It’s Relatively
Simple…
Nick Bremer, Erik Cox, Scott
McKinney, Mike Miller, Logan
Petersen, AJ Schmucker, Nick
Thull
IN REVIEW
• Einstein’s Postulates
• Lorentz Frames
• Minkowski Inner Product
Einstein’s Postulates
1. There exists a Lorentz frame
for Spacetime.
2. A Lorentz transformation of a
Lorentz frame gives a Lorentz
frame.
1st Lorentz Postulate
– For stationary events, Physical Clock Time
and Coordinate Time should agree.
– That is, we assume that stationary
standard clocks measure coordinate time.
nd
2
Lorentz Postulate
• The velocity of light called c = 1.
• Light always moves in straight lines with
unit velocity in a vacuum.
• T  (T , vT  r0 ) , time and spatial position
• Note: Think of the light pulse as a
moving particle.
Minkowski Space
(Geometry of Spacetime)
• The symmetric, non-degenerate bilinear form
of the inner product has the properties
•
•
•
•
<x,y>=<y,x>
<x1 + x2, y> = <x1, y> + <x2, y>
<cx,y> = c<x,y>
The inner product does not have to be positive
definite, which means the product of it with
itself could be negative.
• Non-degenerate meaning only the zero
vector is orthogonal to all other vectors
• Spacetime has it’s own geometry described
by the Minkowski Inner Product.
Minkowski Inner Product
•
•
•
•
•
Defined on R4:
u = (u0,u1,u2,u3)
v = (v0,v1,v2,v3)
<u,v>:=u0v0- u1v1- u2v2- u3v3
<•,•> also called the Lorentz Metric, the
Minkowski metric, and the Metric Tensor
• M = R4 with Minkowski Inner Product
• “•” represents the usual inner product (dot
product) in R3.
• In this case you have an inner product that
allows negative length.
WorldLines
Vector Position Functions and
Worldlines
• In Newtonian physics/calculus, moving
particles are described by functions
t
r(t)
• r(t) = ( x(t) , y(t) , z(t) )
This curve gives the
“history” of the
particle.
Vector position functions and
Worldlines cont…
• View this from R4 perspective
t
( t, r(t) )
• In the above ‘t’ represents time and ‘r(t)’
represents the position.
• This can be thought of as a “curve in R4”,
called the Worldline of the particle.
• A worldline (at +b) is given a non-Euclidian
“geometry” in M by the Minkowski Metric.
A Brief Description of
Relativistic Time Dilation
The Einstein – Langevin Clock
• Time is measured where the period
between light emission and return is
regarded as one unit.
t  2LC
Light
source
Where L = length of tube,
and c = speed of light.
Mirror
The Relativistic Time Dilation
Factor
Let t’ = time of ½ pendulum
Light
Source
L
L = ct’
Mirror
Consider a spaceship with an Einstein-Langevin Clock
onboard. We will look at what happens after time t.
Time Relative to a Stationary
Lorentz Frame
Ship has moved D = vt.
Light
Source
ct
L
vt
Z
Mirror
Z
’
“e”
We’ll let t be the measure of ½ of
the light pendulum according to
the clock of the stationary Lorentz
frame.
Observe:
(ct)2 = L2 + (vt)2
Y
O
O’
X
X’
Y’
Relativistic Dilation Factor
• Solve for
(ct)2 = L2 + (vt)2
And recall: L = ct’
We call
c t  L v t
L
t
c2  v2
ct '
t
2
2
c v
2 2
2
2 2
 (v ) 
c
c v
2
the relativistic dilation factor.
This shows up in many
equations in relativity theory.
Too make things easier, we
often set c = 1.
2
What Is Proper Time?
Proper time is the elapsed time measured by a moving object.
Recall:
t
t'
1 v2
t   (v)t '
( Note :  (v) 
1
1 v
2
)
where t  represents the elapsed time of the moving object in
proper time and t is the elapsed time of the moving object in

coordinate time (using  (v ) as a conversion factor).
However, this formula only applies for constant velocities.
Solution:
Take small time intervals on the worldline, each with
approximately constant velocity.

