Theory of Relativity Formula Calculator

Lorentz factor, time dilation, length contraction and relativistic energy with step-by-step solutions
Find the Lorentz factor and time dilation at v = 0.80c
Find the rest energy of 1.0 g of matter in joules
Find the total energy and kinetic energy of an electron at 0.90c
A 100 m rocket passes at 0.80c. What is its contracted length?

The Lorentz Factor Runs Everything

Special relativity rests on two postulates — the laws of physics are identical in every inertial frame, and cc is the same in every frame — and almost every formula that follows is one dimensionless quantity times a classical expression. That quantity is the Lorentz factor:

γ=11v2/c2=11β2,β=vc\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - \beta^2}}, \qquad \beta = \frac{v}{c}

with vv the relative speed in m/s and c=2.998×108c = 2.998 \times 10^8 m/s. γ1\gamma \ge 1 always, and γ1\gamma \to 1 as β0\beta \to 0, which is why Newtonian mechanics survives at everyday speeds.

The three kinematic results:

Δt=γΔt0,L=L0γ,u=uv1uv/c2\Delta t = \gamma\, \Delta t_0, \qquad L = \frac{L_0}{\gamma}, \qquad u' = \frac{u - v}{1 - uv/c^2}

The assumption people forget: Δt0\Delta t_0 is the proper time, measured by a clock at rest with respect to the event, and L0L_0 is the proper length, measured in the object's own rest frame. Get those two backwards and the effect comes out inverted.

Mass, Energy and Momentum

Mass-energy equivalence is the famous line, but the version you actually compute with is the total energy:

E0=mc2,E=γmc2,K=(γ1)mc2E_0 = mc^2, \qquad E = \gamma m c^2, \qquad K = (\gamma - 1)mc^2

  • mminvariant (rest) mass, kg; the same in every frame
  • E0E_0 — rest energy, joules; EE — total energy, joules
  • KK — relativistic kinetic energy, joules

Momentum picks up the same factor, p=γmvp = \gamma m v in kg·m/s, and the three combine into the frame-independent relation

E2=(pc)2+(mc2)2E^2 = (pc)^2 + (mc^2)^2

which is the most useful equation on this page: it lets you go between energy and momentum without ever computing vv. For a photon m=0m = 0 and it collapses to E=pcE = pc.

Practical units: particle physicists work in electronvolts, with 1 eV=1.602×10191\ \text{eV} = 1.602 \times 10^{-19} J. An electron's rest energy is 0.5110.511 MeV, a proton's is 938.3938.3 MeV.

Common Mistakes to Avoid

  • Treating "relativistic mass" γm\gamma m as a real mass — it is an obsolete bookkeeping device. Modern practice keeps mm invariant and puts γ\gamma on the energy and momentum instead.
  • Using K=12mv2K = \tfrac12 mv^2 at high β\beta — at 0.99c0.99c the classical formula is off by a factor of twelve.
  • Forgetting to square β\beta — the denominator is 1v2/c2\sqrt{1 - v^2/c^2}, so v=0.8cv = 0.8c gives 10.64\sqrt{1 - 0.64}, not 10.8\sqrt{1 - 0.8}.
  • Adding velocities classically0.6c+0.6c0.6c + 0.6c is 0.882c0.882c, not 1.2c1.2c. Nothing crosses cc.
  • Applying special relativity to accelerating or gravitating frames — that needs general relativity.
  • Mixing eV and joules mid-calculation — convert once, at the start or at the end, never in the middle.

示例题目

Step 1: Convert the mass: m=1.0 g=1.0×103 kgm = 1.0\ \text{g} = 1.0 \times 10^{-3}\ \text{kg}
Step 2: E0=mc2E_0 = mc^2 with c=2.998×108 m/sc = 2.998 \times 10^{8}\ \text{m/s}
Step 3: c2=(2.998×108 m/s)2=8.988×1016 m2/s2c^2 = (2.998 \times 10^{8}\ \text{m/s})^2 = 8.988 \times 10^{16}\ \text{m}^2\text{/s}^2
Step 4: E0=(1.0×103 kg)(8.988×1016 m2/s2)=8.99×1013 JE_0 = (1.0 \times 10^{-3}\ \text{kg})(8.988 \times 10^{16}\ \text{m}^2\text{/s}^2) = 8.99 \times 10^{13}\ \text{J}
Answer: E09.0×1013E_0 \approx 9.0 \times 10^{13} J — about the energy of a 21 kilotonne blast

Step 1: β2=(0.80)2=0.64\beta^2 = (0.80)^2 = 0.64, so 1β2=0.361 - \beta^2 = 0.36
Step 2: γ=1/0.36=1/0.60=1.667\gamma = 1/\sqrt{0.36} = 1/0.60 = 1.667
Step 3: Proper time on board: Δt0=Δt/γ=(1.00 yr)÷1.667=0.600 yr\Delta t_0 = \Delta t/\gamma = (1.00\ \text{yr}) \div 1.667 = 0.600\ \text{yr}
Step 4: Contracted length in the Earth frame: L=L0/γ=(100 m)÷1.667=60.0 mL = L_0/\gamma = (100\ \text{m}) \div 1.667 = 60.0\ \text{m}
Answer: γ=1.67\gamma = 1.67; the crew ages 0.6000.600 yr; the hull measures 60.060.0 m

Step 1: 1β2=10.81=0.191 - \beta^2 = 1 - 0.81 = 0.19, so γ=1/0.19=1/0.4359=2.294\gamma = 1/\sqrt{0.19} = 1/0.4359 = 2.294
Step 2: E=γmec2=2.294×0.511 MeV=1.172 MeVE = \gamma m_e c^2 = 2.294 \times 0.511\ \text{MeV} = 1.172\ \text{MeV}
Step 3: K=Emec2=1.172 MeV0.511 MeV=0.661 MeVK = E - m_e c^2 = 1.172\ \text{MeV} - 0.511\ \text{MeV} = 0.661\ \text{MeV}
Step 4: Classical check: 12mev2=12(0.511 MeV)(0.81)=0.207 MeV\tfrac12 m_e v^2 = \tfrac12 (0.511\ \text{MeV})(0.81) = 0.207\ \text{MeV}, a 69% underestimate
Answer: E=1.17E = 1.17 MeV, K=0.661K = 0.661 MeV

常见问题

There is no single formula. Special relativity is built on the Lorentz factor γ = 1/√(1 − v²/c²), which then gives time dilation Δt = γΔt₀, length contraction L = L₀/γ, momentum p = γmv and total energy E = γmc². The E = mc² everyone quotes is the special case v = 0, the rest energy.

Total energy is E = γmc², and it splits into rest energy mc² plus kinetic energy (γ − 1)mc². The frame-independent form E² = (pc)² + (mc²)² is usually more convenient, because it links energy to momentum without needing the speed.

In modern usage, no. Mass m is an invariant property of the object. What grows without bound as v approaches c is the energy γmc² and the momentum γmv, which is why no finite energy can push a massive object to c. Older texts folded γ into a 'relativistic mass' γm; the physics is identical, the bookkeeping is not.

When γ is close enough to 1 for your tolerance. At v = 0.1c, γ = 1.005, a 0.5 percent correction. At 0.5c it is 1.155, and by 0.9c it is 2.29 and classical mechanics is simply wrong.

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