How to Compare Classical and Relativistic Momentum Numerically

An electron travelling at \(0.80c\) gives two different momentum values, depending on which model you use. The classical calculation gives approximately \(2.19\times10^{-22}\ \text{kg m s}^{-1}\), while the relativistic calculation gives \(3.64\times10^{-22}\ \text{kg m s}^{-1}\). That’s a difference of 66.7%.

So, how do you compare classical and relativistic momentum numerically?

Use the same rest mass and the same velocity in both calculations. First calculate classical momentum using \(p=mv\). Then calculate the Lorentz factor, \(\gamma\), and use it to calculate relativistic momentum. Finally, compare the two values using a ratio or percentage difference.

There is one important detail: the mass in both calculations is the particle’s rest mass, measured in its own rest frame. It does not increase when the particle moves faster. Its momentum and total energy, however, do depend on the observer’s frame.

01Why physicists needed a different momentum model

Imagine sending an electron through a magnetic field. Its path curves because the magnetic force changes the direction of its momentum.

At ordinary speeds, the classical model predicts how much the electron should bend. But what happens when the electron approaches the speed of light?

The relativistic model predicts a larger momentum than the classical model at the same speed. Consequently, the electron is harder to deflect by the same magnetic field.

This gave physicists a way to investigate whether the classical model was still accurate for very fast electrons.

The early experimental problem

In the early twentieth century, physicists were investigating whether the behaviour of fast electrons agreed with classical mechanics or the emerging theory of special relativity.

Walter Kaufmann investigated the behaviour of fast electrons using an apparatus that combined electric and magnetic fields.

The arrangement can be pictured from bottom to top:

• A radioactive source at the bottom produced fast electrons through beta decay.
• The electrons travelled upwards through a region containing electric and magnetic fields.
• A capacitor formed part of the electric-field arrangement.
• The fields deflected the electrons from their original paths.
• A photographic plate at the top recorded where the electrons arrived.

The recorded positions provided evidence about how the electrons responded to the applied fields. Their speeds and momenta could then be inferred using the experimental model.

What should the experiment have shown?

If the classical model was sufficient, the electron’s momentum would be \(p=m_0v\). If the relativistic model was correct, its momentum would be \(p=\gamma m_0v\).

At the same speed, the relativistic model predicts greater momentum and therefore a larger radius of curvature in a uniform magnetic field.

Kaufmann’s reported measurements did not agree with Einstein’s relativistic prediction. This contributed to an important debate about how fast electrons behaved and how reliably their properties could be measured.

Bucherer’s revised experiment

Alfred Bucherer subsequently developed a modified experimental arrangement.

His apparatus placed the radioactive source between two circular plates, with photographic film arranged around the wall of a cylindrical instrument. This differed from Kaufmann’s arrangement, in which the source and recording plate occupied opposite ends of the apparatus.

Bucherer’s measurements were presented as experimental support for the Lorentz-Einstein description of fast electrons.

However, the early experiments did not settle every question. Later analyses identified difficulties involving the resolving power of direct electric and magnetic deflection methods, as well as possible electron scattering.

This is an important distinction when discussing experimental evidence: a measurement is not the same thing as the theoretical model used to interpret it.

Historical descriptions of these instruments are available in the Smithsonian Institution Scholarly Press chapter on early electron experiments and the Max Planck Institute for the History of Science’s account of Bucherer’s apparatus. A later 1938 analysis published in Physical Review also illustrates the experimental difficulties involved.

Scale estimate: how different would the electron’s path be?

We can estimate the size of the predicted difference using a simplified magnetic-field arrangement.

Suppose an electron travels at \(0.80c\) through a uniform magnetic field of strength \(0.10\ \text{T}\), perpendicular to its velocity.

The radius of its circular path is related to momentum by:

\[
r=\frac{p}{|q|B}
\]

where:

• \(r\) is the radius of curvature, in metres;
• \(p\) is the magnitude of momentum, in \(\text{kg m s}^{-1}\);
• \(q\) is the electron’s charge, in coulombs;
• \(B\) is magnetic field strength, in tesla.

Use the electron’s charge magnitude \(1.60\times10^{-19}\ \text{C}\).

