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Einstein postulates and limits of classical mechanics

Learning objectives

  • Outcome

    State the two postulates of special relativity and name the classical assumption that each one rejects.

  • Outcome

    Compute a speed with u=u′+v1+u′v/c2u = \frac{u' + v}{1 + u'v/c^2} and compare the result with the Galilean value u′+vu' + v for given u′u' and vv.

  • Outcome

    Use the relative error of Galilean addition to determine for which v/cv/c classical mechanics stays within 1 % error and for which it becomes inapplicable.

  • Outcome

    Explain, using the spectrum of a moving source, why the constancy of cc contradicts Galilean velocity addition.

Understand the idea

Einstein's Postulates and the Limits of Classical Mechanics

Text

Classical mechanics rests on two tacit assumptions: time is absolute (t′=tt' = t) and velocities add simply, u=u′+vu = u' + v (Galilean relativity). Maxwell's electromagnetism, however, predicts waves travelling in vacuum at c=1/ε0μ0c = 1/\sqrt{\varepsilon_0\mu_0}, a speed fixed by constants of nature rather than by the motion of the source; the emission spectrum shows that a moving source shifts the frequency (the lines move) but does not change the propagation speed. Einstein removed the conflict with two postulates: (1) the principle of relativity — all inertial frames are equivalent for every law of physics, not only for mechanics; (2) the constancy of the speed of light — every inertial observer measures the same vacuum value cc, regardless of the motion of source or observer. Together these postulates demand the Lorentz transformation instead of the Galilean one, and hence time dilation, length contraction, and relativistic velocity addition u=u′+v1+u′v/c2u = \dfrac{u' + v}{1 + u'v/c^2}. Classical mechanics is not wrong: it is the low-speed limit v≪cv \ll c, where u′v/c2→0u'v/c^2 \to 0 and the Galilean formula is recovered. The limits appear wherever speeds approach cc: fast particles, precise time synchronisation (GPS), and the propagation of light itself.

Worked examples

Direct

Problem

A spacecraft recedes from Earth at v=0.5cv = 0.5c and launches a probe forward at u′=0.5cu' = 0.5c relative to the spacecraft. Find the probe's speed relative to Earth with the relativistic formula and compare it with the Galilean answer.

Solution

Both velocities lie along the same line, so we use the velocity-addition formula:

u=u′+v1+u′vc2=0.5c+0.5c1+(0.5c)(0.5c)c2=c1+0.25=0.8c.u = \frac{u' + v}{1 + \frac{u'v}{c^2}} = \frac{0.5c + 0.5c}{1 + \frac{(0.5c)(0.5c)}{c^2}} = \frac{c}{1 + 0.25} = 0.8c.

The Galilean count would give

uгал=u′+v=0.5c+0.5c=c.u_{\text{гал}} = u' + v = 0.5c + 0.5c = c.

The relativistic result 0.8c0.8c is below cc, whereas the Galilean one already reaches the limit cc; at any higher speeds it would give u>cu > c, which is impossible for a massive probe. This is a direct consequence of the second postulate: no addition of velocities can exceed cc.

Application

Problem

Two stages accelerate a craft: relative to Earth a stage moves at v=0.3cv = 0.3c, and a module separates from it at u′=0.4cu' = 0.4c in the same direction. Find the module's speed relative to Earth and the relative error that classical mechanics would give. May Galilean addition be used in this problem?

Solution

Relativistic calculation.

u=u′+v1+u′vc2=0.4c+0.3c1+0.12=0.7c1.12=0.625c.u = \frac{u' + v}{1 + \frac{u'v}{c^2}} = \frac{0.4c + 0.3c}{1 + 0.12} = \frac{0.7c}{1.12} = 0.625c.

Galilean calculation: uгал=u′+v=0.7cu_{\text{гал}} = u' + v = 0.7c.

Relative error:

δ=∣uгал−u∣u=0.7c−0.625c0.625c=0.12=12 %.\delta = \frac{|u_{\text{гал}} - u|}{u} = \frac{0.7c - 0.625c}{0.625c} = 0.12 = 12\,\%.

An error of 12 % vastly exceeds the acceptable 1 % threshold, so Galilean addition may not be used here. Note that for equal speeds u′=v=xcu' = v = xc the error is approximately (v/c)2(v/c)^2, so the 1 % threshold requires v≲0.1cv \lesssim 0.1c.

Transfer

Problem

A star recedes from an observer at v=0.3cv = 0.3c and emits light that reaches the observer. In the star's frame the light travels toward the observer at u′=−cu' = -c. What does Galilean addition predict, and what does the relativistic formula give? Is the answer consistent with the spectrum: the star's spectral lines are shifted to longer wavelengths by about a factor 1.36 in wavelength?

Solution

Galilean prediction:

uгал=u′+v=−c+0.3c=−0.7c.u_{\text{гал}} = u' + v = -c + 0.3c = -0.7c.

That is, the observer would see light travelling at 0.7c0.7c — and the "speed of light" would depend on the motion of the source.

Relativistic calculation:

u=u′+v1+u′vc2=−c+0.3c1+(−c)(0.3c)c2=−0.7c1−0.3=−c.u = \frac{u' + v}{1 + \frac{u'v}{c^2}} = \frac{-c + 0.3c}{1 + \frac{(-c)(0.3c)}{c^2}} = \frac{-0.7c}{1 - 0.3} = -c.

The speed magnitude equals cc for any vv: substituting u′=±cu' = \pm c into the formula always gives ∣u∣=c|u| = c. This is exactly what the second postulate asserts.

