Relativistic Momentum
Momentum at speeds near light.
Formula
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How it works
This formula gives the momentum of an object when its speed is high enough that Newton’s simple p = mv no longer stays accurate. The extra factor 1/\sqrt{1 - v^2/c^2} comes from special relativity: as v approaches the speed of light, the momentum increases much more rapidly than linearly, reflecting the fact that no massive object can be accelerated to c. It reduces to ordinary momentum at low speeds because the square-root term is then very close to 1.
When to use it
Use this when a particle or object is moving at a significant fraction of the speed of light, such as in particle accelerators, cosmic-ray physics, or relativistic astrophysics. It is also useful whenever you need the correct relationship between velocity and momentum in special relativity rather than classical mechanics.
Worked example, step by step
A proton has mass m = 1.67 \times 10^{-27} kg and moves at v = 0.80c. Step 1: Write the formula: p = mv / \sqrt{1 - v^2/c^2} Step 2: Compute the speed factor: v^2/c^2 = (0.80)^2 = 0.64 Step 3: Compute the denominator: \sqrt{1 - 0.64} = \sqrt{0.36} = 0.60 Step 4: Compute mv: (1.67 \times 10^{-27})(0.80 \times 3.00 \times 10^8) = 4.008 \times 10^{-19} kg\cdot m/s Step 5: Divide by the denominator: p = (4.008 \times 10^{-19}) / 0.60 = 6.68 \times 10^{-19} kg\cdot m/s Final answer: p \approx 6.7 \times 10^{-19} kg\cdot m/s
Common mistakes to avoid
- Using p = mv even when the speed is close to c
- Forgetting that c must be in the same units as v so that v^2/c^2 is dimensionless
- Putting the square-root factor in the numerator instead of the denominator
Did You Know?
Momentum grows without limit as speed nears light's, which is why no massive object can ever reach it.
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