From our earlier comparison:

\[
p_{\text{classical}}=2.19\times10^{-22}
\]

\[
p_{\text{relativistic}}=3.64\times10^{-22}
\]

The classical prediction is:

\[
r_{\text{classical}}
=
\frac{2.19\times10^{-22}}
{(1.60\times10^{-19})(0.10)}
\]

\[
r_{\text{classical}}=0.0137\ \text{m}
\]

The relativistic prediction is:

\[
r_{\text{relativistic}}
=
\frac{3.64\times10^{-22}}
{(1.60\times10^{-19})(0.10)}
\]

\[
r_{\text{relativistic}}=0.0228\ \text{m}
\]

The predicted radii are therefore approximately 1.37 cm and 2.28 cm.

The relativistic model predicts a larger radius, meaning less curvature. This is exactly what we expect from the greater momentum.

These are illustrative calculations, not measurements from Kaufmann’s or Bucherer’s historical experiments. We have also simplified the apparatus to a magnetic-field-only region.

02The numerical method: comparing both momentum models

You only need a few equations to make a reliable comparison.

1. Identify the rest mass

The rest mass, \(m_0\), is the mass of the particle measured in its own rest frame.

A rest frame is a reference frame in which the particle is stationary.

For example, an electron travelling at \(0.80c\) relative to a laboratory still has the same rest mass as an electron sitting stationary in a laboratory.

The electron’s rest mass is approximately:

\[
m_0=9.11\times10^{-31}\ \text{kg}
\]

Do not replace this with a larger, velocity-dependent mass.

2. Calculate classical momentum

Classical momentum is:

\[
p_{\text{classical}}=m_0v
\]

where:

• \(m_0\) is rest mass, in kilograms;
• \(v\) is velocity, in metres per second.

This model is a good approximation when the speed is much smaller than the speed of light.

3. Calculate the Lorentz factor

The Lorentz factor, represented by \(\gamma\), accounts for relativistic effects.

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

Here, \(c\) is the speed of light in a vacuum:

\[
c=3.00\times10^8\ \text{m s}^{-1}
\]

It is often useful to calculate the speed as a fraction of the speed of light first.

For example, if \(v=0.80c\), then:

\[
\frac{v}{c}=0.80
\]

Therefore:

\[
\gamma=
\frac{1}{\sqrt{1-0.80^2}}
=1.667
\]

Notice that \(\gamma\) is always greater than or equal to 1.

At low speeds, it is very close to 1. As the speed approaches \(c\), it increases rapidly.

4. Calculate relativistic momentum

Once you know \(\gamma\), use:

\[
p_{\text{relativistic}}=\gamma m_0v
\]

The only difference from the classical equation is the additional Lorentz factor.

This is why relativistic momentum is always greater than classical momentum for a particle moving at a non-zero speed.

5. Compare the results

The ratio between the two momentum magnitudes is:

\[
\frac{p_{\text{relativistic}}}
{p_{\text{classical}}}
=\gamma
\]

This is particularly useful because you can compare the models without needing to calculate the particle’s mass.

If a question asks for the percentage by which relativistic momentum exceeds classical momentum, use:

\[
\%\text{ difference}
=
\frac{p_{\text{relativistic}}-p_{\text{classical}}}
{p_{\text{classical}}}
\times100
\]

Substituting the Lorentz factor gives:

\[
\%\text{ difference}=(\gamma-1)\times100
\]

Be careful with the denominator. If a question instead asks for the percentage error in the classical result relative to the relativistic result, divide by relativistic momentum, not classical momentum.

How quickly does the difference grow?

The following table uses illustrative speeds to show how much relativistic momentum exceeds classical momentum.

At \(0.10c\), the two models give almost identical answers.

At \(0.90c\), relativistic momentum is more than twice the classical prediction.

At \(0.99c\), relativistic momentum is approximately 7.09 times the classical value.

This is why you should not assume that classical momentum remains accurate simply because you know the particle’s mass and velocity.

03Rest mass, total energy, and momentum are different quantities

There is another common source of confusion: treating mass and energy as interchangeable quantities in every calculation.

Special relativity distinguishes between invariant rest mass and the energy measured by an observer.

Consider a particle travelling through a laboratory at relativistic speed.

