•Conservation of momentum is one of the most fundamental and most useful concepts of elementary physis •Does it apply in special relativity? •Consider the following collision process. All balls have mass m and speed v before and after collision m Before m m m 1 0, ,0 u v 2 0, ,0 u v 3 ,0,0 u v 4 ,0,0 u v After •Surely this will conserve momentum p mu 2 4 1 3 0, ,0 0, , ,0,0 ,0, 0 0 p mv p m p m mv v v p 2 4 1 3 0,0, 0,0, 0 0 p p p p Momentum is conserve d But what about other frames?
Transcript
•Conservation of momentum is one of the most fundamental and most useful concepts of elementary physis•Does it apply in special relativity?
•Consider the following collision process. All balls have mass m and speed v before and after collision
mBefore
m
mm
1 0, ,0u v
2 0, ,0u v 3 ,0,0u v
4 ,0,0u v
After
•Surely this will conserve momentum
p mu
2
4
1
3
0, ,0
0, ,
,0,0
,0,
0
0
p mv
p m
p m
mv
v
v
p
2
4
1
3
0,0,
0,0,
0
0p p
p p
Momentum is conserved
But what about other frames?
m
m
mm
1 0, ,0u v
2 0, ,0u v 3 ,0,0u v
4 ,0,0u v
Momentum in another frame
x’
y’
v
2
2
,1
,1
xx
x
yy
x
u vu
u v c
uu
u v c
1
1
,1 0
,1 0
x
y
vu v
v vu
4 2 2 2 2
4 2 2
2,
1 1
00,
1
x
y
v v vu
v c v c
uv c
3
2 2
1
4
2
, ,0
, ,0
0,0,0
2 1 ,0,0
u
u v
v
v
c
v
v
u v
u
2 2
3
2
4
1 2 ,0,0
2 1 ,0,0
p p
p mv v c
v
p
m
In this frame, momentum is not conserved!
The problem with momentumHow can we explain this failure of conservation of momentum? Three choices:1. This process is physically impossible; it cannot occur2. Conservation of momentum is not a conservation law in S.R.3. Our formula for momentum is wrong in S.R. p mu
x x
y y
z z
xp mu m
ty
p mu mtz
p mu mt
• Why is t given special privilege over other coordinates?• We want something that is like time, but not coordinate
dependant in the denominator, like proper time t
x
xp m
x
xm mu
t
2 21
mup mu
u c
Does this fix the problem? (1)
2 21
mup
u c
1
2
, ,0
, ,0
u v v
u v v
2 2
1
1 v c
1 2 2
2 2 21x
mvp
v vc c
4 2 2
2 22 2 2
2 1
41 11
x
mvp
v vc c v c
2 2 2
2 2 21 1
mv
v v vc c c
2
21
mv
vc
2m v
22 2
2 2
2
41
mv
v vc c
2 4
2 4
2
21
mv
v vc c
22 mv
2xp
2
2
2
1
mv
vc
3
4 2 2
0,0,0
2,0,0
1
u
vu
v c
Does this fix the problem? (2)
Momentum is conserved
21
22
, ,0
, ,0
p mv mv
p mv mv
3
24
0,0,0
2 ,0,0
p
p mv
2
3 4
21 2 2 ,0,0
2 ,0,0
p
p
p
v
m
p m
v
What’s missing here . . ., ,
x y zp m
•Hint – we’ve got all three dimensions . . .
•But we live in four-dimensional spacetime•Is there a conserved fourth component of momentum?
t
tp m
•These quantities are related by Lorentz boost, just as px and py are related by rotations•If px, py and pz are conserved, then so must be pt
•Is it pt that is conserved, or cpt?•Actually it doesn’t matter; if pt is conserved, so is any multiple of it•We will work with a different multiple
•The ERIC quantity•Eric is conserved
2 2t
tE c p c m
t
2mc
2E mc
What is Eric, really?2E mc m1 m1
m2 m2
1u
Consider low velocity limit:
2u
1u
2u
1 2 1 2E E E E
1 2 2 2 2
1 2 1 2m c m c m c m c
Not careful enough . . .2
21
2
u
c
2 2 2 22 2 2 21 2 1 2
1 2 1 22 2 2 21 1 1 1
2 2 2 2
u u u um c m c m c m c
c c c c
2 2 2 21 1 1 11 1 2 2 1 1 2 22 2 2 2m u m u m u m u E is really
energy
Potential and Kinetic Energy2E mc
•E is the total energy – it includes both potential and kinetic energy•Total energy is always conserved
•Potential energy is the energy something has when it’s not moving•Also called rest energy•More about this later
•Kinetic energy is how much extra energy there is due to motion
20E mc
2kin 0 1E E E mc 2 4 231
2 8mv mv c
A sample problem . . .
20E mc
In 2004, the total world consumption of energy was 15 TW. How much fuel would have to be consumed each hour to power the world if we could extract
100% of its potential energy?
