RELATIVITY
PART III
• Momentum
• Rest Mass
• C is the ultimate speed limit
• Forces
• Energy
Momentum
Is momentum conserved in relativistic collisions?
Yes, if you redefine what you mean by “momentum”!
The classical definition of momentum, p=mv, is not conserved.
Invariant v.s. Conser ved
Invariant quantities are those that remain the same when
calculated in two different reference frames.
examples: c, Δs, q, m
Conserved quantities are those that remain the same during
some process in a given reference frame.
examples: energy, momentum
Momentum
S’
y’
v
A
x’
S
y
B
x
Sam and Sam Prime are
moving relative to each other
with speed v along the x axis.
Momentum
S’
y’
v
A
-u0
S
u0
x’
They each throw a ball with
speed u0 along their y axis
(measured in their rest frame).
y
B
x
Momentum
What Sam sees:
S’
v
Y component of
Sam Prime’s ball is
u yB =
uyB
u yB =
S
u0
u y! 1 " v 2 / c 2
vu !
1 + 2x
c
!u0 1 ! v 2 / c 2
v"0
1+ 2
c
u yB = !u0 1 ! v 2 / c 2
m u yA ! m u yB
Relativistic Momentum
Redefine momentum as:
!
p!
where
!u "
!
mu
1" u / c
2
2
!
= # u mu
1
1 # u 2 / c2
" = 1 then lim p = mv
Notice that since lim
v!0
v!0
Momentum
y momentum of Sam’s Ball:
mu0
pyA !
1 " u02 / c 2
pyB !
y momentum of Sam Prime’s Ball:
pyB =
(
!m u0 1 ! v 2 / c 2
(
1 ! v 2 + u 20 1 ! v 2 / c 2
=
)) / c
2
!m u0 1 ! v 2 / c 2
(1 ! v
2
)(
/ c 2 1 ! u 20 / c 2
=
)
(
m u yB
)
1 " u 2xB + u 2yB / c 2
!m u0 1 ! v 2 / c 2
(
)(
1 ! v 2 / c 2 ! u 20 / c 2 + v 2 / c 2 u 2 / c 2
=
So, we see that momentum is conserved:
)
!m u0
1 ! u 20 / c 2
pyA + pyB = ! u mu0 " ! u mu0 = 0
Mass
We define mass to be invariant. It does not change under a
Lorentz transformation. If you wanted to define m ( u ) ! " u m
as the “relativistic mass” then you could use the classical
!
!
definition of momentum p = m ( u ) u , but Einstein doesn’t think
it is a good idea.
Many texts still write the mass as m0 to distinguish it from the
relativistic masss m (u )
Mass is Rest Mass
“It is not good to introduce the
1/2
concept of mass M = m / (1 ! v 2 / c 2 )
of a body for which no clear
definition can be given. It is better to
introduce no other mass than “the
rest mass” m. Instead of introducing
M, it is better to mention the
expression for the momentum and
energy of a body in motion.”
–Albert Einstein
c is the ultimate speed limit
4
3
p/m
v
1 ! v2 / c2
2
1
v
0
0
0.2
0.4
0.6
0.8
v/c
Relativistic Forces
The standard form of Newtons 2nd Law is NOT relativistic
!
!
F = ma
But the Momentum form is relativistic
! dp!
F=
dt
1
Kinetic Energy
Work-Kinetic Energy Theorem “works” relativistically
Let EK = kinetic energy
u
Ek =
!
u
u
d
= "
(! u mu ) dx
dt
u=0
F dx
u=0
=
" ud (!
u
mu )
u=0
You will show that
#
u2 &
d (! u mu ) = m % 1 " 2 (
c '
$
"3/2
du
plug in
$
u2 '
Ek = " ud (! u mu ) = " m & 1 # 2 )
c (
%
0
0
u
u
#3/2
u du
"
%
1
= mc 2 $
! 1'
2
2
# 1! u / c
&
Kinetic Energy
Relativistic Kinetic Energy
Low velocity limit:
so
Ek = ! u mc 2 " mc 2
#
u2 &
! u = %1 " 2 (
c '
$
1/2
)1+
1 u2
+!
