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The consequences of special relativity
Grant Einstein his two postulates and time dilation, length contraction and a limiting speed of c for massive objects all follow: clocks in flight run slow, lengths shrink along the direction of travel, and, in the relativistic-mass convention this option uses, the inertia grows with speed until no push, however hard, can carry an object up to c. Experimental tests of these predictions have agreed with them.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on The Michelson-Morley experiment and Mass-energy and binding energy.
IN THIS TOPIC
- Use the time dilation equation, and identify which observer measures the proper time.
- Cite muon decay as the evidence, and run the numbers both ways round.
- Use the length contraction equation with proper length.
- Describe how mass and kinetic energy vary with speed, and outline Bertozzi's direct test.
- For Edexcel, say when the relativistic stretching of a particle's lifetime is significant; that board asks for no equation with it.
COMMON MISCONCEPTION
With a strong enough accelerator you could push an electron past the speed of light.
Time dilation
Einstein's two postulates cannot both hold while time stays universal. The first thing to go is the idea that all observers share one clock. The proper time t0 between two events is the time measured in the inertial frame where both events happen at the same place, on a clock riding along with them. Any observer moving at speed v relative to that clock measures a longer time:
The root is smaller than one, so t is bigger than t0. A moving clock runs slow, as judged from the frame it moves through. At everyday speeds the correction is invisibly tiny. Near c it is enormous.
WORKED EXAMPLE
A muon's stretched lifetime
A muon at rest lives about t0 = 2.2 μs before decaying. Find its lifetime as measured by a laboratory it passes at 0.99c.
t = t0/√(1 − v2/c2) = 2.2/√(1 − 0.992) = 2.2/0.141 = 16 μs.
The muon's own clock still reads 2.2 μs at decay; the laboratory watches that clock run seven times slow. Proper time belongs to the muon, because both events, birth and decay, happen at the muon.
The muons that should not arrive
Nature runs this experiment constantly. Cosmic rays striking the upper atmosphere create muons about 10 km up, travelling near c. One undilated lifetime at 0.996c covers 650 m, so the 10 km descent lasts fifteen lifetimes, and since decay is exponential, fewer than three muons in every ten million should survive it. Ground-level detectors measure far more than that. Around a quarter of them arrive.
Time dilation closes the gap exactly. At 0.996c the factor is 11.2, the laboratory-frame lifetime stretches to 25 μs, the descent lasts only 1.4 dilated lifetimes, and the predicted survival is 26 per cent, matching what detectors count. Muon showers are time dilation measured daily, with the sky itself for apparatus.
Length contraction
The same root reaches the metre rule. The proper length l0 of an object is its length measured in its own rest frame; an observer it passes at speed v measures it shorter along the direction of motion:
Nothing is being crushed. Lengths, like durations, are relations between frames. The muon's own frame shows this. In the muon's rest frame its lifetime really is 2.2 μs, and the atmosphere is length-contracted from 10 km to under 900 m. At 0.996c that takes 3.0 μs, the same 1.4 lifetimes the ground observer counted. The two frames describe the descent differently and predict the same survival rate.
GUIDED PRACTICE
The travelling metre
A metre rule flies past at 0.80c, aligned with its motion. Find its measured length, and its measured length if instead it flies at the same speed aligned across its motion.
Show the working
Along the motion, l = 1.00 × √(1 − 0.802) = 1.00 × 0.60 = 0.60 m.
Across the motion, 1.00 m, unchanged. Contraction acts only along the direction of travel, and questions regularly test that.
Mass, energy, and Bertozzi's electrons
Speed changes the mass as well. Push on an object near light speed and the work done produces ever less increase in speed, as though the inertia itself were growing. AQA writes that growth as a relativistic mass above the rest mass m0, m = m0/√(1 − v2/c2), the convention this option examines; modern treatments keep the mass itself invariant and let energy and momentum carry the growth, and both languages describe the same physics. Tying the AQA form to the equivalence of mass and energy gives the total energy of a moving object, and the booklet prints the whole chain as one line:
Subtract off the rest energy m0c2 and what remains is the kinetic energy, which the booklet does not print:
As v approaches c the denominator collapses towards zero, so the mass, and with it the kinetic energy, grows without limit. That is what stops any accelerator, however strong, from pushing an electron past c. It is also why the check question in Cathode rays and the electron broke. eV = ½mv2 assumes all the work becomes kinetic energy of a fixed mass, when past a few tenths of c it increasingly becomes mass-energy instead. An electron's rest energy m0c2 is 8.2 × 10−14 J, about 0.51 MeV, and beyond that most of the energy an accelerator supplies goes into mass-energy rather than speed.