13 131.5 10 W 1.5 10 J/sP
2
Em
c
E Pt 131.5 10 J/s 3600 s 165.40 10 J
16 2 2
28
5.40 10 kg m /s
3.00 10 m/s
0.600 kg
Good vs. Bad Notation 2E mcp mu
Eric’s notation:•m, called the mass, is a constant, and independent of velocity
•It does not increase with velocity•It does not approach infinity as u c
•The famous formula E = mc2 is not true except when u = 0.•Einstein never wrote this formula
•The quantity m is then sometimes called the relativistic mass - but I won’t use this term.
2E mc2
0E mc
Satan’s notation:•m m0 is renamed the rest mass •We then define m = m0 as the mass
•In this notation, mass m goes to infinity as u c•In this notation, E = mc2 is always true
•Do not use this notation/nomenclature in this class
Momentum and Energy
2 21
mup mu
u c
•For small velocities, 1, and the momentum is just the usual formula•As u c, the momentum and energy both go to infinity
•This tells us we can’t reach u = c.
22
2 21
mcE mc
u c
u/c
E/mc2
p/mc
Particle physics units•Forces are usually induced by using electric fields•Charges are usually the same as the electron charge (-e) or simple multiples of this (+e, -2e, etc.)•Electric potentials are measured in volts (V)•Easy to work with units of electron charge times volts, or eV’s (called electron volts)
F qE
191 eV 1.602 10 C 1 V 191.602 10 J •Metric multiples are also common•Momentum has units of energy/velocity, so it is often expressed in units of eV/c
•Mass has units of energy/velocity2, so it is often expressed in units of eV/c2
3
6
9
12
1 keV 10 eV
1 MeV 10 eV
1 GeV 10 eV
1 TeV 10 eV
0.980 TeV /p c
20.938 GeV /m c
A sample problem . . .A Z-particle at rest (mass mZ = 91.19 GeV/c2) decays to a B-meson and an anti-B meson (both mass mB = 5.279 GeV/c2). What is the energy and velocity of the resulting B-mesons?
ZB B•Since the initial Z is at rest, pZ = 0 and EZ
= mZc2 = 91.19 GeV•The B-meson and anti-B-meson must have equal & opposite momenta•Since they have the same mass, this means they have equal speeds•They therefore have equal energies 1
2b ZE E 45.59 GeV2
b bE m c 5.279 GeV
45.59 GeV8.637
5.279 GeV
2 2
1
1 v c
2 2 21 1v c 0.9866
0.9933v c 82.978 10 m/s Note: Mass is not conserved!
Two Useful Formulas2E mcp mu
u pc
c E
Written this way, both sides are dimensionless
Square and subtract them:
2 2 2E c p 22 2 2 2 2mc c m u
2 2 2 2
2 21
m c c u
u c
2 4 2 2
2 2
m c c u
c u
22 2 2 2E c p mc
Divide them:
2
p mu
E mc
2
u
c
Different types of particlesu pc
c E
2 2 2 2 4E c p m c
Must the combination E2 – c2p2 always be positive? Maybe m2 needn’t always be positive
1. If m2 > 0 we call the particle a massive particle• Electrons, protons, rockets, people• E > cp, so u < c: Always slower than light
2. If m2 = 0 we call the particle a massless particle• Photons (particles of light)• E = cp, so u = c: Always at the speed of light
3. If m2 < 0 we call the particle a tachyon• No known particles• E < cp, so u > c: Always faster than light
Massless and nearly massless particles2E mcp mu
u pc
c E
E c p
•First two equations make no sense for massless particles – don’t use them•If mass is very small (mc2 much smaller than E or cp), then often easier to treat as massless•Massless versions of these equations:
22 2 2 2E c p mc
u c
The Large Electron-Positron collider (LEP) accelerates electrons (me = 0.511MeV/c2) to an
energy of E = 100 GeV. How fast are they moving?
u c
0.999999999987u c
Sample problemA Z-particle at rest (mZ = 91.2 GeV/c2) decays to an electron (me =
0.511 MeV/c2) with energy 25.0 GeV, a positron (me) with energy 20.0 GeV moving at 71° angle compared to the electron, and an unknown X particle. What is the momentum, energy, mass, and velocity of X?
Z e-
e+
X
91.2 GeV
0Z
Z
E
p
•Electrons are nearly massless
1 25.0 GeVE 71
2ˆ ˆ20.0 cos 71 sin 71 GeVp c i j
ˆ ˆ6.51 18.9 GeVi j
2 20.0 GeVE
1ˆ25.0 GeVp c i
Conservation of energy:
91.2 25.0 20.0 GeV 1 2X ZE E E E 46.2 GeVXE
Continued . . .
Sample problem continued . . .. . . What is the momentum, energy, mass, and velocity of X?
Suppose I measure the momentum and energy of a particle or collection of particles. What does a moving observer measure?•Far easier than the book makes it out•Difference in coordinates (x, y, z, t) satisfy same equations as (x, y, z, t)
2
x x v t
y y
z z
t t v x c
2x x
y y
z z
x
p p vE c
p p
p p
E E vp
These formulas work for either a single particle or for the sum of all particles
