Relativistic Momentum
The relativistic momentum is defined as $$ \vec{p} = \gamma m \vec{v} $$ where $ \gamma $ is the Lorentz factor, $$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
This relation, $ \vec{p} = \gamma m \vec{v} $, extends the classical definition of momentum ($ \vec{p} = m\vec{v} $) to include the effects predicted by Einstein’s special theory of relativity.
The Lorentz factor $ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $ increases sharply as the velocity $ v $ approaches the speed of light $ c $.
Although the object’s rest mass $ m $ remains constant, the relativistic momentum $ \vec{p} = \gamma m \vec{v} $ grows dramatically with velocity, diverging to infinity as $ v \to c $.
This is why no object with mass can ever reach or exceed the speed of light - doing so would require an infinite amount of energy.
$$ E^2 = p^2 c^2 + m^2 c^4 $$
With this relativistic formulation, momentum is conserved in all inertial reference frames, fully consistent with the principle of relativity.
Note. The relativistic momentum $ \vec{p} = \gamma m \vec{v} $ forms the spatial component of the four-momentum vector: $$ P^\mu = (\gamma m c, \gamma m \vec{v}) = (\gamma m c, \gamma m v_x , \gamma m v_y , \gamma m v_z ) $$ This four-vector unifies energy and momentum into a single invariant quantity, ensuring that both are conserved in every inertial frame.
Explanation and Derivation
In classical mechanics, momentum (or linear momentum) is defined as the product of mass and velocity:
$$ \vec{p} = m \vec{v} $$
However, this expression is not invariant between inertial frames - that is, between observers moving at constant relative velocity.
The classical definition breaks down at speeds approaching the speed of light ($ c $), where space and time are no longer absolute but interdependent.
To remain consistent with Lorentz transformations, momentum must be redefined so that it transforms properly between all inertial observers.
In the framework of special relativity, momentum is part of a single four-dimensional entity: the four-momentum vector:
$$ P^\mu = m U^\mu $$
Here $ m $ denotes the invariant (rest) mass of the object, and $ U^\mu $ is the four-velocity:
$$ U^\mu = (\gamma c, \gamma \vec{v}) = (\gamma c,\, \gamma v_x,\, \gamma v_y,\, \gamma v_z)\, $$
Hence,
$$ P^\mu = m U^\mu = m\, (\gamma c, \gamma \vec{v}) $$
$$ P^\mu = (\gamma m c, \gamma m \vec{v}) $$
$$ P^\mu = (\gamma m c, \gamma m v_x , \gamma m v_y , \gamma m v_z ) $$
The time component $ P^0 = \gamma m c $ of the four-vector corresponds to the total energy $ E = \gamma m c^2 $ divided by $ c $:
$$ P^0 = \frac{E}{c} = \frac{\gamma m c^2}{c} = \gamma m c $$
The spatial components $ P^i = \gamma m v_i $ represent the relativistic momentum:
$$ \vec{p} = \gamma m \vec{v} $$
This definition ensures that momentum transforms consistently between all inertial reference frames and obeys the law of momentum conservation.
Limiting Cases
In the non-relativistic limit, where the velocity $ v $ is much smaller than the speed of light ($ v \ll c $), the Lorentz factor approaches unity ($ \gamma \approx 1 $), and the relativistic momentum reduces to the familiar classical expression:
$$ \vec{p} \approx m\vec{v} $$
In the relativistic limit, as the velocity $ v $ approaches $ c $, the Lorentz factor grows without bound ($ \gamma \to \infty $), causing the momentum to increase indefinitely - even though the velocity itself can never reach $ c $.
Relationship Between Energy and Momentum
The contravariant four-momentum vector is defined as:
$$ P^\mu = (\gamma m c, \gamma m \vec{v}) = (\gamma m c, \gamma m v_x, \gamma m v_y, \gamma m v_z) $$
To obtain its covariant form, $ P_\mu = (\gamma m c, -\gamma m \vec{v}) $, we apply the Minkowski metric with signature $(+,-,-,-)$:
$$ P_\mu = g_{\mu\nu} P^\nu $$
$$ P_\mu = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} P^\nu $$
$$ P_\mu = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} (\gamma m c, \gamma m v_x, \gamma m v_y, \gamma m v_z) $$
$$ P_\mu = (\gamma m c, -\gamma m v_x, -\gamma m v_y, -\gamma m v_z) $$
The scalar product of the two forms of the four-momentum is:
$$ P^\mu P_\mu = (\gamma m c \cdot \gamma m c) + (\gamma m v_x \cdot (-\gamma m v_x)) + (\gamma m v_y \cdot (-\gamma m v_y)) + (\gamma m v_z \cdot (-\gamma m v_z)) $$
$$ P^\mu P_\mu = (\gamma m c)^2 - (\gamma m v_x)^2 - (\gamma m v_y)^2 - (\gamma m v_z)^2 $$
$$ P^\mu P_\mu = (\gamma m c)^2 - \gamma^2 m^2 (v_x^2 + v_y^2 + v_z^2) $$
$$ P^\mu P_\mu = (\gamma m c)^2 - \gamma^2 m^2 v^2 $$
$$ P^\mu P_\mu = \gamma^2 m^2 (c^2 - v^2) $$
Since the Lorentz factor is $ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $, we can write:
$$ P^\mu P_\mu = m^2 \frac{c^2 - v^2}{1 - \frac{v^2}{c^2}} $$
$$ P^\mu P_\mu = m^2 \frac{c^2 - v^2}{\frac{c^2 - v^2}{c^2}} $$
$$ P^\mu P_\mu = m^2 (c^2 - v^2) \frac{c^2}{c^2 - v^2} $$
$$ P^\mu P_\mu = m^2 c^2 $$
This proves that the scalar product is invariant under Lorentz transformations - it has the same value in all inertial reference frames.
