Four-vector

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

Mathematics of four-vectors

A point in Minkowski space is called an "event" and is described by a set of four coordinates such as

$x^{\mu }:=\left(ct,x,y,z\right)$

for $\mu$ = 0, 1, 2, 3, where c is the speed of light. 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:

$\Delta x^{\mu }:=\left(\Delta ct,\Delta x,\Delta y,\Delta z\right)$

(Note that the position vector is the displacement vector when one of the two events is the origin of the coordinate system). The inner product of two four-vectors $u$ and $v$ is defined (using Einstein notation) as

$u\cdot v=u^{\mu }\eta _{\mu \nu }v^{\nu }=\left({\begin{matrix}u^{0}&u^{1}&u^{2}&u^{3}\end{matrix}}\right)\left({\begin{matrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{matrix}}\right)\left({\begin{matrix}v^{0}\\v^{1}\\v^{2}\\v^{3}\end{matrix}}\right)=u^{0}v^{0}-u^{1}v^{1}-u^{2}v^{2}-u^{3}v^{3}$

where η is the Minkowski metric. Sometimes this inner product is called the Minkowski inner product. The four-vectors are arrows on the spacetime diagram or Minkowski diagram.

Four-vectors may be classified as either spacelike, timelike or null. In this article, four-vectors will be referred to simply as vectors. Spacelike, timelike, and null vectors are ones whose inner product with themselves is greater than, less than, and equal to zero respectively.

Examples of four-vectors in dynamics

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 (τ) 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:

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

where γ is the Lorentz factor. Important four-vectors in relativity theory can now be defined, such as the four-velocity defined by:

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

where

$u^{i}={\frac {dx^{i}}{dt}}$

for i = 1, 2, 3. Notice that

$U^{\mu }U_{\mu }=c^{2}\,$

The four-acceleration is given by:

$A^{\mu }={\frac {dU^{\mu }}{d\tau }}=\left(\gamma {\dot {\gamma }}c,\gamma {\dot {\gamma }}\mathbf {u} +\gamma ^{2}\mathbf {\dot {u}} \right)={\frac {\partial U_{\mu }}{\partial x^{\nu }}}U^{\nu }=U_{\mu ,\nu }U^{\nu }$

Since the magnitude of $U^{\mu }$ is a constant

$0={\frac {\partial U^{\mu }U_{\mu }}{\partial x^{\nu }}}=2U_{\mu ,\nu }U^{\mu }\,$

multiplying both sides by $U^{\nu }$ yields the identity

$A^{\mu }U_{\mu }=0\,$

which is true for all velocity distributions.

The four-momentum is for a massive particle is given by:

$p^{\mu }:=mU^{\mu }=\left(\gamma mc,\mathbf {p} \right)$

where m is the invariant mass of the particle.

The four-force is defined by:

$F^{\mu }:=mA^{\mu }=\left(\gamma {\dot {\gamma }}mc,\gamma \mathbf {f} \right)$

where

$\mathbf {f} ={\frac {d\mathbf {p} }{dt}}=m{\dot {\gamma }}\mathbf {u} +m\gamma \mathbf {\dot {u}}$.

Physics of four-vectors

The power and elegance of the four-vector formalism may be demonstrated by deriving some important relations between the physical quantities energy, mass and momentum.

Deriving E = mc²

Here, an expression for the total energy of a particle will be derived. The kinetic energy (K) of a particle is defined analogously to the classical definition, namely as

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

with f as above. Noticing that $F^{\mu }U_{\mu }=0$ and expanding this out we get

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

Hence

${\frac {dK}{dt}}=c^{2}{\frac {d\gamma m}{dt}}$

which yields

$K=\gamma mc^{2}+S\,$

for some constant S. When the particle is at rest (u = 0), we take its kinetic energy to be zero (K = 0). This gives

$S=-mc^{2}\,$

Thus, we interpret the total energy E of the particle as composed of its kinetic energy K and its rest energy m c^{2. Thus, we have}

$E=\gamma mc^{2}\,$

Deriving E² = p²c² + m²c⁴

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

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

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

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

i.e.

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

Hence

$E^{2}=p^{2}c^{2}+m^{2}c^{4}.$

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

$J^{\mu }:=\left(\rho c,\mathbf {j} \right)$

formed from the current density j and charge density ρ, and the electromagnetic four-potential defined by

$\phi ^{\mu }:=\left(\psi ,\mathbf {A} c\right)$

formed from the vector potential A and the scalar potential φ.

A plane electromagnetic wave can be described by the four-frequency defined as

$N^{\mu }:=\left(\nu ,\nu \mathbf {n} \right)$

where $\nu$ is the frequency of the wave and n is a unit vector in the travel direction of the wave. Notice that

$N^{\mu }N_{\mu }=\nu ^{2}\left(n^{2}-1\right)=0$

so that the four-frequency is always a null vector.

Deriving Planck's law

It is often assumed that Planck's law relating the energy and frequency of a photon must necessarily come from quantum mechanics. However, Planck's law can be obtained purely within the formalism of special relativity. In analogy with the definition for the four-momentum of a particle, the photon four-momentum is defined by

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

where:

$E\,$ is the photon's energy

$\mathbf {p} =p\mathbf {n}$ is the photon's momentum

$\mathbf {n}$ a unit vector in the direction of motion of the photon.

Note that $p^{\mu }p_{\mu }=0$ by virtue of the relation $E=pc\,$ which comes from electromagnetic theory. Given that $p^{\mu }$ and $N^{a}$ are both null vectors (with each one clearly non-zero, and noting that $p^{\mu }N_{\mu }=0$, this means that $p^{\mu }$ and $N^{\mu }$ must be proportional, i.e.

$p^{\mu }=sN^{\mu }$

for some real number s. Multiplying the above relation by ${\frac {\partial x^{'\nu }}{\partial x^{\mu }}}$ gives

$p^{'\nu }=sN^{'\nu }.$

Considering the 0-th component of the last two relations shows that the ratio of a photon's energy to its frequency is the same in any two inertial reference frames, i.e.

$E=h\nu \,$

which is Planck's law, the constant traditionally being denoted by $h$ and called Planck's constant. In fact, by employing the same definitions for the photon four- momentum and the four-frequency in general relativity, the above proof appears to be valid in GR.

Note that by combining $E=pc\,$ with Planck's law, the momentum of a photon may be written as the famous de Broglie equation:

$p={\frac {h}{\lambda }}$

See also

• four-velocity
• four-acceleration
• four-momentum
• four-force
• four-current
• electromagnetic four-potential
• Basic introduction to the mathematics of curved spacetime

References

• Rindler, W. Introduction to Special Relativity (2nd edn.) (1991) Clarendon Press Oxford ISBN 0-19-853952-5