research30 min studyLesson 993 in Physics

Limiting cases and correspondence principles

Recover Newtonian kinetic energy with declared variables, truncation errors and the rest-energy subtraction.

On this page

What you’ll understand

  • Derive a low-speed expansion after subtracting rest energy.
  • Evaluate a truncation against an exact finite comparison.
  • Explain why fixed-speed and fixed-momentum corrections have different signs.

A correspondence principle connects descriptions in an overlapping regime, but it must specify the quantity compared and the variables held fixed. Relativistic total energy does not approach Newtonian kinetic energy directly: the rest energy must first be subtracted. At fixed speed the first relativistic correction is positive; at fixed momentum the corresponding correction is negative. Neither sign is paradoxical. They concern different experimental specifications. An effective research check makes this bookkeeping explicit and tests whether the chosen expansion parameter is small enough for the observable and accuracy being claimed.

Subtract the quantity that survives at rest

For a free massive particle in an inertial frame, total energy equals gamma times rest mass times c squared. The mass is the invariant rest mass, and speed is the frame-dependent three-speed. Subtracting the rest energy isolates kinetic energy. Expanding the inverse square root for beta much smaller than one yields the Newtonian term followed by powers of beta squared. The speed of light supplies the conversion between mass and energy, not an adjustable fit parameter.

Taking beta to zero while examining total energy mostly tests its rest-energy term. To recover Newtonian motion, compare the much smaller difference above that term. This distinction matters computationally because subtracting two nearly equal floating-point numbers can lose relative precision. Algebraic rearrangement or an appropriately controlled series may evaluate the small difference better. Analytical correspondence and numerical stability are related checks, but one does not replace the other.

β=v/c,K(v)mc2=11−β2−1=β22+3β48+O(β6)\beta=v/c,\qquad \frac{K(v)}{mc^2}=\frac{1}{\sqrt{1-\beta^2}}-1=\frac{\beta^2}{2}+\frac{3\beta^4}{8}+O(\beta^6)
m is rest mass in kg, c and v have m/s units, and K is kinetic energy in joules. The power series is a low-speed expansion at fixed v; its leading neglected term after the quartic term is 5β⁶/16.↔ Scroll wide equations horizontally.

Estimate error in the stated observable

Relative to the Newtonian kinetic term, the first speed-based correction is three quarters of beta squared. This estimates an accuracy regime only when higher terms are smaller. At beta equal to six tenths the expansion parameter is no longer especially small, although the exact expression is perfectly regular. Keeping an additional term can improve the answer without making a low-order truncation a rigorously bounded prediction at every larger speed. Near light speed all orders become important.

The relevant error definition also needs a denominator. An error quoted relative to total energy can look tiny while the same error is substantial relative to kinetic energy. A measurement designed to resolve the kinetic correction therefore cannot use rest energy to dilute its accuracy requirement. When a theory is checked against an older theory, report agreement of matching observables in the overlapping regime, together with the size of the first discrepancy and the conditions under which it was estimated.

q=p/(mc),K(p)mc2=1+q2−1=q22−q48+O(q6)q=p/(mc),\qquad \frac{K(p)}{mc^2}=\sqrt{1+q^2}-1=\frac{q^2}{2}-\frac{q^4}{8}+O(q^6)
This second expansion holds momentum p fixed instead of speed. q is dimensionless and the Newtonian expression is p²/(2m). Its negative correction is compatible with the positive fixed-speed correction because p=γmv in relativity.↔ Scroll wide equations horizontally.

A limit can fail to be uniform

A small parameter does not necessarily guarantee a small change in every derived prediction. If a frequency has a small fractional correction, its accumulated phase difference grows with the observation duration. Likewise, a small correction to an energy difference can dominate a nearly canceled residual. One must specify which times, lengths and near-degeneracies are allowed to vary during the limit. Correspondence at fixed finite time need not justify extrapolation over a duration that diverges as the parameter shrinks.

These cautions are constructive. A limiting calculation supplies a benchmark, a leading correction and a way to plan a measurement that resolves that correction. A failure to recover a known overlap can expose a missed constant, a different definition of momentum, or an uncontrolled approximation before extensive simulation begins. Successful recovery establishes consistency in that regime; it does not prove the more general theory or its empirical applicability outside the shared domain. The worked and practice comparisons deliberately hold different physical inputs fixed.

Test a speed expansion at a finite speed

For v=0.60c, calculate K/(mc²) exactly, its Newtonian value, and its value through the quartic speed term. Express each approximation’s shortfall as a percentage of the exact kinetic energy. Use √0.64=0.8.

  1. Here β²=0.36 and γ=1/0.8=1.25. After subtracting rest energy, the exact dimensionless kinetic energy is 0.25.

  2. The Newtonian value is β²/2=0.18. Its shortfall is 0.07, or 0.07/0.25=28% of the exact kinetic energy. Comparing with total energy would answer a different accuracy question.

  3. The quartic term is (3/8)(0.36)²=0.0486. Adding it gives 0.2286, short by 0.0214. Relative to the exact kinetic energy that is 8.56%. The improvement is real, but these errors do not support a high-precision low-speed approximation at this speed.

The result

The exact, Newtonian and quartic values are 0.25, 0.18 and 0.2286 in units of mc². Their comparison tests a controlled limiting idea using a finite case where the truncation error is visible.

Explore the model

Interactive lab

Interactive lab · function

Isolate the leading speed term

Interpret x as β and y as the Newtonian K/(mc²). Reset and set both controls to Coefficient a=0.5 and Position x=0.6; predict y=0.18. For a fresh comparison, Reset and set both controls to a=0.5 and x=0.1; predict y=0.005. The leading relative speed correction scales with x².

Isolate the leading speed termThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis quadratic instrument displays only the leading term. It does not evaluate gamma, exact relativistic energy, the quartic correction or a fixed-momentum comparison.

Check your understanding

Why is the quartic correction positive at fixed speed but negative at fixed momentum?

Try it yourself

Instead specify p=0.10mc. Use √1.01=1.00498756 to calculate the exact K/(mc²), the Newtonian fixed-momentum value, and the approximation through q⁴. Does the Newtonian estimate lie above or below the exact answer?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the momentum expansion, not the speed expansion with β replaced by q. Here q=p/(mc).

Compare with the solution

The exact value is √1.01−1=0.00498756. The Newtonian fixed-momentum value is q²/2=0.005, which lies above the exact value by 0.00001244, about 0.2494% of the exact kinetic energy. Including −q⁴/8 subtracts 0.0000125 and gives 0.0049875. Its remaining shortfall is about 0.00000006 in units of mc², consistent with a higher-order correction. The opposite Newtonian error sign from the fixed-speed example follows from fixing momentum instead of speed, not from choosing a different energy law.

Take it with you

  • Compare matching observables after removing constants such as rest energy.
  • A correspondence claim includes its fixed variables and truncation accuracy.
  • Small-parameter limits can become nonuniform for long times or small residuals.

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