HomeLearning HubIB DP PhysicsA.5 Galilean and special relativity
A.5

Galilean and special relativity

Theme A · Space, time and motion · HL only

This was Option A. It is now core HL content, examined on the same papers as everything else.

🎯What you need to be able to do

  • Use reference frames and Galilean transformations, and state why they fail at high speed.
  • State the two postulates of special relativity and work out what follows from them.
  • Use the Lorentz transformations and the invariant spacetime interval.
  • Calculate time dilation and length contraction, and identify proper time and proper length correctly.
  • Explain the relativity of simultaneity, and use muon decay as evidence.
  • Read and draw spacetime diagrams.

📚The physics

Galilean relativity is the common-sense version and it works perfectly well at everyday speeds. Velocities simply add: walk forward at 2 m s\(^{-1}\) on a train doing 30 m s\(^{-1}\) and the ground sees 32 m s\(^{-1}\). The transformations are \( x' = x - vt \) and \( t' = t \) — note that second equation, because it quietly assumes everyone shares one universal clock. That assumption is the one Einstein throws away.

The problem is that Maxwell’s equations predict a specific speed for light, with no mention of who is measuring it. Under Galilean addition, chasing a light beam at half its speed should make it recede at half speed. Experiment says otherwise.

The two postulates. First, the laws of physics are identical in all inertial frames. Second, the speed of light in a vacuum is the same for all inertial observers, regardless of the motion of the source or the observer. Everything else in this topic is a consequence — and each consequence is strange precisely because the second postulate is being taken seriously.

Time dilation. A moving clock runs slow:

\[ \Delta t = \gamma\,\Delta t_0 \qquad \text{where} \qquad \gamma = \frac{1}{\sqrt{1 - v^{2}/c^{2}}} \]

and \( \Delta t_0 \) is the proper time — the interval measured in the frame where the two events happen at the same place. Getting proper time right is most of the battle. Ask yourself which observer sees both events at one location; that observer measures the proper time, and every other observer measures something longer.

Length contraction. A moving object is shortened along its direction of motion: \( L = L_0/\gamma \), where proper length \(L_0\) is measured in the frame where the object is at rest. Note the asymmetry with time: proper time is the shortest interval, proper length is the longest length. Contraction happens only along the direction of motion; transverse dimensions are unaffected.

Relativity of simultaneity. Two events that are simultaneous in one frame are generally not simultaneous in another. This is not a measurement error or a signal-delay effect — it is a statement about time itself. It is also the resolution of most apparent paradoxes in this topic: when two observers disagree about lengths or times, they are usually disagreeing about which events counted as happening “at once”.

The spacetime interval is what everyone does agree on. While \(\Delta x\) and \(\Delta t\) differ between frames, the quantity \( (c\Delta t)^{2} - (\Delta x)^{2} \) is invariant. Relativity is less a theory about things being subjective than a theory about which quantity is the real one.

Muons are the standard evidence. Created high in the atmosphere, they have a half-life so short that classically almost none should reach the ground. Far more arrive than that. From our frame, their clocks run slow so they live long enough; from the muon’s frame, its lifetime is perfectly ordinary but the atmosphere is length-contracted to a fraction of its thickness. Both frames predict the same number of muons at the detector, which is the point — the physics agrees even when the descriptions do not.

✏️Worked example

A spacecraft passes Earth at \(0.80c\) heading for a star 6.0 light-years away as measured from Earth.

First find \(\gamma\). \( \gamma = 1/\sqrt{1 - 0.80^{2}} = 1/\sqrt{0.36} = 1/0.60 = 1.67 \).

(a) How long does the trip take in the Earth frame? Distance 6.0 ly at \(0.80c\) gives \( t = 6.0/0.80 = 7.5 \) years.

(b) How long does it take according to the crew? The crew are present at both departure and arrival at the same place — in their own ship — so they measure proper time: \( \Delta t_0 = 7.5/1.67 = 4.5 \) years.

(c) How far is the star according to the crew? In their frame the star rushes towards them and the distance is contracted: \( L = 6.0/1.67 = 3.6 \) light-years.

Check the consistency. The crew travel 3.6 ly at \(0.80c\), which takes \( 3.6/0.80 = 4.5 \) years — exactly the answer to (b). The two frames disagree about distance and about duration, but they agree about speed and about the crew’s age on arrival. If your numbers do not close like this, something is wrong.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Calculate the Lorentz factor \(\gamma\) for a speed of \( 0.60c \).
\( \gamma = 1/\sqrt{1 - 0.60^{2}} = 1/\sqrt{0.64} = 1/0.80 = 1.25 \).
2. A muon has a proper lifetime of 2.2 µs. It travels at \( 0.995c \), for which \( \gamma = 10.0 \). How long does it live in the Earth frame?
\( \Delta t = \gamma\Delta t_0 = 10.0 \times 2.2 = 22 \) µs. The proper time is the 2.2 µs, because in the muon’s own frame it is created and decays at the same place.
3. A spacecraft of proper length 120 m passes Earth at \( 0.80c \) (\( \gamma = 1.67 \)). How long does it appear to an observer on Earth?
\( L = L_0/\gamma = 120/1.67 = 72 \) m. Proper length is measured in the frame where the object is at rest, and every other frame measures something shorter.
4. In the muon experiment, explain how the muon’s own frame accounts for so many muons reaching the ground, given that its lifetime in that frame is perfectly ordinary.
In the muon’s frame its clock runs normally, so time dilation cannot be the explanation. Instead the atmosphere is rushing past at close to \(c\) and is length-contracted to a fraction of its thickness. The muon therefore has far less distance to cover, and covers it within an ordinary lifetime. Both frames predict the same number of muons at the detector, which is the whole point — the descriptions differ, the physics does not.
5. Two events are separated by \( \Delta x = 3.0 \times 10^{8} \) m and \( \Delta t = 2.0 \) s in one frame. Calculate the invariant spacetime interval \( (c\Delta t)^{2} - (\Delta x)^{2} \).
\( c\Delta t = 3.0 \times 10^{8} \times 2.0 = 6.0 \times 10^{8} \) m. So the interval is \( (6.0 \times 10^{8})^{2} - (3.0 \times 10^{8})^{2} = 3.6 \times 10^{17} - 9.0 \times 10^{16} = 2.7 \times 10^{17} \) m\(^{2}\). Every inertial observer gets this same number, however much they disagree about \(\Delta x\) and \(\Delta t\) separately.
6. State the two postulates of special relativity, and explain which one makes Galilean velocity addition fail.
First: the laws of physics are identical in all inertial frames. Second: the speed of light in a vacuum is the same for all inertial observers, regardless of the motion of the source or observer. The second is the one that breaks Galilean addition — under Galileo, chasing a light beam at \( 0.5c \) should make it recede at \( 0.5c \), but the postulate (and experiment) says it still recedes at \(c\). Keeping that forces the abandonment of the universal clock hidden in \( t' = t \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Hypertextbook — special relativity
  • HyperPhysics — relativity, time dilation and the muon experiment