Physics Lorentz Transformations March
Lorentz Transformations
This handout summarizes the lecture discussion of Lorentz transforma
tions
Introduction
When two observers are moving relative to each other we use Lorentz trans
formations to relate their observations
Four vectors
Lorentz transformations op erate on four vectors more generally ten
sors with four vector indices An example of a four vector is the space
time co ordinate formed from the time co ordinate t and the three dimensional
vector space co ordinate x We multiply the time by the velocity of light
ct
x
x
We can lab el the comp onents of the four vector as x with ranging from
0
to The sup erscript is conventional Then the time comp onent is x ct
Another common four vector is the four momentum
E
p
pc
0
The comp onent p is the energy which must include the rest energy The
four momentum of a particle of mass m at rest is just
2
mc
p
The Lorentz transformation is the central feature of sp ecial relativity that
was adopted in order to account for the remarkable observation that the
sp eed of a light ray measured by an observer was apparently the same
regardless of how fast the observer was moving
0
Supp ose observers in frames S and S are moving with a velocity v
relative to each other To b e more precise let s align the spatial co ordinate
0
systems for each observer so S is moving along the z axis of observer S and
0 0
the x and y axes are parallel with the x and y axes Supp ose also that
b oth observers set their clo cks so that when their origins are on top of each
other their clo cks read also
Then if an observer in frame S sees an event at the space time co ordinate
0 0
x and an observer S sees the same event at space time co ordinate x the
co ordinates are related by the Lorentz transformation
00 0 3
x x x
10 1
x x
20 2
x x
30 3 0
x x x
We have introduced
v c
p
2
In terms of t x y and z the Lorentz transformation is
0
ct ct z
0
x x
0
y y
0
z z ct
As we mentioned this transformation assures that if b oth observers see
the same light ray and measure its sp eed they b oth get the same result
namely c A simple way to check this is to supp ose that the light ray leaves
the origin at time t Later on the observer S notices that it has reached
0
the space time co ordinate x while observer S notices that it has reached
0
x The relationship b etween these observations is given by the Lorentz
transformation It is easy to check the Lorentz invariance prop erty
0 2 1 2 2 2 3 2 00 2 10 2 20 2 30 2
x x x x x x x x or
2 2 0 2 0 2
ct r ct r
p
2 2 2
where we have introduced the distance to the origin r x y z for
p
0 0
0 2 0 2 0 2
y z for observer S For a light ray observer S and r x
moving with velocity c we must have r ct So b oth sides of the equation
ab ove must b e zero More gnerally Both observers always get the same
0 0
value for any Lorentz invariant quantity So we conclude that also r ct
and b oth observers agree that the sp eed is c
If the relative sp eed v is much less than the sp eed of light we have
and
0
ct ct
0
x x
0
y y
0
z z v t
which is the Galilean relativity that we learned in our rst course in classical
Time dilation and Lorentz Fitzerald contrac
tion
As an application of a space time Lorentz transformation consider how
time intervals are recorded by di erent observers As a sp eci c application
supp ose a particle is placed at rest at the origin of observer S at time
0 0
t t It decays after a p erio d of time Observer S also sees the
decay According to his her clo cks what is the elapsed time
0
The answer is easily found from the relation ct ct z which we
apply to the decay event Notice that to use this relation we have to sp ecify
z Since the decaying particle is still at the S origin we have z and
0
t So fast moving particles app ear to have longer lifetimes according
to the Lorentz factor This is the p eculiar phenomenon of time dilation
As a second application consider how distance measures are recorded
by di erent observers Supp ose that observer S places a meter stick along
0
the z axis and observer S measures its length What will the result b e
As always we answer by considering the relationship b etween a pair of
0
events In this case the events are S measuring the two ends of the meter
0
stick which she must do at the same instant t Otherwise she would
have to correct for the motion of the meter stick We can use the Lorentz
transformation equations given ab ove without mo di cation if we supp ose
0 0 0
that S makes the measurement at t when the S and S origins coincide
0
Both observers agree that the left end of the meter stick is at z z
0
when t t S would say the right end of the meter stick is always
0 0
at z and S records it as z What seems p eculiar is that they will
disagree ab out the simultaneity of the events of measuring b oth ends If we
0 0
lo ok at the relation ct ct z we see that t requires ct z so
only for an event at z or for for vanishingly low velocities v c ! do
0
b oth observers agree on the time So when observer S measures the right
0
end of the meter stick at time t observer S says that measurement
happ ens at time ct z with z meter But from the fourth relation
0 0
z ct observer S must b e doing this measurement at p osition z
0 2 0
z z z z So to S the meter stick is short by a factor
This is the Lorentz Fitzgerald contraction
Note that b oth observers always agree on lengths p erp endicular to their
relative velocity The contraction is only along the direction of relative
motion
Lorentz transformations and other four vectors
The four momentum transforms just like the space time co ordinate under a
Lorentz transformation
