How to Calculate Speed from the Lorentz Factor

A muon created high in Earth’s atmosphere has an average lifetime of about \(2.2\,\mu\text{s}\) when measured in its own rest frame. Even travelling at the speed of light, it would cover only about 660 m during that time. Yet muons created kilometres above the ground are regularly detected near Earth’s surface. How can they survive the journey?

The Lorentz factor explains the difference. More importantly, if you already know the Lorentz factor, you can calculate the particle’s speed directly:

\[
\boxed{v=c\sqrt{1-\frac{1}{\gamma^2}}}
\]

Here, \(v\) is the particle’s speed relative to an observer, \(c\) is the speed of light in a vacuum (\(3.00\times10^8\,\text{m s}^{-1}\)), and \(\gamma\) is the Lorentz factor.

The main challenge is making sure you have calculated \(\gamma\) from the correct quantities and in the correct reference frame. Let’s work through the method and see how experimental evidence supports it.

01Finding speed from the Lorentz factor

Start with the Lorentz factor equation:

\[
\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
\]

You want to make \(v\) the subject.

Step 1: Square both sides.

\[
\gamma^2=\frac{1}{1-\frac{v^2}{c^2}}
\]

Step 2: Take the reciprocal.

\[
\frac{1}{\gamma^2}=1-\frac{v^2}{c^2}
\]

Step 3: Rearrange to isolate the speed term.

\[
\frac{v^2}{c^2}=1-\frac{1}{\gamma^2}
\]

Step 4: Multiply by \(c^2\), then take the square root.

\[
v^2=c^2\left(1-\frac{1}{\gamma^2}\right)
\]

Therefore:

\[
\boxed{v=c\sqrt{1-\frac{1}{\gamma^2}}}
\]

That’s the equation you need.

Notice that the Lorentz factor is dimensionless. It tells you how much relativistic effects differ from their classical equivalents, but it isn’t itself a speed.

What should you expect from the answer?

The relationship has some useful properties.

Three checks are particularly useful:

• If \(\gamma=1\), the object is stationary relative to the observer.
• If \(\gamma\) increases, the speed increases towards \(c\).
• A massive object cannot have \(\gamma<1\), or reach \(c\) at any finite Lorentz factor.

If your calculated speed is greater than or equal to \(c\), check your substitution and algebra.

02Get the reference frame right

Before substituting numbers, identify who is measuring the motion.

Imagine a spacecraft travelling past Earth. There are two observers:

• Observer A is on Earth and measures the spacecraft moving.
• Observer B is inside the spacecraft and measures themselves as stationary.

These observers do not measure the same elapsed time for the same pair of events.

Proper time versus laboratory time

The time measured in the rest frame of the clock is called the proper time, represented by \(\Delta\tau\).

For the spacecraft, this is the time measured by its onboard clock between two events occurring at the same position inside the spacecraft.

The Earth observer measures a longer interval, \(\Delta t\).

The relationship is:

\[
\Delta t=\gamma\Delta\tau
\]

Rearranging:

\[
\boxed{\gamma=\frac{\Delta t}{\Delta\tau}}
\]

This gives you a second way to find the Lorentz factor before calculating speed.

For example, suppose the spacecraft clock measures \(4.0\,\mu\text{s}\), while Earth clocks measure \(10.0\,\mu\text{s}\).

\[
\gamma=\frac{10.0}{4.0}=2.50
\]

You can now substitute \(\gamma=2.50\) into the speed equation.

The important point is that you must not reverse the time ratio. The laboratory measures the longer interval, so it belongs in the numerator.

For a fuller explanation of this relationship, see Time Dilation in HSC Physics: Using the Lorentz Factor.

Does direction affect the Lorentz factor?

The Lorentz factor depends on speed, not direction.

If a particle travels east at \(0.90c\), it has the same Lorentz factor as an otherwise identical particle travelling west at \(0.90c\), relative to the same laboratory frame.

This is because speed is the magnitude of velocity.

For motion along the x-axis:

• Motion in the positive x direction has \(v_x=+v\).
• Motion in the negative x direction has \(v_x=-v\).

However, the Lorentz factor uses \(v^2\), so both directions give the same result.

For motion in two or three dimensions, use the magnitude of the complete velocity vector:

\[
v=\sqrt{v_x^2+v_y^2+v_z^2}
\]

Do not calculate the Lorentz factor using only one component unless the particle is moving entirely along that axis.

