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Massless and low-speed energy limits

Key idea: The energy–momentum relation gives E = pc for zero rest mass. For a massive particle at low speed, total energy approaches mc² + ½mv²; the rest energy remains.

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  • GCE A-Level H3 Physics 2027
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The energy–momentum relation, E² = (pc)² + (mc²)², covers both massive and massless particles. Its two required limits use different conditions: zero rest mass, or a massive particle moving slowly compared with light. Keep that distinction visible throughout each derivation.

Here E is total energy, p ≥ 0 is momentum magnitude and m is invariant rest mass. Use one inertial frame’s energy and momentum. The vacuum light speed is c; energy is in joules and momentum in kg m s⁻¹.

Check your understanding 1 Predict first

A massive particle moves at v≪ c. What does E² = (pc)² + (mc²)² give for its total energy?

Total energy at low speed
Show answer

About mc² + (1/2)mv². At low speed pc≪ mc², so the rest energy dominates and the motion adds a small kinetic term. The section on low speed below derives this.

Massless means zero rest mass, not zero energy

Set the rest mass to zero in the energy–momentum relation:

E² = (pc)².

Taking the positive-energy root, with momentum magnitude p ≥ 0, gives

E = pc.

This result is exact for a massless particle, such as a photon. It is not a low-speed approximation. A photon in vacuum travels at c and has no rest frame; the massive-particle expressions involving γ m cannot be evaluated by multiplying zero mass by an infinite Lorentz factor.

For a massive particle, E > pc. At sufficiently large momentum pc≫ mc², E ≈ pc can be an approximation even though m is still non-zero. State the condition rather than describing such a particle as massless.

Guided practice 1

A photon moving along the negative axis

About 5 min

Problem

A photon has momentum component pₓ = -2.4 × 10⁻²⁷ kg m s⁻¹. On paper, start from the energy–momentum relation and derive the limiting form. Find the energy, then compare your derivation with the solution.

Try this before viewing the solution

Form of the energy–momentum relation
Unit: J

Hints

Hint 1

direction and magnitude

The energy relation uses p = |pₓ| here. The negative component tells you the direction.
Hint 2

choose the physical square root

Set m = 0 and take positive energy after taking the square root.
Show solution step by step

  1. Apply zero rest mass

    Method

    Set m = 0 in the energy–momentum relation.

    Reason

    The photon is massless; a rest-frame Lorentz factor is not available.

    Working

    Answer

    E² = (pc)².

  2. Take the positive root

    Method

    Use p = |pₓ| and E = pc.

    Reason

    Energy is positive regardless of the chosen direction of the axis.

    Working

    Answer

    p = 2.4 × 10⁻²⁷ kg m s⁻¹.

  3. Calculate

    Method

    Multiply momentum magnitude by the vacuum light speed.

    Reason

    Use c = 3.00 × 10⁸ m s⁻¹ and round the final answer to two significant figures.

    Working

    Answer

    E = (2.4 × 10⁻²⁷)(3.00 × 10⁸) = 7.2 × 10⁻¹⁹ J.

Low speed retains rest energy

Now let m > 0 and v≪ c. Factor out the rest-energy scale:

E = mc² square root of (1 + (p/mc)²).

For a small dimensionless quantity ε, the binomial approximation is square root of (1 + ε) ≈ 1 + ε/2. Here

ε = (p/mc)² = γ²v²/c²≪1,

because γ ≈ 1 at low speed. Substitution gives

E ≈ mc²[1 + (1/2)(p/mc)²] = mc² + p²/2m.

Using p ≈ mv in this same regime,

E ≈ mc² + (1/2)mv².

The first term is rest energy; only the second is the leading kinetic-energy term. Subtracting mc² recovers K ≈ (1/2)mv². Do not divide by m in this derivation and then put m = 0: it was derived for a massive particle.

Worked example 2

A small momentum correction

Problem

A massive particle has p/(mc) = 0.10. Express its approximate total and kinetic energies in units of mc², starting from the energy–momentum relation.
Show full solution

  1. Factor out rest energy

    Method

    Write the energy as a dimensionless square root.

    Reason

    The small quantity must be a ratio with no units.

    Working

    Answer

    E/(mc²) = square root of (1 + 0.10²).

  2. Keep the leading correction

    Method

    Use square root of (1 + ε) ≈ 1 + ε/2 with ε = 0.010.

    Reason

    The expansion parameter is the squared momentum ratio.

    Working

    Answer

    E/(mc²) ≈ 1 + 0.010/2 = 1.005.

  3. Separate kinetic energy

    Method

    Subtract the rest-energy ratio one.

    Reason

    Total energy and kinetic energy are different quantities.

    Working

    Answer

    K/(mc²) ≈ 0.0050.

Spot the mistake 3

Low-speed check

About 5 min

Learner claim

A learner says K = (γ-1)mc² cannot agree with K = (1/2)mv² because it contains rest-energy factors. Use the low-speed expansion to correct the claim.

