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In relativity, a four-vector is a vector in a four-dimensional real vector space, called Minkowski space, whose components transform like the space and time coordinates ${\displaystyle \left(x,y,z,t\right)}$ under spatial rotations and boosts (a change by a constant velocity to another inertial reference frame). The set of all such rotations and boosts, called Lorentz transformations and described by 4×4 matrices, forms the Lorentz group.

Mathematics of four-vectors

A point in Minkowski space is called an "event" and is described by the position four-vector defined as:

${\displaystyle x^{a}=\left(x,y,z,ct\right)}$ (${\displaystyle a=1,2,3,4}$)

where c is the speed of light.

When considering physical phenomena, differential equations arise naturally; however, when considering space and time derivatives of functions, it is unclear which reference frame these derivatives are taken with respect to. It is agreed that time derivatives are taken with respect to the proper time (${\displaystyle \tau }$) in the given reference frame. It is then important to find a relation between this time derivative and another time derivative (taken in another inertial reference frame). This relation is provided by the time transformation in the Lorentz transformations and is:

${\displaystyle {\frac {d\tau }{dt}}={\frac {1}{\gamma }}}$

where ${\displaystyle \gamma }$ is the gamma factor of relativity. #

The inner product of two four-vectors x and y is defined as:

${\displaystyle x\cdot y=x^{a}\eta _{ab}y^{b}\left({\begin{matrix}x^{1}&x^{2}&x^{3}&x^{4}\end{matrix}}\right)\left({\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{matrix}}\right)\left({\begin{matrix}y_{1}\\y_{2}\\y_{3}\\y_{4}\end{matrix}}\right)=x^{1}y_{1}+x^{2}y_{2}+x^{3}y_{3}-x^{4}y_{4}}$

where ${\displaystyle \eta }$ is the Minkowski metric.

Examples of four-vectors in dynamics

Important four-vectors in the relativity theory can now be defined, such as the four-velocity defined by:

${\displaystyle U^{a}:={\frac {dx^{a}}{d\tau }}={\frac {dx^{a}}{dt}}{\frac {dt}{d\tau }}=\left(\gamma \mathbf {u} ,\gamma c\right)}$

where ${\displaystyle u^{i}={\frac {dx^{i}}{dt}}}$ (${\displaystyle i=1,2,3}$). Note that ${\displaystyle U^{a}U_{a}=-c^{2}}$. The four-acceleration is defined by:

${\displaystyle A^{a}:={\frac {dU^{a}}{d\tau }}=\left(\gamma {\dot {\gamma }}\mathbf {u} +\gamma ^{2}\mathbf {\dot {u}} ,\gamma {\dot {\gamma }}c\right)}$

Note that by direct calculation, it is always true that ${\displaystyle A^{a}U_{a}=0}$. The four-momentum is defined by:

${\displaystyle P^{a}=m_{o}U^{a}=\left(\mathbf {p} ,mc\right)}$

where ${\displaystyle m_{o}}$ is the rest mass of the particle ${\displaystyle m=\gamma m_{o}}$ and ${\displaystyle \mathbf {p} =m\mathbf {u} }$.

An important relation can be obtained by calculating the inner product of the four-momentum with itself in two different ways:

${\displaystyle p^{2}-m^{2}c^{2}=P^{a}P_{a}=m_{o}^{2}U^{a}U_{a}=-m_{o}^{2}c^{2}}$

The four-force is defined by:

${\displaystyle F^{a}=m_{o}A^{a}=\left(\gamma \mathbf {f} ,\gamma {\dot {m}}c\right)}$

where ${\displaystyle \mathbf {f} =m_{o}{\dot {\gamma }}\mathbf {u} +m_{0}\gamma \mathbf {\dot {u}} }$.

Physics of four-vectors

It is possible to derive some important relations between energy, mass and momentum by using the formalism of four-vectors.

Deriving ${\displaystyle E=mc^{2}}$

Using the formalism of four-vectors, it is possible to derive an expression for the total energy of a particle. The kinetic energy of a particle (${\displaystyle K}$) is defined analogously to the classical definition, namely as:

${\displaystyle {\frac {dK}{dt}}=\mathbf {f} \cdot \mathbf {u} }$

with ${\displaystyle \mathbf {f} }$ as above. Note that ${\displaystyle F^{a}U_{a}=0}$ and expanding this out we get:

${\displaystyle \gamma ^{2}\left(\mathbf {f} \cdot \mathbf {u} -{\dot {m}}c^{2}\right)=0}$

Hence,

${\displaystyle {\frac {dK}{dt}}=c^{2}{\frac {dm}{dt}}\Rightarrow K=mc^{2}+S}$

for some constant ${\displaystyle S}$. When the particle is at rest (${\displaystyle u=0}$), we take it's kinetic energy to be zero (${\displaystyle K=0}$); this gives,

${\displaystyle S=-m_{0}c^{2}}$

Thus, we interpret the total energy of the particle (${\displaystyle E}$) as composed of it's kinetic energy ${\displaystyle K}$ and it's rest energy ${\displaystyle m_{0}c^{2}}$. Thus, we have:

${\displaystyle E=mc^{2}}$

Deriving ${\displaystyle E^{2}=p^{2}c^{2}+m_{0}^{2}c^{4}}$

Using the relation ${\displaystyle E=mc^{2}}$, we can write the four-momentum as:

${\displaystyle P^{a}=\left(\mathbf {p} ,{\frac {E}{c}}\right)}$.

Taking the inner product of the four-momentum with itself in two different ways, we obtain the relation:

${\displaystyle p^{2}-{\frac {E^{2}}{c^{2}}}=P^{a}P_{a}=m_{o}^{2}U^{a}U_{a}=-m_{o}^{2}c^{2}}$

i.e.

${\displaystyle p^{2}-{\frac {E^{2}}{c^{2}}}=-m_{o}^{2}c^{2}}$

Hence,

${\displaystyle E^{2}=p^{2}c^{2}+m_{0}^{2}c^{4}}$

This last relation is useful in many areas of physics.

Examples of four-vectors in electromagnetism

Examples of four-vectors in electromagnetism include the four-current defined by

${\displaystyle J^{a}=\left(j,\rho c\right)}$

formed from the current and charge densities (${\displaystyle j}$ and ${\displaystyle \rho }$, respectively) and the electromagnetic four-potential defined by

${\displaystyle {\tilde {A}}^{a}=\left(\mathbf {A} ,{\frac {\phi }{c}}\right)}$ formed from the vector and scalar potentials (${\displaystyle A}$ and ${\displaystyle \phi }$, respectively).

The laws of physics are also postulated to be invariant under Lorentz transformations. An object in an inertial reference frame will perceive the universe as if the universe were Lorentz-transformed so that the perceiving object is stationary.

See also

• four-velocity
• four-acceleration
• four-momentum
• four-force

• Relativity