00 0 3
p p p
10 1
p p
20 2
p p
30 3 0
p p p
so
0
E E p c
z
0
p c p c
x
x
0
c p c p
y
y
0
p c p c E
z
z
In fact a four vector is de ned to b e any quantity that transforms in this way
Velocity transformation
Let s see how velocities transform under a Lorentz transformation We con
sider a particle moving along the z axis with z w t according to observer
0
S He says it has velocity w Observer S moving with velocity v relative
0 0 0
to S says it has velocity w z t We want to nd it Now we are dealing
with three velocities the velocities of the particle as seen by the observers
and the relative velocity of the observers For simplicity let s do this in only
one dimension so supp ose all velocities are along the z axis Then a Lorentz
transformation relates the p osition of the particle as seen by b oth observers
0
z z ct w v t
0
ct ct z c v w c t
We divide the equations to get
w v
0
w
2
v w c
0
In the low velocity limit we recover the familiar Galilean result w w v
0
But at the other extreme if w c we get w c so all observers agree on
the same velocity for a particle moving at the sp eed of light
Lorentz invariants
We have already seen that the quantity
2 0 2 1 2 2 2 3 2 2 2
x x x x x ct x
is invariant under a Lorentz transformation By the same token the quantity
2 0 2 1 2 2 2 3 2 2 2 2 2 4
p p p p p E p c m c
is also invariant since it transforms the same way This result is fortunate
since then all observers agree on the same mass This relativistic relationship
b etween energy momentum and mass is sometimes called the mass shell
condition
More general invariants can b e constructed from any pair of four vectors
For example if we have two four momenta k and p the combination
0 0 1 1 2 2 3 3
p k p k p k p k p k
is also invariant Notice that this expression de nes an inner pro duct rather
like a dot pro duct but the signs metric are di erent from the familiar
Cartesian dot pro duct The familiar dot pro duct of three dimensional vec
tors is similarly invariant under rotations
Photon four momentum
A photon has zero mass If it has frequency then its wavelength is c
and if it is moving in the z direction its energy and momentum are given
by
E
B C
B C
p
B C
A
pc
where E h and p h E c This result is consistent with the
statement that
2 2 2
p E cp
meaning the photon has zero mass
Relativistic kinematics
A moving particle has kinetic energy In Newtonian mechanics we thought
of kinetic energy as b eing the primary quantity and then tacked on the
rest mass when we b egan talking ab out low velocity nuclear physics In
sp ecial relativity we treat the total energy as the primary quantity given
p
2 4 2 2
m c p c and the kinetic energy K is then derived from it by E
using
2
K E mc
It is a useful exercise to check that the relativistic mass energy momentum
relation ab ove turns into the standard nonrelativistic result for sp eeds much
less than the sp eed of light
q
2 2
2 4 2 2
K m c p c mc p m
Two bo dy collisions
Consider a two bo dy reaction of the type
!
Let the four momenta of the particles b e p p p and p resp ectively
1 2 3 4
Notice we are now using subscripts to lab el the particle Each of these
terms represents a full four vector Let the masses similarly b e m m
1 2
m and m Energy and momentum must b e conserved The conservation
3 4
condition can b e written concisely in terms of four vectors as
p p p p
1 2 3 4
Supp ose that particle is at rest and particle arrives along the p ositive z
axis Then the initial four momenta are
2
E m c
2
B B C C
B B C C
p p
B B C C
1 2
A A
pc
4 2 2 2 2
c so there is really only one Of course we must also have E p c m
1
kinematic parameter for the incoming particle Let the scattering plane b e
the xz plane Then the nal four momenta are
E E
3 4
B C B C
p c sin p c sin
B C B C
3 3
p p
B C B C
3 4
A A
p c cos p p cos c
3 3
where we have already arranged for three momentum conservation The
energies must also b e related through the energy conservation condition
2
E m c E E
2 3 4
This constraint together with the mass shell condition for particles and
determines the outcome completely in terms of the one free parameter
namely the scattering angle
Compton Scattering
Let s supp ose that we are considering Compton scattering so that particles
and are light particles and particles and are an electron A light
particle has zero mass so we have E pc and E p c The mass shell
3 3
condition for particle then b ecomes
2 2 2 2 4
E E sin E E cos m c
3 3
4
which together with energy conservation and a little algegra gives for the
energy of the scattered photon
2
E mc
E
3
2
E cos mc
Relating lab and c m frames
It is often helpful to analyze collisions in the c m frame We can use Lorentz
invariants to help relate quantities in the two frames For example consider
the two bo dy reaction with any mass In the c m frame let s write four
momenta of particles and this way
E E
1c 2c
B B C C
B B C C
p p
B B C C
1 2
A A
p c p c
c c
The total energy in the c m frame is
W E E
1c 2c
The total four momentum in the c m frame is then
W
B C
B C
P p p
B C
1 2
A
2
We can use a trick with the Lorentz invariant P to relate the total energy
W in the c m to the energy E in the lab oratory frame We have
2 2 2 2 2
W P p p p p p p
1 2 1 2
1 2
Since p p is a Lorentz invariant we can evaluate it in the lab oratory frame
1 2
2
where it is just E m c So we get
2
2 2 4 2 4 2
W m c m c E m c
2
1 2
To get the momentum of each particle in the c m requires a bit more algebra
We just quote the result
q
4 4 2 2 2 2
4 8 8 2 4 2 4 8
p c W m c m c W m c W m c m m c W
c
1 2 1 2 1 2
From the momentum we can then get the energies E and E for each
1c 2c particle