A rotation of your coordinate axes changes the velocity components, but not the actual speed or Lorentz factor.

03How experiments revealed relativistic motion

The Lorentz factor isn’t just a mathematical feature of special relativity. Experiments involving cosmic-ray muons provide evidence for the time dilation it describes.

One particularly useful historical experiment was conducted by David Frisch and James Smith in 1963.

Their problem was straightforward: if muons have such short lifetimes, why do so many reach the Earth’s surface?

The experimental setup

Muons are produced when cosmic rays interact with particles in the upper atmosphere. They travel towards Earth at speeds close to the speed of light.

Frisch and Smith compared muons detected at two locations:

• Near the top of Mount Washington, New Hampshire.
• Near sea level in Cambridge, Massachusetts.

At Mount Washington, the apparatus included a thick iron absorber above a plastic scintillator.

The iron acted as a filter. Muons with insufficient energy were stopped before reaching the scintillator. Faster muons could pass through the iron and, depending on their energy, through the detector.

A scintillator produces flashes of light when charged particles pass through it. A photomultiplier converted these flashes into electrical signals.

The signals were sent to an oscilloscope, which displayed them as pulses.

When a muon stopped inside the scintillator, it eventually decayed. The resulting electron produced another flash and therefore another electrical pulse.

The time between the two pulses measured how long the stopped muon survived. Repeating this process for many muons allowed the researchers to determine their average lifetime at rest.

What did the original model predict?

The measured mean lifetime of muons at rest was approximately:

\[
\tau_0=2.2\,\mu\text{s}
\]

The distance between the two experimental locations was approximately \(1907\,\text{m}\).

The selected muons had an average speed of approximately \(0.9952c\).

First, calculate how long the journey would take according to an observer on Earth.

\[
\Delta t=\frac{d}{v}
\]

Substituting:

\[
\Delta t=
\frac{1907}{0.9952(3.00\times10^8)}
\]

\[
\Delta t\approx6.39\times10^{-6}\,\text{s}
\]

So the journey takes approximately \(6.4\,\mu\text{s}\) in the Earth frame.

If there were no time dilation, the muons would have to survive for nearly three times their average rest-frame lifetime.

Using the exponential decay model, the expected surviving fraction would be:

\[
P=e^{-\Delta t/\tau_0}
\]

\[
P=e^{-6.39/2.2}
\]

\[
P\approx0.055
\]

Only about 5.5% of the original population would survive.

For an initial sample of 568 muons, this predicts approximately 31 survivors.

What did the experiment actually observe?

The researchers recorded 412 decay events in their corresponding one-hour run at Cambridge.

That is dramatically more than the approximately 31 expected without time dilation.

The observation supports the prediction of special relativity: moving muons have a longer measured lifetime in the Earth frame.

How did the researchers improve the comparison?

There was an important experimental complication.

Muons lose energy as they pass through the atmosphere. Therefore, simply comparing two detectors at different altitudes could mean comparing different speed distributions.

Frisch and Smith addressed this by adjusting the iron absorber at the Cambridge detector. They removed material equivalent to the additional atmospheric energy loss, helping ensure that the muons being compared belonged to the same selected speed range at the mountain altitude.

They also measured the decay-time distribution of stopped muons to establish their rest-frame lifetime.

Their full analysis reported an observed time-dilation factor of \(8.8\pm0.8\), consistent with their geometry-adjusted prediction of \(8.4\pm2\).

The effective factors account for the real detector geometry and spread of particle speeds, rather than describing one perfectly identical muon.

The original experiment is described in Frisch and Smith’s 1963 research paper.

The key connection to our calculation is that once a Lorentz factor has been established, the speed can be found by reversing the time-dilation relationship.

For a single particle travelling at constant speed:

\[
\gamma=\frac{\Delta t}{\Delta\tau}
\]

Then:

\[
v=c\sqrt{1-\frac{1}{\gamma^2}}
\]

04Finding the Lorentz factor from energy

Sometimes an examination question won’t give you a Lorentz factor or a pair of time measurements.

Instead, it will provide the particle’s total energy and rest energy.

This is where you need to distinguish between invariant mass, rest energy, total energy, and kinetic energy.

Invariant mass and rest energy

The invariant mass, represented by \(m\), is the mass of an object measured in its rest frame. It has the same value for all inertial observers.

