Relativistic Mechanics
Mass, Energy, and Momentum
General Relativity
“In science one tries to tell people, in such a way as to be understood by
everyone, something that no one ever knew before. But in poetry, it's the exact
opposite.”—P. A. M. Dirac
Relativistic Momentum
We believe very strongly that momentum is conserved. Let’s
see what effect a relativistic calculation has on momentum.
Imagine this collision: person A on the ground throws a ball
straight at the side of a railroad flatcar moving past him.
Person B moving with the flatcar throws an identical ball
straight out at the same speed (relative to the flatcar). The
two balls meet head-on and rebound.
B
v
flatcar’s velocity
B throws ball in this direction
B’s ball also has a velocity component
in this direction…
A throws ball in
this direction
…so that B’s ball has a total velocity
along this direction and follows this path
A
collision! momentum is (??) conserved
B
v
flatcar’s velocity
B throws ball in this direction
B’s ball also has a velocity component
in this direction…
A throws ball in
this direction
…so that B’s ball has a total velocity
along this direction and follows this path
A
collision! momentum is (??) conserved
Read Beiser for a more detailed description of this thought
experiment.
Because of time dilation, A and B measure different travel
times for ball thrown from the flatcar. They will calculate
different initial and final momenta. They will not agree that
momentum is conserved.
Non-conservation of momentum is an alarming idea. What
can we do to fix this situation?
It turns out that if the ball thrown by the moving* flatcar
observer has a greater mass than the ball thrown by the
stationary observer, momentum is conserved:
m(v) =
m
1- v
,
2
c2
where m is the mass of the ball at rest, and m(v) is the mass it
needs to have when it is moving.
*So we are taking the point of view of the observer on earth.
In the old days* we then said “OK, m is the mass of the
object at rest and m(v) is its mass when it is moving. Let’s
call m0 the rest mass and m the mass when it is moving
(‘relativistic mass’).” This notation is consistent with our
equations for time dilation and length contraction, so we
have
m0
m=
.
2
1- v 2
c
Mass depends on speed. This was Einstein’s original
approach, but later he said it is “not good.”
* when I studied relativity in college. Also, in the previous edition of our
modern physics text.
The “new” approach is to say “Look, mass is mass. We
believe it is something fundamental. If we believe in
conservation of momentum, we had better change our
definition of momentum.”
If we define momentum as p = mv where  =
1
1- v
,
2
c2
then “mass is mass,” momentum is conserved in our thought
experiment (and in real life), and relativistic momentum
reduces to classical (Newtonian) momentum in the limit
v0.*
*More satisfying than saying “mass changes with velocity.”
In the old days, rest mass was relativistically invariant. Now
mass is relativistically invariant. Same reality, just different
use of words.
Let’s make this new notation official.
OSE :
=
1
2
v
1-
,
OSE :
p = mv
c2
A consequence of this new definition of momentum:
F=
dp
= ma
dt
F=
dp d
=  mv 
dt dt
More consequences of this new definition of momentum:

2
F
v
a=
1c2
m

3/2
relativistic momentum increases
without bound as vc; c is the
“speed limit”
For finite F,
a0 as vc.*
No finite force can
accelerate an object
having nonzero
mass up to the
speed of light!
*example 1.5
classical momentum
increases linearly
with velocity; no limit
to classical velocity
“When can I use classical mechanics, and when do I have to
use relativistic mechanics?”
1
.
The correction factor for momentum is  =
2
1- v 2
c
Here’s a table showing the correction factor for different
objects:
Object
v
v/c
m(v)/m
jogger
10 km/h .000000009
≈1
space
shuttle
104 m/s
electron
106 m/s
0.0033
1.001
electron
108 m/s
0.333
1.061
0.000033 1.0000001
subtract 1, multiply by 100 to get % error
Mass and Energy
From your first-semester physics course:
d  mv 
KE   F  ds  
 ds.
dt
Use the definition of  and integrate by parts to get
KE = mc 2 - mc 2 =   - 1  mc 2
mc2 = mc2 + KE.
Assuming potential energy is zero (we can always choose
coordinates to do this), we interpret mc2 as total energy.
E = mc 2 + KE.
The “box” indicates an OSE.
When an object is at rest KE = 0, and any energy that
remains is interpreted as the object’s rest energy E0.
E0 = mc 2 .
When an object is moving, its total energy is
E = mc 2 =
mc 2
2
v
1-
.
c2
This is really just a
variation of the OSE
on the previous slide.
This is the closest you’ll come to seeing E=mc2 in
this class. In the “old days,” E=mc2 would have
been written E=mc2.
These equations have a number of interesting implications.
Mass and energy are two different aspects of the same “thing.”
Conservation of energy is actually conservation of massenergy.
The c2 in E0=mc2 means a little mass is “worth” a lot of
energy.
Your lunch: an example of relativity at work in “everyday life.”
Total energy is conserved but not relativistically invariant.
Rest (or proper) mass is relativistically invariant.
Mass is not conserved! (But it is for the purposes of chemistry.)
Example: when 1 kg (how much is that?) of dynamite
explodes, it releases 5.4x106 joules of energy. How much
mass disappears?
This is actually a conservation of mass-energy problem. If
this material were presented in Physics 23, I would make
you start with your conservation of mass-energy OSE and
derive the appropriate equations from there.
For now, it is sufficient to realize that the problem is just
asking “what is the mass equivalent of 5.4x106 joules of
energy?”
E0
2
E0 = mc
m= 2
c
m=
5.4 ×106
3×10 
8 2
= 6 ×10-11 kg.
Conservation of mass is a
very good approximation!
If we are to claim relativistic mechanics as a replacement
theory for Newtonian mechanics, then relativistic mechanics
had better reduce to Newtonian mechanics in the limit of
small relative velocities.
KE = mc 2 - mc 2 =
mc 2
2
v
1-
Our text shows that for v<<c,
1
KE  mv 2 .
2
- mc 2 .
c2
When can I get away with using KE = mv2/2, and when do
I have to use KE = mc2 - mc2?
Use Newtonian KE every time you can get away with it!
Use relativistic KE only when you must!
If v = 1x107 m/s (fast!) then mv2/2 is off by only 0.08%.
Probably OK to use mv2/2. If v = 0.5 c, then mv2/2 is off
by 19%. Better use relativity.
Energy and Momentum
Total energy and magnitude of momentum are given by
E=
mc 2
1- v
mv
p=
2
c
1- v
2
.
2
c2
With a bit of algebra, you can show
E - p c = mc
2
2 2
2

2
.
The quantities on the LHS and RHS of the above equation
are relativistically invariant (same for all inertial observers).
Rearranging:
E = mc
2
2

2
2 2
+p c .
See p. 31 for comment
on particle vs. system
of particles.
E=
mc 2
1- v
p=
2
c2
mv
2
v
1-
c2
Is it possible for a particle to have no mass? If m = 0, what
are E and p?
For a particle with m = 0 and v < c, then E=0 and
p=0. A “non-particle.” No such particle.`
But if m = 0 and v = c, then the two equations above are
indeterminate. We can’t use them to say anything about our
proposed particle.
If m = 0 and v = c, we must use E2 = mc 2  + p2c 2 .
2
The energy of such a particle is E = pc. We could detect this
particle! It could exist!
Do you know of any massless particles?
 photon
 neutrino*
 graviton**
graviton is to gravity as photon is to E&M field
*Doubtful. Nobel prize for you if you prove mneutrino = 0 or mneutrino ≠ 0.
**Maybe. Nobel prize for its discoverer. Problem: gravitational fields are
much, much weaker than E&M fields.
 Particles having KE >> E0 (or pc >> mc2) become more
photon-like and behave more like waves.
 The momentum carried by massless particles is nonzero
(E = pc).
Quickly concealing
his pocket laser
pointer?
Could you stop a freight train with a flashlight?
Could you stop a beam of atoms with a laser beam?
A note on units.
We will use the electron volt (eV) as an energy unit
throughout this course.

