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Four-vector

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 (t, x, y, z) 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.

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Mathematics of four-vectors

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

$x^a := \left(ct, x, y, z \right)$

for a = 0, 1, 2, 3, where c is the speed of light.

The inner product of two four-vectors x and y is defined (using Einstein notation) as

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

where η is the Minkowski metric. Sometimes this inner product is called the Minkowski inner product. The four-vectors make up 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 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 gamma factor of relativity. Important four-vectors in relativity theory can now be defined, such as the four-velocity defined by:

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

where

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

for i = 1, 2, 3. Notice that

UaUa = - c2.

The four-acceleration is defined by:

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

Note that by direct calculation, it is always true that

AaUa = 0.

The four-momentum is defined by:

$P^a :=m_0 U^a = \left(mc, \mathbf{p} \right)$

where m0 is the rest mass of the particle (with m = γm0) and p = mu.

The four-force is defined by:

$F^a := m_0 A^a = \left(\gamma \dot{m} c, \gamma \mathbf{f} \right)$

where

$\mathbf{f} = m_0 \dot{\gamma} \mathbf{u} + m_0 \gamma \mathbf{\dot{u}}$.

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 FaUa = 0 and expanding this out we get

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

Hence

$\frac{dK}{dt} = c^2 \frac{dm}{dt}$

which yields

K = mc2 + 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 = - m0c2

Thus, we interpret the total energy E of the particle as composed of its kinetic energy K and its rest energy m0c2. Thus, we have

E = mc2

Deriving E2 = p2c2 + m02c4

Using the relation E = mc2, we can write the four-momentum as

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

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

$p^2 - \frac{E^2}{c^2} = P^a P_a = m_0^2 U^a U_a = -m_0^2 c^2$

i.e.

$p^2 - \frac{E^2}{c^2} = -m_0^2 c^2$

Hence

$E^2 = p^2 c^2 + m_0^2 c^4.$

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

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

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

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

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

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

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

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

$N^a N_a = \nu ^2 \left(n^2 - 1 \right) = 0$

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

Deriving Planck's relation

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

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

where E is the photon's energy and $\mathbf{p}=p \mathbf{n}$ the photon's momentum with $\mathbf{n}$ a unit vector in the direction of motion of the photon. Note that $\tilde{P}^a\tilde{P}_a=0$ by virtue of the relation E = pc which comes from electromagnetic theory. Given that $\tilde{P}^a$ and Na are both null vectors (with each one clearly non-zero, and noting that $\tilde{P}^aN_a=0$, this means that $\tilde{P}$ and Na must be proportional, i.e.

$\tilde{P}^a=sN^a$

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

$\tilde{P}^{'a}=sN^{'a}.$

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 = hν

which is Planck's relation , the constant traditionally being denoted by h and called Planck's constant. Note that by combining E = pc with Planck's relation, the momentum of a photon may be written as:

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

the famous de Broglie equation.