Try this before viewing the solution

Show solution step by step

  1. Expand gamma

    Method

    For a massive particle at v≪ c, retain the leading correction.

    Reason

    Expand (1-v²/c²)^(-1/2) to first order in v²/c².

    Working

    Answer

    γ ≈ 1 + (1/2)v²/c²

  2. Subtract rest contribution

    Method

    γ-1 removes the leading one.

    Reason

    Kinetic energy excludes rest energy.

    Working

    Answer

    γ-1 ≈ (1/2)v²/c²

  3. Cancel factors

    Method

    The c² factors cancel.

    Reason

    The classical kinetic-energy scale remains.

    Working

    Answer

    K ≈ ((1/2)v²/c²)mc² = (1/2)mv²

Decide whether the approximation is accurate enough

For a massive particle, write β = v/c and divide kinetic energy by the non-zero scale mc²:

Kᵣₑₗ/mc² = 1/(square root of (1-β²))-1, K_Newtonian/mc² = (1/2)β².

Exact and Newtonian kinetic energy

Both curves begin at zero and agree closely at low speed. The exact kinetic-energy ratio rises increasingly above the Newtonian curve as speed approaches c. The normalisation requires m greater than zero.

Scroll across the figure to read all labels.

Both curves begin at zero and agree closely at low speed. The exact kinetic-energy ratio rises increasingly above the Newtonian curve as speed approaches c. The normalisation requires m greater than zero.Both curves begin at zero and agree closely at low speed. The exact kinetic-energy ratio rises increasingly above the Newtonian curve as speed approaches c. The normalisation requires m greater than zero.
Calculated kinetic-energy ratios for a massive particle, using K/(mc²) = 1/sqrt(1 − β²) − 1 and its Newtonian approximation β²/2. Here β = v/c; samples extend to 0.95c. These are model values, not measurements.
Open full-size graph
What to noticeAt 0.80c, the exact ratio is 2/3 and the Newtonian ratio is 0.32. Low speed is an approximation condition, not a condition of zero rest mass.
View figure data
Figure data: Exact and Newtonian kinetic energy
Speed ratio v/c (1)RelativisticNewtonian approximation
0.0000.000000000.0000000
0.0100.000050000.0000500
0.0200.000200060.0002000
0.0300.000450300.0004500
0.0400.000800960.0008000
0.0500.001252350.0012500
0.0600.001804870.0018000
0.0700.002459040.0024500
0.0800.003215440.0032000
0.0900.004074770.0040500
0.1000.005037820.0050000
0.1100.006105460.0060500
0.1200.007278710.0072000
0.1300.008558630.0084500
0.1400.009946450.0098000
0.1500.011443470.0112500
0.1600.013051120.0128000
0.1700.014770940.0144500
0.1800.016604600.0162000
0.1900.018553890.0180500
0.2000.020620730.0200000
0.2100.022807180.0220500
0.2200.025115460.0242000
0.2300.027547910.0264500
0.2400.030107050.0288000
0.2500.032795560.0312500
0.2600.035616280.0338000
0.2700.038572240.0364500
0.2800.041666670.0392000
0.2900.044902990.0420500
0.3000.048284840.0450000
0.3100.051816080.0480500
0.3200.055500830.0512000
0.3300.059343430.0544500
0.3400.063348530.0578000
0.3500.067521030.0612500
0.3600.071866160.0648000
0.3700.076389470.0684500
0.3800.081096880.0722000
0.3900.085994640.0760500
0.4000.091089450.0800000
0.4100.096388410.0840500
0.4200.101899110.0882000
0.4300.107629600.0924500
0.4400.113588510.0968000
0.4500.119785020.1012500
0.4600.126228960.1058000
0.4700.132930840.1104500
0.4800.139901880.1152000
0.4900.147154140.1200500
0.5000.154700540.1250000
0.5100.162554940.1300500
0.5200.170732260.1352000
0.5300.179248560.1404500
0.5400.188121140.1458000
0.5500.197368680.1512500
0.5600.207011370.1568000
0.5700.217071090.1624500
0.5800.227571540.1682000
0.5900.238538500.1740500
0.6000.250000000.1800000
0.6100.261986640.1860500
0.6200.274531850.1922000
0.6300.287672250.1984500
0.6400.301448020.2048000
0.6500.315903390.2112500
0.6600.331087170.2178000
0.6700.347053340.2244500
0.6800.363861810.2312000
0.6900.381579250.2380500
0.7000.400280080.2450000
0.7100.420047740.2520500
0.7200.440976040.2592000
0.7300.463170940.2664500
0.7400.486752580.2738000
0.7500.511857890.2812500
0.7600.538643640.2888000
0.7700.567290310.2964500
0.7800.598006930.3042000
0.7900.631037090.3120500
0.8000.666666670.3200000
0.8100.705233720.3280500
0.8200.747141390.3362000
0.8300.792874930.3444500
0.8400.843024450.3528000
0.8500.898315990.3612500
0.8600.959654500.3698000
0.8701.028184790.3784500
0.8801.105379800.3872000
0.8901.193172320.3960500
0.9001.294157340.4050000
0.9011.305108350.4059005
0.9021.316230130.4068020
0.9031.327527060.4077045
0.9041.339003710.4086080
0.9051.350664820.4095125
0.9061.362515270.4104180
0.9071.374560170.4113245
0.9081.386804790.4122320
0.9091.399254620.4131405
0.9101.411915350.4140500
0.9111.424792910.4149605
0.9121.437893450.4158720
0.9131.451223370.4167845
0.9141.464789340.4176980
0.9151.478598300.4186125
0.9161.492657470.4195280
0.9171.506974390.4204445
0.9181.521556900.4213620
0.9191.536413190.4222805
0.9201.551551820.4232000
0.9211.566981700.4241205
0.9221.582712160.4250420
0.9231.598752950.4259645
0.9241.615114270.4268880
0.9251.631806780.4278125
0.9261.648841660.4287380
0.9271.666230630.4296645
0.9281.683985940.4305920
0.9291.702120500.4315205
0.9301.720647810.4324500
0.9311.739582090.4333805
0.9321.758938260.4343120
0.9331.778732060.4352445
0.9341.798980010.4361780
0.9351.819699560.4371125
0.9361.840909090.4380480
0.9371.862628000.4389845
0.9381.884876790.4399220
0.9391.907677110.4408605
0.9401.931051910.4418000
0.9411.955025460.4427405
0.9421.979623500.4436820
0.9432.004873350.4446245
0.9442.030804030.4455680
0.9452.057446400.4465125
0.9462.084833310.4474580
0.9472.112999780.4484045
0.9482.141983150.4493520
0.9492.171823350.4503005
0.9502.202563080.4512500