Its corresponding rest energy is:

\[
E_0=mc^2
\]

The total energy measured in a frame where the object is moving is:

\[
E=\gamma mc^2
\]

Since \(E_0=mc^2\), this becomes:

\[
\boxed{E=\gamma E_0}
\]

Rearranging:

\[
\boxed{\gamma=\frac{E}{E_0}}
\]

This means you can find the Lorentz factor by dividing the total energy by the rest energy.

What if the question gives kinetic energy instead?

This is a common source of incorrect answers.

Total energy is made up of rest energy and kinetic energy:

\[
E=E_0+K
\]

where \(K\) is the relativistic kinetic energy.

Combining this with \(E=\gamma E_0\):

\[
\gamma E_0=E_0+K
\]

Therefore:

\[
\boxed{\gamma=1+\frac{K}{E_0}}
\]

Notice the extra 1.

Consider two different situations:

Situation A: A particle has total energy equal to four times its rest energy.

\[
E=4E_0
\]

Therefore:

\[
\gamma=4
\]

Situation B: A particle has kinetic energy equal to four times its rest energy.

\[
K=4E_0
\]

Therefore:

\[
\gamma=1+4=5
\]

These are not the same situation.

If you confuse total energy with kinetic energy, you will calculate the wrong Lorentz factor and therefore the wrong speed.

Does the particle’s mass increase?

You may encounter older explanations describing an increase in relativistic mass.

For modern calculations, use invariant mass and total energy instead.

The particle’s invariant mass remains constant. Its total energy increases as its speed relative to the observer increases.

The Lorentz factor describes the relationship between the total energy and rest energy, not an actual change in invariant mass.

05Worked examples

06Common mistakes when calculating relativistic speed

Before attempting the practice questions, check these common errors.

One final limitation: this method applies to massive objects with a rest frame. Photons travel at \(c\), but they have no rest frame, so you cannot use this equation to calculate their speed from a finite Lorentz factor.

07Questions and solutions

Question 1

A particle has a Lorentz factor of \(1.25\).

Calculate its speed as:

a) A fraction of the speed of light.

b) A value in \(\text{m s}^{-1}\).

Solution 1

The particle travels at \(0.600c\), equivalent to \(1.80\times10^8\,\text{m s}^{-1}\).

Start with:

\[
v=c\sqrt{1-\frac{1}{\gamma^2}}
\]

Substitute \(\gamma=1.25\):

\[
v=c\sqrt{1-\frac{1}{1.25^2}}
\]

\[
v=c\sqrt{1-\frac{1}{1.5625}}
\]

\[
v=c\sqrt{0.36}
\]

\[
\boxed{v=0.600c}
\]

For part (b), substitute the speed of light:

\[
v=0.600(3.00\times10^8)
\]

\[
\boxed{v=1.80\times10^8\,\text{m s}^{-1}}
\]

The result is less than \(c\), as expected for a massive particle.

Question 2

An unstable particle travels in the positive x direction through a laboratory.

Its mean lifetime measured in its own rest frame is \(2.20\,\mu\text{s}\). Its mean lifetime measured by laboratory observers is \(11.0\,\mu\text{s}\).

Assume that the particle travels at constant speed.

a) Calculate its Lorentz factor.

b) Determine its speed in \(\text{m s}^{-1}\).

c) Calculate the mean distance travelled before decay, as measured in the laboratory.

Solution 2

The particle has a Lorentz factor of \(5.00\), a speed of \(2.94\times10^8\,\text{m s}^{-1}\), and a mean decay distance of approximately \(3.23\,\text{km}\).

Part a: Calculate the Lorentz factor.

The proper lifetime is measured in the particle’s rest frame:

\[
\tau_0=2.20\,\mu\text{s}
\]

The laboratory lifetime is:

\[
\Delta t=11.0\,\mu\text{s}
\]

Therefore:

\[
\gamma=\frac{\Delta t}{\tau_0}
\]

\[
\gamma=\frac{11.0}{2.20}
\]

\[
\boxed{\gamma=5.00}
\]

Part b: Calculate the speed.

Use the Lorentz factor equation:

\[
v=c\sqrt{1-\frac{1}{\gamma^2}}
\]

\[
v=3.00\times10^8
\sqrt{1-\frac{1}{25}}
\]

\[
v=3.00\times10^8\sqrt{0.96}
\]

\[
\boxed{v=2.94\times10^8\,\text{m s}^{-1}}
\]

Part c: Calculate the mean distance travelled.

