Four-vector

Four-vectors in a real-valued basis

A four-vector A is a vector with a "timelike" component and three "spacelike" components, and can be written in various equivalent notations:^[4] $${\displaystyle {\begin{aligned}\mathbf {A} &=\left(A^{0},\,A^{1},\,A^{2},\,A^{3}\right)\\&=A^{0}\mathbf {E} _{0}+A^{1}\mathbf {E} _{1}+A^{2}\mathbf {E} _{2}+A^{3}\mathbf {E} _{3}\\&=A^{0}\mathbf {E} _{0}+A^{i}\mathbf {E} _{i}\\&=A^{\alpha }\mathbf {E} _{\alpha }\end{aligned}}}$$ where A^α is the component multiplier and E_α is the basis vector; note that both are necessary to make a vector, and that when A^α is seen alone, it refers strictly to the components of the vector.

The upper indices indicate contravariant components. Here the standard convention is that Latin indices take values for spatial components, so that i = 1, 2, 3, and Greek indices take values for time and space components, so α = 0, 1, 2, 3, used with the summation convention. The split between the time component and the spatial components is a useful one to make when determining contractions of one four vector with other tensor quantities, such as for calculating Lorentz invariants in scalar products (examples are given below), or raising and lowering indices.

In special relativity, the spacelike basis E₁, E₂, E₃ and components A¹, A², A³ are often Cartesian basis and components: $${\displaystyle {\begin{aligned}\mathbf {A} &=\left(A_{t},\,A_{x},\,A_{y},\,A_{z}\right)\\&=A_{t}\mathbf {E} _{t}+A_{x}\mathbf {E} _{x}+A_{y}\mathbf {E} _{y}+A_{z}\mathbf {E} _{z}\\\end{aligned}}}$$ although, of course, any other basis and components may be used, such as spherical polar coordinates $${\displaystyle {\begin{aligned}\mathbf {A} &=\left(A_{t},\,A_{r},\,A_{\theta },\,A_{\phi }\right)\\&=A_{t}\mathbf {E} _{t}+A_{r}\mathbf {E} _{r}+A_{\theta }\mathbf {E} _{\theta }+A_{\phi }\mathbf {E} _{\phi }\\\end{aligned}}}$$ or cylindrical polar coordinates, $${\displaystyle {\begin{aligned}\mathbf {A} &=(A_{t},\,A_{r},\,A_{\theta },\,A_{z})\\&=A_{t}\mathbf {E} _{t}+A_{r}\mathbf {E} _{r}+A_{\theta }\mathbf {E} _{\theta }+A_{z}\mathbf {E} _{z}\\\end{aligned}}}$$ or any other orthogonal coordinates, or even general curvilinear coordinates. Note the coordinate labels are always subscripted as labels and are not indices taking numerical values. In general relativity, local curvilinear coordinates in a local basis must be used. Geometrically, a four-vector can still be interpreted as an arrow, but in spacetime - not just space. In relativity, the arrows are drawn as part of Minkowski diagram (also called spacetime diagram). In this article, four-vectors will be referred to simply as vectors.

It is also customary to represent the bases by column vectors: $${\displaystyle \mathbf {E} _{0}={\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}\,,\quad \mathbf {E} _{1}={\begin{pmatrix}0\\1\\0\\0\end{pmatrix}}\,,\quad \mathbf {E} _{2}={\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}\,,\quad \mathbf {E} _{3}={\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}}$$ so that: $${\displaystyle \mathbf {A} ={\begin{pmatrix}A^{0}\\A^{1}\\A^{2}\\A^{3}\end{pmatrix}}}$$

The relation between the covariant and contravariant coordinates is through the Minkowski metric tensor (referred to as the metric), η which raises and lowers indices as follows: $${\displaystyle A_{\mu }=\eta _{\mu \nu }A^{\nu }\,,}$$ and in various equivalent notations the covariant components are: $${\displaystyle {\begin{aligned}\mathbf {A} &=(A_{0},\,A_{1},\,A_{2},\,A_{3})\\&=A_{0}\mathbf {E} ^{0}+A_{1}\mathbf {E} ^{1}+A_{2}\mathbf {E} ^{2}+A_{3}\mathbf {E} ^{3}\\&=A_{0}\mathbf {E} ^{0}+A_{i}\mathbf {E} ^{i}\\&=A_{\alpha }\mathbf {E} ^{\alpha }\\\end{aligned}}}$$ where the lowered index indicates it to be covariant. Often the metric is diagonal, as is the case for orthogonal coordinates (see line element), but not in general curvilinear coordinates.