2 c2
"
%
1 u2
1
Ek ! mc 2 $ 1 +
+ !' ( mc 2 ! mu 2
2
2c
#
&
2
= ! u mc 2 " mc 2
Total Energy & Rest Energy
Ek = ! u mc 2 " mc 2
Relativistic Kinetic Energy
Total Energy:
Total Energy
Rest Energy
E = ! u mc 2
E = mc 2
E = Ek + mc 2
Low velocity limit v << c:
E!
1 2
mu + mc 2
2
Energy & Momentum
Classical Physics:
Relativity:
Ek =
p2
2m
(
E 2 = ( pc ) + mc 2
2
Rewrite as:
Compare to spacetime interval:
)
2
( mc )
2 2
= E 2 ! ( pc )
2
!s 2 = ( c!t ) " ( !x )
2
2
Units
Atomic Physics:
electron Volt
High Energy Physics:
1 eV = 1.602 ! 10 "19 Joules
MeV = 10 6 eV
GeV = 10 9 eV
TeV = 1012 eV
Masses of fundamental particles are usually quoted in terms of
energy m = E / c 2
Rest Masses
Photon (γ)
0
a.m.u. (u)
Electron (e–)
0.5110 MeV/c 2
Deuteron (2H) 1875.613 MeV/c 2
Muon (μ–)
105.7 MeV/c 2
Hydrogen (H)
Proton (p)
938.272 MeV/c 2
Helium (He)
Neutron (n)
939.565 MeV/c 2
931.494 MeV/c 2
938.783 MeV/c 2
3727.379 MeV/c 2
Nuclear Masses
When a particle is composed of smaller units bound together
by attractive forces, the rest mass of the particle is given by:
Rest Mass of System = Σ(rest masses) + Σ(PE) + Σ(KE)
–
+
–
!##
#"###
$
PEeff
Since the binding energy (i.e. PE) is negative, the rest mass of
particles is usually less than the rest mass of the composite particle.
This means that energy is released when particles combine, and
that energy is required to break them apart.
Mass of Deuteron
mass of Proton (p)
938.272 MeV/c 2
+ mass of Neutron (n) 939.565 MeV/c 2
Total
1878.348 MeV/c 2
mass of Deuteron (2H) 1875.613 MeV/c 2
mass difference
2.735 MeV/c 2
Hydrogen Atom
Binding energy of electron to proton13.6 eV has a mass
equivalent of 0.0000136 MeV/c2 or 1.46x10–8 u
mass of Proton (p)
938.272 MeV/c 2
+ mass of Electron (e–)
Total
mass of Hydrogen (H)
0.5110 MeV/c 2
938.783 MeV/c 2
938.783 MeV/c 2
Massless Par ticles
The idea of a “massless particle” in classical physics makes no sense:
KE =
1 2
mv ! 0
2
p = mv ! 0
Fg =
GmM
!0
r2
A classical particle can not “function” without mass.
Massless Par ticles
In relativity a massless particle can still have energy & momentum:
E=
( pc )2 + ( mc 2 )
2
E = pc
0
Derive the speed of a massless particle:
p = ! mv =
mv
1" v / c
2
2
=
E
c
Since the numerator goes to zero, the denominator must be zero as
well for the ratio to be finite.
v=c
Par ticles & Antipar ticles
P. Dirac showed that relativity and quantum mechanics predict the
existence of antimatter.
Antiparticles have the same mass, but opposite charges of their
counterparts. But, when they meet, the annihilate each other,
converting their mass into energy.
positronium:
electron (e–)
highly unstable
lifetime ~10-10s
positron (e+)
Par ticles & Antipar ticles
The electron and the positron will annihilate to form two oppositely
moving photons (gamma rays), each with 0.511 MeV of energy
0.511 MeV photon
0.511 MeV photon
Pair Production
Pair production is the opposite process of annihilation: mass is
created from energy (usually high energy photons).
In order for pair production to occur, a nearby particle must be
present to carry away some momentum.
moving electron (e–)
E1
p1
innocent electron (e–)
at rest
E = mc 2
p=0
Pair Production
After the reaction:
positron/electron pair
v
innocent electron (e–)
Each has:
v
v
moving electron (e–)
E2
p2
v
Minimum initial speed for pair production occurs when all 4 particles
have the same final velocity.
Show that the KE of the original moving electron must be = 6mc2