At 0.10c, the ratios are about 0.005038 and 0.005000: the exact kinetic energy is about 0.76% above the Newtonian value. At 0.80c, they are 2/3 and 0.32, a substantial difference. The normalisation lets the same graph describe any non-zero rest mass.

There is no single speed cutoff for every calculation. “Low speed” means that the omitted terms are small enough for the precision required. The graph is sampled below c; a massive particle cannot have a finite-energy point at v = c.

Check your understanding 2 How good is ½mv² at half the speed of light?

At v = 0.50c, find Kᵣₑₗ/(mc²) and K_Newtonian/(mc²). By what percentage is the exact kinetic energy larger?

Show the steps Optional working
Show the steps Optional working
Unit: %
Show answer

γ = 1/square root of 0.75 = 1.1547, so Kᵣₑₗ/(mc²) = 0.155. The Newtonian ratio is 0.125. The exact value is about 24% larger, so (1/2)mv² is not accurate at 0.50c for most purposes.

Check your understanding 3 Choose the condition, not just the shorter formula

A proton and a photon have the same momentum magnitude. Which has the larger total energy? Is the proton’s energy exactly pc if its speed is close to c?

Larger total energy
Near c, the proton's energy is
Show answer

The proton has non-zero rest mass, so its total energy is square root of ((pc)² + (mc²)²) > pc. The photon’s energy is exactly pc. For the proton, E ≈ pc can be a good approximation when pc≫ mc², but the rest-mass term remains non-zero and the equality is not exact.

Try it yourself 4

Choose from particle data

Minimal support

Problem

Three particles have these data: A: m = 0 and p > 0; B: m > 0 and v = 0.020c; C: m > 0 and v = 0.80c. Choose an expression for the total energy of A and B, and find C’s total energy as a multiple of mc². Explain on paper where an approximation is made and whether the rest-energy term is present; compare with the solution.

Try this before viewing the solution

Particle A
Particle B

Hints

Hint 1

read the regime first

Separate a condition on rest mass from a condition on speed. Then decide whether the square-root relation can be simplified exactly or approximately.

Show solution step by step

  1. Particle A

    Method

    Use E = pc exactly.

    Reason

    The rest mass is zero, while the stated momentum is non-zero.

    Working

    Answer

    mc² = 0.

  2. Particle B

    Method

    Use E ≈ mc² + (1/2)mv² if leading-order low-speed accuracy is sufficient.

    Reason

    v²/c² = 0.0004 is small.

    Working

    Answer

    K ≈ (1/2)mv²; total energy still includes mc².

  3. Particle C

    Method

    Use the exact relation, or equivalently E = γ mc².

    Reason

    v/c = 0.80 is not a low-speed regime.

    Working

    Answer

    E = (5/3)mc², including rest energy.

Common mistakes

  • Treating E = pc as a low-speed result; it is exact only for a massless particle.
  • Dropping the rest energy mc² when taking the low-speed limit of the total energy.
  • Putting m = 0 into γ mc² for a photon.
  • Calling a fast massive particle massless because E ≈ pc.
  • Assuming a single speed below which Newtonian kinetic energy is always accurate enough.

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