The bases can be represented by row vectors: $${\displaystyle {\begin{aligned}\mathbf {E} ^{0}&={\begin{pmatrix}1&0&0&0\end{pmatrix}}\,,&\mathbf {E} ^{1}&={\begin{pmatrix}0&1&0&0\end{pmatrix}}\,,\\[1ex]\mathbf {E} ^{2}&={\begin{pmatrix}0&0&1&0\end{pmatrix}}\,,&\mathbf {E} ^{3}&={\begin{pmatrix}0&0&0&1\end{pmatrix}},\end{aligned}}}$$ so that: $${\displaystyle \mathbf {A} ={\begin{pmatrix}A_{0}&A_{1}&A_{2}&A_{3}\end{pmatrix}}}$$

The motivation for the above conventions are that the scalar product is a scalar, see below for details.

Lorentz transformation

Given two inertial or rotated frames of reference, a four-vector is defined as a quantity which transforms according to the Lorentz transformation matrix Λ: $${\displaystyle \mathbf {A} '={\boldsymbol {\Lambda }}\mathbf {A} }$$

In index notation, the contravariant and covariant components transform according to, respectively: $${\displaystyle {A'}^{\mu }=\Lambda ^{\mu }{}_{\nu }A^{\nu }\,,\quad {A'}_{\mu }=\Lambda _{\mu }{}^{\nu }A_{\nu }}$$ in which the matrix Λ has components Λ^μ_ν in row μ and column ν, and the matrix (Λ⁻¹)^T has components Λ_μ^ν in row μ and column ν.

For background on the nature of this transformation definition, see tensor. All four-vectors transform in the same way, and this can be generalized to four-dimensional relativistic tensors; see special relativity.

Pure boosts in any direction

For two frames moving at constant relative three-velocity v (not four-velocity, see below), it is convenient to denote and define the relative velocity in units of c by: $${\displaystyle {\boldsymbol {\beta }}=(\beta _{1},\,\beta _{2},\,\beta _{3})={\frac {1}{c}}(v_{1},\,v_{2},\,v_{3})={\frac {1}{c}}\mathbf {v} \,.}$$

Then without rotations, the matrix Λ has components given by:^[6] $${\displaystyle {\begin{aligned}\Lambda _{00}&=\gamma ,\\\Lambda _{0i}=\Lambda _{i0}&=-\gamma \beta _{i},\\\Lambda _{ij}=\Lambda _{ji}&=(\gamma -1){\frac {\beta _{i}\beta _{j}}{\beta ^{2}}}+\delta _{ij}=(\gamma -1){\frac {v_{i}v_{j}}{v^{2}}}+\delta _{ij},\\\end{aligned}}}$$ where the Lorentz factor is defined by: $${\displaystyle \gamma ={\frac {1}{\sqrt {1-{\boldsymbol {\beta }}\cdot {\boldsymbol {\beta }}}}}\,,}$$ and δ_ij is the Kronecker delta. Contrary to the case for pure rotations, the spacelike and timelike components are mixed together under boosts.

For the case of a boost in the x-direction only, the matrix reduces to;^[7][8] $${\displaystyle {\begin{pmatrix}A'^{0}\\A'^{1}\\A'^{2}\\A'^{3}\end{pmatrix}}={\begin{pmatrix}\cosh \phi &-\sinh \phi &0&0\\-\sinh \phi &\cosh \phi &0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}{\begin{pmatrix}A^{0}\\A^{1}\\A^{2}\\A^{3}\end{pmatrix}}}$$

Where the rapidity ϕ expression has been used, written in terms of the hyperbolic functions: $${\displaystyle \gamma =\cosh \phi .}$$

This Lorentz matrix illustrates the boost to be a hyperbolic rotation in four dimensional spacetime, analogous to the circular rotation above in three-dimensional space.

