LECTURE 3
1. Recap: Lorentz Transforms
Recall the two axioms of Special Relativity:
1.) There are no experiments which can differentiate between a moving and stationary frame.
2.) The speed of light is the same for all observers
Einstein made these axioms the foundation of his theory, but they weren’t arbitrary. They were born of
the symmetries inherent in equations describing electric and magnetic fields, and of observations in the
laboratory. In short, you can consider them experimental fact. All Einstein did was work out the
implications of these lab-based observations. The first three we have seen several times:
a. Moving clocks tick slower (time dilation)
b. Moving rulers measure shorter (length contraction)
c. When moving with respect to one another, different people won’t agree on whether events at
different locations happened at the same time (loss of simultaneity)
The implication is that ideas of time and space are relative, and measurements of these things (and many
other things, we will find) depend on who’s looking. What people do agree on is the existence of events.
(A ship blew up. A person threw a ball. A clock output a time.) Just not where or when these things
occurred.
The mathematics describing how these things vary between frames is captures in the Lorentz transform
equations:
 Focus on events occurring at specific locations AND times in a particular frame: (tA,xA), (tB,xB)
 Define Δx = xB-xA, Δt = tB-tA
 Then the equivalent quantities in a frame moving at speed v, Δx = xB-xA, Δt = tB-tA, can be
found via
Δx = γ (Δx - βc Δt)
Δt = γ (Δt - β Δx/c)
and/or the equivalent inverse equations
Δx = γ (Δx + βc Δt)
Δt = γ (Δt + β Δx/c)
Where β = v/c and γ = (1- β2)-1/2
Key points:
 Strange implications (a)-(c) are contained in these equations as special cases.
 Loss of simultaneity occurs when Δx differs from zero. This affects the definition of when. (e.g.
Mr Thompkins and Eustace in HW#2.)
 This is not an indication of measurement error! LT equations describe actual differences in the
definition of space and time in different frames. Finite speed of light has additional and different
effects. (e.g. dancing cow in HW#2, measurement of star’s speed in HW#1.)
Example: Discussion 3, Problem 3(d) (equivalent to train paradox problem):
The Nostromo crew has a clock at the front end of the ship and another one at the back end. Naturally,
they keep these clocks synchronized [using some experimental technique whose details are irrelevant].
According to the vacationers, when the aft Nostromo clock reads 1:00 pm, what does the forward clock
read?
What are the events? A- the back clock outputs a time
B- the front clock outputs a time
Strategy box: Δt = 0 -we are interested in what the vacationers think the readings are at a point in time
Δx = 600ft -clocks are on opposite ends of the ship
Δt = ? We want to know what the times on the clocks are reading
Pick most useful LT equation:
Δt = γ (Δt + β Δx/c)
0 = γ (Δt + 4/5 * 600 ft/ 1ft/ns)
Δt = -480 ns
Which clock is slower? → Definition of positive implies front.
Δx = xfront – xback if we want to use +600ft (as we did). Therefore, Δt= tfront-tback.
Additional Note: - Notice the careful wording of the question. What we just calculated is what the clock
on the Nostromos would read in vacationer’s frame. This is different than saying this is what the
vacationers see. The latter statement implies some measurement (e.g. the light getting to the eyes of the
vacationers, or some such nonsense.) The LT describes what the time is once measurement errors have
been corrected for.
2. Derivation of Lorentz Transform
Using only the axioms of SR, one can show that the LT equations are the only set of equations
which relate two inertial frames:
Define two frames S, S, where the prime frame is moving, by convention, in the +x direction at speed v.
We are seeking general functions, f,g, where:
Δx = f(Δt, Δx, v) and Δt = g(Δt, Δx, v)
But I am lazy and don’t like writing Δ all the time, so without loss of generality, let’s just define our
origins such that (t, x) = (t, x) = (0, 0) at some synchronizing event. Then,
x = f(t, x, v) and t = g(t, x, v)
What are f and g? Let’s start applying conditions.
I. An object moving as constant speed in the S frame is seen as moving as a constant speed in the S
frame:
x = x0 + u t ↔ x = x0 + u t
If this weren’t true, then one could distinguish which frame was moving by looking for the frame
in which the object was accelerating!
Mathematically, this is the same as saying that f and g describe a linear transform
x = Ax +Bt
t = Cx+Dt
II. The origin of x moves at speed v in the S frame, by definition:
x = 0
↔
x=vt
Plug and play:
0 = Avt + Bt
B = -Av
→Now, see how this conclusion modifies our equations:
x = Ax – Avt
t = Cx + Dt
III. Likewise, the origin of x moves at velocity –v in the S frame:
x=0
↔
x = -v t
Plug and play:
-vt = 0 -Avt
t = 0 + Dt
Together these imply D = A. (Solve first equation for t, and set equal to RHS of second equation.)
→Now
x = Ax – Avt
t = Cx + At
IV. A beam of light will be seen as moving at speed c in both frames:
x=ct
↔
x = c t
Plug and play:
ct = Act -Avt
t = Cct + At
Together these imply (set expressions for t/t equal to one another):
A - Av/c = Cc + A
C = -Av/c2
→Now
x = A (x – vt) = A (x – βct)
t = A (t – vx/c2) = A (t – βt/c)
This should start looking familiar. We are getting close!
V. By first axiom of SR, the equations must be true if we interchange the definition of S and S and
reverse the sign of v. Otherwise the two frames would not be identical. (S would look
different to S than S looked to S.)
Therefore, we can say
x = A (x + vt)
t = A (t + vx/c2)
Solve first equation for At:
At = (x – Ax)/v
Plug into second equation and solve for x:
t = (x – Ax)/v + Avx/c2
vt = x – Ax + Av2x/c2
x-vt = Ax (1-v2/c2)
x = (x-vt)/A(1-v2/c2)
But if we compare this result to the most recent form of our equations (after IV), this must mean:
A = 1/A(1-v2/c2)
A2 = 1/(1-v2/c2)
A = (1-v2/c2)-1/2  γ
Thus, the only way we can consistently apply all these constraints (I-V) is if
x = γ (x + vt)
t = γ (t + vx/c2)
which are the Lorentz transform equations.
QED.