SME
University Physics III: Waves, Optics & Modern · Problem type
The Lorentz transformation
Use this for “A rocket passes Earth at 0.8c. An event happens 1,200 m away along its path, 2,000 m of ct after they meet. Where and when is it in the rocket’s frame?”
Diagram
γ
=
1
÷
√
1 −
β
2
=
1
÷
√
1 −
0.6
2
=
1.25
x′ = γ(x − βct)
x′ = 1.25 × (900 − 0.6 × 600) = 675 m
ct′ = γ(ct − βx)
ct′ = 1.25 × (600 − 0.6 × 900) = 75 m
s
2
=
(ct)
2
−
x
2
=
600
2
−
900
2
s
2
=
−450,000
m²
(ct′)² − (x′)² = 75² − 675² = −450,000 m²: the same in both frames.
s² < 0: spacelike, so no signal links the event to the origin.
The S′ axes tilt by tan⁻¹β = 30.96°.
This is an example. Type your own values to start a new problem.
x′ (
)
=
γ
×
(
x (
)
−
β
×
ct (
)
)
ct′ (
)
=
γ
×
(
ct (
)
−
β
×
x (
)
)
s²
Interval in S
worked out (not typed)
s′²
Interval in S′
worked out (not typed)
Formulas
γ
=
1
÷
√
1 −
β
2
1.25
=
1
÷
√
1 −
0.6
2
x
′
=
γ
×
(
x
−
β
×
ct)
675 = 1.25 × (900 − 0.6 × 600)
ct′
=
γ
×
(ct
−
β
×
x
)
75 = 1.25 × (600 − 0.6 × 900)
x
=
γ
×
(
x
′
+
β
×
ct′)
900 = 1.25 × (675 + 0.6 × 75)
ct
=
γ
×
(ct′
+
β
×
x
′)
600 = 1.25 × (75 + 0.6 × 675)
s
2
=
ct²
−
x
2
−450,000
=
600
2
−
900
2
s′²
=
ct′²
−
x
′²
−450,000
=
75
2
−
675
2
Assumptions
•
S′ moves at β = v ÷ c along +x past S, and the origins meet at ct = ct′ = 0; a negative β moves toward −x.
•
Time is written as ct, in meters, so x and ct share a unit: x′ = γ(x − β·ct), ct′ = γ(ct − βx), and back with +β.
•
The interval s² = (ct)² − x² is the same in both frames: positive is timelike, negative spacelike.
Step-by-step
WE KNOW
β = 0.6, x = 900 m, ct = 600 m
FIND
Lorentz factor (γ), event position in S′ (x′), event time in S′, as ct′ (ct′), interval in S (s²), interval in S′ (s′²)
Step 1 · Find Lorentz factor (γ)
Use γ = 1 ÷ √(1 − β²)
The
Lorentz
factor
for
this
speed:
1
at
rest,
growing
without
limit
as
β
nears
1.
γ
=
1
÷
√
1 −
0.6
2
γ
=
1
÷
√
1 − 0.36
γ
=
1
÷
√
0.64
γ
=
1
0.8
γ
=
1.25
Step 2 · Find event position in S′ (x′)
Use x′ = γ × (x − β × ct)
In
S′
the
event
sits
x
−
β
·ct
from
the
moving
origin,
stretched
by
γ
.
x
′
=
1.25
×
(900
−
0.6
×
600)
x
′
=
1.25
×
(900
−
360)
x
′
=
1.25
×
540
x
′
=
675
m
Step 3 · Find event time in S′, as ct′ (ct′)
Use ct′ = γ × (ct − β × x)
Clocks in S′ disagree with ours by βx: events farther along happen earlier for S′.
ct′ = 1.25 × (600 − 0.6 × 900)
ct′ = 1.25 × (600 − 540)
ct′ = 1.25 × 60
ct′ = 75 m
Step 4 · Find interval in S (s²)
Use s² = ct² − x²
The
interval
in
S:
(ct)
2
−
x
2
,
worked
as
(ct
−
x
)(ct
+
x
).
s
2
=
(ct
−
x
)
×
(ct
+
x
)
s
2
=
(600
−
900)
×
(600
+
900)
s
2
=
−300
×
1,500
s
2
=
−450,000
m²
Step 5 · Find interval in S′ (s′²)
Use s′² = ct′² − x′²
The
same
in
S′:
it
matches
s
2
,
since
the
interval
is
the
same
in
every
frame.
s′²
=
(ct′
−
x
′)
×
(ct′
+
x
′)
s′² = (75 − 675) × (75 + 675)
s′² = −600 × 750
s′² = −450,000 m²
Check
Put the numbers back in. Both sides should match.
1.25
=
1
÷
√
1 −
0.6
2
✓
675 = 1.25 × (900 − 0.6 × 600) ✓
75 = 1.25 × (600 − 0.6 × 900) ✓
900 = 1.25 × (675 + 0.6 × 75) ✓
600 = 1.25 × (75 + 0.6 × 675) ✓
−450,000
=
600
2
−
900
2
✓
−450,000
=
75
2
−
675
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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