Relativistic Speed Calculator

Recover v from the Lorentz factor, energy or voltage, and add velocities relativistically
At what speed does a clock run at half its proper rate?
Find the speed of an electron accelerated through 2.00 MV
Muons at 0.995c have a proper lifetime of 2.20 microseconds. How far do they travel?
Add 0.60c and 0.60c relativistically

Going Backwards From γ to v

Most relativity problems hand you vv and ask for γ\gamma. The harder and more common direction in practice is the reverse — you measure a dilation, an energy or a decay length, and you want the speed. Invert the Lorentz factor:

γ=11β2β=vc=11γ2\gamma = \frac{1}{\sqrt{1-\beta^2}} \quad\Longrightarrow\quad \beta = \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}}

with c=2.998×108c = 2.998 \times 10^8 m/s. The three routes into γ\gamma:

  • From a time dilation or lifetime: γ=Δt/Δt0\gamma = \Delta t / \Delta t_0
  • From a total energy: γ=E/(mc2)\gamma = E/(mc^2)
  • From a kinetic energy or an accelerating voltage: γ=1+K/(mc2)\gamma = 1 + K/(mc^2), with K=qVK = qV in eV for a charge accelerated through VV volts

The assumption people forget: β\beta approaches 1 but never reaches it. Doubling the accelerator voltage roughly doubles γ\gamma but barely moves vv — from 0.979c0.979c to 0.995c0.995c — which is why beam energy, not beam speed, is the number accelerator physicists quote.

Adding Relativistic Speeds

Velocities do not simply add. If SS' moves at vv relative to SS, and an object moves at uu' in SS', then in SS:

u=u+v1+uv/c2u = \frac{u' + v}{1 + u'v/c^2}

all speeds in m/s along the same axis. The denominator is the whole story: it is close to 1 at everyday speeds, so the classical sum survives, and it grows just fast enough to keep u<cu < c whenever uu' and vv are.

Check the limits. Put u=cu' = c and the expression returns exactly cc for any vv — the second postulate falls out of the algebra. Put u=v=0.6cu' = v = 0.6c and you get 0.882c0.882c, not 1.2c1.2c.

A practical consequence is the decay length of an unstable particle. Its lifetime in the lab is γτ0\gamma\tau_0, so it travels

d=βcγτ0d = \beta c \gamma \tau_0

metres before decaying — the reason atmospheric muons reach the ground at all.

When it applies: inertial frames and collinear motion. Perpendicular components need the full Lorentz transformation.

Common Mistakes to Avoid

  • Solving γ\gamma for β\beta without squaring — it is β=11/γ2\beta = \sqrt{1 - 1/\gamma^2}, not 11/γ\sqrt{1 - 1/\gamma}.
  • Using K=12mv2K = \tfrac12 mv^2 to get a speed — at MeV energies this returns values above cc, which is the giveaway that the classical formula has broken down.
  • Forgetting that γ=1+K/mc2\gamma = 1 + K/mc^2 needs the rest energy in the same units0.5110.511 MeV for an electron, 938.3938.3 MeV for a proton.
  • Applying time dilation with the wrong clockΔt0\Delta t_0 is the proper time, measured where the two events happen at the same place.
  • Adding velocities classically and getting a result above cc.
  • Quoting a speed to four figures near cc0.9999c0.9999c and 0.99999c0.99999c differ enormously in energy but almost not at all in speed.

示例题目

Step 1: Half rate means Δt=2Δt0\Delta t = 2\Delta t_0, so γ=2\gamma = 2
Step 2: β=11/γ2=11/4=0.750\beta = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/4} = \sqrt{0.750}
Step 3: β=0.8660\beta = 0.8660
Step 4: v=0.8660×(2.998×108 m/s)=2.596×108 m/sv = 0.8660 \times (2.998 \times 10^{8}\ \text{m/s}) = 2.596 \times 10^{8}\ \text{m/s}
Answer: v0.866c=2.60×108v \approx 0.866c = 2.60 \times 10^{8} m/s

Step 1: Kinetic energy gained: K=eV=2.00 MeVK = eV = 2.00\ \text{MeV}
Step 2: γ=1+K/(mec2)=1+(2.00 MeV)÷(0.511 MeV)=1+3.914=4.914\gamma = 1 + K/(m_ec^2) = 1 + (2.00\ \text{MeV}) \div (0.511\ \text{MeV}) = 1 + 3.914 = 4.914
Step 3: γ2=24.15\gamma^2 = 24.15, so 1/γ2=0.041411/\gamma^2 = 0.04141
Step 4: β=10.04141=0.95859=0.9791\beta = \sqrt{1 - 0.04141} = \sqrt{0.95859} = 0.9791
Step 5: v=0.9791×(2.998×108 m/s)=2.935×108 m/sv = 0.9791 \times (2.998 \times 10^{8}\ \text{m/s}) = 2.935 \times 10^{8}\ \text{m/s}
Answer: v0.979c=2.94×108v \approx 0.979c = 2.94 \times 10^{8} m/s

Step 1: β2=(0.995)2=0.990025\beta^2 = (0.995)^2 = 0.990025, so 1β2=0.0099751 - \beta^2 = 0.009975
Step 2: γ=1/0.009975=1/0.09987=10.01\gamma = 1/\sqrt{0.009975} = 1/0.09987 = 10.01
Step 3: Lab-frame lifetime: Δt=γτ0=10.01×2.20 μs=22.03 μs\Delta t = \gamma\tau_0 = 10.01 \times 2.20\ \mu\text{s} = 22.03\ \mu\text{s}
Step 4: v=0.995×2.998×108=2.983×108 m/sv = 0.995 \times 2.998 \times 10^{8} = 2.983 \times 10^{8}\ \text{m/s}
Step 5: d=vΔt=(2.983×108 m/s)(2.203×105 s)=6.57×103 md = v\,\Delta t = (2.983 \times 10^{8}\ \text{m/s})(2.203 \times 10^{-5}\ \text{s}) = 6.57 \times 10^{3}\ \text{m}
Answer: d6.57d \approx 6.57 km — against only 656656 m if time dilation is ignored

常见问题

Invert it: β = v/c = √(1 − 1/γ²), then multiply by c = 2.998 × 10⁸ m/s. A γ of 2 corresponds to 0.866c, a γ of 10 to 0.995c, and a γ of 100 to 0.99995c.

Conventionally anything above about 0.1c, where γ = 1.005 and classical formulas are already half a percent out. By 0.5c the error in kinetic energy is around 19 percent, and above 0.9c the classical results are meaningless.

No object with mass can. Its momentum γmv and energy γmc² both diverge as v approaches c, so reaching c would take infinite energy. The relativistic velocity addition formula enforces the same limit: 0.6c added to 0.6c gives 0.882c, never more than c.

Time dilation. At 0.995c a muon's 2.20 μs proper lifetime becomes 22.0 μs in the ground frame, so it covers about 6.6 km instead of 656 m. In the muon's own frame nothing is dilated — instead the atmosphere is length-contracted by the same factor of 10.

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