Four-Momentum Vector

The four-momentum is one of the key quantities in special relativity. It unifies, in a single mathematical object, two concepts that in classical mechanics are treated separately: the momentum $\vec{p}$ and the energy $E$.

The scalar product between the contravariant and covariant forms of the four-momentum is a Lorentz invariant, meaning it has the same value in all inertial reference frames:

$$ P^\mu P_\mu = m^2 c^2 $$

From this relation, we can derive the fundamental energy-momentum relation of relativity:

$$ E^2 = p^2 c^2 + m^2 c^4 $$

In this way, the four-momentum provides a unified geometric framework for energy and momentum, ensuring that the conservation laws retain their covariant form in every inertial frame.

Explanation and Derivation

The contravariant four-momentum vector has four components:

$$ P^\mu =( \gamma mc, \gamma m v_x,\ \gamma m v_y,\ \gamma m v_z ) $$

The first, or temporal, component is $P^0 = \gamma m c$, which 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 remaining three spatial components form the relativistic momentum:

$$ \vec{p} = \gamma m \vec{v} $$

Thus the four-momentum can be written compactly as:

$$ P^\mu = \left(\frac{E}{c}, \vec{p}\right) $$

The covariant form of the four-momentum is obtained by applying the Minkowski metric, which changes the sign of the spatial components:

$$ P_\mu = g_{\mu\nu} P^\nu $$

$$ P_\mu = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} P^\nu $$

$$ P_\mu = ( \gamma mc, -\gamma m v_x, -\gamma m v_y, -\gamma m v_z ) $$

The scalar product between the contravariant and covariant forms is invariant under Lorentz transformations, meaning it has the same value in every inertial frame:

$$ P^\mu P_\mu = (\gamma mc)^2 - (\gamma m v_x)^2 - (\gamma m v_y)^2 - (\gamma m v_z)^2 $$

$$ P^\mu P_\mu = \left(\frac{E}{c}\right)^2 - p^2 = m^2 c^2 $$

Grouping the spatial terms together yields:

$$ P^\mu P_\mu = (\gamma m c)^2 - (\gamma m)^2 (v_x^2 + v_y^2 + v_z^2) $$

Since $v_x^2 + v_y^2 + v_z^2 = v^2$, this becomes:

$$ P^\mu P_\mu = (\gamma m)^2 (c^2 - v^2) $$

Recalling that $ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $, we can simplify:

$$ P^\mu P_\mu = \left( \frac{1}{\sqrt{1 - v^2/c^2}} \right)^2 m^2 (c^2 - v^2) $$

$$ P^\mu P_\mu = \frac{1}{1 - v^2/c^2} m^2 (c^2 - v^2) $$

$$ P^\mu P_\mu = m^2 c^2 $$

This confirms that the scalar product of the four-momentum is a relativistic invariant, valid in all inertial frames.

The Fundamental Energy-Momentum Relation

The fundamental energy-momentum relation is: $$ E^2 = p^2 c^2 + m^2 c^4 $$

Derivation

The contravariant four-momentum is given by:

$$ P^\mu = \left( \gamma mc , \gamma m v_x, \gamma m v_y, \gamma m v_z \right) $$

Using the Minkowski metric $(+,-,-,-)$, the covariant form becomes:

$$ P_\mu = \left( \gamma mc, -\gamma m v_x , -\gamma m v_y , -\gamma m v_z \right) $$

The temporal component corresponds to the total energy divided by $c$: $ \gamma mc = \frac{E}{c} = \frac{\gamma m c^2}{c} $

$$ P^\mu = \left( \frac{E}{c}, \gamma m v_x, \gamma m v_y, \gamma m v_z \right) $$

Since $ \vec{v} = (v_x, v_y, v_z) $, we can write:

$$ P^\mu = \left( \frac{E}{c}, \gamma m \vec{v} \right) $$

and, using $ \vec{p} = \gamma m \vec{v} $, we obtain the compact form:

$$ P^\mu = \left( \frac{E}{c}, \vec{p} \right) $$

or equivalently:

$$ P^\mu = \left( \frac{E}{c}, p_x, p_y, p_z \right) $$

The corresponding covariant vector is: $$ P_\mu = \left( \frac{E}{c}, -p_x, -p_y, -p_z \right) $$

Their scalar product is therefore:

$$ P^\mu P_\mu = \left( \frac{E}{c} \right)^2 - (p_x^2 + p_y^2 + p_z^2) $$

$$ P^\mu P_\mu = \frac{E^2}{c^2} - p^2 $$

Since this scalar product equals the invariant $ P^\mu P_\mu = m^2 c^2 $, we can write:

$$ m^2 c^2 = \frac{E^2}{c^2} - p^2 $$

Multiplying through by $ c^2 $ gives:

$$ m^2 c^4 = E^2 - p^2 c^2 $$

Rearranging terms, we arrive at the final form:

$$ E^2 = p^2 c^2 + m^2 c^4 $$

This is the fundamental energy-momentum relation of special relativity. It holds for any material particle ($m>0$) and shows that the total energy $E$ includes both the rest energy $m c^2$ and the energy associated with motion, $p c$.

And so on.