Properties

Derivatives and differentials

In special relativity (but not general relativity), the derivative of a four-vector with respect to a scalar λ (invariant) is itself a four-vector. It is also useful to take the differential of the four-vector, dA and divide it by the differential of the scalar, dλ: $${\displaystyle {\underset {\text{differential}}{d\mathbf {A} }}={\underset {\text{derivative}}{\frac {d\mathbf {A} }{d\lambda }}}{\underset {\text{differential}}{d\lambda }}}$$ where the contravariant components are: $${\displaystyle d\mathbf {A} =\left(dA^{0},dA^{1},dA^{2},dA^{3}\right)}$$ while the covariant components are: $${\displaystyle d\mathbf {A} =\left(dA_{0},dA_{1},dA_{2},dA_{3}\right)}$$

In relativistic mechanics, one often takes the differential of a four-vector and divides by the differential in proper time (see below).

Four-position

A point in Minkowski space is a time and spatial position, called an "event", or sometimes the position four-vector or four-position or 4-position, described in some reference frame by a set of four coordinates: $${\displaystyle \mathbf {R} =\left(ct,\mathbf {r} \right)}$$ where r is the three-dimensional space position vector. If r is a function of coordinate time t in the same frame, i.e. r = r(t), this corresponds to a sequence of events as t varies. The definition R⁰ = ct ensures that all the coordinates have the same dimension (of length) and units.^{[9][10][11][12]} These coordinates are the components of the position four-vector for the event.

The displacement four-vector is defined to be an "arrow" linking two events: $${\displaystyle \Delta \mathbf {R} =\left(c\Delta t,\Delta \mathbf {r} \right)}$$

For the differential four-position on a world line we have, using a norm notation: $${\displaystyle \|d\mathbf {R} \|^{2}=\mathbf {dR\cdot dR} =dR^{\mu }dR_{\mu }=c^{2}d\tau ^{2}=ds^{2}\,,}$$ defining the differential line element ds and differential proper time increment dτ, but this "norm" is also: $${\displaystyle \|d\mathbf {R} \|^{2}=(cdt)^{2}-d\mathbf {r} \cdot d\mathbf {r} \,,}$$ so that: $${\displaystyle (cd\tau )^{2}=(cdt)^{2}-d\mathbf {r} \cdot d\mathbf {r} \,.}$$

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 ${\displaystyle \tau }$. As proper time is an invariant, this guarantees that the proper-time-derivative of any four-vector is itself a four-vector. It is then important to find a relation between this proper-time-derivative and another time derivative (using the coordinate time t of an inertial reference frame). This relation is provided by taking the above differential invariant spacetime interval, then dividing by (cdt)² to obtain $${\displaystyle \left({\frac {cd\tau }{cdt}}\right)^{2}=1-\left({\frac {d\mathbf {r} }{cdt}}\cdot {\frac {d\mathbf {r} }{cdt}}\right)=1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}={\frac {1}{\gamma (\mathbf {u} )^{2}}}\,,}$$ where u = dr/dt is the coordinate 3-velocity of an object measured in the same frame as the coordinates x, y, z, and coordinate time t, and $${\displaystyle \gamma (\mathbf {u} )={\frac {1}{\sqrt {1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}}}$$ is the Lorentz factor. This provides a useful relation between the differentials in coordinate time and proper time: $${\displaystyle dt=\gamma (\mathbf {u} )d\tau \,.}$$

This relation can also be found from the time transformation in the Lorentz transformations.

Important four-vectors in relativity theory can be defined by applying this differential ⁠d/dτ⁠.

Four-gradient

Considering that partial derivatives are linear operators, one can form a four-gradient from the partial time derivative ∂/∂t and the spatial gradient operator ∇. Using the standard basis, in index and abbreviated notations, the contravariant components are: $${\displaystyle {\begin{aligned}{\boldsymbol {\partial }}&=\left({\frac {\partial }{\partial x_{0}}},\,-{\frac {\partial }{\partial x_{1}}},\,-{\frac {\partial }{\partial x_{2}}},\,-{\frac {\partial }{\partial x_{3}}}\right)\\&=(\partial ^{0},\,-\partial ^{1},\,-\partial ^{2},\,-\partial ^{3})\\&=\mathbf {E} _{0}\partial ^{0}-\mathbf {E} _{1}\partial ^{1}-\mathbf {E} _{2}\partial ^{2}-\mathbf {E} _{3}\partial ^{3}\\&=\mathbf {E} _{0}\partial ^{0}-\mathbf {E} _{i}\partial ^{i}\\&=\mathbf {E} _{\alpha }\partial ^{\alpha }\\&=\left({\frac {1}{c}}{\frac {\partial }{\partial t}},\,-\nabla \right)\\&=\left({\frac {\partial _{t}}{c}},-\nabla \right)\\&=\mathbf {E} _{0}{\frac {1}{c}}{\frac {\partial }{\partial t}}-\nabla \\\end{aligned}}}$$

