PHYS 3313 – Section 001
Lecture #8
Monday, Feb. 16, 2015
Dr. Jaehoon Yu
•
•
•
•
•
Relativistic Energy
Relationship Between Relativistic
Quantities
Binding Energy
Quantization
Discovery of the X-ray and the Electron
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
1
Announcements
• Reading assignments: CH3.9
• Homework #2
– CH3 end of the chapter problems: 2, 19, 27, 36, 41, 47
and 57
– Due Wednesday, Feb. 25
• Quiz #2 Monday, Feb. 23
– Beginning of the class
– Covers CH1.1 – what we finish this this Wednesday,
Feb. 18
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
How do we keep momentum conserved in a relativistic case?
Redefine the classical momentum in the form:
p = G ( u ) mu =
1
1- u 2 c 2
mu
This (u) is different than the factor since it uses the particle’s speed u
 What? How does this make sense?
 Well the particle itself is moving at a relativistic speed, thus that
must impact the measurements by the observer in the rest frame!!
Now, the agreed form of the momentum in all frames is ( is the proper
time):
1
dr
dr dt
mu
p=m
=m
= mug =
2
2
dt
dt dt
1- u c
p = mg u
Resulting in the new relativistic definition of the momentum:
When u0, this formula becomes that of the classical.
What can the speed u be to maintain the accuracy of the classical
momentum
to 2015
1%?
Monday, Feb. 16,
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
3
Relativistic Energy
• Due to the new idea of relativistic mass, we must now
redefine the concepts of work and energy.
– Modify Newton’s second law to include our new definition of linear
momentum, and force becomes:
æ
ö
d
mu
dp
d
ç
÷
F=
=
g mu =
2
dt çè 1- u 2 c ÷ø
dt
dt
(
)
• The work W done by a force F to move a particle from rest to
d
a certain kinetic energy is
W =K=ò
g mu ) ×udt
(
dt
• Resulting relativistic kinetic energy becomes
gu
K = ò um × d (g u ) =g mc - mc =(g -1) mc
2
2
2
0
• Why doesn’t this look anything like the classical KE?
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
4
Big note on Relativistic KE
• Only K = (g -1) mc is right!
2
•
1 2
1
2
K = mu and K = g mu are
2
2
wrong!
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
5
Total Energy and Rest Energy
Rewriting the relativistic kinetic energy:
g mc2 =
mc 2
1- u c
2
2
= K + mc 2
The term mc2 is called the rest energy and is denoted by E0.
E0 = mc
2
The sum of the kinetic energy and rest energy is interpreted
as the total energy of the particle. (note that u is the speed of
the particle)
ETot = g mc =
2
Monday, Feb. 16, 2015
mc 2
1- u 2 c 2
=
E0
1- u c
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
2
= K + E0
6
Relativistic and Classical Kinetic Energies
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
7
Relationship of Energy and Momentum
p = g mu =
mu
1- u 2 c 2
We square this formula, multiply by c2, and rearrange the terms.
2
æ
ö
u
2 2
2 2 2 2
2 2 4
p c = g m u c = g m c ç 2 ÷ = g 2 m2c4 b 2
èc ø
æ
1ö
b = 1- 2 Þ p c = g m c ç 1- 2 ÷ = g 2 m 2c 4 - m2 c 4
g
è g ø
2
1
Rewrite
Rewrite
Monday, Feb. 16, 2015
2 2
2
2 4
p c =E - E
2 2
2
2
0
E = p c +E = p c +m c
2
2 2
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
0
2 2
2 4
8
Massless Particles have a speed
equal to the speed of light c
• Recall that a photon has “zero” rest mass and
the equation from the last slide reduces to: E =
pc and we may conclude that:
E = g mc = pc = g muc
2
• Thus the speed, u, of a massless particle must
be c since, as m ®0, g ® and it follows
that: u = c.
¥
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
9
Units of Work, Energy and Mass
• The work done in accelerating a charge through a
potential difference V is W = qV.
– For a proton, with the charge e = 1.602 × 10−19 C being
accelerated across a potential difference of 1 V, the
work done is
1 eV = 1.602 × 10−19 J
W = (1.602 × 10−19)(1 V) = 1.602 × 10−19 J
•eV is also used as a unit of energy.
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
10
Other Units
1)
Rest energy of a particle:
Example: Rest energy, E0, of proton
(
)(
)
2
E0 ( proton ) = m p c = 1.67 ´10 -27 kg × 3.00 ´10 8 m s = 1.50 ´10 -10 J
1eV
-10
8
= 1.50 ´10 J ×
=
9.38
´10
eV
-19
1.602 ´10 J
2)
Atomic mass unit (amu): Example: carbon-12
M(
12
12 g mole
C atom ) =
6.02 ´10 23 atoms mole
= 1.99 ´10 -23 g atom
M ( 12C atom ) = 1.99 ´10 -26 kg atom = 12u atom
What is 1u in eV?
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
11
Binding Energy
• The potential energy associated with the force
keeping a system together  EB.
• The difference between the rest energy of the
individual particles and the rest energy of the
combined bound system.
M bound system c + EB = å mi c
2
EB = å mi c - M bound system c
2
2
i
2
i
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
12
Examples 2.13 and 2.15
• Ex. 2.13: A proton with 2-GeV kinetic energy hits
another proton with 2 GeV KE in a head on
collision. (proton rest mass = 938MeV/c2)
• Compute v, , p, K and E for each of the initial
protons
• What happens to the kinetic energy?
• Ex. 2.15: What is the minimum kinetic energy the
protons must have in the head-on collision in the
reaction p+p++d, in order to produce the
positively charged pion (139.6MeV/c2 ) and a
deuteron.(1875.6MeV/c2).
Monday, Feb. 16, 2015
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
13