Time Dilation: Introduction Revision Notes | A Level AQA Physics

1. The Problem With Time

Here is a question that troubled physicists for decades: if two observers move relative to each other, whose clock is correct? The answer — and this is the genuinely strange part — is that both clocks are correct, and they genuinely disagree about how much time has passed.

This is not a measurement error or a trick of perception. Time itself passes at different rates depending on how fast you are moving. This is time dilation, and it is one of the most experimentally confirmed predictions in all of physics.

To understand why this happens, you need to accept two postulates that Einstein proposed in 1905. First, the laws of physics are identical in all inertial reference frames. Second, the speed of light in a vacuum, $c = 3.0 \times 10^8$ m/s, is the same for all observers regardless of their motion. These two ideas, taken together, force time dilation to be real.

2. Proper Time — The Clock That Travels With the Events

Before you can calculate anything, you need to be precise about which time interval you are measuring. Picture a muon — a subatomic particle — created high in the atmosphere and travelling downward at close to the speed of light. The muon carries its own internal clock, ticking away as it travels. The time measured by that internal clock, from the moment the muon is created to the moment it decays, is called the proper time, $t_0$.

The key phrase is at rest relative to the events. The muon's own clock is at rest relative to the muon's decay — it is present at both the start and the end. An observer on the ground, however, sees the muon fly past. To that observer, the two events (creation and decay) happen at different locations, and their clock records a longer time interval, $t$.

This is the core asymmetry of time dilation: the observer who is stationary relative to both events always measures the shortest time, $t_0$. Any observer in relative motion measures a longer time, $t > t_0$. The moving clock runs slow as seen from the rest frame.

3. The Time Dilation Equation

Now that proper time is pinned down, the relationship between $t_0$ and the dilated time $t$ measured by a relatively moving observer follows directly from the constancy of the speed of light. The result is one of the most important equations in A-level physics.

The denominator $\sqrt{1 - \frac{v^2}{c^2}}$ is always less than or equal to 1, which means $t$ is always greater than or equal to $t_0$. At everyday speeds, $v \ll c$, so $\frac{v^2}{c^2} \approx 0$ and $t \approx t_0$ — time dilation is completely negligible. The effect only becomes significant when $v$ is an appreciable fraction of $c$.

The quantity $\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ appears so frequently in special relativity that it has its own symbol: the Lorentz factor, $\gamma$. You can write the equation compactly as $t = \gamma t_0$. Since $\gamma \geq 1$ always, the dilated time is always at least as large as the proper time.

4. Working With the Equation — Guided Calculations

The equation looks straightforward, but the arithmetic inside the square root catches many students out. The worked examples below build from a standard calculation to a more demanding one. Work through each step yourself before revealing the next.

Worked example: Time Dilation Calculation — Muon Decay

A muon travels at $v = 0.98c$ relative to a detector on Earth. In the muon's own rest frame, its lifetime (proper time) is $t_0 = 2.2 \times 10^{-6}$ s. Calculate the lifetime of the muon as measured by the Earth-based detector. [3 marks]

1. $t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}$ — [M1] Selecting the correct equation and identifying $t_0 = 2.2 \times 10^{-6}$ s and $v = 0.98c$.
2. $\frac{v^2}{c^2} = \frac{(0.98c)^2}{c^2} = 0.98 \times 0.98 = 0.9604$ — Because $v = 0.98c$, the $c^2$ terms cancel exactly. Always expand the square: $0.98 \times 0.98 = 0.9604$, not 0.98.
3. $\sqrt{1 - 0.9604} = \sqrt{0.0396} \approx 0.199$ — $1 - 0.9604 = 0.0396$. Taking the square root: $\sqrt{0.0396} \approx 0.199$. This is the Lorentz factor denominator.
4. $t = \frac{2.2 \times 10^{-6}}{0.199} \approx 1.1 \times 10^{-5}$ s — [M1] Correct substitution. [A1] $t \approx 1.1 \times 10^{-5}$ s. The Earth detector measures a lifetime roughly five times longer than the muon's own proper time.
5. $t > t_0$? Yes: $1.1 \times 10^{-5} > 2.2 \times 10^{-6}$ ✓ — The dilated time must always be greater than or equal to the proper time. If your answer gives $t < t_0$, you have inverted the fraction — recheck.