Its rest mass remains constant, but its total energy increases.

The relevant quantities are:

The distinction is important.

The rest energy is not the same as the total energy of a moving particle. Total energy includes both rest energy and kinetic energy.

A common mistake is to say that a particle’s rest mass increases as its speed increases. Older physics texts sometimes describe this using the term relativistic mass.

For modern calculations, keep the invariant rest mass \(m_0\) constant and account for the change through \(\gamma\).

Momentum also has a direction

Momentum is a vector quantity. It has both magnitude and direction.

Suppose a particle travels to the right along the positive \(x\)-axis. Its momentum is positive.

If it travels to the left, its momentum is negative.

The Lorentz factor does not change this direction because it is always positive.

For example:

\[
v=+0.80c
\quad\Rightarrow\quad
p=+\gamma m_0(0.80c)
\]

But:

\[
v=-0.80c
\quad\Rightarrow\quad
p=-\gamma m_0(0.80c)
\]

The magnitude is identical, but the directions are opposite.

Energy behaves differently. Total energy is a scalar quantity, so it does not become negative simply because a particle travels to the left.

04Worked examples

05Questions and solutions

Question 1

An electron has a rest mass of \(9.11\times10^{-31}\ \text{kg}\) and travels at \(0.60c\) in the positive \(x\)-direction.

Calculate:

a) Its classical momentum.

b) Its relativistic momentum.

c) The percentage by which relativistic momentum exceeds classical momentum.

Use \(c=3.00\times10^8\ \text{m s}^{-1}\).

Solution 1

The classical momentum is \(+1.64\times10^{-22}\ \text{kg m s}^{-1}\), while the relativistic momentum is \(+2.05\times10^{-22}\ \text{kg m s}^{-1}\). The relativistic result is 25.0% greater.

Part a: Classical momentum

First calculate velocity:

\[
v=0.60(3.00\times10^8)
\]

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

Use:

\[
p_{\text{classical}}=m_0v
\]

\[
=(9.11\times10^{-31})(1.80\times10^8)
\]

\[
p_{\text{classical}}
=+1.64\times10^{-22}\ \text{kg m s}^{-1}
\]

The positive sign indicates motion in the positive \(x\)-direction.

Part b: Relativistic momentum

Calculate the Lorentz factor:

\[
\gamma=
\frac{1}{\sqrt{1-0.60^2}}
\]

\[
\gamma=\frac{1}{0.80}=1.25
\]

Then:

\[
p_{\text{relativistic}}=\gamma m_0v
\]

\[
=(1.25)(9.11\times10^{-31})
(1.80\times10^8)
\]

\[
p_{\text{relativistic}}
=+2.05\times10^{-22}\ \text{kg m s}^{-1}
\]

Part c: Percentage difference

\[
\%\text{ difference}
=
\frac{2.05-1.64}{1.64}\times100
\]

\[
=25.0\%
\]

The relativistic model predicts greater momentum because the Lorentz factor is greater than one.

Question 2

An ion has a rest mass of \(3.20\times10^{-26}\ \text{kg}\) and travels at \(0.75c\) in the negative \(x\)-direction.

a) Calculate its classical momentum.

b) Calculate its relativistic momentum.

c) Calculate the percentage error if the classical result is used instead of the relativistic result. Take the relativistic result as the reference value.

Solution 2

The classical momentum is \(-7.20\times10^{-18}\ \text{kg m s}^{-1}\), while the relativistic momentum is approximately \(-1.09\times10^{-17}\ \text{kg m s}^{-1}\). Using the classical result produces an error of approximately 33.9% relative to the relativistic value.

Part a: Classical momentum

The velocity is:

\[
v=-0.75(3.00\times10^8)
\]

\[
v=-2.25\times10^8\ \text{m s}^{-1}
\]

Therefore:

\[
p_{\text{classical}}=m_0v
\]

\[
=(3.20\times10^{-26})
(-2.25\times10^8)
\]

\[
p_{\text{classical}}
=-7.20\times10^{-18}\ \text{kg m s}^{-1}
\]

Part b: Relativistic momentum

Calculate:

\[
\gamma=
\frac{1}{\sqrt{1-0.75^2}}
\]

\[
\gamma=
\frac{1}{\sqrt{0.4375}}
\]

\[
\gamma=1.512
\]

Now calculate momentum:

\[
p_{\text{relativistic}}=\gamma m_0v
\]

\[
=(1.512)(3.20\times10^{-26})
(-2.25\times10^8)
\]

\[
p_{\text{relativistic}}
=-1.09\times10^{-17}\ \text{kg m s}^{-1}
\]

Part c: Percentage error

The question specifies that the relativistic value is the reference.