x (t  t )

x (t )
Average Velocity



x (t  t )  x (t ) dx 


v
t
dt
for small t
With constant velocity, we can use:
t2
2
t '  1  v t
t1
As t  0 , summing the intervals
together yields the following:
t2
lim 
t 0
t1
t2
2
2
1  v t   1  v dt
t1
Proper
Time!
Proper Time
t

t , x (t )   

1 v (t )
2
dt
a
Note: T is Tau
where t-a is the elapsed time measured on the
stationary clock and the resulting integral is the
elapsed time measured on the moving clock.

t  a    1 v (t )
t
Note:
a
2
dt
Parameterize the Worldline
Using Coordinate Time
Recall: Spacetime = E = the set of all possible events e

where E is modeled by M  4 , ,
Worldline:

s  e( s )  E

where e(s)  t (s), x (s) and s  

4
Parameterize using coordinate time: z (t )  t , x (t )   

dz  dx 

Differentiate:
 1,   1, v (t ) 
dt  dt 
Fact About Proper Time
Proper time represents a non-Euclidian arc-length.
Proof:
dz

dt
 1, v (t ) 
dz dz


,
 1, v (t ) , 1, v (t ) 
dt dt


 1  v (t )  v (t )
 2
 1  v (t )
dz dz
 2
,
 1  v (t )
dt dt
t

a
 
dz dz
,
dt  
dt dt

 
t
dz dz
,
dt  
dt dt
a

1 v (t )
2

1 v (t )

dt t , x (t ) 
Notice the similarities between the common arclength formula and the expression above:
x = x(t), y = y(t)
t=b
z(t) = ( x(t),y(t) )
t=a
b

a
2
2
 dx   dy 
     dt  
 dt   dt 
a
b
dz dz
,
dt
dt dt
This shows that Proper time is a non-euclidean arc-length.
2
dt
Since Proper Time is similar to arc-length, can the worldline be
parameterized by proper time? Yes, provided the quantity called
proper time is “independent of parameter”.
Change parameter of the worldline:
For any curve, t  e(t ) we can make a change of
parameter using a function s  s (t ) to get s  e( s (t ))
Differentiate z with respect to s:
dz dz dt

ds dt ds
By Chain Rule
dz dz
dz dt dz dt
 dt 
,

,
 
ds ds
dt ds dt ds
 ds 
dz dz
 dt 
,
 
ds ds
 ds 
dz dz
,
dt dt
2
dz dz
,
dt dt
Pick b so t(b) = a
s

b
(i.e. if s = b, then t = a)
dz dz  dt 
,
 ds
dt dt  ds 
s
dz dz
,
ds  
ds ds
b
Make a substitution:
t (s)

t (b )
t
dz dz
,
dt  
dt dt
a
t  t (s )
dt
dt  ds
ds
dz dz

,
dt  (t ( s), x (t ( s )))
dt dt
Thus showing proper time is independent of parameter.
Four-velocity
What is four-velocity?
4-dimensional, relativistic analog of traditional threevelocity, represented by:
dz
or U ()
d
Four-velocity and Threevelocity
How is four-velocity related to traditional three-velocity?
Recall that we can parameterize worldlines by proper time:

z  z ()  (t (), x (t ()))

dz  dt dx 
U () 
 , 
d  d d 

  dx dt 
   (v ),

dt
d




  dx  
   (v ),  (v ) 
dt


 
  (v )1, v 
dt



(v
) shown later.
Note:
d

Factor out  (v )
Important Fact About Fourvelocity
Four-velocity is a timelike unit vector
Proof: Show
dz dz
,
1
d d
t
Recall:

(t , x (t ))  

