, kinematics requires the use of coordinate
systems. However, the choice of a reference system seems
to prefer one coordinate system over the other. The third
axiom of classical mechanics, the Galilean principle of
special relativity, states that this must not be the case:
the equations of motion must have the same functional
form in all inertial reference frames. A frame that moves
with constant velocity uwith respect to an inertial frame
also is an inertial frame. The necessity of inertial frames
is overcome in the theory of general relativity.
What remains unspecified in the axiom is the transfor-
mation of coordinates between two such inertial frames.
Maxwell’s equations describe the propagation of light.
They are form-invariant under coordinate transforma-
tions if time and space transform according to the
Lorentz transformation formulas. Lorentz transforma-
tions guarantee that the relativistic distance between two
events is the same in all coordinate systems. The coordi-
nates of an event Pis described in Kwith the coordinates
(t, x, y, z) and in K′with coordinates (t′, x′, y′, z ′). The
invariant distance between two events P1and P2is given
by
s2
12 =c2(t2−t1)2−(x2−x1)2−(y2−y1)2−(z2−z1)2,(3)

3
and s2
12 = (s′
12)2must hold with
(s′
12)2=c2(t′
2−t′
1)2−(x′
2−x′
1)2−(y′
2−y′
1)2−(z′
2−z′
1)2,
(4)
where we use the coordinates of the two events P1and
P2in the two different reference frames. When the two
events are infinitesimally close, (ds)2=c2(dt)2−(dx)2−
(dy)2−(dz)2is the invariant distance. Note that the
velocity of light is the same in all reference frames, c′=c
(Einstein’s axiom of the invariance of the speed of light).
To simplify the discussion, we assume that the two ref-
erence frames Kand K′coincide at time t=t′= 0, and
K′moves with velocity u=uex. Then, the coordinates
of an event Pis described in Kwith the coordinates
(t, x, y, z) and in K′with coordinates (t′, x′, y′, z ′). The
Lorentz transformation provides the relation between the
coordinates,
x=x′+ut′
p1−u2/c2, y =y′, z =z′, t =t′+ux′/c2
p1−u2/c2.
(5)
For infinitesimal distances, one simply has to re-
place (t, x, y, z) by (dt, dx, dy, dz) and (t′, x′, y′, z′) by
(dt′,dx′,dy′,dz′).
The velocities of a particle are given by v= dr/(dt) in
Kand v′= dr′/(dt′) in K′. Therefore, they transform
according to
vx=v′
x+u
1 + v′
xu/c2, vy=v′
yp1−u2/c2
1 + v′
xu/c2,
vz=v′
zp1−u2/c2
1 + v′
xu/c2,(6)
when we use the Lorentz transformation for the coordi-
nates (5) in their infinitesimal form.
C. Doppler effect
Light is described by electromagnetic plane waves
with frequency ωand wave vector kas solutions of the
Maxwell equations in the absence of external sources,
e.g.,
E(r, t) = E0cos(ωt −k·r) (7)
for the vector of the electric field where E0is a three-
dimensional vector with real components. Apparently,
E(r, t) has extrema and zeros when the phase
φ(r, t) = ωt −k·r(8)
is a multiple of π/2. The number of zeros or the num-
ber of maxima/minima are independent of the reference
frame so that the phase must be a Lorentz scalar. This
implies that (ω, k) form a relativistic four-vector in the
same way as (t, r) so that its components transform anal-
ogously to eq. (5),
kx=k′
x+uω′/c2
p1−u2/c2, ky=k′
y, kz=k′
z,
ω=ω′+uk′
x
p1−u2/c2(9)
when Kand K′move with constant velocity u=uex
relative to each other. When the light also travels along
the x-axis to the right, we have ky=kz= 0 and kx=
k=ω/c from the dispersion relation
ω=|k|c . (10)
Therefore, we obtain the relativistic Doppler formula for
the frequency shift,
ω=ω′s1 + u/c
1−u/c (11)
between the frequencies measured in Kand K′. When
the light travels to the left (ky=kz= 0 and kx=−k=
−ω/c), the signs ‘+’ and ‘−’ in eq. (11) swap their places.
III. DYNAMICS: LAGRANGE FORMALISM
The ultimate goal in classical mechanics is to derive
the motion of particles from basic principles, i.e., to for-
mulate equations from which the particle trajectory r(t)
can be deduced. Newton’s original formulation was su-
perseded by the Lagrange and Hamilton formulation be-
cause the underlying Hamilton principle of least action
constitutes the basis of present-day theoretical physics.
A. Particle mass
To describe particle dynamics, Newton assigns a sec-
ond defining property to a point particle, namely its (in-
ertial) mass m. Below we shall assume that
1. the non-relativistic momentum and (kinetic) en-
ergy of a particle are proportional to the mass m;
2. the mass is a scalar under Lorentz transformations;
3. a particle with mass Mcan decay into two particles
with mass m≤M/2.
Property 1 seems self-evident because it requires twice
the force to push two mugs of beer over a counter com-
pared to pushing a single one. Moreover, we tacitly as-
sume that we do not gain or lose liquid when looking
at the mug from different reference frames (property 2),
and we know that we can split a liquid into equal volumes
without loosing any (property 3).

4
On a more fundamental level, the generation of iner-
tial mass requires an understanding of the interaction
of relativistic particle fields with the Higgs field. Even
more intricate is the notion of a gravitational mass and
its equivalence to the inertial mass. We will not dwell
into these fundamental issue here but move on to the
equations of motions that govern the particle dynamics.
B. Euler-Lagrange equations
Modern physics is based on the principle of least ac-
tion. For a point particle in classic mechanics, the ac-
tion Salong a path R(t) with velocity ˙
R(t) within the
time interval [t1, t2] reads
S=Zt2
t1
dtL(R,˙
R, t).(12)
Here, Lis the Lagrange function that depends only on
the particle coordinates R(t), velocities ˙
R(t), and time t.