Therefore:

\[
\%\text{ error}
=
\frac{|p_{\text{relativistic}}|
-|p_{\text{classical}}|}
{|p_{\text{relativistic}}|}
\times100
\]

Using unrounded values:

\[
=\frac{1.0885\times10^{-17}
-7.20\times10^{-18}}
{1.0885\times10^{-17}}\times100
\]

\[
=33.9\%
\]

The classical result underestimates the magnitude of momentum by 33.9%.

Be careful: this is not the same as saying relativistic momentum is 33.9% greater than classical momentum. The reference value changes the denominator.

Question 3

Two identical electrons, each with a rest mass of \(9.11\times10^{-31}\ \text{kg}\), travel along the same horizontal axis.

Electron A travels at \(0.80c\) in the positive \(x\)-direction.

Electron B travels at \(0.80c\) in the negative \(x\)-direction.

Calculate:

a) The relativistic momentum of each electron.

b) The total momentum of the two-electron system.

c) The total energy of the system.

d) Explain why the total momentum can be zero while the total energy is not zero.

Solution 3

The total momentum is zero, but the total energy is approximately \(2.73\times10^{-13}\ \text{J}\). The momenta cancel because they point in opposite directions, whereas the energies add because energy is a scalar.

Part a: Individual relativistic momenta

Both electrons have the same speed, so:

\[
\gamma=
\frac{1}{\sqrt{1-0.80^2}}
=1.667
\]

The magnitude of momentum for either electron is:

\[
|p|=\gamma m_0v
\]

\[
=(1.667)(9.11\times10^{-31})
(2.40\times10^8)
\]

\[
|p|=3.64\times10^{-22}\ \text{kg m s}^{-1}
\]

Their signed momenta are:

\[
p_A=+3.64\times10^{-22}
\]

\[
p_B=-3.64\times10^{-22}
\]

Part b: Total momentum

Momentum is a vector, so we add the signed values:

\[
p_{\text{total}}=p_A+p_B
\]

\[
=3.64\times10^{-22}
-3.64\times10^{-22}
\]

\[
\boxed{p_{\text{total}}=0}
\]

The two momenta cancel exactly.

Part c: Total energy

First calculate the rest energy of one electron:

\[
E_0=m_0c^2
\]

\[
=(9.11\times10^{-31})(3.00\times10^8)^2
\]

\[
E_0=8.20\times10^{-14}\ \text{J}
\]

The total energy of one moving electron is:

\[
E=\gamma m_0c^2
\]

\[
=(1.667)(8.20\times10^{-14})
\]

\[
E=1.37\times10^{-13}\ \text{J}
\]

For two electrons:

\[
E_{\text{total}}=2E
\]

\[
=2(1.37\times10^{-13})
\]

\[
\boxed{E_{\text{total}}=2.73\times10^{-13}\ \text{J}}
\]

Part d: Explain the difference

Momentum and energy are different quantities.

The electrons have equal momentum magnitudes but opposite directions. Their vector sum is zero.

However, both electrons have positive total energy. Energy is added, not cancelled by reversing direction.

The system therefore has zero total momentum but a substantial amount of energy.

This distinction becomes particularly important when investigating collisions and the mass of a combined system.

Question 4

Two particles, A and B, travel in opposite directions along the same horizontal axis in a laboratory frame.

Their properties are:

A student calculates their classical momenta and notices that they have equal magnitudes and opposite directions.

The student concludes that the total relativistic momentum must also be zero.

Determine whether this conclusion is correct.

Calculate:

a) The classical momentum of each particle.

b) The relativistic momentum of each particle.

c) The total relativistic momentum.

Explain why the classical and relativistic predictions differ.

