Relativistic Kinetic Energy Calculator
Compute relativistic kinetic energy using \(K = (\gamma - 1)mc^2\). Compare with classical kinetic energy, see the Lorentz factor, and explore how energy grows as velocity approaches the speed of light.
Relativistic kinetic energy calculator
You can enter rest mass in SI units or in energy units (MeV/c², GeV/c²). Proton/electron mass are filled automatically.
The calculator automatically checks that \( v < c \approx 2.998×108 \) m/s.
Results
Relativistic kinetic energy formula
At everyday speeds, kinetic energy is well described by the classical formula
\[ K_\text{classical} = \frac{1}{2}mv^2 \]
But when an object moves at a significant fraction of the speed of light \( c \), special relativity must be used. The relativistic kinetic energy is
\[ K_\text{rel} = (\gamma - 1)\,mc^2 \]
\[ \gamma = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} \]
where:
- \( K_\text{rel} \) is the relativistic kinetic energy,
- \( m \) is the rest mass of the object,
- \( v \) is its speed,
- \( c \approx 2.99792458 \times 10^8 \,\text{m/s} \) is the speed of light,
- \( \gamma \) is the Lorentz factor.
Total energy and rest energy
In relativity, the total energy of a particle is
\[ E = \gamma mc^2 \]
The rest energy is
\[ E_0 = mc^2 \]
Relativistic kinetic energy is simply the difference between total and rest energy:
\[ K_\text{rel} = E - E_0 = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2. \]
How this calculator works (step by step)
-
Convert mass to kilograms.
- If you enter kg or g, the value is converted to kg.
- If you enter MeV/c² or GeV/c², the calculator uses \( 1\,\text{eV}/c^2 \approx 1.78266192\times 10^{-36}\,\text{kg} \).
- Choosing “proton mass” or “electron mass” fills in their known rest masses.
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Determine the speed.
- If you enter \( \beta = v/c \), the calculator sets \( v = \beta c \).
- If you enter \( v \) directly, it checks that \( v < c \).
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Compute the Lorentz factor.
\[ \gamma = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} \]
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Compute energies.
- Rest energy: \( E_0 = mc^2 \)
- Total energy: \( E = \gamma mc^2 \)
- Relativistic KE: \( K_\text{rel} = (\gamma - 1)mc^2 \)
- Classical KE: \( K_\text{classical} = \tfrac12 mv^2 \)
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Compare classical vs relativistic.
The percent difference shown is
\[ \%\ \text{difference} = \frac{K_\text{rel} - K_\text{classical}}{K_\text{classical}} \times 100\%. \]
When do you need relativistic kinetic energy?
For small speeds \( v \ll c \), the relativistic formula reduces to the classical one. Using a Taylor expansion of \( \gamma \) for small \( v/c \):
\[ \gamma = \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} \approx 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \dots \]
\[ K_\text{rel} = (\gamma - 1)mc^2 \approx \frac{1}{2}mv^2 + \frac{3}{8}m\frac{v^4}{c^2} + \dots \]
The leading term is exactly the classical kinetic energy. The higher-order terms become important when \( v \) is a significant fraction of \( c \).
- For \( v \lesssim 0.1c \), the error of using \( \tfrac12 mv^2 \) is typically below 1%.
- By \( v \approx 0.5c \), the classical formula underestimates KE by tens of percent.
- Near \( v \approx 0.9c \) and above, the classical formula fails badly; relativistic KE is several times larger.
Worked example: 1 kg mass at 0.8c
Suppose a 1 kg object moves at \( v = 0.8c \).
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Compute \( \gamma \):
\[ \gamma = \frac{1}{\sqrt{1 - (0.8)^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} \approx 1.6667 \]
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Rest energy:
\[ E_0 = mc^2 = (1\,\text{kg})(2.998\times 10^8\,\text{m/s})^2 \approx 8.99\times 10^{16}\,\text{J} \]
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Relativistic kinetic energy:
\[ K_\text{rel} = (\gamma - 1)mc^2 = (1.6667 - 1)\,E_0 \approx 0.6667 \times 8.99\times 10^{16} \approx 6.0\times 10^{16}\,\text{J} \]
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Classical kinetic energy:
\[ K_\text{classical} = \frac{1}{2}mv^2 = \frac{1}{2}(1)(0.8c)^2 = 0.32\,c^2 \approx 2.88\times 10^{16}\,\text{J} \]
Here, the classical formula underestimates the kinetic energy by more than a factor of 2.
Relativistic kinetic energy in particle physics
In high‑energy physics, masses and energies are often expressed in electronvolts (eV), MeV, or GeV. A common convention is to set \( c = 1 \), so that mass and energy share the same units.
- Electron rest mass: \( m_e c^2 \approx 0.511\,\text{MeV} \)
- Proton rest mass: \( m_p c^2 \approx 938.27\,\text{MeV} \)
In these units, the relativistic kinetic energy is still \( K_\text{rel} = (\gamma - 1)mc^2 \), but you can treat \( mc^2 \) directly as “mass in MeV”. This calculator supports that workflow via the MeV/c² and GeV/c² mass options and shows energies in both joules and eV.
Relativistic kinetic energy – FAQ
What is relativistic kinetic energy?
Relativistic kinetic energy is the kinetic energy predicted by Einstein’s special relativity. It accounts for the fact that no object with mass can reach or exceed the speed of light, and that energy grows without bound as speed approaches \( c \). The formula is \( K_\text{rel} = (\gamma - 1)mc^2 \).
When do I need to use the relativistic formula instead of \( \tfrac12 mv^2 \)?
As a rule of thumb, if \( v \lesssim 0.1c \), the classical kinetic energy is accurate to better than about 1%. For speeds above \( 0.1c \), especially in particle accelerators or astrophysics, you should use the relativistic formula.
Why does kinetic energy blow up as \( v \to c \)?
The Lorentz factor \( \gamma = 1/\sqrt{1 - v^2/c^2} \) diverges as \( v \) approaches \( c \). Because \( K_\text{rel} = (\gamma - 1)mc^2 \), the kinetic energy also diverges. This reflects the fact that it would take infinite energy to accelerate a massive particle to the speed of light.
Can relativistic kinetic energy be negative?
No. For any physical speed \( 0 \le v < c \), the Lorentz factor satisfies \( \gamma \ge 1 \), so \( K_\text{rel} = (\gamma - 1)mc^2 \ge 0 \). It is zero only when the object is at rest in the chosen frame.
How does this relate to \( E = mc^2 \)?
The famous equation \( E = mc^2 \) refers to rest energy. The full relativistic energy is \( E = \gamma mc^2 \). The kinetic part is the difference between total and rest energy: \( K_\text{rel} = E - mc^2 \).