Consequently, the particle acceleration must be a func-
tion of the particle position and velocity only, i.e., the
particle motion is deterministic.
To find the realized trajectory r(t), the principle of
least action states that Sis stationary with respect to
small variations of the realized trajectory whereby all
trajectories start and end at the points P1=r(t1) and
P2=r(t2), respectively. For this reason, the Lagrange
function is not unique. For example, the variation does
not change when we add a constant Cto L, i.e., using
˜
L=L+Cin Sleads to the same realized trajectory as
using L.
As shown in textbooks, the realized trajectory r(t) ful-
fills the Euler-Lagrange equations
d
dt
∂L
∂˙
RR=r,˙
R=˙
r
=∂L
∂RR=r,˙
R=˙
r
.(13)
The equations (13) constitute Newton’s second law.
For example, in non-relativistic mechanics, a single
point-particle has the Lagrange function
Lnr(R,˙
R, t) = m
2˙
R2.(14)
When inserted in eq. (13), the Euler-Lagrange equations
read
d
dt(m˙
r) = 0,(15)
i.e., a free particle moves along a straight line (Newton’s
first law).
C. Energy and momentum
One big advantage of the Lagrange formalism over
Newton’s formulation of classical mechanics lies in the
fact that conserved quantities like momentum and en-
ergy are well defined. As shown by Noether [10], when
the Lagrange function is invariant under translations in
space (time), there is a conserved quantity called momen-
tum (energy). Thus, these objects and their conservation
laws result from the homogeneity of space and time.
The simplest example is the non-relativistic point par-
ticle in Sec. III B where eq. (15) expresses that the mo-
mentum
pnr =mv,(16)
with ˙
r=von the realized trajectory, is a conserved
quantity for a non-relativistic free particle, i.e., pnr does
not change in time.
Apparently, momentum conservation is based on the
fact that the Lagrange function does not depend on r.
The theory reflects the fact that space is homogeneous
so that Ldoes not change under translations. Therefore,
for a general Lagrange function L=L(˙
R) the conserved
momentum is defined by
p=∂L(˙
R)
∂˙
R˙
R=˙
r
.(17)
We shall derive the relativistic Lagrange function in the
next subsection III D and thus readily find the expression
for the relativistic momentum of a point particle.
Homogeneity in time implies that Ldoes not explicitly
depend on time, L=L(R,˙
R). The resulting conserved
quantity is the energy (or Jacobi integral),
E=˙
r·∂L(R,˙
R)
∂˙
RR=r,˙
R=˙
r
−L(r,˙
r).(18)
For the non-relativistic point particle in Sec. III B we thus
find ( ˙
r=v)
Enr =˙
r·(m˙
r)−m
2˙
r2=m
2v2≡Tnr ,(19)
the well-known expression for the (kinetic) energy of a
non-relativistic point particle.
D. Action for a particle in Minkowski space
Axiomatically, the action Sis a scalar under Lorentz
transformations. The only infinitesimal scalar for a single
particle on the world line from P1to P2is the infinites-
imal distance dsbetween two point on the world-line.
Therefore,
S=−αZP2
P1
ds=−αc Zt2
t1
dtq1−˙
R2/c2.(20)
Thus, we know the relativistic Lagrange function up to a
constant α > 0 that is determined from the comparison

5
with the non-relativistic limit. Indeed, we can read off
the relativistic Lagrange function from eq. (20),
L(R,˙
R, t)≡L(˙
R) = −αcq1−˙
R2/c2.(21)
For small velocities |˙
R| ≪ c, the Taylor expansion leads
to
L(˙
R)≈ −αc 1−˙
R2
2c2!=−αc +α
2c˙
R2.(22)
The comparison with the non-relativistic Lagrange func-
tion in eq. (14) shows that we must set α=mc to arrive
at L(˙
R)≈Lnr(˙
R) + C.
E. Energy and momentum for a single particle
With the relativistic Lagrange function from eq. (21),
L(˙
R) = −mc2q1−˙
R2/c2(23)
and the results from Sec. III C we can readily determine
the conserved momentum of a single point particle with
mass m, see eq. (17),
p=mv
p1−v2/c2=γmv(24)
with the relativistic factor
γ=1
p1−v2/c2.(25)
The particle’s conserved energy reads, see eq. (18),
E=v·mv
p1−v2/c2+mc2p1−v2/c2=γmc2.(26)
For a particle at rest, v=0, the energy is finite,
E0≡E(v=0) = mc2,(27)
the famous Einstein formula for the rest energy of a par-
ticle.
Equations (24) and (26) constitute the main results
that need to be proven using the pedestrian approach
outlined in the next section.
IV. PEDESTRIAN DERIVATION
The derivation in Sect. III is concise and elegant but it
uses a number of concepts in theoretical physics that were
not common knowledge at the beginning of the 20th cen-
tury nor are they familiar to newcomers nowadays. For
this reason, we collect the main ideas to derive compre-
hensively the Einstein formula using only basic concepts
of classical mechanics as outlined in Secs. II and III.
We pursue the following strategy.
A. Derive the relativistic momentum from elastic scat-
tering of two particles;
B. Derive the relativistic kinetic energy from elastic
scattering of two particles;
C. Derive the mass defect formula from fission of a
heavy particle into two equal light particles;
D. Derive the Einstein energy-mass relation.
We only invoke the concepts of relativistic classical me-
chanics and do not refer to classical electrodynamics
(electromagnetic waves) or concepts of quantum mechan-
ics (photons).
FIG. 1. Collision of two identical particles with velocities ±v
in the center-of-mass frame.