1 v (t )
2
dt
a
Differentiate using
Fund. Thm. Calculus:
Using Chain Rule and
inverses, we get:
d d
 
dt dt a

1 v (t )
dt

d
1
t

1 v (t )
2

2
dt 

1 v (t )
1
dz dz
,
dt dt
2

  (v )




dz  dx 
dz dz dt  dx 
1
Recall:
 1, 

 1, 
dt  dt 
d dt d  dt  dz dz
 dt , dt







 
 
  dx 

dz dz
1
1
 dx 
,
 1, 
,
1
,
  dt 

d d
dt

 dz dz  
 dz dz 
 dt , dt 
 dt , dt 








 dx   dx 


1, , 1, 


1
d
x
d
x
 1, , 1,    dt   dt   1



 dz dz   dt   dt 
d
x
d
x




,
1
,
,
1
,





dt
dt
 dt dt 



Thus showing
dz dz
,
 1 and proving four-velocity
d d
is a timelike unit vector.







Classical
Momentum
Conservation of Momentum
Recall the Classical Law of Conservation of
Momentum




m1v1  m2v2  m1v1  m2v2
It can be shown, however, that this Law is NOT
Lorentz invariant for inelastic collisions.
Due to the principle of covariance, the laws of
physics should hold for any Lorentz Frame.
Something must not be quite right!
So what do we do?
Question: If the Classical Law isn’t
quite right, how do we fix it?
Answer: Would a 4-dimensional analog work?
Gearing up
What if we use 4-velocity[U () ]in place of
standard velocity?

dt dx
Recall:U ()  ( , ) and
d d

dx
 2
 1 | v | (space component of U () )
dt


 d x dx

Note that for small | v |,
dt d
Relativistic Momentum
Our first analog of the Classical Law will look like:
m1U1  m2U 2  m1U1  m2U 2
We’ll call this equation *. Let’s consider the time
and space components of *.
Time Component
dt

  (v )
The time component of U(T) is simply:
d
Which gives:





m1 (v1 )  m2 (v2 )  m1 (v1 )  m2 (v2 )

As you may have noticed, this seems trivial since for small v
all it says is that m1  m2  m1  m2 . However, we will
see that this will be a new law analogous to the Classical
Law of Conservation of Energy.
Space Component
Now let’s examine the space part of *. Recall that:

dx

dt
 2 (space component of U () )
1 v


dx
 2 dx
 1 v
dt
d
1
2
1 v


dx dx

dt d
 
a.k.a.  (v )v
Space Component
Now we can write the space component of *
as:
 
 
 
 
m1 (v1 )v1  m2 (v2 )v2  m1 (v1)v1  m2 (v2 )v2



(
v
)
Notice that when v is small,
1
2
1 v
1
and we get the Classical Law of Conservation of
Momentum.
Philosophy
It behooves us to stop and think here for a moment. It
seems that the space component of * is very close to the
Classical Law of Conservation of Momentum. Does this
make sense?
It does. When moving from 3 dimensions to 4, we
added time. If we look at the space part of our 4dimensional analog, it seems reasonable to see things
that were developed in 3-dimensional space.
So now what?
 
 
 
 
m1 (v1 )v1  m2 (v2 )v2  m1 (v1)v1  m2 (v2 )v2
is very close to the Classical Law of
Conservation of Momentum. If we could
somehow redefine mass as some sort of rest