In the laboratory frame, the mean decay time is \(11.0\,\mu\text{s}\).

Use:

\[
d=v\Delta t
\]

Convert microseconds into seconds:

\[
\Delta t=11.0\times10^{-6}\,\text{s}
\]

Substitute:

\[
d=(2.939\times10^8)(11.0\times10^{-6})
\]

\[
d=3233\,\text{m}
\]

Therefore:

\[
\boxed{d=3.23\,\text{km}}
\]

The particle travels an average of approximately \(3.23\,\text{km}\) before decaying, as measured in the laboratory.

This is a mean distance, not a guaranteed distance for every individual particle. Radioactive decay is statistical.

Question 3

A proton travels through a particle accelerator.

Its invariant rest energy is \(0.938\,\text{GeV}\), and its total energy in the laboratory frame is \(7.504\,\text{GeV}\).

a) Calculate the Lorentz factor.

b) Determine its speed as a fraction of \(c\).

c) Calculate its relativistic kinetic energy.

d) Explain whether the proton’s invariant mass changes as its speed increases.

Solution 3

The proton has a Lorentz factor of \(8.00\), travels at approximately \(0.992c\), and has a kinetic energy of \(6.566\,\text{GeV}\). Its invariant mass does not change.

Part a: Calculate the Lorentz factor.

Use the total-energy relationship:

\[
\gamma=\frac{E}{E_0}
\]

Substitute:

\[
\gamma=\frac{7.504}{0.938}
\]

\[
\boxed{\gamma=8.00}
\]

Part b: Determine the speed.

\[
v=c\sqrt{1-\frac{1}{\gamma^2}}
\]

\[
v=c\sqrt{1-\frac{1}{64}}
\]

\[
v=c\sqrt{\frac{63}{64}}
\]

\[
v\approx0.9922c
\]

Therefore:

\[
\boxed{v\approx0.992c}
\]

In metres per second:

\[
v=0.9922(3.00\times10^8)
\]

\[
\boxed{v\approx2.98\times10^8\,\text{m s}^{-1}}
\]

Part c: Calculate the kinetic energy.

Total energy is the sum of rest energy and kinetic energy.

\[
K=E-E_0
\]

\[
K=7.504-0.938
\]

\[
\boxed{K=6.566\,\text{GeV}}
\]

Part d: Explain the change in mass.

The proton’s invariant mass does not increase.

Its rest energy remains \(0.938\,\text{GeV}\), while its total energy increases because of its motion relative to the laboratory.

The relationship is:

\[
E=\gamma mc^2
\]

The increase is in total energy, not invariant mass. Confusing these quantities can lead to incorrect explanations of relativistic motion.

Question 4

A particle travels through a laboratory with the following velocity components:

• \(0.36c\) towards the east.
• \(0.48c\) towards the north.

a) Calculate the particle’s total speed.

b) Determine its Lorentz factor.

c) A student calculates the Lorentz factor using only the eastward component. Explain why this is incorrect.

d) Suppose the coordinate axes are rotated so that the new x-axis points directly along the particle’s path. What are its new velocity components, and does its Lorentz factor change?

Solution 4

The particle’s speed is \(0.600c\), and its Lorentz factor is \(1.25\). Rotating the coordinate axes changes the velocity components but does not change the Lorentz factor.

Part a: Calculate the total speed.

The particle has two perpendicular velocity components.

Therefore, use Pythagoras:

\[
v=\sqrt{v_x^2+v_y^2}
\]

Substitute:

\[
v=\sqrt{(0.36c)^2+(0.48c)^2}
\]

\[
v=c\sqrt{0.1296+0.2304}
\]

\[
v=c\sqrt{0.36}
\]

\[
\boxed{v=0.600c}
\]

Part b: Calculate the Lorentz factor.

Use the total speed, not either individual component.

\[
\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
\]

\[
\gamma=\frac{1}{\sqrt{1-0.600^2}}
\]

\[
\gamma=\frac{1}{\sqrt{0.64}}
\]

\[
\boxed{\gamma=1.25}
\]

Part c: Explain the student’s error.

The student has ignored the northward component.

The Lorentz factor depends on the magnitude of the complete velocity vector, not just one direction of travel.