Note the basis vectors are placed in front of the components, to prevent confusion between taking the derivative of the basis vector, or simply indicating the partial derivative is a component of this four-vector. The covariant components are: $${\displaystyle {\begin{aligned}{\boldsymbol {\partial }}&=\left({\frac {\partial }{\partial x^{0}}},\,{\frac {\partial }{\partial x^{1}}},\,{\frac {\partial }{\partial x^{2}}},\,{\frac {\partial }{\partial x^{3}}}\right)\\&=(\partial _{0},\,\partial _{1},\,\partial _{2},\,\partial _{3})\\&=\mathbf {E} ^{0}\partial _{0}+\mathbf {E} ^{1}\partial _{1}+\mathbf {E} ^{2}\partial _{2}+\mathbf {E} ^{3}\partial _{3}\\&=\mathbf {E} ^{0}\partial _{0}+\mathbf {E} ^{i}\partial _{i}\\&=\mathbf {E} ^{\alpha }\partial _{\alpha }\\&=\left({\frac {1}{c}}{\frac {\partial }{\partial t}},\,\nabla \right)\\&=\left({\frac {\partial _{t}}{c}},\nabla \right)\\&=\mathbf {E} ^{0}{\frac {1}{c}}{\frac {\partial }{\partial t}}+\nabla \\\end{aligned}}}$$

Since this is an operator, it does not have a "length", but evaluating the scalar product of the operator with itself gives another operator: $${\displaystyle \partial ^{\mu }\partial _{\mu }={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\nabla ^{2}={\frac {{\partial _{t}}^{2}}{c^{2}}}-\nabla ^{2}}$$ called the D'Alembert operator.

Four-velocity

The four-velocity of a particle is defined by: $${\displaystyle \mathbf {U} ={\frac {d\mathbf {X} }{d\tau }}={\frac {d\mathbf {X} }{dt}}{\frac {dt}{d\tau }}=\gamma (\mathbf {u} )\left(c,\mathbf {u} \right),}$$

Geometrically, U is a normalized vector tangent to the world line of the particle. Using the differential of the four-position, the magnitude of the four-velocity can be obtained: $${\displaystyle \|\mathbf {U} \|^{2}=U^{\mu }U_{\mu }={\frac {dX^{\mu }}{d\tau }}{\frac {dX_{\mu }}{d\tau }}={\frac {dX^{\mu }dX_{\mu }}{d\tau ^{2}}}=c^{2}\,,}$$

In short, the magnitude of the four-velocity for any object is always a fixed constant: $${\displaystyle \|\mathbf {U} \|^{2}=c^{2}}$$

The norm is also: $${\displaystyle \|\mathbf {U} \|^{2}={\gamma (\mathbf {u} )}^{2}\left(c^{2}-\mathbf {u} \cdot \mathbf {u} \right)\,,}$$ so that: $${\displaystyle c^{2}={\gamma (\mathbf {u} )}^{2}\left(c^{2}-\mathbf {u} \cdot \mathbf {u} \right)\,,}$$ which reduces to the definition of the Lorentz factor.

Units of four-velocity are m/s in SI and 1 in the geometrized unit system. Four-velocity is a contravariant vector.

Four-acceleration

The four-acceleration is given by: $${\displaystyle \mathbf {A} ={\frac {d\mathbf {U} }{d\tau }}=\gamma (\mathbf {u} )\left({\frac {d{\gamma }(\mathbf {u} )}{dt}}c,{\frac {d{\gamma }(\mathbf {u} )}{dt}}\mathbf {u} +\gamma (\mathbf {u} )\mathbf {a} \right).}$$ where a = du/dt is the coordinate 3-acceleration. Since the magnitude of U is a constant, the four acceleration is orthogonal to the four velocity, i.e. the Minkowski scalar product of the four-acceleration and the four-velocity is zero: $${\displaystyle \mathbf {A} \cdot \mathbf {U} =A^{\mu }U_{\mu }={\frac {dU^{\mu }}{d\tau }}U_{\mu }={\frac {1}{2}}\,{\frac {d}{d\tau }}\left(U^{\mu }U_{\mu }\right)=0\,}$$ which is true for all world lines. The geometric meaning of four-acceleration is the curvature vector of the world line in Minkowski space.

