Relativity III
I.
Collisions
Linear momentum is still conserved in collisions of high speed particles. Modern
physicists use this knowledge and extremely large particle accelerators to study
the ultimate questions of the universe (1. What is matter?; 2. What are the forces
that exist?; 3. How did the universe begin?; etc..). You should work through all of
the “Test Your Knowledge Problems” and make sure that you can do these
problems as they require you to synthesize many of the concepts that we have
covered.
II.
Work-Energy Theorem and Special Relativity
Einstein also considered the effect of Special relativity upon classical concepts of
energy!
A.
Kinetic Energy
The work-energy theorem in PHYS1224 related the work done by the net external
force to the change in an object's kinetic energy! We now calculate the net work
done on a body moving at relativistic speeds in the x-direction by
f
ΔK  
i
f
ΔK   dp
i
dp
dx
dt
f
dx
 v dp
dt i
Using integration by parts, we have
f
ΔK  pv i   p dv  mv
f
2 f
i
f

i
i
mo v
v
1-  
c
2
dv
For a body whose initial speed was zero, the equation reduces to
v
K  mv  m o 
2
0
v
v
1-  
c
2
dv .
2
-c
v
To evaluate the integral, we define η  1    . Thus, v dv 
dη .
2
c
2
Substituting into our integral, we have that

v
v
1-  
c
2
dv 
- c2
2
η
-1/2
dη   c 2 η1/2
We now evaluate our equation by applying the limits
K  mv  m o c η
2
2
v
1/2 1  c 
1
2
2
v
K  mv 2  m o c 2 1     m o c 2
c
We can simplify this equation using our mass formula and a little bit of algebra
v
K  mv 2  m o c 2 1   
c



mo
K  mv 2  
2

v
1

 