mass times  (v ), then this equation would
match the Classical Law.
Mass
Can Mass Change?
If so, what happens
if Mass is not constant?
How to Define Rest Mass
•Fix a Lorentz coordinatization
•Arbitrarily choose a particle at rest
•Define the rest mass of that particle to be one
In other words,
The initial mass and velocity
Equals the post collision mass and velocities
Than by solving for m1 and than looking as v1 approaches zero
Rest mass is than defined
as just the mass of an object
having velocity zero
Relativistic Momentum
also called 4- Momentum
or energy momentum
p = (Rest mass)(4-velocity)
The break down of p
The time component in P is related to
the Newtonian concept of Kinetic energy.
While the space component in P is related to
the Classical Law of Conservation of Momentum.
Substitution and Distribution
Recall,
Inspired by the tendency of Newtonian
concepts to appear in spatial components of
relativistic momentum. We determine
relativistic mass to equal
M=(rest mass)(time dilation factor)
Hence,
p can now be defined as
a vector with relation to M.
Observations on
Kinetic Energy
Background on Kinetic Energy
• Kinetic energy is the
energy of motion
• Kinetic energy is a
scalar quantity; it does
not have a direction.
• The Kinetic energy of
an object is completely
described by magnitude
alone.
KE  1 2m v
v
= speed of the object
m
= mass of the object
2
Setup for Conversion to
Newton
Newtonian Perspective versus the classical
definition of momentum


p  mv
Newton’s 2nd Law of Motion
Explains how an object will
change velocity if it is
pushed or pulled upon


dp
F :
dt
Information for the Proof
The Kinetic Energy’s rate of change is
particularly interesting:
Claim:


d
dv  dp   
2
(1 2 m v )  m  v 
v  F v
dt
dt
dt
Conversion to Newtonian KE
Breakup the
2
v using the product rule for dot-products
d
d
2
 
(1 2m v ) 
(1 2 mv  v )
dt
dt
Pullout the constant
1 2m
d
d  
 
(1 2 mv  v )  (1 2 m) (v  v )
dt
dt
Conversion (cont.)
Distribute the dot-product


d  
dv   dv
(1 2m) (v  v )  (1 2m) (
v  v 
)
dt
dt
dt
Combine the terms



dv   dv
 dv
(1 2m) (
v  v 
)  (1 2 m)( 2(v 
))
dt
dt
dt
Reduce


 dv
 dv
2(v  )(1 2 m)  v  m
dt
dt
Conclusion



dv
 dv  
 F , so we know that v  m
 vF
Recall that m
dt
dt
So…
 
d
( KE )  F  v
dt
Before we move on, this equation can be used to define
Relativistic Energy as:
dE  
 F v
dt
Relativistic
Energy
E=
2
mc
Working out the equation…
• Let m = rest mass
m 2  m 2 1
Recall: unit vector of four velocity: <U, U>
 m 2 U ,U
 mU, mU
Recall: Four momentum p = mU
 p, p


 ( po , p), ( po , p)
Continuing on…
From the previous slide we had:


 ( po , p), ( po , p)
 po
2
 
 p p
Recall: Relativistic Mass M = po
2
 
 M  p p
This is also the Minkowski Inner Product definition
Which gives us the following equation:
 
m  M  p p
2
2
Differentiating both sides with
respect to t…

 
d
2
0
M  p  p
dt

 dp
dM
0  2M
2p
dt
dt
By Product Rule for Dot Product
Recall:

  dp 
dM
0  2M
 2 Mv 

dt
dt 

Recall:

 
dM
0  2M
 2M F  v 
dt

p  Mv


dp
F
dt
by Newton’s II
Continuing on…
From the previous slide we had…

 
dM
0  2M
 2M F  v 
dt
 
dM
0
 F v
dt

dM
dE
0

dt
dt
dE
dM

dt
dt

dE
dt 
dt

dM
dt
dt
By factoring out 2M we get…
Recall: Relativistic Energy
dE  
 F v
dt
Almost there…
From the previous slide…

dE
dt 
dt

dM
dt
dt
E  M C
If we consider smaller and smaller masses that get closer
and closer to zero, the energy will also get smaller and
smaller; therefore not allowing for any remainder C. So
we will claim C is zero.
Which brings us to…
E = M!!!
proving the equation E = mc2 where
c (speed of light) is set to 1
Special Thanks
Thanks to Dr. Deckelman for his guidance
on the creation of this presentation.