Four-momentum

For a massive particle of rest mass (or invariant mass) m₀, the four-momentum is given by: $${\displaystyle \mathbf {P} =m_{0}\mathbf {U} =m_{0}\gamma (\mathbf {u} )(c,\mathbf {u} )=\left({\frac {E}{c}},\mathbf {p} \right)}$$ where the total energy of the moving particle is: $${\displaystyle E=\gamma (\mathbf {u} )m_{0}c^{2}}$$ and the total relativistic momentum is: $${\displaystyle \mathbf {p} =\gamma (\mathbf {u} )m_{0}\mathbf {u} }$$

Taking the scalar product of the four-momentum with itself: $${\displaystyle \|\mathbf {P} \|^{2}=P^{\mu }P_{\mu }=m_{0}^{2}U^{\mu }U_{\mu }=m_{0}^{2}c^{2}}$$ and also: $${\displaystyle \|\mathbf {P} \|^{2}={\frac {E^{2}}{c^{2}}}-\mathbf {p} \cdot \mathbf {p} }$$ which leads to the energy–momentum relation: $${\displaystyle E^{2}=c^{2}\mathbf {p} \cdot \mathbf {p} +\left(m_{0}c^{2}\right)^{2}\,.}$$

This last relation is useful in relativistic mechanics, essential in relativistic quantum mechanics and relativistic quantum field theory, all with applications to particle physics.

Four-force

The four-force acting on a particle is defined analogously to the 3-force as the time derivative of 3-momentum in Newton's second law: $${\displaystyle \mathbf {F} ={\frac {d\mathbf {P} }{d\tau }}=\gamma (\mathbf {u} )\left({\frac {1}{c}}{\frac {dE}{dt}},{\frac {d\mathbf {p} }{dt}}\right)=\gamma (\mathbf {u} )\left({\frac {P}{c}},\mathbf {f} \right)}$$ where P is the power transferred to move the particle, and f is the 3-force acting on the particle. For a particle of constant invariant mass m₀, this is equivalent to $${\displaystyle \mathbf {F} =m_{0}\mathbf {A} =m_{0}\gamma (\mathbf {u} )\left({\frac {d{\gamma }(\mathbf {u} )}{dt}}c,\left({\frac {d{\gamma }(\mathbf {u} )}{dt}}\mathbf {u} +\gamma (\mathbf {u} )\mathbf {a} \right)\right)}$$

An invariant derived from the four-force is: $${\displaystyle \mathbf {F} \cdot \mathbf {U} =F^{\mu }U_{\mu }=m_{0}A^{\mu }U_{\mu }=0}$$ from the above result.

Four-heat flux

The four-heat flux vector field, is essentially similar to the 3-d heat flux vector field q, in the local frame of the fluid:^[13] $${\displaystyle \mathbf {Q} =-k{\boldsymbol {\partial }}T=-k\left({\frac {1}{c}}{\frac {\partial T}{\partial t}},\nabla T\right)}$$ where T is absolute temperature and k is thermal conductivity.

Four-baryon number flux

The flux of baryons is:^[14] $${\displaystyle \mathbf {S} =n\mathbf {U} }$$ where n is the number density of baryons in the local rest frame of the baryon fluid (positive values for baryons, negative for antibaryons), and U the four-velocity field (of the fluid) as above.

Four-entropy

The four-entropy vector is defined by:^[15] $${\displaystyle \mathbf {s} =s\mathbf {S} +{\frac {\mathbf {Q} }{T}}}$$ where s is the entropy per baryon, and T the absolute temperature, in the local rest frame of the fluid.^[16]

Four-current

The electromagnetic four-current (or more correctly a four-current density)^[17] is defined by $${\displaystyle \mathbf {J} =\left(\rho c,\mathbf {j} \right)}$$ formed from the current density j and charge density ρ.

Four-potential

The electromagnetic four-potential (or more correctly a four-EM vector potential) defined by $${\displaystyle \mathbf {A} =\left({\frac {\phi }{c}},\mathbf {a} \right)}$$ formed from the vector potential a and the scalar potential ϕ.

The four-potential is not uniquely determined, because it depends on a choice of gauge.