1 eV  1.6  10 19 C
  1 V   1.6  10
19
J
Variations on the eV:
1 meV = 10-3 eV (milli)
1 keV = 103 eV (kilo)
1 MeV = 106 eV (mega)
1 GeV = 109 eV (giga)
There are a few rare cases where mixing eV units in
calculations with SI units leads to numerical errors. I will
warn you when we come to those cases.
Because mass and energy are convenient, we sometimes
write masses in “energy units.”
An electron has a rest mass of 9.11x10-31 kg. If you plug
that mass into E0 = mc2, you get an energy of 511,000 eV, or
511 keV, or 0.511 MeV, or 0.511x10-3 GeV.
We sometimes write the electron mass as 0.511 MeV/c2.
It is also possible to express momentum in “energy units.”
An electron might have a momentum of 0.3 MeV/c.
If you are making a calculation with an equation like
E = mc
2
2

2
+ p2 c 2
and you want to use 0.511 MeV/c2 for the electron mass,
please do. It often simplifies the calculation. But watch
out…
What is the total energy of an electron that has a momentum
of 1.0 MeV/c?
E = mc
2
2

2
+ p2 c 2
2
2
 0.511 MeV
 1.0 MeV  2
2
2
E =
×c  +
 c
2
c
c




E2 =  0.511 MeV  + 1.0 MeV 
2
2
E2 = 1.26 MeV 2 
E =1.12 MeV
Notice the convenient cancellation of the c’s in the 2nd step.
Avoid the common mistake: don’t divide by an extra c2 or
multiply by an extra c2 in the 2nd step.
General Relativity
Something to think about. Is the mass that goes in F =
ma (or the relativistic version) the same “thing” as the
mass that goes in F = Gm1m2/r2?
Not necessarily!
Experimentally, the two “kinds” of mass are the same to
within better than one part in 1012, and most of us believe
they are the same anyway, so…
“An observer in a closed laboratory cannot distinguish
between the effects of a gravitational field or an acceleration
of the lab.”
The principle of equivalence.
The principle of equivalence leads one to conclude that light
must be deflected by a gravitational field.
Experimental observation of this effect in 1919 was one of
Einstein’s great triumphs.
We will hear more about light and gravity in the next
chapter.
Before I give the first quiz, I would like to work a few problems
in class. I’ve already done time dilation and length contraction
problems. I should work some problems with energy and
momentum.
"If A equals success, then the formula is: A=X+Y+Z. X is work. Y is play.
Z is keep your mouth shut.“—A. Einstein
Problem 1.29. At what speed does the kinetic energy of a
particle equal its rest energy?
Kinetic energy:
K  mc 2  mc 2
Rest energy:
E0  mc 2
When the two are equal E0=K so that:
mc 2  mc 2  mc 2
2
mc  2mc
2
 2
2
1
2
v
1
c2
Is it a good idea to
combine several algebraic
steps in one “line?”
Only if you can guarantee
that you will never make a
mistake.
2
v
1
2
v
1
v2
1
2 
c
2
1
2 
c
4
1
2 1
c
4
v2
3
2 
c
4
3 2
v  c
4
2
v
3
c
2
cross-multiply
square both sides
solve for v2/c2
simplify
solve for v2
Problem 1.33. A particle has a kinetic energy 20 times its rest
energy. Find the speed of the particle in terms of c.
K  mc 2  mc 2
mc 2  K  mc 2
mc 2
2
v
1
 K  mc 2
rearrange
plug in expression for 
c2
2
mc 2
v
 1
2
2
c
K  mc
algebra
2
mc 2
v
 1
2
2
c
K  mc
1
K
1
mc 2
2
v
 1
1
K 