A. Elastic scattering: momentum conservation
We start our investigation with an elastic scattering of
two identical classical particles, as shown in Fig. 1. The
particles approach each other on the x-axis and move
along the y-axis after the scattering. Elastic scattering
means that there is no loss of energy and the particles
remain the same so that the speed of each particle after
the impact will remain the same (v=|v|), only its direc-
tion will change. It is intuitively clear that the laws of
momentum and energy conservation are fulfilled: the to-
tal momentum is zero before and after the collision, and
the total energy of the two particles remains the same.
We now demand that momentum conservation is also
fulfilled in a different frame of reference. Let there be
two reference frames, one ‘primed’ where all values will
be marked with primes, the other ‘unprimed’. Let the
primed frame move relative to the unprimed one in the
direction of the x-axis with velocity u=uex. Then, the
connection between the ‘primed’ particle velocity v′=
(v′
x, v′
y, v′
z) and its ‘unprimed’ velocity v= (vx, vy, vz) is
given by eq. (6) in Sect. II.
As ‘primed frame’ we choose the center-of-mass frame
as in Fig. 1 so that the ‘unprimed frame’ is the laboratory
frame in which the observer may be viewed at rest. A

6
short calculation convinces the reader that the momen-
tum of a particle cannot be given by pnr =mvbecause
mv1+mv2=mv3+mv4. Since the system is rotational
invariant, the modulus of the momentum of a particle
must be a function of the modulus of the velocity,
|p(v)|=p(v),(28)
where p=qp2
x+p2
y+p2
zand v=qv2
x+v2
y+v2
zso that
px=vx
vp(v), py=vy
vp(v), pz=vz
vp(v),(29)
with the Cartesian components, vx,y,z =ex,y,z ·v,
px,y,z =ex,y,z ·p.
FIG. 2. Elastic collision in the ‘primed’ center-of-mass refer-
ence frame (left) and in the ‘unprimed’ laboratory reference
frame (right). The ‘primed’ frame moves to the right with
velocity u=uex, relative to the ‘unprimed’ one.
To find the unknown function p(v), we use the conser-
vation of momentum in a collision,
p1+p2=p3+p4(30)
for which we only need to consider the x-projection of this
equality because the other components just give 0 = 0.
We consider the elastic scattering event in the laboratory
frame, see Fig. 2. We note that
p1,x =p1=p(v1) = pv+u
1 + vu/c2(31)
because the velocity v1is just directed along the x-axis.
Next,
p2,x =−p2=−p(v2) = −pv−u
1−vu/c2.(32)
To be definite, we assume that uis less than vso that
the velocity v2is directed towards the negative x-axis,
and hence p2,x =−p2. Lastly,
p3,x =v3,x
v3
p(v3) =
=u
pv2+u2−v2u2/c2ppv2+u2−v2u2/c2,(33)
and p4,x =p3,x. When we insert these expressions into
p1,x +p2,x =p3,x +p4,x, we obtain
pv+u
1 + vu/c2−pv−u
1−vu/c2=
= 2 u
pv2+u2−v2u2/c2ppv2+u2−v2u2/c2.
(34)
This is the equation for the unknown function p(v). The
unique solution for p(v) for all 0< u < v < c is
p(v) = Cp
v
p1−v2/c2(35)
with some constant Cpwhich is readily achieved after
some algebra, or by using Mathematica [18]. A deriva-
tion of eq. (35) is given in appendix A.
It remains to determine the constant Cpin eq. (35).
It is known that, for small velocities, |v| ≪ c, we have
the non-relativistic relation pnr =mvfor the relation
between particle velocity and its momentum, see eq. (16).
The expansion of eq. (35) for small vgives p(v≪c)≈
Cpvso that the constant is the mass of the particle, Cp=
m. Thus, p(v) = mv/p1−v2/c2,p=γmv, or, in vector
form,
p=γmv.(36)
In sum, we re-derived the relativistic momentum as a
function of the particle velocity as given in eq. (24).
B. Elastic scattering: kinetic energy conservation
We reconsider the scattering experiment in Figs. 1
and 2 in the context of energy conservation. The collision
in this experiment is elastic, i.e., the internal states of
the particles do not change. The only difference between
the states of the particles results from their velocities.
Therefore, the energy Eof each particle is a function of
its velocity vonly, E=E(v), again assuming rotational
invariance. Energy conservation thus implies
E(v1) + E(v2) = E(v3) + E(v4) (37)
in the scattering event in both reference frames.
It is convenient to divide the particle’s energy Einto
its internal energy (also called ‘potential energy’ or ‘rest
energy’) E0, and its kinetic energy T,
E(v) = E0+T(v).(38)
The internal energy is nothing else but the energy at
zero velocity, E0≡E(0). Since the internal energy does
not change in elastic collisions, we can rewrite the en-
ergy conservation law for our scattering experiment as
conservation of the kinetic energy,
T(v1) + T(v2) = T(v3) + T(v4).(39)

7
Apparently, the value of E0cannot be determined using
elastic scattering.
Now, we derive the function T(v) from the condi-
tion (39). In the ‘primed’ reference frame (center-of-mass
system, left part of Fig. 2), it is evident that the kinetic
energy is conserved, T(v′
1) + T(v′
2) = T(v′
3) + T(v′
4), be-
cause the velocity moduli are all equal, v′
1=v′
2=v′
3=
v′
4=v.
For the laboratory frame, things are a little bit more
complicated. The velocity moduli are
v1=v+u
1 + vu/c2, v2=v−u
1−vu/c2,
v3=v4=rv2+u2−v2u2
c2.(40)
As for the case of the momentum, the non-relativistic
expression, Tnr(v) = mv2/2 in eq. (19), does not fulfill
eq. (39) when eq. (40) is used.
