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Spacetime diagrams IB DP Physics Study Notes

Spacetime diagrams IB DP Physics Study Notes - 2025 Syllabus

Spacetime diagrams IB DP Physics Study Notes

Spacetime diagrams IB DP Physics Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

  • space-time diagrams

  • that the angle between the world line of a moving particle and the time axis on a space-time diagram is related to the particle’s speed as given by $\tan \theta=\frac{v}{c}$

  • that muon decay experiments provide experimental evidence for time dilation and length contraction.

Standard level and higher level: There is no Standard level content
Additional higher level: 8 hours

IB DP Physics 2025 -Study Notes -All Topics

Space-Time Diagrams

Space-time diagrams are a graphical tool used in special relativity to visualize events, motion, and relativistic effects such as time dilation and length contraction.

Basic Components of Space-Time Diagrams

1. Axes:

  • Vertical axis (\( ct \)): Represents time, multiplied by \( c \) (speed of light) so that both axes have the same units.
  • Horizontal axis (\( x \)): Represents spatial position in one dimension.
  •  Events are plotted as points in this coordinate system.

2. World Lines:

  • Stationary objects: Have vertical world lines, since their position does not change over time.
  • Moving objects: Have slanted world lines, with the slope representing velocity.
  • Light: Travels at 45° since it moves at speed \( c \) in all reference frames.

Interpreting World Lines

The slope (\( v/c \)) of a world line is linked to the object’s velocity \( v \):
\[
\tan{\theta} = \frac{v}{c}
\]
\( \theta = 0^\circ \): The object is at rest (world line is vertical).
\( \theta \) increases as velocity increases.
\( \theta = 45^\circ \) for light (\( v = c \)).
As \( v \to c \), the world line becomes nearly horizontal, showing that high-speed objects cover more space in less time.

Applications of Space-Time Diagrams

Simultaneity: Events that are simultaneous in one frame may not be simultaneous in another frame.
Time Dilation: A moving clock ticks slower, visible by comparing proper time intervals along different world lines.
Length Contraction: The spatial separation of events differs based on the observer’s frame.

 

Muons – An Example of Time Dilation and Length Contraction

Muons are unstable particles created when cosmic rays interact with the upper atmosphere. They move at very high velocities $(\beta \sim 0.9999)$ and have very short lifetimes, $\tau=2 \times 10^{-6} \mathrm{~s}$, as measured in the lab.

Do muons reach the ground, given an atmospheric “thickness” of about 10 km ?

“Classical” answer: distance $=$ velocity $\times$ time $=0.9999 \mathrm{c} \times 2 \times 10^{-6} \mathrm{~s} \cong 0.6 \mathrm{~km}$
$\therefore$ conclude that muons will never reach the ground.

However, they do! What is wrong? Because muons move so quickly, relativistic effects are important. The classical answer mixes up reference frames.

  • $\tau$ – refers to lifetime in the muon reference frame (lab where $\tau$ is measured)
  • atmospheric thickness – refers to length in the Earth’s reference frame

(1) Time dilation approach – from Earth frame

$\rightarrow$ treat the muons as a clock in $\mathrm{A}^{\prime}$
$\mathrm{t}^{\prime}=0$ muons are created
$\mathrm{t}^{\prime}=\tau$ muons are destroyed
$\Delta t^{\prime}=\tau$
The creation and destruction of muons occurs at the same place in $\mathrm{A}^{\prime}$.

What is the time interval as measured in $A$ ?

$$
\Delta t=\gamma \Delta t^{\prime}=\gamma \tau=1.4 \times 10^{-4} s \text { i.e., it takes } 70 \text { times longer }
$$

$$
\therefore \text { distance }=0.9999 \mathrm{c} \times 1.4 \times 10^{-4} \mathrm{~s} \cong 42 \mathrm{~km}
$$

Since $42 \mathrm{~km}>10 \mathrm{~km}$, the muons will reach the ground.

(2) Length contraction approach – from muon frame

In reference frame A (now with the muons)

$$
\text { lifetime }=\tau=2 \times 10^{-6} \mathrm{~s}
$$

velocity of ground $=0.9999 \mathrm{c}$

So the distance that the ground travels before the muon decays is 0.6 km . But what is the thickness of the atmosphere that the muon sees? proper length of atmosphere $=10 \mathrm{~km}$ length of atmosphere in muon frame is

$$
\Delta x=\Delta x^{\prime} \sqrt{1-v^2 / c^2}=10 \sqrt{1-0.9999^2}=0.14 \mathrm{~km}
$$

i.e., the atmosphere that the muon sees is 70 times thinner $0.6 \mathrm{~km}>0.14 \mathrm{~km}$ and so the ground will reach the muon.

Thus, length contraction and time dilation are real!

IB Physics Spacetime diagrams Exam Style Worked Out Questions

Question

A spaceship leaves Earth and travels at a speed of $0.60 \mathrm{c}$ relative to the Earth to a point $P$.
$P$ is 3.0 lightyears from Earth.
The spaceship then returns to Earth. Ignore the time taken to reverse the direction of the spaceship.
What is the time taken for the total journey to and from $\mathrm{P}$ as measured by an observer in the spaceship?
A. 6.3 years
B. 5.0 years
C. 8.0 years
D. 10 years

▶️Answer/Explanation

Ans:C

Question

A spaceship is travelling at $0.60 \mathrm{c}$ from Earth when it launches a probe at $0.10 \mathrm{c}$ relative to the spaceship and away from Earth.

What is the speed of the probe relative to Earth?
A. $0.50 c$
B. $0.66 c$
C. $0.75 c$
D. $0.90 \mathrm{c}$

▶️Answer/Explanation

Ans:B

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