Using only the eastward component would give:

\[
\gamma_x=
\frac{1}{\sqrt{1-0.36^2}}
\]

\[
\gamma_x\approx1.07
\]

This is incorrect because the particle is also moving north.

Part d: Consider the rotated axes.

When the new x-axis is aligned with the particle’s direction of motion, its velocity components become:

\[
v_x’=0.600c
\]

\[
v_y’=0
\]

The total speed is still:

\[
v=0.600c
\]

Therefore:

\[
\boxed{\gamma’=1.25}
\]

A coordinate rotation changes how we describe the velocity, not the physical motion itself.

This distinction is important: rotating coordinate axes is not the same as changing to an observer moving relative to the particle.

Question 5

A particle initially at rest decays into two identical daughter particles.

The parent particle has a rest energy of \(5E_0\), where \(E_0\) is the rest energy of either daughter particle.

Assume:

• The parent is initially stationary in the laboratory frame.
• Only two identical daughter particles are produced.
• There are no additional particles, photons, or energy losses.
• Both daughter particles move in opposite directions along the same straight line.

Determine:

a) The total energy of each daughter particle.

b) The Lorentz factor of each daughter particle.

c) The speed of each daughter particle.

d) Explain why the total momentum of the two daughter particles is zero even though neither particle is stationary.

A student argues that each daughter must have a Lorentz factor of 5 because the parent had five times the rest energy of either daughter. Evaluate this argument.

Solution 5

Each daughter particle has a Lorentz factor of \(2.50\) and travels at approximately \(0.917c\). The student’s argument is incorrect because the parent’s total energy is shared between both daughter particles.

This question requires you to combine conservation of energy, conservation of momentum, and the relationship between total energy and rest energy.

Part a: Calculate the total energy of each daughter.

Initially, the parent is stationary.

Its total energy is therefore equal to its rest energy:

\[
E_{\text{parent}}=5E_0
\]

Energy is conserved.

Since the two daughter particles are identical and travel in opposite directions, they have equal energies.

Therefore:

\[
E_1+E_2=5E_0
\]

Since:

\[
E_1=E_2
\]

Each daughter receives:

\[
E_1=E_2=\frac{5E_0}{2}
\]

Therefore:

\[
\boxed{E_{\text{daughter}}=2.50E_0}
\]

Part b: Calculate the Lorentz factor.

The total energy of each daughter is \(2.50E_0\).

Its rest energy is \(E_0\).

Using:

\[
\gamma=\frac{E}{E_0}
\]

We obtain:

\[
\gamma=\frac{2.50E_0}{E_0}
\]

\[
\boxed{\gamma=2.50}
\]

Part c: Calculate the speed.

Use the Lorentz factor equation:

\[
v=c\sqrt{1-\frac{1}{\gamma^2}}
\]

Substitute:

\[
v=c\sqrt{1-\frac{1}{2.50^2}}
\]

\[
v=c\sqrt{0.84}
\]

\[
\boxed{v=0.917c}
\]

Each daughter moves at approximately \(91.7\%\) of the speed of light, in opposite directions.

Part d: Explain the momentum.

The total momentum is zero because the daughter particles have equal and opposite momenta.

One travels in the positive x direction, while the other travels in the negative x direction.

Their momentum vectors cancel:

\[
\vec p_1+\vec p_2=0
\]

However, zero total momentum does not mean that each particle has zero momentum.

Both particles are moving rapidly. Their individual momenta are simply equal in magnitude and opposite in direction.

The student’s original argument also confuses the parent’s total energy with the energy of each daughter.

The parent has \(5E_0\) of energy, but that energy is divided between two particles. Assigning \(\gamma=5\) to each daughter would give a combined energy of \(10E_0\), violating conservation of energy.

This is the key idea: a Lorentz factor belongs to a particle’s motion relative to a particular reference frame. It is not automatically inherited from the object that produced it.

07Where to go next

Once you can calculate speed from the Lorentz factor, you can use the same information to investigate relativistic momentum.

The relativistic momentum equation is:

\[
p=\gamma mv
\]

Here, \(m\) is the invariant mass, \(v\) is the speed relative to the observer, and \(\gamma\) is the Lorentz factor.

Unlike the classical momentum equation \(p=mv\), this expression accounts for the increasingly significant relativistic effects as a particle approaches the speed of light.

Continue with How to Calculate Relativistic Momentum from Mass and Speed to connect speed, invariant mass, and momentum in particle-motion problems.