We find the correct expression for T(v) as for the mo-
mentum function p(v). For general 0 < u < v < c, we
must solve eq. (39) for the velocities given in eq. (40),
Tv+u
1 + vu/c2+Tv−u
1−vu/c2=
= 2T rv2+u2−v2u2
c2!.(41)
The unique solution that reproduces the non-relativistic
limit, T(v≪c)≈mv2/2, obeys
T(v)=(γ−1)mc2.(42)
This is shown explicitly in appendix B. The proof that
T(v) from eq. (42) requires only some basic algebra, or
can be done using Mathematica [18].
Eq. (42) provides the relativistic formula for the kinetic
energy T(v) of a particle with mass mand velocity |v|=
v. We recall that the mass is considered as independent of
the velocity. The velocity vappears here in the γ-factor
only, γ= 1/p1−v2/c2.
C. Particle fission: mass defect
Thus far, we used elastic collisions as a convenient tool
to derive the relativistic formulas for the momentum and
the kinetic energy. But elastic processes, by definition,
do not change internal states of particles. Therefore, we
need to address inelastic processes to get access to the in-
ternal energy. The simplest example process is the fission
of a particle of mass Minto two lighter identical particles
of mass m≤M/2. This process is shown in Fig. 3 in two
different reference frames. When we reverse the arrow of
time, the process describes particle fusion.
The left part of Fig. 3 shows the particle decay in the
center-of-mass frame. The particle with mass Msplits
FIG. 3. Decay of a particle of mass Minto two identical
particles of mass m≤M/2. The left part depicts this process
in the (primed) center-of-mass reference frame, and the right
part in the unprimed frame, in which the center of mass moves
to the right with velocity u(laboratory frame).
into two particles that fly away in opposite directions
with velocities v=±vex. Before the decay, the kinetic
energy of the heavy particle is equal to zero. After the
decay, the total kinetic energy is equal to 2mc2(γ−1),
where γ= 1/p1−v2/c2. Hence, due to energy conser-
vation, the total internal energy decreases by the same
amount,
∆E0(m)=2mc2(γ−1) .(43)
This decrease in internal energy is proportional to the
mass defect as we shall show next.
To this end, we apply the momentum conservation law
in the unprimed (laboratory) frame. The right part of
Fig. 3 shows the same fission process in a reference frame,
where the center-of-mass frame moves to the right with
velocity u < v < c. In this frame, the heavy particle
moves to the right with velocity u, one light particles
moves to the right with velocity
v1=v+u
1 + vu/c2,(44)
and the other light particle moves to the left with velocity
v2=v−u
1−vu/c2.(45)
The momentum conservation law along the x-axis reads
Mu
p1−u2/c2=mv1
p1−v2
1/c2−mv2
p1−v2
2/c2.(46)
Therefore, the heavy mass reads
M=mp1−u2/c2
u v1
p1−v2
1/c2−v2
p1−v2
2/c2!.
(47)
We substitute v1,2from eqs. (44) and (45) and find after
some algebra
M=2m
p1−v2/c2.(48)
We see that the mass Mof the heavy particle is larger
than the total mass of the debris particles 2m, assuming
that the decay particles move away from each other, v >

8
0. Due to the decay, the total mass decreases by
∆M(m) = M−2m= 2m 1
p1−v2/c2−1!= 2m(γ−1) ,
(49)
which is sometimes called the mass defect.
Now we can compare the mass defect ∆M(m) in
eq. (49) with the decrease of internal energy ∆E0(m)
in eq. (43), which leads to
∆E0(m)=∆M(m)c2.(50)
Apparently, every change of the total internal energy
∆E0(m) of the system is accompanied by a change of
its total mass ∆M(m), such that eq. (50) holds.
D. Energy-mass formula
A point particle has no internal degrees of freedom like,
e.g., a molecule. Point particles in classical physics are
solely characterized by their mass. Therefore, its internal
energy can only depend on m,E0≡E0(m). We apply
this Ansatz to eq. (43) and write explicitly
∆E0(m) = E0(M)−2E0(m)=2mc2(γ−1) .(51)
Note that the velocity vwas never specified because, in a
real experiment, it is measured and used to interpret the
fission process and the internal structure of the particle
of mass M.
In our thought experiment, we can consider the ex-
treme case that the heavy particle just splits into two
halves that do not separate from each other, v= 0.
Therefore, the internal energy must obey the relation
∆E0(m) = E0(2m)−2E0(m) = 0 (52)
because there is no mass defect in this case, M= 2m, as
∆M(m) = 0, see eq. (49). Since eq. (52) must hold for
all m, it follows that
E0(m) = C0m , (53)
and the constant must be C0=c2in view of eq. (50).
Therefore, we finally arrive at the desired result
E0(m) = mc2(54)
for the internal energy of a point particle of mass m.
The total energy for a moving particle is then
E(v) = γmc2=mc2
p1−v2/c2,(55)
according to eqs. (38) and (42), and in agreement with
the expression from the Lagrange formalism outlined in
Sec. III, see eq. (26).
V. PHOTON DYNAMICS
So far we considered collisions of massive particles. In
this section, we study the collision of a massive particle
with a photon, i.e., Compton scattering, and the emission
of two photons by a massive particle.
A. Compton scattering
First, we show that energy and momentum conserva-
tion in the Compton scattering event provides enough
information to prove
Eph =ℏω , Eph =|pph |c(56)
for photons. In this sense, particle scattering alone also
proves Einstein’s second famous formula (56), without
resorting to the photoelectric effect.
Our derivation is based on the assumption that the
photon energy is some (yet unknown) function f(ω) of
its (angular) frequency ω, and that the absolute value of
its momentum is given by another function g(ω),
Eph(ω) = f(ω),|pph |=g(ω).(57)
The latter also depends only on ωdue to the dispersion
relation (10).