c

2
v
1  
c
2
v
1  
c
2
 moc2


   v 2 
c 2  1      m o c 2
   c  


  v 2 
K  mv 2  mc 2 1      m o c 2
 c 


K  mv 2  mc 2  mv 2  m o c 2
K  mc 2  m o c 2
K  m  mo  c 2  Δ mc 2
This amazing result states that adding kinetic energy to an object is the SAME
as increasing the object's mass!! So if you go to the lab and heat the gas in a
balloon, you actually make it weigh more!!
Note : We obtained a totally different result for kinetic energy using the workenergy theorem than for Newtonian Physics. In relativity,
B.
K  1 M v2
2
!!
Einstein's Mass-Energy Equivalence
Our amazing result in part A yields more surprises!! First, we rewrite the equation
as
m c 2  K  mo c 2 .
By dimensional analysis, we know that every term in our equation must have the
units of energy!
Therefore, mass appears to be a type of energy or equivalently energy is a type of
mass. This means that we need to change our notions about the independence of
mass and energy.
Classical Physics - Two Conservation Laws That Must Be Obeyed
1)
The mass of the universe is conserved.
2)
The total energy of the universe is conserved.
Special Relativity - One Mass-Energy Conservation Law
1)
The total mass-energy of the universe is conserved.
Since mass is just a type of energy, it can be converted into other types of energy
including kinetic! Thus, mass may not be conserved in a system. Energy can be
converted into mass (kinetic energy converted into particles in particle
accelerators) or mass into energy (stars, power plants, atomic bombs, etc.).
C.
Kinetic Energy
In part A, we found a formula for calculating the kinetic energy of a body due to
its change in mass (mass defect). However, we also need to remember the basic
definition of kinetic energy.
Kinetic energy is defined as the energy a body has due to its motion.
Thus, we could rewrite our previous equation as
m c 2  motion energy mo c 2 .
D.
Rest Energy - Eo
If the body is at rest (K=0), both sides of our equation reduce to moc2. Thus, this
quantity represents a type of energy that the body has at rest due to its intrinsic
mass!!
Therefore, a body that would have no energy in Classical Physics still has a
minimum amount of intrinsic energy due to its intrinsic mass. Einstein called this
quantity the rest energy.
Eo  mo c2
E.
Total Energy - E
We can now rewrite our amazing energy equation as
m c 2  motion energy   rest energy  .
Thus, the quantity on the left-hand side of the equation must represent the total
energy of the body. Thus, the true (relativistic) mass tells us the total energy of
the body. Mass and energy are equivalent!!
E  m c2
We have now derived the most famous equation in all of modern physics!!
Aside: You will often hear people on TV complaining that Einstein and physicist
shouldn't have been trying to make bombs or that physics is a useless degree
unless you want to be a bomb builder. These comments are due to a lack of
science literacy among the general public. Even extremely well educated people
generally have little understanding about one practices science. Consider the
individual steps in our journey over the past year,
1)
Galileo proposes the law of inertia by rolling balls down inclined planes.
2)
Equations are needed to compare measurements between different
observers (Galilean Transformation). It seems necessary that the rolling
ball's motion have some reality and not be a hallucination of a particular
observer.
3)
Newton develops the Laws of Mechanics that describes the cause of the
ball motion and allows us if we know the cause (net external force) to
predict the acceleration, velocity, and position of the ball at each instant of
time.
4)
Maxwell developed the laws of Electricity and Magnetism.
5)
Experiments including the Michelson-Morely experiment indicate that we
have the wrong equations for describing the motion of a ball moving at
high speeds.
6)
Einstein and Lorentz obtain the correct equations (as far as our
experiments allow us to determine) for a fast ball moving down a plane.
What physicist was trying to build an atomic bomb or weapon?
From the apparently harmless observation of balls rolling down planes, we are
inevitably led to an equation that states: If a ball's intrinsic desire to travel in a
straight line at constant speed is reduced then a tremendous amount of energy will
be released. Galileo invented the concept of inertia while doing his experiments
and the concept of energy developed 300 years later. How could he have know
what Einstein would discover.
This is true of all basic research. To ask a scientist doing basic research the
question "what will you discover" or "what will you create from your research"
is to not understand science! History suggests that the over all pay-off from basic
research will be tremendous although the importance of any particular project is
difficult or impossible to determine over the short range.
The building of the atomic bomb was a great engineering achievement and not a
great physics achievement. This is also true of going to the moon. The physics
was known! The question was how to apply the physics to make a bomb. It was a
series of technical engineering challenges. Why didn't engineers run the
Manhattan bomb project! The science was new and hadn't entered into the
engineering curriculum. Most devices based upon new scientific discoveries are
built by physicists (laser, transistor, radar, NMR, x-rays, etc.). Once the device
becomes important, the information is disseminated into the engineering
education. Engineers that apply their special talents of designing and constructing
improved, reliable, and cost effective devices and physicists go back to doing
basic scientific discovery.
Fact: The time between a scientific discovery and its impact on the market
place is shrinking. A basic research project in genetic engineering or laser physics
becomes a new bioengineering or information technology product in a couple of
years. Thus, there is a need for individuals who transcend the engineer/physicist
divide at all education levels (BS, MS, and PhD).
IV.
Total Energy and Linear Momentum
A.
Classically
p2
1
2
K  mo v 
2
2 mo
This equation is not valid in relativity since its derivation assumed that mass was
constant!
B.
Relativity
Einstein also considered the fact that all inertial observers must agree that energy
is conserved in elastic collisions. This led him to discover the following formula
relating linear momentum and the total energy of an object
E 2  p c 2  mo c 2   p c2  E o 2
2
Note: Kinetic energy does not appear in the equation!! It is the total energy that
appears in the equation!!
WARNING
E  E o 2  E  E o 2
2
C.
Special Case of Massless Particles (mo = 0)
We know from PHYS2424 that light carries momentum as well as energy. In
Phys2424, we found that Maxwell's equations relate the energy of the light to its
linear momentum by the equation
Epc
Since Special Relativity is suppose to be compatible with Maxwell's
electromagnetic theory, Einstein suggested that the equation for light is a special
case of his more general equation in part B with the rest mass of light equal to
zero! Furthermore, light would have no rest energy but does have total energy due
to its motion (kinetic energy).
We will see later that Einstein also developed a theory for light in which light is
composed of particles called photons. The paper was published at the same time
and in the same journal as his relativity paper. Thus, he felt that any massless (ie
no rest mass) particle must obey his energy equations:
Massless (Zero Rest Mass) Particles
E  m c2
Epc
However, our previous formula relating relativistic mass to rest mass is invalid
since it was derived assuming mo  0.
For Massless Particles
m
mo
1  v 
c
2
Questions:
1)
Since light has mass, does it feel the force of gravity?
2)
Is the mass in Relativity (inertial mass) the same as the source of gravity
(gravitational mass)?
Einstein considered these questions in developing his General Theory of Relativity. It is
an extremely complicated mathematical theory whose primary application is in
astrophysics. Therefore, it is left for graduate study except for some small discussion in
the book. Thus, I may discuss some of its ideas in class, but I have not included them in
theses notes.
V.
Modern Physics Units
It is neither practical nor useful to solve most Modern Physics problems in SI units. The
following units should be used in solving relativity problems.
A.
Energy - MeV
B.
Mass - MeV/c2
C.
Linear Momentum - MeV/c
D.
Speed – c
VI.
Linear Momentum and Velocity
Velocity is a poor quantity to characterize a system since an extremely small change in
the speed of an object traveling near the speed of light can greatly change the object’s
mass, momentum, and energy. Thus, we usually find it better to characterize an object by
its linear momentum and energy.
Example: An electron has a kinetic energy of 0.100 MeV. Find its speed according to
classical and relativistic mechanics.
Given:
Find:
A.
Classical Physics
B.
Relativity