In 1964 William Bertozzi tested this about as directly as anyone could. Accelerate electrons through a known pd, so the kinetic energy is known from eV, then time their flight over 8.4 m to measure the speed outright. Classically v2 should climb in proportion to the energy forever. His electrons instead saturated just below c, their measured speed flattening while the energy, delivered as heat to the target, kept rising exactly as supplied. Kinetic energy varies with speed the relativistic way, by direct measurement.
INDEPENDENT PRACTICE
An impossible speed, corrected
An electron is given 2.0 MeV of kinetic energy. Show that classical physics predicts an impossible speed, and find the true speed (electron rest energy 0.51 MeV).
Show the working
Classically ½m0v2 = 2.0 MeV gives v = 8.4 × 108 m s−1, which is 2.8 times the speed of light and therefore forbidden.
Relativistically the total energy is E = 2.0 + 0.51 = 2.51 MeV, so E/m0c2 = 4.9, and v/c = √(1 − 1/4.92) = 0.98: v = 2.9 × 108 m s−1, just under c, exactly where Bertozzi's electrons sat.
ASSESSMENT FOCUS
- Choose the proper quantity first, and say why you chose it. Proper time is measured in the frame where the two events happen at the same place. Proper length is measured in the object's rest frame. Most lost marks in this topic are proper-quantity errors.
- Sanity-check the direction of every answer. Moving clocks read less time, so t0 is always the smallest. Moving lengths measure shorter, so l0 is always the longest.
- The muon answer needs its numbers in a chain. Lifetime dilated by the factor, descent time compared with it, survival transformed from negligible to substantial, and the prediction matched against what detectors count.
- Do the muon the other way round too, since papers ask for it. In the muon's own frame the lifetime is unchanged and the atmosphere is contracted instead.
- Sketching mass or kinetic energy against speed, start at the rest value, stay near it out to a few tenths of c, then rise steeply towards a vertical asymptote at c, never touching it.
- Bertozzi scores on the method's directness. Kinetic energy known from the accelerating pd and checked calorimetrically, speed measured by time of flight, and the measured v2 flattening below c.
- This page sits in an AQA option, but one strip of it is core Edexcel. That board asks when the relativistic stretching of a particle's lifetime matters, and says outright that the equations are not required, so the muon section is the part to read. Two situations answer it. Cosmic-ray muons reach the ground when an undilated lifetime would never get them there, and short-lived particles made in accelerators survive long enough to cross a detector and be recorded. Both share the condition worth stating: the speed is a large fraction of c, and the unstable particle's natural lifetime is short compared with the time the trip takes. At everyday speeds the stretch is there and far too small to notice.
CHECK YOURSELF
A spacecraft passes Earth at 0.60c. Its on-board clock records 100 s between two ticks, and its proper length is 50 m. Find the time between those ticks and the craft's length as measured from Earth, and state which measurement each observer would call "correct".
Show a hint
The root is √(1 − 0.36) = 0.80. Decide which quantity dilates and which contracts.
Show the answer
The ticks happen at the ship, so 100 s is the proper time, and Earth measures t = 100/0.80 = 125 s. The moving clock runs slow.
For length, 50 m is the proper length, and Earth measures l = 50 × 0.80 = 40 m, contracted along the motion.
Both are correct. Each observer's measurements are right in their own inertial frame. The point of relativity here is that these quantities are frame-dependent, with the proper values belonging to the frame riding with the clock and the rule.
Moving clocks run slow and moving lengths shrink, by the same root, and the muons prove it.
Proper time is measured where both events happen in one place. Proper length is measured at rest with the object.
Mass grows with speed, E = mc² says by how much, and c is a limiting speed no amount of energy can reach.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the consequences of special relativity questions page.
CHECK YOUR PROGRESS
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- Use the time dilation equation, and identify which observer measures the proper time.
- Cite muon decay as the evidence, and run the numbers both ways round.
- Use the length contraction equation with proper length.
- Describe how mass and kinetic energy vary with speed, and outline Bertozzi's direct test.
- For Edexcel, say when the relativistic stretching of a particle's lifetime is significant; that board asks for no equation with it.
Open the full revision checklist to track your progress across the whole unit.