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 ╬│ to v

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

╬│=11тИТ╬▓2тЯ╣╬▓=vc=1тИТ1╬│2\gamma = \frac{1}{\sqrt{1-\beta^2}} \quad\Longrightarrow\quad \beta = \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}}

with c=2.998├Ч108c = 2.998 \times 10^8 m/s. The three routes into ╬│\gamma:

  • From a time dilation or lifetime: ╬│=╬Фt/╬Фt0\gamma = \Delta t / \Delta t_0
  • From a total energy: ╬│=E/(mc2)\gamma = E/(mc^2)
  • From a kinetic energy or an accelerating voltage: ╬│=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: ╬▓\beta approaches 1 but never reaches it. Doubling the accelerator voltage roughly doubles ╬│\gamma but barely moves vv тАФ from 0.979c0.979c to 0.995c0.995c тАФ which is why beam energy, not beam speed, is the number accelerator physicists quote.

Adding Relativistic Speeds

Velocities do not simply add. If SтА▓S' moves at vv relative to SS, and an object moves at uтА▓u' in SтА▓S', then in SS:

u=uтА▓+v1+uтА▓v/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 uтА▓u' and vv are.

Check the limits. Put uтА▓=cu' = c and the expression returns exactly cc for any vv тАФ the second postulate falls out of the algebra. Put uтА▓=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 ╬│╧Д0\gamma\tau_0, so it travels

d=╬▓c╬│╧Д0d = \beta c \gamma \tau_0

metres before decaying тАФ 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 ╬│\gamma for ╬▓\beta without squaring тАФ it is ╬▓=1тИТ1/╬│2\beta = \sqrt{1 - 1/\gamma^2}, not 1тИТ1/╬│\sqrt{1 - 1/\gamma}.
  • Using K=12mv2K = \tfrac12 mv^2 to get a speed тАФ at MeV energies this returns values above cc, which is the giveaway that the classical formula has broken down.
  • Forgetting that ╬│=1+K/mc2\gamma = 1 + K/mc^2 needs the rest energy in the same units тАФ 0.5110.511 MeV for an electron, 938.3938.3 MeV for a proton.
  • Applying time dilation with the wrong clock тАФ ╬Фt0\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 cc тАФ 0.9999c0.9999c and 0.99999c0.99999c differ enormously in energy but almost not at all in speed.

Examples

Step 1: Half rate means ╬Фt=2╬Фt0\Delta t = 2\Delta t_0, so ╬│=2\gamma = 2
Step 2: ╬▓=1тИТ1/╬│2=1тИТ1/4=0.750\beta = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/4} = \sqrt{0.750}
Step 3: ╬▓=0.8660\beta = 0.8660
Step 4: v=0.8660├Ч(2.998├Ч108┬аm/s)=2.596├Ч108┬аm/sv = 0.8660 \times (2.998 \times 10^{8}\ \text{m/s}) = 2.596 \times 10^{8}\ \text{m/s}
Answer: vтЙИ0.866c=2.60├Ч108v \approx 0.866c = 2.60 \times 10^{8} m/s

Step 1: Kinetic energy gained: K=eV=2.00┬аMeVK = eV = 2.00\ \text{MeV}
Step 2: ╬│=1+K/(mec2)=1+(2.00┬аMeV)├╖(0.511┬аMeV)=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: ╬│2=24.15\gamma^2 = 24.15, so 1/╬│2=0.041411/\gamma^2 = 0.04141
Step 4: ╬▓=1тИТ0.04141=0.95859=0.9791\beta = \sqrt{1 - 0.04141} = \sqrt{0.95859} = 0.9791
Step 5: v=0.9791├Ч(2.998├Ч108┬аm/s)=2.935├Ч108┬аm/sv = 0.9791 \times (2.998 \times 10^{8}\ \text{m/s}) = 2.935 \times 10^{8}\ \text{m/s}
Answer: vтЙИ0.979c=2.94├Ч108v \approx 0.979c = 2.94 \times 10^{8} m/s

Step 1: ╬▓2=(0.995)2=0.990025\beta^2 = (0.995)^2 = 0.990025, so 1тИТ╬▓2=0.0099751 - \beta^2 = 0.009975
Step 2: ╬│=1/0.009975=1/0.09987=10.01\gamma = 1/\sqrt{0.009975} = 1/0.09987 = 10.01
Step 3: Lab-frame lifetime: ╬Фt=╬│╧Д0=10.01├Ч2.20┬а╬╝s=22.03┬а╬╝s\Delta t = \gamma\tau_0 = 10.01 \times 2.20\ \mu\text{s} = 22.03\ \mu\text{s}
Step 4: v=0.995├Ч2.998├Ч108=2.983├Ч108┬аm/sv = 0.995 \times 2.998 \times 10^{8} = 2.983 \times 10^{8}\ \text{m/s}
Step 5: d=vтАЙ╬Фt=(2.983├Ч108┬аm/s)(2.203├Ч10тИТ5┬аs)=6.57├Ч103┬аmd = 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: dтЙИ6.57d \approx 6.57 km тАФ against only 656656 m if time dilation is ignored

Frequently Asked Questions

Invert it: ╬▓ = v/c = тИЪ(1 тИТ 1/╬│┬▓), then multiply by c = 2.998 ├Ч 10тБ╕ m/s. A ╬│ of 2 corresponds to 0.866c, a ╬│ of 10 to 0.995c, and a ╬│ of 100 to 0.99995c.

Conventionally anything above about 0.1c, where ╬│ = 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 ╬│mv and energy ╬│mc┬▓ 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 ╬╝s proper lifetime becomes 22.0 ╬╝s in the ground frame, so it covers about 6.6 km instead of 656 m. In the muon's own frame nothing is dilated тАФ instead the atmosphere is length-contracted by the same factor of 10.

Related Solvers

Try AI-Math for Free

Get step-by-step solutions to any math problem. Upload a photo or type your question.

Start Solving