Solution 4

The student’s conclusion is incorrect. The classical momenta cancel, but the total relativistic momentum is approximately \(+5.72\times10^{-20}\ \text{kg m s}^{-1}\).

The reason is that the two particles have different speeds and therefore different Lorentz factors.

Part a: Classical momenta

For particle A:

\[
p_{A,\text{classical}}=m_Av_A
\]

\[
=(2.00\times10^{-27})(0.60)
(3.00\times10^8)
\]

\[
p_{A,\text{classical}}
=+3.60\times10^{-19}
\]

For particle B:

\[
p_{B,\text{classical}}=m_Bv_B
\]

\[
=(3.00\times10^{-27})(-0.40)
(3.00\times10^8)
\]

\[
p_{B,\text{classical}}
=-3.60\times10^{-19}
\]

Therefore:

\[
p_{\text{total,classical}}=0
\]

The student’s cancellation is correct for the classical model.

Part b: Relativistic momenta

First calculate the Lorentz factor for particle A:

\[
\gamma_A=
\frac{1}{\sqrt{1-0.60^2}}
\]

\[
\gamma_A=1.25
\]

Its relativistic momentum is:

\[
p_A=\gamma_A m_Av_A
\]

\[
=(1.25)(2.00\times10^{-27})
(1.80\times10^8)
\]

\[
p_A=+4.50\times10^{-19}
\ \text{kg m s}^{-1}
\]

Now calculate the Lorentz factor for particle B:

\[
\gamma_B=
\frac{1}{\sqrt{1-0.40^2}}
\]

\[
\gamma_B=\frac{1}{\sqrt{0.84}}
\]

\[
\gamma_B=1.091
\]

Its relativistic momentum is:

\[
p_B=\gamma_Bm_Bv_B
\]

\[
=(1.091)(3.00\times10^{-27})
(-1.20\times10^8)
\]

\[
p_B=-3.93\times10^{-19}
\ \text{kg m s}^{-1}
\]

Part c: Total relativistic momentum

Add the signed values:

\[
p_{\text{total}}=p_A+p_B
\]

\[
=4.50\times10^{-19}
-3.93\times10^{-19}
\]

\[
\boxed{p_{\text{total}}
=+5.72\times10^{-20}\ \text{kg m s}^{-1}}
\]

The net momentum is in the positive \(x\)-direction.

Part d: Explain the difference

The classical model uses:

\[
p=m_0v
\]

The two classical momenta cancel because the masses and speeds were chosen to satisfy:

\[
m_A(0.60c)=m_B(0.40c)
\]

However, relativistic momentum includes an additional factor:

\[
p=\gamma m_0v
\]

Particle A travels faster, so its Lorentz factor is larger:

\[
\gamma_A=1.25
\]

\[
\gamma_B=1.091
\]

The relativistic correction is therefore different for each particle.

The original cancellation no longer holds.

The important lesson is that equal and opposite classical momenta do not necessarily produce equal and opposite relativistic momenta.

Question 5

Two identical particles each have a rest mass of \(1.50\times10^{-27}\ \text{kg}\).

They approach each other along the same horizontal axis, travelling at equal speeds of \(0.80c\) in opposite directions.

They collide and form a single composite object that remains at rest in the laboratory frame.

Assume:

• The collision is isolated, so total momentum and total energy are conserved.
• No energy escapes as radiation or otherwise.
• All initial kinetic energy is retained as internal energy of the composite object.
• The final object remains intact.

A student argues:

“The total momentum before the collision is zero, so the final object must simply have a mass equal to the sum of the two original rest masses.”

Another student argues:

“The particles are travelling very quickly, so each particle’s rest mass has increased.”

Both arguments contain a mistake.

Determine:

a) The initial total relativistic momentum.

b) The initial total energy.

c) The final velocity of the composite object.

d) The invariant mass of the final composite object.

e) Explain why the final object’s mass is greater than the sum of the two original rest masses, even though the individual particles’ rest masses do not increase because of their motion.

Solution 5

The initial total momentum is zero, the total energy is \(4.50\times10^{-10}\ \text{J}\), and the final composite object has an invariant mass of \(5.00\times10^{-27}\ \text{kg}\).