Answer: $t \approx 1.1 \times 10^{-5}$ s (to 2 significant figures)

Notice how dramatically the lifetime extends at $0.98c$. This is not a theoretical curiosity — muons created by cosmic rays at altitudes of around 10–15 km should, based on their proper lifetime, decay long before reaching sea level. Yet detectors at ground level routinely record them. Time dilation is the explanation.

Worked example: Finding Speed from a Known Time Dilation

An astronaut on a spacecraft measures a journey time of $t_0 = 4.0$ years. Mission control on Earth measures the same journey as taking $t = 5.0$ years. Calculate the speed of the spacecraft as a fraction of $c$. [4 marks]

1. $t_0 = 4.0$ years (astronaut's clock, at rest relative to the journey events in their frame); $t = 5.0$ years (Earth clock) — The astronaut is present at both departure and arrival in their own frame, so they measure proper time. Earth observers see the spacecraft move between two locations, so they measure the dilated time.
2. $t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}$ \Rightarrow \sqrt{1 - \frac{v^2}{c^2}} = \frac{t_0}{t} = \frac{4.0}{5.0} = 0.80$ — [M1] Rearranging correctly. Squaring both sides: $1 - \frac{v^2}{c^2} = 0.80 \times 0.80 = 0.64$.
3. $\frac{v^2}{c^2} = 1 - 0.64 = 0.36$ — [M1] Correct isolation of the velocity term.
4. $\frac{v}{c} = \sqrt{0.36} = 0.60$, so $v = 0.60c$ — [A1] $v = 0.60c = 1.8 \times 10^8$ m/s. Always take the positive root — speed is a magnitude.

Answer: $v = 0.60c$ (i.e. $1.8 \times 10^8$ m/s)

5. Physical Interpretation and Experimental Evidence

Time dilation is not a philosophical puzzle — it has been confirmed experimentally to high precision. The muon example above is one of the most compelling pieces of evidence. Cosmic-ray muons are produced at altitudes of roughly 10–15 km and travel at speeds close to $0.998c$. Their proper lifetime is about $2.2 \times 10^{-6}$ s, which would classically allow them to travel only around 660 m before decaying. Yet they arrive at sea level in large numbers.

From the Earth frame, the muon's clock runs slow — its lifetime is extended by the Lorentz factor to around $35 \times 10^{-6}$ s at $0.998c$, giving it enough time to traverse the full 15 km. The Hafele–Keating experiment of 1971 provided further confirmation: atomic clocks flown around the world on aircraft returned showing slightly less elapsed time than identical clocks left on the ground, consistent with relativistic predictions.

GPS satellites also require relativistic time dilation corrections. Without them, positional errors would accumulate at around 38 microseconds per day from the combined special and general relativistic effects — enough to cause navigation errors of roughly 10 km per day. Time dilation is built into the software of every GPS receiver you have ever used.

6. The Lorentz Factor in Depth

It is worth building intuition for how $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ behaves across different speeds. At $v = 0.10c$, $\gamma \approx 1.005$ — a 0.5% effect, completely undetectable in everyday life. At $v = 0.90c$, $\gamma \approx 2.29$, meaning a moving clock runs at less than half the rate of a stationary one. At $v = 0.999c$, $\gamma \approx 22.4$.

Two features of this graph are worth noting for exam questions. First, $\gamma$ rises slowly at low speeds — the curve is nearly flat below $0.5c$. Second, $\gamma$ increases steeply and approaches infinity as $v \to c$. This is why no object with mass can ever reach $c$: it would require infinite energy to accelerate to that speed.

7. Exam Technique and Common Errors

Time dilation questions on AQA papers follow predictable patterns. The most common task is a straightforward substitution into $t = \frac{t_0}{\sqrt{1 - v^2/c^2}}$, but the marks are often lost in the arithmetic rather than the physics. The guidance below targets the errors that appear most frequently in mark scheme reports.

For extended-response questions asking you to explain time dilation, a full-marks answer needs three components: a statement of the postulate (constancy of $c$), a description of what proper time is and who measures it, and a clear statement that the moving clock runs slow as observed from the rest frame. Linking to experimental evidence such as muon survival strengthens an evaluation answer.

The time-dilation equation required by AQA is t = t0 / √(1 − v^2/c^2), where t0 is the proper time measured in the frame where the clock is at rest. Because the denominator is less than 1 for any non-zero speed, the observed time interval t is greater than t0.