Physics/Option Modules/Turning Points in Physics/Relativistic Energy
Physics · Turning Points in Physics · A2 Level

Relativistic Energy

Every time an accelerator like CERN's Large Hadron Collider pushes a beam of particles a little faster, it gets a little more expensive to squeeze out that last increment of speed — and no matter how much energy is poured in, the particles never quite reach the speed of light. That isn't an engineering shortfall. It falls directly out of putting rest energy and relativistic momentum together into a single expression for total energy.

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Total relativistic energy

The previous two lessons built up the two pieces needed here: relativistic momentum, $p=\gamma mv$, and rest energy, $E_0=mc^2$. Combining the idea that a moving object carries more energy than a stationary one leads to the full expression for an object's total relativistic energy:

$E = \gamma mc^2$
where $\gamma=1/\sqrt{1-v^2/c^2}$ is, once again, the same Lorentz factor controlling every relativistic effect in this module, and $m$ is the object's fixed rest mass.

This total energy splits naturally into two parts. At $v=0$, $\gamma=1$, and $E$ reduces to exactly the rest energy $E_0=mc^2$ from the previous lesson — the energy an object has purely by virtue of its mass, even standing still. Any additional speed adds extra energy on top of that rest energy, and this extra part is the relativistic kinetic energy:

$E_k = E - E_0 = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2$
So total energy is rest energy plus kinetic energy, exactly as intuition demands — the relativistic treatment just makes both terms precise and shows how they're linked through the single factor $\gamma$.

DerivationRecovering the classical ½mv² at low speed
1
For $v\ll c$, the ratio $x=v^2/c^2$ is small, so $\gamma=(1-x)^{-1/2}$ can be expanded using the binomial approximation $(1+u)^n\approx1+nu$ for small $u$, with $u=-x$ and $n=-\tfrac12$.
2
This gives $\gamma \approx 1 + \left(-\tfrac12\right)(-x) = 1 + \tfrac12 x = 1 + \tfrac12\dfrac{v^2}{c^2}$.
3
Substitute into $E_k=(\gamma-1)mc^2$: $E_k \approx \left(1+\tfrac12\dfrac{v^2}{c^2}-1\right)mc^2 = \tfrac12\dfrac{v^2}{c^2}\times mc^2$.
4
The $c^2$ cancels: $E_k \approx \tfrac12 mv^2$.
The familiar Newtonian kinetic energy formula is recovered exactly as the low-speed limit of the full relativistic expression — it was never "wrong", only ever an excellent approximation that quietly assumes $v\ll c$, which covers essentially every everyday situation but breaks down badly as $v$ approaches a significant fraction of $c$.
▪ derived
Worked exampleRelativistic vs classical KE for a fast electron
1
An electron ($m = 9.11\times10^{-31}\,\text{kg}$) in a particle accelerator travels at $v = 0.90c$. Find its relativistic kinetic energy and compare it with the classical (Newtonian) prediction.
2
Find $\gamma$ at $v=0.90c$: $\gamma = \dfrac{1}{\sqrt{1-0.90^2}} = \dfrac{1}{\sqrt{1-0.81}} = \dfrac{1}{\sqrt{0.19}} \approx \dfrac{1}{0.4359} \approx 2.294$.
3
Relativistic KE: $E_k = (\gamma-1)mc^2 = (2.294-1)\times9.11\times10^{-31}\times(3.00\times10^8)^2$.
4
$E_k = 1.294 \times 9.11\times10^{-31} \times 9.00\times10^{16} \approx 1.294 \times 8.199\times10^{-14} \approx 1.061\times10^{-13}\,\text{J}$ (about $663\,\text{keV}$).
5
Classical (Newtonian) KE for comparison: $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}\times9.11\times10^{-31}\times(0.90\times3.00\times10^8)^2 \approx \tfrac{1}{2}\times9.11\times10^{-31}\times7.29\times10^{16} \approx 3.32\times10^{-14}\,\text{J}$ (about $207\,\text{keV}$).
The classical formula underestimates the true kinetic energy by more than a factor of 3 at this speed — a huge discrepancy. At $v=0.90c$, the electron is moving fast enough that the low-speed approximation used to derive $\tfrac12mv^2$ has broken down badly, and the full relativistic formula is essential.
▪ answer
Graph of kinetic energy on the vertical axis against speed v as a fraction of the speed of light c on the horizontal axis, running from 0 to 1. Two curves are drawn from the same origin: a dashed curve labelled classical, one half m v squared, following a gentle upward parabola that continues smoothly and only moderately steeply across the full width of the graph; a solid curve labelled relativistic, gamma minus one m c squared, that closely follows the dashed classical curve at small v/c but then bends sharply upward after about v/c equals 0.5, rising steeply and diverging further and further above the dashed curve as v/c approaches 1, with a dashed vertical asymptote line at v/c equals 1 that the solid curve approaches but never touches.
Classical kinetic energy (½mv²) and relativistic kinetic energy ((γ−1)mc²) agree closely at low speed, then diverge sharply as v approaches c — the relativistic curve rises without bound.
Speed (v/c)Lorentz factor γClassical Eₖ = ½mv² (relative)Relativistic Eₖ = (γ−1)mc² (relative)
0.101.0050.00500.0050
0.501.1550.1250.155
0.902.2940.4051.294
0.997.0890.4906.089
0.99922.370.499521.37

Kinetic energy values above are given as multiples of $mc^2$ for direct comparison. The two formulas agree closely below about $v=0.3c$, then diverge rapidly — classical KE approaches a finite limit as $v\to c$, while relativistic KE grows without bound.

