12. A spaceship moves away from Earth at a constant velocity. On the spacetime diagram, the
𝑐𝑡-axis is the worldline of Earth and the 𝑐𝑡′-axis is the worldline of the spaceship.
a. Calculate the speed of the spaceship relative to Earth.
𝑣
tan 𝜃 = ∴ 𝑣 = 𝑐 × tan 𝜃
𝑐
tan 𝜃 is the gradient f the shifted 𝑐𝑡′ axis
𝑦 −𝑦
4−0
2
Gradient 𝑥2−𝑥1 = 6−0 = 3
2
1
2
2
Hence, 𝑣 = 3 × 𝑐 = (3 × 108 ) × 3 = 2 × 108 𝑚𝑠 −1
A radio pulse is emitted from the spaceship towards Earth. In the reference frame of Earth, the
emission occurs at a distance of 2.00 × 10 6 𝑚 from Earth. The emission of the pulse is event P
and its detection is event Q.
b. Construct
events
P
and
Q
on
the
spacetime
diagram.
Emission occurs at E. time 𝑡 = 0, in spaceship’s FOR.
2
Since spaceship travels at 𝑣 = 3 𝑐, it travels a distance
𝑑 = 2 × 106 (𝑚) = (𝑣)𝑡
2
∴ 2 × 106 (𝑚) = ( 𝑐) 𝑡
3
2 × 106
∴ 𝑐𝑡 =
2
3
∴ 𝑐𝑡 = 𝟑 × 106
Explanation: Signal travels backwards to earth and is detected at displacement
𝑥 = 0 (according to earth). The worldline of photon is always at 𝟒𝟓 °, i.e, (mod of)
the gradient should be 1. Hence gradient of photon’s worldline is -1, with line
starting from the coordinates (2,3) to (0,5) [𝑏𝑜𝑡ℎ × 106 𝑚] in earth’s FOR.
c. Calculate:
i.
the proper time interval between the launch of the spaceship and event P.
Proper time is the time as recorded by the spaceship, since the event of measuring
time occurs in the spaceship.
This interval will be shorter than the time as measured from the earth, since from
earth (external observer), time appears longer than the proper time.
1
1
𝛾=
=
= 1.341
2
2
𝑣
2
√1 − ( 2 ) √1 − ( )
𝑐
3
Since,
Δ𝑡 = 𝛾 Δ𝑡 ′
∴ 𝑐 Δ𝑡 = 𝛾 (𝑐Δ𝑡 ′ )
From the graph, 𝑐 Δ𝑡 = 3 × 106 𝑚, therefore,
3 × 106 = 1.341(𝑐 Δ𝑡′)
Hence,
3
) × 106 = 2.237 × 106 (𝑚)
∴ 𝑐Δt ′ = (
1.341
2.237 × 106
Δ𝑡 =
= 7.45 × 10−3 (𝑠)
3 × 108
′
ii.
the coordinates (𝑐𝑡′, 𝑥′) of event Q in the reference frame of the spaceship.
Referring to figure below,
Draw a (light blue) line parallel to 𝑥′ axis passing through Event Q and wherever it
meets the 𝑐𝑡′ axis, read the 𝑐𝑡 (not 𝑐𝑡′) coordinate of this point.
Here, the (light blue) line meets the 𝑐𝑡′ axis at 9 when read from the 𝑐𝑡 axis.
Divide
this
𝑐𝑡
value
by
the
Lorentz
factor
𝛾
.
Hence,
9
𝑐𝑡 ′ =
= 6.71 (× 106 𝑚)
1.341
Similarly, to obtain 𝑥′, draw a (light red) line parallel to 𝑐𝑡′ axis passing through Event
Q and wherever it meets the 𝑥′ axis, divide that by the Lorentz factor 𝛾 .
Here, the (light red) line meets the 𝑐𝑡′ axis at -6.
Hence,
−6
𝑥′ =
= −4.47 (× 106 𝑚)
1.341
d. State the distance travelled by the radio pulse according to:
Distance travelled is difference between the 𝑥 coordinates in each frame of reference
respectively.
ii.
an observer on Earth,
For an earth observer it is pretty straightforward, since we only need to take the
difference between the 𝑥 coordinates of the two events P and Q, the usual way.