In the wave equation for the electromagnetic field:

• In vacuum, $${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})\mathbf {A} =0}$$
• With a four-current source and using the Lorenz gauge condition ${\displaystyle ({\boldsymbol {\partial }}\cdot \mathbf {A} )=0}$, $${\displaystyle ({\boldsymbol {\partial }}\cdot {\boldsymbol {\partial }})\mathbf {A} =\mu _{0}\mathbf {J} }$$

Four-frequency

A photonic plane wave can be described by the four-frequency, defined as $${\displaystyle \mathbf {N} =\nu \left(1,{\hat {\mathbf {n} }}\right)}$$ where ν is the frequency of the wave and ${\displaystyle {\hat {\mathbf {n} }}}$ is a unit vector in the travel direction of the wave. Now, $${\displaystyle \|\mathbf {N} \|=N^{\mu }N_{\mu }=\nu ^{2}\left(1-{\hat {\mathbf {n} }}\cdot {\hat {\mathbf {n} }}\right)=0}$$ so the four-frequency of a photon is always a null vector.

Four-wavevector

The quantities reciprocal to time t and space r are the angular frequency ω and angular wave vector k, respectively. They form the components of the four-wavevector or wave four-vector: $${\displaystyle \mathbf {K} =\left({\frac {\omega }{c}},{\vec {\mathbf {k} }}\right)=\left({\frac {\omega }{c}},{\frac {\omega }{v_{p}}}{\hat {\mathbf {n} }}\right)\,.}$$

The wave four-vector has coherent derived unit of reciprocal meters in the SI.^[18]

A wave packet of nearly monochromatic light can be described by: $${\displaystyle \mathbf {K} ={\frac {2\pi }{c}}\mathbf {N} ={\frac {2\pi }{c}}\nu \left(1,{\hat {\mathbf {n} }}\right)={\frac {\omega }{c}}\left(1,{\hat {\mathbf {n} }}\right)~.}$$

The de Broglie relations then showed that four-wavevector applied to matter waves as well as to light waves: $${\displaystyle \mathbf {P} =\hbar \mathbf {K} =\left({\frac {E}{c}},{\vec {p}}\right)=\hbar \left({\frac {\omega }{c}},{\vec {k}}\right),}$$ yielding ${\displaystyle E=\hbar \omega }$ and ${\displaystyle {\vec {p}}=\hbar {\vec {k}}}$, where ħ is the Planck constant divided by 2π.

The square of the norm is: $${\displaystyle \|\mathbf {K} \|^{2}=K^{\mu }K_{\mu }=\left({\frac {\omega }{c}}\right)^{2}-\mathbf {k} \cdot \mathbf {k} \,,}$$ and by the de Broglie relation: $${\displaystyle \|\mathbf {K} \|^{2}={\frac {1}{\hbar ^{2}}}\|\mathbf {P} \|^{2}=\left({\frac {m_{0}c}{\hbar }}\right)^{2}\,,}$$ we have the matter wave analogue of the energy–momentum relation: $${\displaystyle \left({\frac {\omega }{c}}\right)^{2}-\mathbf {k} \cdot \mathbf {k} =\left({\frac {m_{0}c}{\hbar }}\right)^{2}~.}$$

Note that for massless particles, in which case m₀ = 0, we have: $${\displaystyle \left({\frac {\omega }{c}}\right)^{2}=\mathbf {k} \cdot \mathbf {k} \,,}$$ or ‖k‖ = ω/c. Note this is consistent with the above case; for photons with a 3-wavevector of modulus ω / c , in the direction of wave propagation defined by the unit vector ${\displaystyle {\hat {\mathbf {n} }}~.}$

Four-probability current

In quantum mechanics, the four-probability current or probability four-current is analogous to the electromagnetic four-current:^[19] $${\displaystyle \mathbf {J} =(\rho c,\mathbf {j} )}$$ where ρ is the probability density function corresponding to the time component, and j is the probability current vector. In non-relativistic quantum mechanics, this current is always well defined because the expressions for density and current are positive definite and can admit a probability interpretation. In relativistic quantum mechanics and quantum field theory, it is not always possible to find a current, particularly when interactions are involved.

Replacing the energy by the energy operator and the momentum by the momentum operator in the four-momentum, one obtains the four-momentum operator, used in relativistic wave equations.