The key is that momentum is a vector, while energy is a scalar. Opposite momenta cancel, but the kinetic energies add. Because no energy escapes, the combined system retains that energy as internal energy.

Part a: Initial total momentum

Both particles have:

\[
\gamma=
\frac{1}{\sqrt{1-0.80^2}}
\]

\[
\gamma=1.667
\]

The magnitude of each relativistic momentum is:

\[
|p|=\gamma m_0v
\]

\[
=(1.667)(1.50\times10^{-27})
(2.40\times10^8)
\]

\[
|p|=6.00\times10^{-19}
\ \text{kg m s}^{-1}
\]

Because the particles travel in opposite directions:

\[
p_{\text{total}}=
+6.00\times10^{-19}
-6.00\times10^{-19}
\]

\[
\boxed{p_{\text{total}}=0}
\]

Part b: Initial total energy

The total energy of one particle is:

\[
E=\gamma m_0c^2
\]

\[
=(1.667)(1.50\times10^{-27})
(3.00\times10^8)^2
\]

\[
E=2.25\times10^{-10}\ \text{J}
\]

There are two identical particles, so:

\[
E_{\text{total}}=2E
\]

\[
\boxed{E_{\text{total}}
=4.50\times10^{-10}\ \text{J}}
\]

Notice that energy adds even though momentum cancels.

Part c: Final velocity

The initial total momentum is zero.

Momentum is conserved, and the collision is isolated.

Therefore, the final composite object must also have zero momentum.

Since it has non-zero mass, its velocity must be:

\[
\boxed{v_{\text{final}}=0}
\]

The final object is stationary in the laboratory frame.

Part d: Final invariant mass

The final object is at rest, so its total energy is its rest energy:

\[
E_{\text{final}}=M c^2
\]

where \(M\) is the invariant mass of the composite object.

Energy conservation gives:

\[
Mc^2=4.50\times10^{-10}
\]

Rearranging:

\[
M=\frac{4.50\times10^{-10}}
{(3.00\times10^8)^2}
\]

\[
\boxed{M=5.00\times10^{-27}\ \text{kg}}
\]

This is the invariant mass of the combined system.

Part e: Why is the final mass greater?

The sum of the original rest masses is:

\[
m_{\text{original}}=2m_0
\]

\[
=2(1.50\times10^{-27})
\]

\[
m_{\text{original}}=3.00\times10^{-27}\ \text{kg}
\]

But the final composite mass is:

\[
M=5.00\times10^{-27}\ \text{kg}
\]

The difference is:

\[
\Delta M=M-2m_0
\]

\[
=5.00\times10^{-27}
-3.00\times10^{-27}
\]

\[
\boxed{\Delta M=2.00\times10^{-27}\ \text{kg}}
\]

Where did this additional mass come from?

It comes from the kinetic energy that was retained inside the composite object.

The total initial kinetic energy is:

\[
K_{\text{total}}
=2(\gamma-1)m_0c^2
\]

\[
=2(1.667-1)(1.50\times10^{-27})c^2
\]

\[
K_{\text{total}}=1.80\times10^{-10}\ \text{J}
\]

The equivalent mass contribution is:

\[
\frac{K_{\text{total}}}{c^2}
=
\frac{1.80\times10^{-10}}
{9.00\times10^{16}}
\]

\[
=2.00\times10^{-27}\ \text{kg}
\]

This matches the additional mass calculated above.

The individual particles did not acquire greater invariant rest masses simply because they were moving.

Instead, the final composite system has a greater invariant mass because it retains the energy associated with the particles’ initial motion.

This is the important distinction: zero total momentum does not mean zero energy, and the invariant mass of a system depends on its total energy in its own centre-of-momentum frame.

06Where to go next

Comparing classical and relativistic momentum is more than substituting values into two equations. You need to identify the correct rest mass, calculate the Lorentz factor, preserve the direction of momentum, and distinguish momentum from energy.

Once you can do that, the next useful step is understanding how forces change momentum.

The impulse and change in momentum explanation develops that connection.

For the broader explanation of why momentum increases so rapidly as a massive particle approaches the speed of light, continue with Relativistic Momentum: How HSC Physics Explains the Speed Limit.