1  mc 2 


2
c
2
v
1
rewriting equation, so
you can see the next step
÷ LHS top and bottom by mc2
2
c
v2
1

1

2
c2
K 

1  mc 2 


2
square both sides
solve for v2/c2
v
1
 1
2
c
K


1


2 
mc


for the exam,
you should have
this equation
v  c 1
solve for v; not “official,”
but often useful
1
K 

1  mc 2 


take  of both sides
2
K MUST be the relativistic
kinetic energy. K=½mv2
is wrong here!
By now you’ve forgotten what the problem asked, so here it is
again: “A particle has a kinetic energy 20 times its rest
energy. Find the speed of the particle in terms of c.”
But the rest energy is just mc2, and the problem says use
K=20mc2.
v  c 1
1

20mc 
1 
2 
mc


v  c 1
2
1
1  20 
1
v  c 1
441
2
2
use K=20mc2
algebra
math
v  0.9989 c
Important conceptual result: if an object’s kinetic energy (or total energy) is much
greater (“much” meaning a factor of 5 or 10 or so) than its rest energy, the object is
going almost the speed of light.
Problem 1.43. Find the momentum (in MeV/c) of an electron
whose speed is 0.6c.
“I thought momentum had units of [mass]·[velocity]?”
It does! And MeV/c works out to [mass]·[velocity], but is a
much more convenient unit to use when working with
relativistic particles. Physicists who deal with high-energy
particles use MeV/c for momentum units, and MeV/c2 for mass
units.
p  mv
p
mv
1 v
OSE
plug in expression for 
2
c2
The electron mass in “energy units” is 0.511 MeV/c2.
MeV
0.511 2  0.6c
c
p
2
 0.6c 
1
c2
MeV
0.511 2  0.6c
c
p
2
 0.6c 
1
c2
MeV
0.511  0.6
c
p
1  0.36
substitute numerical values
notice how conveniently the
units lead to cancellation
don’t forget to square the 0.6
MeV
p  0.383
c
My Mathcad solution is rather klunky. For the quiz or exam,
work the problem the way I did here.
“I don’t trust these funny “energy units” for mass and
momentum. Would I get the correct result if I work the
problem in SI units?”
How many different approaches are there to solving any
physics problem? How many right answers are there to any
physics problem?
p  mv
p
p
mv
2
v
1
c2
9.11  10 31 kg  0.6  3  10 8 m / s
0.6  3  10 m / s 

1
8
p
2
 3  10 m / s 
1.64  1022 kg  m / s
1  0.36
8
2
p  2.05  1022 kg  m / s
“Is this answer correct?”
If this were a multiple choice quiz problem and I asked for an
answer in MeV/c, you would not see the above answer as a
choice. If this were a “work-it-out-for-partial-credit” problem,
I would take off a few points (one point for “math,” plus a few
more points, depending on how annoyed I happened to be
when I was grading).
“Can you show me how to convert to energy units?”
Sure. Keep in mind that in algebra, you can always multiply
or divide by 1, or add or subtract 0, without changing your
answer. (You may change the form of the answer, but not the
value.)
m
3  10
kg

m
s
p  2.05  10 22

s
c
1
kg  m2
2
14
s
p  6.15  10
c
8
p  6.15  1014
p  6.15  10 14
joules
c
joules
eV
MeV


c
1.6  10 19 joules 106 eV
1
1
6.15
14 196 MeV
p
 10
1.6
c
MeV
p  3.84  10
c
1
MeV
p  0.384
c
roundoff error crept into the
3rd decimal place
Which way would you rather work this problem?
Keep in mind that every time you punch a button on your
calculator, and every time you write a symbol on paper, you
have increased the chance that you made a mistake.*
*Also true, I suppose, when you show all your steps, but if each step is simple, the chance for
error goes way down, plus I can give more partial credit if I see where you made your mistake.