Four-spin

The four-spin of a particle is defined in the rest frame of a particle to be $${\displaystyle \mathbf {S} =(0,\mathbf {s} )}$$ where s is the spin pseudovector. In quantum mechanics, not all three components of this vector are simultaneously measurable, only one component is. The timelike component is zero in the particle's rest frame, but not in any other frame. This component can be found from an appropriate Lorentz transformation.

The norm squared is the (negative of the) magnitude squared of the spin, and according to quantum mechanics we have $${\displaystyle \|\mathbf {S} \|^{2}=-|\mathbf {s} |^{2}=-\hbar ^{2}s(s+1)}$$

This value is observable and quantized, with s the spin quantum number (not the magnitude of the spin vector).

Four-vectors in the algebra of physical space

A four-vector A can also be defined in using the Pauli matrices as a basis, again in various equivalent notations:^[20] $${\displaystyle {\begin{aligned}\mathbf {A} &=\left(A^{0},\,A^{1},\,A^{2},\,A^{3}\right)\\&=A^{0}{\boldsymbol {\sigma }}_{0}+A^{1}{\boldsymbol {\sigma }}_{1}+A^{2}{\boldsymbol {\sigma }}_{2}+A^{3}{\boldsymbol {\sigma }}_{3}\\&=A^{0}{\boldsymbol {\sigma }}_{0}+A^{i}{\boldsymbol {\sigma }}_{i}\\&=A^{\alpha }{\boldsymbol {\sigma }}_{\alpha }\\\end{aligned}}}$$ or explicitly: $${\displaystyle {\begin{aligned}\mathbf {A} &=A^{0}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}+A^{1}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}+A^{2}{\begin{pmatrix}0&-i\\i&0\end{pmatrix}}+A^{3}{\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\\&={\begin{pmatrix}A^{0}+A^{3}&A^{1}-iA^{2}\\A^{1}+iA^{2}&A^{0}-A^{3}\end{pmatrix}}\end{aligned}}}$$ and in this formulation, the four-vector is represented as a Hermitian matrix (the matrix transpose and complex conjugate of the matrix leaves it unchanged), rather than a real-valued column or row vector. The determinant of the matrix is the modulus of the four-vector, so the determinant is an invariant: $${\displaystyle {\begin{aligned}|\mathbf {A} |&={\begin{vmatrix}A^{0}+A^{3}&A^{1}-iA^{2}\\A^{1}+iA^{2}&A^{0}-A^{3}\end{vmatrix}}\\[1ex]&=\left(A^{0}+A^{3}\right)\left(A^{0}-A^{3}\right)-\left(A^{1}-iA^{2}\right)\left(A^{1}+iA^{2}\right)\\[1ex]&=\left(A^{0}\right)^{2}-\left(A^{1}\right)^{2}-\left(A^{2}\right)^{2}-\left(A^{3}\right)^{2}\end{aligned}}}$$

This idea of using the Pauli matrices as basis vectors is employed in the algebra of physical space, an example of a Clifford algebra.

Four-vectors in spacetime algebra

In spacetime algebra, another example of Clifford algebra, the gamma matrices can also form a basis. (They are also called the Dirac matrices, owing to their appearance in the Dirac equation). There is more than one way to express the gamma matrices, detailed in that main article.

The Feynman slash notation is a shorthand for a four-vector A contracted with the gamma matrices: $${\displaystyle \mathbf {A} \!\!\!\!/=A_{\alpha }\gamma ^{\alpha }=A_{0}\gamma ^{0}+A_{1}\gamma ^{1}+A_{2}\gamma ^{2}+A_{3}\gamma ^{3}}$$

The four-momentum contracted with the gamma matrices is an important case in relativistic quantum mechanics and relativistic quantum field theory. In the Dirac equation and other relativistic wave equations, terms of the form: $${\displaystyle {\begin{aligned}\mathbf {P} \!\!\!\!/=P_{\alpha }\gamma ^{\alpha }&=P_{0}\gamma ^{0}+P_{1}\gamma ^{1}+P_{2}\gamma ^{2}+P_{3}\gamma ^{3}\\[4pt]&={\dfrac {E}{c}}\gamma ^{0}-p_{x}\gamma ^{1}-p_{y}\gamma ^{2}-p_{z}\gamma ^{3}\\\end{aligned}}}$$ appear, in which the energy E and momentum components (p_x, p_y, p_z) are replaced by their respective operators.