For simplicity, we direct the photon momentum along
its wave vector. As in the previous sections, we imply
that the functions f(ω) and g(ω) permit Taylor expan-
sions at any ω > 0.
Let us consider a collision of a photon with a massive
particle (an electron for brevity) in the center-of-mass
frame where the net momentum is equal to zero. To be
definite, we let the particles travel towards each other
along the x-axis before the collision, the electron to the
right with some velocity v=vex, and the photon to the
left. The condition of zero net momentum relates the
electron velocity vto the photon frequency ω
γ(v)mv =g(ω),(58)
where mis the electron mass, and γ(v) = (1−v2/c2)−1/2
as in Sect. IV. The conservation laws permit that both
particle fly apart also along the xaxis after the collision,
the electron to the left with velocity −vex, and the pho-
ton to the right with the same frequency ωas before the
collision. This process is schematically depicted in the
left part of Fig. 4.
Now we look at this very process from another refer-
ence frame K, see Fig. 4, right part, that moves to the
left along x-axes with some velocity u, as seen from the
center-of-mass frame K′. In the frame K, the initial and
the final electron velocities are
v1=v+u
1 + vu/c2exand v3=−v−u
1−vu/c2ex,(59)
correspondingly, as follows from the relativistic velocity
addition rules (6). The photon frequency undergoes the

9
FIG. 4. Compton scattering process as seen in the center-of-mass frame K′(left part) and in another reference frame Kthat
moves to the left relative to K′with velocity u(right part).
Doppler shift when switching from frame K′to frame K,
see eq. (11),
ω2=ωs1−u/c
1 + u/c and ω4=ωs1 + u/c
1−u/c (60)
before and after the collision, respectively.
The energy conservation law in the laboratory frame
Ktherefore reads
γv+u
1 + vu/c2mc2+f ωs1−u/c
1 + u/c!=
=γv−u
1−vu/c2mc2+f ωs1 + u/c
1−u/c!.(61)
Similarly, the momentum conservation law in the frame
Khas the following form when projected onto the x-axis
L(v, u, ω) = R(v , u, ω),
L(v, u, ω) =
=γv+u
1 + vu/c2mv+u
1 + vu/c2−g ωs1−u/c
1 + u/c!,
R(v, u, ω) =
=−γv−u
1−vu/c2mv−u
1−vu/c2+g ωs1 + u/c
1−u/c!.
(62)
Equations (61) and (62) must be fulfilled for all positive
values ωand for all values of u, provided that electron
velocity vis related to ωby eq. (58). They permit to
determine the functions f(ω) and g(ω), up to a few pa-
rameters.
As in the previous section, we expand eqs. (61)
and (62) in a power series in uwhich provides differ-
ential equations for the functions f(ω) and g(ω), see Ap-
pendix C. Their solution leads to the following expres-
sions for the photon energy f(ω),
f(ω) = C1ω−C2
ω+C3,(63)
and for the photon momentum g(ω),
g(ω) = C1ω+C2
ω1
c,(64)
where C1,C2, and C3are some constants. To fix these
constants, we note that the energy and pressure of light
are always positive, thence f(ω)>0 and g(ω)>0 for all
positive frequencies ω. Therefore,
C1>0, C2= 0,and C3≥0.(65)
Also, knowing that photons can have arbitrarily small
energies, we conclude that C3= 0. Hence, the only free
constant is C1so that from f(ω) = cg(ω) we find that
Eph =|pph|c(66)
holds. Moreover, the result
Eph(ω) = C1ω(67)
is nothing but Planck’s fundamental assertion that light
comes in quanta (called photons) with energy Eph(ω) =
ℏω, i.e., the constant C1was first named by Planck,
C1=ℏ. Using this identification, eqs. (66) and (67)
prove eq. (56) entirely.
B. Two-photon emission
An alternative approach to the mass-defect for-
mula (49) modifies Einstein’s original thought experi-
ment [5] by making it independent of the knowledge of

10
electrodynamics. In the original setting, a body at rest
with mass Memits two equal bunches of light in opposite
directions, see Fig. 5a. Rohrlich [2] argues that even the
nineteen-century non-relativistic physics, supplied by the
photons’ energy and momentum formulas, makes it possi-
ble to derive the mass-defect formula (49) by an analysis
of the experiment depicted in Fig. 5a. Feigenbaum and
Mermin [19] suggest to replace the light by massive par-
ticles which makes the problem purely mechanical, and
we are left with the problem analyzed in Sec. IV C.
FIG. 5. (a) Einstein’s seminal thought experiment [5] modi-
fied later by Feigenbaum and Mermin [19] and by Rohrlich [2];
(b, c) its simplified version, where a particle annihilates into
two photons.
Here, we retrace the thought experiment of particle
annihilation into two photons. Note that we make use of
the information that a photon of frequency ωhas energy
E=ℏωand momentum |p|=ℏω/c.
In the frame K′where the particle is at rest, see
Fig. 5b, both photons must have the same absolute value
of momentum and, consequently, the same frequency ω.
The energy balance in K′therefore reads
E0= 2ℏω , (68)
where E0is the particle’s internal energy, and ℏωis the
energy of one photon. In another frame K, see Fig. 5c,
that moves with speed uto the left relative to K′, the
photons undergo the Doppler shift and acquire the fre-
quencies
ω1=ωs1 + u/c
1−u/c and ω2=ωs1−u/c
1 + u/c ,(69)
according to eq. (11). Next, we apply momentum conser-
vation in the frame K. The particle’s momentum γ(u)mu
turns into the photon momenta, ℏω1/c and ℏω2/c, which
are directed to the right and to the left, respectively.