Here, it is simply
Δ𝑥𝐸 = 𝑥𝑄 − 𝑥𝑃 = 0 − 2 = −2 (× 106 𝑚)
-ve sign implies displacement is towards it (-ve x-direction by convention)
Disregarding the sign, the distance is 2 × 106 𝑚
i.
a spaceship observer
For the spaceship observer, use the same logic
[Remember, length contracts/shrinks when measured by an external observer
(earth here)]
Hence
Δ𝑥𝑠 = 𝑥𝑃′ − 𝑥𝑄′ = 0 − 4.47 = 4.47 (× 106 𝑚)
13. A spaceship is sent from Earth towards a distant planet. The diagram shows the spacetime
axes (𝑥, 𝑐𝑡) of the reference frame of Earth and the worldline 𝑐𝑡′ of the spaceship. The
spaceship leaves Earth at 𝑡 = 0 and moves at a constant velocity relative to Earth, in the
direction of the negative 𝑥-axis. Coordinate axes are scaled in light years (𝑙𝑦).
When 𝑡 = 2.00 𝑦𝑒𝑎𝑟𝑠, a communication signal travelling at the speed of light is sent from
Earth towards the spaceship.
a. Copy the spacetime diagram and draw on it:
i.
the space axis 𝑥′ for the reference frame of the spaceship
Space axis 𝑥′ is titled above the x axis by the angle 𝜃, and
𝑣
1
tan 𝜃 = =
𝑐 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑜𝑟𝑙𝑑𝑙𝑖𝑛𝑒
Since, 𝜃 is the angle between the ct (y) axis and the worldline.
Hence,
𝑥2 − 𝑥 −2 − 0
tan 𝜃 =
=
= −0.5
𝑦 − 𝑦1
4−0
∴ 𝜃 = tan−1 0.5 = −26.6 °
Thus, 𝑥′ axis will be at an angle of 𝜃 = 26.6 ° below the 𝑥-axis
ii.
the worldline of the communication signal.
The signal is emitted from the earth at time 2 𝑌,
hence, 𝑡 = 2 𝑦, i.e, 𝑐𝑡 = 2 𝑙𝑦, and 𝑥 = 0. Call this event P
Thus, the signal was emitted at P(0,2) ; 𝑥𝑃 = 0 , 𝑐𝑡𝑃 = 2
Signal is electromagnetic wave; hence it moves at speed of light and has
worldline with (mod of) gradient = 1.
Signal moves ahead in time (towards top of graph) and in -ve x direction (left of
graph).
Hence, draw a line starting from P (0,2) that has gradient -1 as the signal’s
worldline.
b. Estimate, using the diagram, the time at which the communication signal is received:
i.
according to the clock in the spaceship
Signal will be received by the spaceship when the worldline of the signal meets
the 𝑥 ′ line. Call this point Q. (At point P, 𝑥′ = 0).
To get this time 𝑐𝑡 ′ , read the value of ct corresponding to this point Q and then
divide the ct value by the Lorentz factor 𝛾
1
1
𝛾=
=
= 1.155
2
(−0.52 )
√1
𝑣
−
√1 − 2
𝑐
draw a line parallel to the 𝑥′ axis through the event Q
To get that, draw a line parallel to the 𝑥′ axis and read it’s 𝑐𝑡 coordinate.
Multiply this 𝑐𝑡 coordinate by the Lorentz factor 𝛾 to get the value of 𝑐𝑡′.
Here, the (light red) line meets the 𝑐𝑡 axis at 𝑐𝑡 = 1.5 𝑙𝑦.
Divide this by 𝛾 =
1
𝑣2
√(1−( 2 ) )
𝑐
1
= √1−0.52 = 1.155
since (tan 𝜃 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑜𝑟𝑙𝑑 𝑙𝑖𝑛𝑒 = −0.5
Hence 𝑐𝑡 ′ =
ii.
4
1.155
= 3.46 𝑙𝑦, that is 𝑡 ′ = 3.46 𝑌 as seen from the spaceship.
according to the clock on Earth
Just read the 𝑐𝑡 value for the same even Q.
Thus,
𝑡 =4𝑌
c. The speed of the spaceship is 0.500c. Determine, using the Lorentz transformation, the
spacetime coordinates in the reference frame of the spaceship of the event when the
communication signal is received.
Read the coordinates 𝑐𝑡’ and 𝑥’ of the event Q
𝑐𝑡 ′ = 3.46 𝑙𝑦 𝑎𝑛𝑑 𝑥 ′ = 0