Returning to the earlier form,
$$ P^\mu P_\mu = (\gamma m c)^2 - \gamma^2 m^2 v^2 $$
and recalling that $ \gamma m c = \frac{E}{c} $, we obtain:
$$ P^\mu P_\mu = \left( \frac{E}{c} \right)^2 - \gamma^2 m^2 v^2 $$
$$ P^\mu P_\mu = \frac{E^2}{c^2} - \gamma^2 m^2 v^2 $$
Since $ P^\mu P_\mu = m^2 c^2 $, we have:
$$ m^2 c^2 = \frac{E^2}{c^2} - \gamma^2 m^2 v^2 $$
Multiplying both sides by $ c^2 $ gives:
$$ m^2 c^4 = E^2 - \gamma^2 m^2 v^2 c^2 $$
$$ m^2 c^4 = E^2 - (\gamma m v)^2 c^2 $$
Since $ \vec{p} = \gamma m \vec{v} $, whose magnitude is $ |\vec{p}| = \gamma m v $, we finally obtain:
$$ m^2 c^4 = E^2 - p^2 c^2 $$
$$ E^2 = m^2 c^4 + p^2 c^2 $$
This is the fundamental energy-momentum relation, a Lorentz-invariant equation.
The term $ m^2 c^4 $ represents the rest energy $ E_0 = m c^2 $, while $ p^2 c^2 $ corresponds to the kinetic contribution due to motion. Their sum gives the total energy of a relativistic particle.
Limiting Cases
The general equation $ E^2 = p^2 c^2 + m^2 c^4 $ applies to both massive and massless particles. Let’s examine two key special cases:
- Massive particle at rest
When a particle is at rest relative to the observer, its momentum is zero ($ p = 0 $). The relation simplifies to:
$$ E^2 = 0 + m^2 c^4 $$ $$ E^2 = m^2 c^4 $$ $$ E = m c^2 $$ In this case, the total energy equals the rest energy $ E_0 $, the intrinsic energy a body possesses even in the absence of motion - Einstein’s celebrated equation. - Massless particle (e.g., photon)
For a particle with zero rest mass ($ m = 0 $), the equation becomes:
$$ E^2 = p^2 c^2 + 0 \cdot c^4 $$ $$ E^2 = p^2 c^2 $$ $$ E = p c $$ This describes photons and all massless particles, which can never be at rest and always travel at the speed of light $ c $. Their energy depends solely on their momentum-or equivalently, on their frequency: $$ E = h\nu = \frac{h c}{\lambda} $$
These two limiting cases illustrate how the relativistic energy-momentum relation unifies the dynamics of matter and radiation into a single, elegant framework. For particles with mass, it yields the rest energy $ E_0 = m c^2 $; for massless particles, it describes pure kinetic energy associated with motion at light speed. In both situations, energy and momentum remain bound by the same Lorentz-invariant law, valid for all observers in every inertial reference frame.
Numerical Example
Consider a particle with mass $ m = 1\ \text{kg} $ moving at $ v = 0.6c $.
The Lorentz factor is:
$$ \gamma = \frac{1}{\sqrt{1 - (0.6c)^2 / c^2}} = 1.25 $$
The particle’s momentum is:
$$ p = \gamma m v = 1.25 \times 1\ \text{kg} \times 0.6c $$
$$ p = (0.75\ \text{kg}) \cdot c $$
Using $ c = 3\times10^8\ \text{m/s} $ (approximately 300,000 km/s):
$$ p = 0.75\ \text{kg} \times 3 \times 10^8\ \text{m/s} $$
$$ p = 2.25 \times 10^8\ \text{kg·m/s} $$
The total energy is:
$$ E = \gamma m c^2 = 1.25 \times 1\ \text{kg} \times (3\times10^8\ \text{m/s})^2 $$
$$ E = 1.25 \times 9\times10^{16}\ \text{kg·m}^2\text{/s}^2 $$
$$ E = 1.125 \times10^{17}\ \text{J} $$
This numerical example confirms the internal consistency of the relativistic energy-momentum relation and shows how both energy and momentum increase together as velocity approaches the speed of light.