Therefore,
γ(u)mu =ℏω
cs1 + u/c
1−u/c −ℏω
cs1−u/c
1 + u/c .(70)
We divide both sides by γ(u) = 1/p(1 + u/c)(1 −u/c)
and obtain
mu =2ℏωu
c2.(71)
This gives ℏω=mc2/2 so that substituting into eq. (68)
finally gives
E0=mc2.(72)
Note that, instead of the momentum conservation in
frame K, we can also consider the energy conservation
in this frame which leads to the same conclusion.
VI. RELATION TO OTHER WORK
In the literature, one can find many clever designs
that aim at the construction or derivation of the rela-
tivistic momentum, the kinetic energy, or the mass-to-
energy relation from thought experiments with collisions,
including a version by Einstein himself, dated from the
year 1935 [1]. Many of them are cited and discussed
by Hu [20] who also suggests two additional collision
schemes. Here, we briefly relate to those that employ
particle collisions.
A. Relativistic momentum
Lewis and Tolman [11] consider the elastic collision of
two identical particles that approach each other at an
infinitely small angle θrelative to the x-axis, and move
at the same angle θafter the collision, see Fig. 6a. When
we look at this process from the reference frame K1that
moves to the right relative to the center-of-mass frame K′
along with the lower particle, see Fig. 6b, one recognizes
that the lower particle moves up and down at some small,
non-relativistic speed w. The higher particle moves at
some speed v. In the frame K2that moves to the left
along with the upper particle, see Fig. 6c, the particles
exchange their roles: the upper one moves at the low
speed w, and the lower one at the high speed v. The
comparison of Figs. 6b and 6c shows that the vertical
projection of the upper particle’s velocity in frame K1
is equal to ±w/γ(v), where the denominator arises from
the relativistic time dilation. Therefore the momentum
conservation law in the projection to the vertical axis
reads
mw −p(v)w/γ(v)
v=−mw +p(v)w/γ(v)
v,(73)
where mis the particle mass. Resolving this equation
with respect to the particle’s momentum p(v), one readily
FIG. 6. Thought experiment by Lewis and Tolman [11–
13] that proves the relativistic formula for the momentum,
eq. (24).

11
recovers the relativistic formula p(v) = γ(v)mv, eq. (24).
A number of schemes use the following convenient
rule for the transformation of the γ-factor between the
‘primed’ and ‘unprimed’ reference frames,
γ(v) = γ(v′)γ(u)1 + v′·u
c2,(74)
where uis the velocity of the ‘primed’ frame relative to
the ‘unprimed’ one. Eq. (74) is a direct consequence of
the velocity transformation rule (6). Using eq. (74) one
can rewrite the transformation rules for the components
of the velocity in the y, z-directions, assuming that uis
parallel to the x-axis,
vy=v′
y
γ(v′)
γ(v), vz=v′
z
γ(v′)
γ(v).(75)
FIG. 7. Thought experiment adopted from the paper by Fin-
kler [21] that proves the relativistic formula for the momen-
tum, eq. (24).
Finkler [21] considers two identical particles that move
in the xy-plane with opposite velocities v′
1and v′
2=−v′
1
relative to the center-of-mass frame K′, see Fig. 7a. In an
‘unprimed’ frame K, see Fig. 7b, that moves along the x-
axis with respect to K′, the y-component of the total mo-
mentum p1y+p2yis equal to zero. Indeed, one may imag-
ine that the particles originally move along the y-axis in
frame K′, as shown by dashed lines, and were scattered
elastically. Then, in frame K, the y-components of their
momenta before the collision compensated each other by
symmetry, as seen in Fig. 7b. We write the y-component
of a particle’s momentum as py=p(v)vy/v, and express
the vanishing of the total y-momentum in frame Kas
p(v1)v1y
v1
=−p(v2)v2y
v2
.(76)
The y-components of the particle velocities v1and v2in
the ‘unprimed’ frame Kcan be transformed to frame K′
using the rule (75),
v1y=v′
1y
γ(v′
1)
γ(v1), v2y=v′
2y
γ(v′
2)
γ(v2).(77)
Since v′
1=v′
2and v′
1y=−v′
2yit follows from eq. (77)
that
γ(v1)v1y=−γ(v2)v2y.(78)
Finally, dividing each side of eq. (76) by the correspond-
ing side of eq. (78) we find
p(v1)
γ(v1)v1
=p(v2)
γ(v2)v2
.(79)
Varying the conditions of this thought experiment with
velocities and inclination angles in frame K′, and the
velocity of frame Krelative to K′, we can independently
vary v1and v2in eq. (79). Therefore, eq. (79) implies
that both sides are constants,
F=p(v)
γ(v)v(80)
is independent of v. In other words, it is proven that
the relativistic momentum p(v) is proportional to γ(v)v.
Using the non-relativistic limit it is seen that F≡mis
the particle mass.
B. Relativistic kinetic energy
In his unpublished lectures in the 1920s, Langevin uses
the collision scheme as in Fig. 2 to derive the relativistic
formula for the kinetic energy. Later, his arguments were
reconstructed by Penrose, Rindler, and Ehlers [22, 23].
Here, we only provide a brief summary of their argu-
ments.
FIG. 8. Kinetic energy as a function of the γ-factor.