Consistency with the spectrum. The red shift of the lines (fobs=fsrc1−β1+βf_{\text{obs}} = f_{\text{src}}\sqrt{\frac{1 - \beta}{1 + \beta}} with β=0.3\beta = 0.3 gives 0.7/1.3≈0.73\sqrt{0.7/1.3} \approx 0.73, i.e. a wavelength stretch of about 1.36) reveals the motion of the source but not a slowing of light: the frequency changes while the propagation speed stays cc. Hence the classical rule u=u′+vu = u' + v cannot be applied to light and must be replaced by the relativistic one.

Practice

Text

  1. Open an available interactive special-relativity simulation (a light clock or a relativistic velocity adder), or a spreadsheet.
  2. Build a table: with u′=0.5cu' = 0.5c fixed and v/cv/c from 0.010.01 to 0.990.99 in steps of 0.010.01, compute

uрел=u′+v1+u′vc2,uгал=u′+v,δ=∣uгал−uрел∣uрел.u_{\text{рел}} = \frac{u' + v}{1 + \frac{u'v}{c^2}}, \qquad u_{\text{гал}} = u' + v, \qquad \delta = \frac{|u_{\text{гал}} - u_{\text{рел}}|}{u_{\text{рел}}}.

  1. Plot δ\delta against v/cv/c; mark the point where δ=1 %\delta = 1\,\% and the point where uгал=cu_{\text{гал}} = c. Check that as v→cv \to c the value uрел→cu_{\text{рел}} \to c, while uгал→1.5cu_{\text{гал}} \to 1.5c (it grows linearly with vv and exceeds cc for v>0.5cv > 0.5c).
  2. State a conclusion: up to which v/cv/c the classical formula keeps the error below 1 %, and why it cannot be used for light.

Safety. A numerical experiment and a simulation are sufficient. Do not run experiments with radiation sources, high-voltage vacuum tubes, or fast particles.

Visual reference

Purpose

Show where Galilean velocity addition violates the second postulate and how the relativistic formula keeps every speed below the limit cc.

Representation

A graph: horizontal axis u′/cu'/c over the range −1-1 to 11, vertical axis u/cu/c for fixed v=0.5cv = 0.5c. Two curves — the straight line uгал=u′+vu_{\text{гал}} = u' + v (rising above the level cc for u′>0.5cu' > 0.5c) and the curve u=u′+v1+u′v/c2u = \frac{u' + v}{1 + u'v/c^2}, which reaches u=cu = c at u′=cu' = c (and u=−cu = -c at u′=−cu' = -c) and stays within the light-speed limit. Mark the example points: u′=0.5cu' = 0.5c gives u=0.8cu = 0.8c, while u′=cu' = c gives u=cu = c for any vv.

Real-world examples

GPS satellite clocks

Scenario

A GPS satellite orbits at an altitude of 20 200 km with a speed of about 3.9×103 m/s3.9\times 10^3\ \mathrm{m/s} relative to the ground. Its radio signal covers the 2.02×107 m2.02\times 10^7\ \mathrm{m} to your phone in t=d/c≈0.067 st = d/c \approx 0.067\ \mathrm{s}, and the receiver turns that travel time into a distance to within a few metres.

Connection

The second postulate is at work here: in the receiver's frame the signal travels at exactly c=2.998×108 m/sc = 2.998\times 10^8\ \mathrm{m/s}, whatever the satellite's motion. The Galilean rule u′=c±3.9×103 m/su' = c \pm 3.9\times 10^3\ \mathrm{m/s} would shift the range by ±3.9×103×0.067 s≈260 m\pm 3.9\times 10^3 \times 0.067\ \mathrm{s} \approx 260\ \mathrm{m} — far more than the few metres the travel-time measurement actually delivers.

Headlight and horn: where velocity addition breaks

Scenario

A train moves at 30 m/s30\ \mathrm{m/s} (108 km/h) towards a horn sounding in still air. A passenger measures the sound arriving at 343+30=373 m/s343 + 30 = 373\ \mathrm{m/s} — the ordinary Galilean sum, an 8.7%8.7\% change. The same passenger then measures the light from a lamp: classical addition would give c+30 m/sc + 30\ \mathrm{m/s}, a shift of only 1×10−71\times 10^{-7} of cc, too small to see. The decisive test used the Earth's orbital motion, v=3.0×104 m/sv = 3.0\times 10^4\ \mathrm{m/s}, that is v/c=1.0×10−4v/c = 1.0\times 10^{-4}, which a Michelson interferometer could resolve — and no change in the speed of light was found.

Connection

Classical mechanics is built on adding velocities, and for sound the rule holds: the passenger's 373 m/s373\ \mathrm{m/s} is exactly what Galilean relativity predicts. The second postulate says light does not obey it: a lamp and a star give the same cc for every inertial observer, so no frame exists in which light is at rest — and with that frame the classical picture of an absolute space in which light propagates loses its footing.

Muons that should not reach the ground

Scenario

Cosmic rays create muons about 10 km10\ \mathrm{km} above the ground. A muon flies at 0.998c=2.99×108 m/s0.998c = 2.99\times 10^8\ \mathrm{m/s} and has a mean lifetime of 2.2 μs2.2\ \mu\mathrm{s}. Classical kinematics gives a mean path of 2.99×108×2.2×10−6≈660 m2.99\times 10^8 \times 2.2\times 10^{-6} \approx 660\ \mathrm{m}, so only a fraction e−10000/660≈3×10−7e^{-10000/660} \approx 3\times 10^{-7} should survive the trip. In fact about one muon crosses each square centimetre of ground every minute.

Connection

This is the limit of classical mechanics in a single measurement. Under Galilean transformations neither the 2.2 μs2.2\ \mu\mathrm{s} lifetime nor the 10 km10\ \mathrm{km} of air changes between frames, so classical kinematics has no way to account for the observed flux; 'time' and 'length' themselves must depend on the frame, and the postulates are the minimal assumptions that deliver that dependence.