Problem :
Show that the if we have
tanhθ = v/c
, the Lorentz transformations can be written as:
Δx = Δx'coshθ + cΔt'sinhθ




Δct = Δx'sinhθ + cΔt'coshθ




Since
tanhθ = sinhθ/coshθ = v/c
. The first transformation says (dropping the
Δ
s):
x = +
. Now the
γ
factor is
= = coshθ
. Thus the first term
becomes
x'coshθ
and the second
vt'coshθ
. But
v = c tanhθ
, so the second term
becomes
c tanhθt'cosθ = ct'sinhθ
. Similarly, the second transformation says
ct = +
. The first term becomes
ct'coshθ
and the second
vx'coshθ/c = x'tanhθcoshθ = x'sinhθ
.
Problem :
The Lorentz transformation expressed in the problem above may be represented in matrix form by:
Show that is you apply one Lorentz Transformation with
tanhθ
_{1} = v
_{1}/c
followed by another
Lorentz transformation with
tanhθ
_{2} = v
_{2}/c
, the result is also a Lorentz transformation with
tanh(θ
_{1} + θ
_{2}) = v
.
To show this all we need to do is multiply the matrix with
θ
_{1}
s by the matrix with
θ
_{2}
s:
This is the Lorentz Transformation with
v/c = tanh(θ
_{1} + θ
_{2})
as desired. This is an
important result: it means that when we transform from frame A to frame B and then from frame B to frame
C, there exists a single transform of the same form which takes straight from A to C.
Problem :
In the reference frame of an outside observer two particles move towards each other, both with velocity
v
. The angle between them is
2θ
as shown in the figure below. What is the speed of one
of the particles as viewed by the other?
Two particle approaching each other at an angle
2θ
.
Consider the frame moving along with the particles along the dashed line midway between them. Call this
frame
F'
. It has a speed
v cosθ
as given by the horizontal projection of the particles' speeds. The
γ
factor between
F'
and the frame of an outside observer is
γ =
. In
F'
the particle both move directly towards the fixed midpoint with speed
u = γsinθ
. The
γ
factor comes from the fact that clocks run slower in
F'
than in the
outside observer's frame so the speed is greater in
F'
by a factor
γ
than it would be in the lab.
We then use the velocity addition formula to find the speed of one particle with respect to another as:
w = =


Problem :
Show that the angle between the
x
and
x'
axes, as shown in , on a
Minkowski diagram is given by
tanθ
_{2} = v/c
. Also, determine the size of one unit on the
x'
axis.
The point
(x', ct') = (1, 0)
lies on the
x'
axis one unit from the origin. This Lorentz transforms to
the point
(x, ct) = (γ, vγ/c)
. The tangent of the angle
θ
_{2}
is given by
ct/x = = v/c
. Hence
tanθ
_{2} = v/c
. This point is a distance
= γ1+v
^{2}/c
^{2}
. This is the size of one unit on the
x'
axis.
Problem :
Use a Minkowski diagram to solve the following problem. Frame
F'
moves at a speed
v
with respect to
frame
F
along the
x
direction. A 1meter stick (as measured in
F'
) lies along the
x'
axis, at rest
in
F'
. An observer in
F
measures the length of the stick. What is the result?
Meterstick in two different frames.
The solution to this problem is easy using Lorentz transformations, but we are not going to do it that
way. We can choose the left end of the stick to be at the origin of
F'
, and as such the worldlines of
the ends of the stick are the
ct'
axis and another line parallel to it. The distance between these lines, as
measured parallel to the
x'
axis must be 1 meter (shown as AC in ). But in
terms of
F
, the segment AC has length
(the length of one unit
in
F'
). Someone in
F
measures the length by writing down the coordinates at the same time: for the
sake of simplicity lets take
t = 0
. At this time the length of the stick is the length of the line segment AB.
Now all is a matter of geometry. The
x'
angle is tilted by an angle
θ
, where
tanθ = v/c
. The
length of CD is
(AC)sinθ
and the angle
âà BCD = θ
so
BD = (CD)tanθ = (AC)sinθtanθ
. Now,
AB = AD  BD = (AC)cosθ  (AC)sinθtanθ = (AC)cosθ(1  tan^{2}
θ) =
(1  v
^{2}/c
^{2}) =
. Thus the observer in
F
sees the stick shorter by
, as
we already knew.
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