Energy conservation during the elastic collision in the
’unprimed’ frame, see the right part of Fig. 2, states that
T(v1) + T(v2) = T(v3) + T(v4) where T(v) denotes the
kinetic energy. Since v3=v4by symmetry, we have
T(v3) = T(v1) + T(v2)
2.(81)
Then, applying the rule (74) to the velocities v1,v2,v3,
and v4, one obtains
γ(v1) = γ(v)γ(u)1 + vu
c2,(82)
γ(v2) = γ(v)γ(u)1−vu
c2,(83)
γ(v3) = γ(v4) = γ(v)γ(u),(84)

12
whence
γ(v3) = γ(v1) + γ(v2)
2.(85)
When we plot the kinetic energy T(v) versus γ(v),
see Fig. 8, eqs. (81) and (85) imply that the point
γ(v3), T (v3)in the plot is always located exactly
in the middle between the points γ(v1), T (v1)and
γ(v2), T (v2). Since v1and v2can be varied indepen-
dently, the only possible shape of the curve T(v) as a
function of γ(v) is a straight line,
T(v) = c1γ(v) + c2.(86)
Here, c1and c2are some coefficients to be found from
comparison with the non-relativistic expression T(v≪
c)≈mv2/2, namely, c1=mc2=−c2. Using these
coefficients, eq. (86) reduces to the relativistic expression
T(v) = mc2(γ(v)−1) for the relativistic kinetic energy,
see eq. (42).
C. Energy-mass relation
With relativistic formulas for the momentum and the
kinetic energy in our hands, we can go ahead and find
the relation between mass and energy. In the Feynman
Lectures on Physics [13] this is done in the following way.
FIG. 9. Thought experiment considered in the Feynman Lec-
tures on Physics [13] for obtaining the energy-mass relation.
Imaging that two equal masses mapproach each other
along the x-axis with equal speeds |v|=v. When they
meet, they coalesce into a larger body of some mass M,
see Fig. 9a . Let us look at this process from another
reference frame K, see Fig. 9b, that moves along the y-
axis relative to the center-of-mass frame K′with a very
small, non-relativistic speed u. Conservation of the y-
component of the momentum in the frame Kreads
2γ(˜v)mu =M u , (87)
where ˜vis the speed of the initial masses in the frame K.
Neglecting the difference between vand ˜v, we find the
fused mass Mfrom eq. (87),
M= 2γ(v)m . (88)
Therefore, the fused mass exceeds the original particle
masses by
∆m=M−2m= 2m(γ(v)−1) .(89)
At the same time, the sum of the internal energies of the
particles have increased by
∆E0= 2mc2(γ(v)−1) (90)
because all the kinetic energy T(v) = 2mc2(γ(v)−1) of
the two masses, as seen from the center-of-mass frame
K′, transformed into internal energy. From comparison
between equations (89) and (90) we come to the conclu-
sion that
∆E0= ∆mc2,(91)
the mass defect formula (49).
VII. CONCLUSIONS
In this work we used relativistic kinematics of point
particles and two-particle collisions to derive the energy-
mass relation that reduces to Einstein’s famous for-
mula (1) for a particle at rest. Our derivation offers sev-
eral advantages over other derivations briefly discussed
in Sec. VI.
1. Since it does not appeal to electrodynamics, it is
conceptually simpler than Einstein’s original ar-
gument that involves electromagnetic radiation.
Moreover, using the relativistic Doppler formula,
we can derive the energy of photons as massless
particles and Planck’s formula (67) from an analy-
sis of Compton scattering or from particle annihi-
lation into two photons.
2. We address very simple geometries for the two-
particle scattering, see Fig. 2, which already pro-
vides us with the relativistic expressions for the rel-
ativistic expressions for particle momentum and its
kinetic energy.
3. The mass defect formula straightforwardly follows
from the decay of a massive particle at rest into
two identical particle, see Fig. 3. With the plain
assumption that the internal energy of a point par-
ticle can only depend on its mass, Einstein’s for-
mula (1) readily follows.
The price of the conceptual simplicity is the use of
some standard elements of calculus, namely (second-
order) Taylor expansion and ordinary first-order differ-
ential equations to find the unique solutions of eqs. (34),
(41), (61), and (62). The proof that the given expressions
solve these equations requires only elementary mathe-
matics.
Students who are scared off by calculus and prefer
physically motivated shortcuts may resort to the liter-
ature, some of which we briefly reviewed above. We
presume that the most straightforward ‘derivation’ of
Einstein’s energy formula starts from the four-vector of
the relativistic velocity. Scattering thought experiments

13
show that the three spatial components are conserved in a
scattering experiment, i.e., they must be the particle mo-
mentum up to a mass factor. Therefore, the ‘Babylonian
approach’ inspires the notion that the zero component
must be the particle energy, up to a mass factor [1, 9].
In this work we argue that the pedestrian but still Eu-
clidean way to Einstein’s formula is neither short nor
simple. Instead, it requires the detailed analysis of scat-
tering experiments and some calculus to extract the cor-
rect formulas for the relativistic momentum and kinetic
energy.
Appendix A: Derivation of the momentum modulus
In this appendix we derive eq. (35). We expand eq. (34)
in a Taylor series in uaround u= 0 up to first order in
u. First, we find
pv+u
1 + vu/c2=p(v) + (1 −v2/c2)p′(v)u+. . . , (A1)
where the ellipsis denotes further terms that are propor-
tional to u2,u3, and so on. The prime denotes the deriva-
tive with respect to v. Next,
pv−u
1−vu/c2=p(v)−(1 −v2/c2)p′(v)u+. . . (A2)
and
2u
pv2+u2−v2u2/c2ppv2+u2−v2u2/c2=
=2u
vp(v) + . . . . (A3)
We collect the terms to first order in uin eq. (34) and
find the condition
2(1 −v2/c2)p′(v) = 2
vp(v).(A4)
Writing p′(v) = dp/dv, this differential equation can be
solved by separation of variables,
dp
p=dv
v(1 −v2/c2),(A5)
so that the integration of both sides leads to
ln p(v) = ln "v
p1−v2/c2#+ const ,(A6)
or
p(v) = Cp
v
p1−v2/c2,(A7)
which proves eq. (35).