Why nothing with mass can reach the speed of light

Putting total energy and kinetic energy together gives the clearest possible derivation of the relativistic speed limit already previewed in the earlier lessons on time dilation and momentum.

DerivationThe energy cost of approaching c
1
As an object's speed $v$ approaches $c$, the ratio $v^2/c^2$ approaches 1, so the quantity under the square root in $\gamma = 1/\sqrt{1-v^2/c^2}$ approaches zero.
2
Dividing by a quantity approaching zero means $\gamma$ itself grows without bound: as $v\to c$, $\gamma\to\infty$.
3
Since kinetic energy is $E_k = (\gamma-1)mc^2$, and $\gamma\to\infty$ as $v\to c$, the kinetic energy required to accelerate any object with non-zero rest mass all the way to $v=c$ would itself have to be infinite.
4
No real source can supply infinite energy, so no object with mass can ever actually be accelerated to reach — let alone exceed — the speed of light. $c$ acts as an unreachable asymptotic limit, approached ever more closely (and at ever greater energy cost) but never attained.
▪ derived
Purpose, method, result, conclusion — Bertozzi's experiment

Purpose: to test directly, by measuring actual electron speeds, whether kinetic energy and speed really follow the relativistic relationship $E_k=(\gamma-1)mc^2$ rather than the classical $E_k=\tfrac12mv^2$ at high energy. Method: William Bertozzi, in 1964, used a Van de Graaff accelerator to give electrons a series of increasingly large, precisely known kinetic energies, then measured each batch's actual speed directly and independently — by timing how long the electrons took to cross a fixed, accurately measured distance (a time-of-flight measurement), with no assumption about the relationship between energy and speed built into the measurement itself. Result: at low kinetic energy, the measured speeds matched both the classical and relativistic predictions equally well, but as kinetic energy increased, the measured speeds climbed ever more slowly and asymptoted toward $c$, tracking the relativistic prediction closely — the classical prediction, by contrast, implied speeds exceeding $c$ at the higher energies used, which was never observed. Conclusion: kinetic energy and speed are related by the relativistic formula, not the classical one; however much kinetic energy is supplied, speed approaches but never reaches $c$ — direct, quantitative experimental confirmation of the energy-speed relationship derived above, independent of the muon-decay evidence for time dilation covered earlier in this module.

Why accelerators need ever more energy for diminishing returns

This is exactly the practical situation facing large particle accelerators like CERN's Large Hadron Collider. As a beam of particles is accelerated closer and closer to $c$, each further increase in speed requires a disproportionately larger input of energy than the last — because kinetic energy grows roughly as $\gamma$, which itself grows increasingly steeply as $v\to c$. Doubling a particle's energy at low speed roughly doubles its speed; doubling a highly relativistic particle's energy barely nudges its speed any closer to $c$ at all, since it's already so close. This is why accelerator design focuses on reaching higher and higher energies (measured in electronvolts, GeV or TeV) rather than talking about speed at all once particles are highly relativistic — energy, not speed, is the quantity that keeps usefully increasing with more input power.

Speed limit, not an energy limit

It's worth being precise about what is and isn't limited here. There is no upper limit on an object's energy or momentum — both can be increased without bound by supplying more energy. What's limited is speed: no matter how much energy is added, $v$ can only ever get closer to $c$, never reach or exceed it. This is also why photons, which travel at exactly $c$, must have zero rest mass — the derivation above shows that reaching $v=c$ with any non-zero rest mass would require infinite energy, so anything that does travel at exactly $c$ can only do so by having no rest mass at all.

Key takeaways

  • Total relativistic energy is $E=\gamma mc^2$, splitting into rest energy $E_0=mc^2$ (at $v=0$, where $\gamma=1$) plus relativistic kinetic energy $E_k=(\gamma-1)mc^2$.
  • A binomial expansion of $\gamma$ at low speed ($\gamma\approx1+\tfrac12v^2/c^2$) shows $E_k\approx\tfrac12mv^2$ is recovered exactly as the low-speed limit of the full relativistic formula.
  • At $v=0.90c$ an electron's relativistic KE (≈663 keV) is more than three times the classical prediction (≈207 keV); the two formulas diverge sharply above about $v\approx0.3c$.
  • As $v\to c$, $\gamma\to\infty$, so the kinetic energy needed to reach $c$ would be infinite — this is why nothing with non-zero rest mass can reach or exceed the speed of light.
  • Bertozzi's experiment (1964) measured electron speeds directly by time-of-flight across a range of kinetic energies, confirming speeds asymptote toward $c$ exactly as the relativistic formula predicts, never approaching the classical prediction's speeds above $c$.
  • Particle accelerators like CERN's LHC need ever-larger energy inputs for diminishing increases in speed as $v\to c$, which is why accelerator performance is measured in energy (GeV/TeV) rather than speed.
  • Only particles with zero rest mass, like photons, can travel at exactly $c$ — anything with mass approaches but never reaches it.
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