Appendix B: Derivation of the kinetic energy
Eq. (41) defines the yet unknown function T(v) for the
kinetic energy. As in appendix A, we expand each term
of this equation in a Taylor series around u= 0 to second
order in u,
Tv+u
1 + vu/c2≈T(v) + u1−v2
c2T′(v)
+u2
21−v2
c2−2v
c2T′(v) + 1−v2
c2T′′(v),(B1)
Tv−u
1−vu/c2≈T(v)−u1−v2
c2T′(v)
+u2
21−v2
c2−2v
c2T′(v) + 1−v2
c2T′′(v),(B2)
T rv2+u2−v2u2
c2!≈T(v) + u2
21−v2
c2T′(v)
v,(B3)
where T′(v) and T′′(v) are first and second derivatives of
function T(v), and higher orders in the Taylor expansion
were ignored. When inserted into eq. (41), the constant
and linear terms drop out, and the quadratic terms lead
to the differential equation
−2v
c2T′(v) + 1−v2
c2T′′(v) = T′(v)
v.(B4)
We temporarily denote T′(v) as f, and T′′ (v) as df/dv
and find the first-order differential equation
−2v
c2f+1−v2
c2df
dv =f
v(B5)
that is solved by separation of variables. It has the solu-
tion
f=f0
v
(1 −v2/c2)3/2,(B6)
where f0is an arbitrary constant. We recall that f=
T′(v), and integrate once more to find
T(v) = Zv
dx T ′(x) = Zv
dx f0
x
(1 −x2/c2)3/2=
=C1
1
p1−v2/c2+C2,(B7)
or
T(v) = γC1+C2,(B8)
where C1and C2are some constants.
We can fix C2by considering the particle at rest, when
v= 0, T= 0, and γ= 1,
0 = C1+C2,(B9)
hence C2=−C1and
T(v)=(γ−1)C1.(B10)

14
To determine C1, we consider small but non-zero veloci-
ties v/c ≪1. The expansion of the γ-factor near v= 0
gives γ(v)≈1 + v2/(2c2) so that
T(v≪c)≈C1
2c2v2.(B11)
The non-relativistic formula reads Tnr =mv2/2, see
eq. (19), so that we must set C1=mc2. Hence,
T(v)=(γ−1)mc2,(B12)
as given in eq. (42).
Appendix C: Derivation of photon energy and
momentum
For better readability, we set the speed of light c
to unity within this appendix. Hence, equations (61)
and (62) take the simpler forms
γv+u
1 + vu m+f ωr1−u
1 + u!=
=γv−u
1−vu m+f ωr1 + u
1−u!,(C1)
and
γv+u
1 + vu mv+u
1 + vu −g ωr1−u
1 + u!=
=−γv−u
1−vu mv−u
1−vu +g ωr1 + u
1−u!,(C2)
where γ(v) = (1 −v2)−1/2.
The Taylor expansions in eq. (C1) at u= 0 up to and
including linear terms in urequire
γv±u
1±vu m=γ(v)m1±uv +O(u2),(C3)
f ωr1±u
1∓u!=f(ω)±uωf ′(ω) + O(u2).(C4)
Substituting these expansions into eq. (C1), we see that
the constant terms cancel each other, and the linear
terms lead to
γ(v)mv =ωf ′(ω).(C5)
The left-hand side of eq. (C5) can be replaced with g(ω),
see eq. (58). Hence, from eqs. (C1) and (58) we obtain a
differential relation between the photon energy f(ω) and
its momentum g(ω),
ωf ′(ω) = g(ω).(C6)
Note that, for a massive particle traveling with some
velocity v, the momentum-to-energy ratio fulfills p/E =
γmv/γmc2=v/c2. Into the latter relation we might
insert v=cfor a photon traveling with velocity c.
Then, g(ω)/f(ω)≡p/E = 1/c immediately follows, i.e.,
g(ω) = f(ω) when c= 1. Therefore, eq. (C6) takes the
form ωf ′(ω) = f(ω) that immediately gives rise to the
conclusion that the photon energy f(ω) is proportional to
the frequency ω. However, we shall not follow this short-
cut. Instead, we will employ the momentum conservation
law, eq. (C2).
The Taylor expansion of the terms contributing to
eq. (C2) at u= 0 require
γv±u
1±vu mv±u
1±vu =γ(v)mv±u+u2v
2+O(u3)
(C7)
and
g ωr1±u
1∓u!=
=g(ω)±uωg′(ω) + u2
2ωg′(ω) + ω2g′′ (ω)+O(u3)
(C8)
up to and including second order in u. Substituting these
expansions into eq. (C2), we see that the constant terms
reproduce the known eq. (58) whereas the terms pro-
portional to ucancel each other. Collecting the terms
proportional to u2we find
γ(v)mv =ωg′(ω) + ω2g′′ (ω).(C9)
In eq. (C9), the left-hand side can be replaced by g(ω)
due to eq. (58). Therefore, we arrive at a differential
equation for the function g(ω),
ω2g′′(ω) + ωg′(ω)−g(ω) = 0 .(C10)
This is a homogeneous second-order linear equation with
the two independent solutions g1(ω) = ωand g2(ω) =
ω−1. Hence, the general solution is
g(ω) = C1ω+C2
ω,(C11)
where C1and C2are arbitrary constants. Substituting
this solution into eq. (C6) and integrating over ωleads
to the function f(ω),
f(ω) = C1ω−C2
ω+C3,(C12)
where C3is yet another constant. When we return to
physical units, we have to divide the right-hand side of
eq. (C11) by cto account the difference between the unit
of energy and that of momentum. In this way, we obtain
eqs. (63) and (64) of the main text.

15
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