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Relativistic mechanics

Lorentz invariance

Newtonian physics assumes that absolute time and space exist outside of any observer; this gives rise to the Galilean invariance described earlier. It also results in a prediction that the speed of light can vary from one reference frame to another. This is contrary to observation. In the special theory of relativity, Einstein keeps the postulate that the equations of motion do not depend on the reference frame, but assumes that the speed of light

c

is invariant. As a result, position and time in two reference frames are related by the Lorentz transformation instead of the Galilean transformation.^[25]
Consider, for example, a reference frame moving relative to another at velocity

v

in the

x

direction. The Galilean transformation gives the coordinates of the moving frame as

$\begin{align} t' &= t \\ x' &= x - v t \end{align}$

while the Lorentz transformation gives^[26]

$\begin{align} t' &= \gamma \left( t - \frac{v x}{c^2} \right) \\ x' &= \gamma \left( x - v t \right)\,, \end{align}$

where

γ

is the Lorentz factor:

$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\,.$

Newton's second law, with mass fixed, is not invariant under a Lorentz transformation. However, it can be made invariant by making the inertial mass

m

of an object a function of velocity:

$m = \gamma m_0\,;$

m 0

is the object's invariant mass.^[27]
The modified momentum,

$\boldsymbol{p} = \gamma m_0 \boldsymbol{v}\,,$

obeys Newton's second law:

$\boldsymbol{F} = \frac{d \boldsymbol{p}}{dt}\,.$

Within the domain of classical mechanics, relativistic momentum closely approximates Newtonian momentum: at low velocity,

γm 0 v

is approximately equal to

m 0 v

, the Newtonian expression for momentum.

Four-vector formulation

Main article: Four-momentum

In the theory of relativity, physical quantities are expressed in terms of four-vectors that include time as a fourth coordinate along with the three space coordinates. These vectors are generally represented by capital letters, for example

R

for position. The expression for the four-momentum depends on how the coordinates are expressed. Time may be given in its normal units or multiplied by the speed of light so that all the components of the four-vector have dimensions of length. If the latter scaling is used, an interval of proper time,

τ

, defined by^[28]

$c^2d\tau^2 = c^2dt^2-dx^2-dy^2-dz^2\,,$

is invariant under Lorentz transformations (in this expression and in what follows the (+ − − −) metric signature has been used). Mathematically this invariance can be ensured in one of two ways: by treating the four-vectors as Euclidean vectors and multiplying time by the square root of

-1

; or by keeping time a real quantity and embedding the vectors in a Minkowski space.^[29] In a Minkowski space, the scalar product of two four-vectors

U = (U 0, U 1, U 2, U 3)

and

V = (V 0, V 1, V 2, V 3)

is defined as

$\boldsymbol{U} \cdot \boldsymbol{V} = U_0 V_0 - U_1 V_1 - U_2 V_2 - U_3 V_3\,.$

In all the coordinate systems, the (contravariant) relativistic four-velocity is defined by

$\boldsymbol{U} \equiv \frac{d \boldsymbol{R}}{d \tau} = \gamma \frac{d \boldsymbol{R}}{dt}\,,$

and the (contravariant) four-momentum is

$\boldsymbol{P} = m_0\boldsymbol{U}\,,$

where

m 0

is the invariant mass. If

R = (ct,x,y,z)

(in Minkowski space), then^{[note 1]}

$\boldsymbol{P} = \gamma m_0 \left(c,\boldsymbol{v}\right) = (m c,\boldsymbol{p})\,.$

Using Einstein's mass-energy equivalence,

E = mc 2

, this can be rewritten as

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

Thus, conservation of four-momentum is Lorentz-invariant and implies conservation of both mass and energy.
The magnitude of the momentum four-vector is equal to

m 0 c

$||\boldsymbol{P}||^2 = \boldsymbol{P}\cdot\boldsymbol{P} = \gamma^2m_0^2(c^2-v^2) = (m_0c)^2\,,$

and is invariant across all reference frames.
The relativistic energy–momentum relationship holds even for massless particles such as photons; by setting

m 0 = 0

it follows that

$E = pc\,.$

In a game of relativistic "billiards", if a stationary particle is hit by a moving particle in an elastic collision, the paths formed by the two afterwards will form an acute angle. This is unlike the non-relativistic case where they travel at right angles.^[30]

Classical electromagnetism

In Newtonian mechanics, the law of conservation of momentum can be derived from the law of action and reaction, which states that the forces between two particles are equal and opposite. Electromagnetic forces violate this law. Under some circumstances one moving charged particle can exert a force on another without any return force.^[31] Moreover, Maxwell's equations, the foundation of classical electrodynamics, are Lorentz-invariant. However, momentum is still conserved.

Vacuum

In Maxwell's equations, the forces between particles are mediated by electric and magnetic fields. The electromagnetic force (Lorentz force) on a particle with charge

q

due to a combination of electric field

E

and magnetic field (as given by the "B-field"

B

) is

$\boldsymbol{F} = q(\boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B}).$

This force imparts a momentum to the particle, so by Newton's second law the particle must impart a momentum to the electromagnetic fields.^[32]
In a vacuum, the momentum per unit volume is

$\boldsymbol{g} = \frac{1}{\mu_0 c^2}\boldsymbol{E}\times\boldsymbol{B}\,,$

where

μ 0

is the vacuum permeability and

c

is the speed of light. The momentum density is proportional to the Poynting vector

S

which gives the directional rate of energy transfer per unit area:^[32]^[33]

$\boldsymbol{g} = \frac{\boldsymbol{S}}{c^2}\,.$

If momentum is to be conserved in a volume

V

, changes in the momentum of matter through the Lorentz force must be balanced by changes in the momentum of the electromagnetic field and outflow of momentum. If

P mech

is the momentum of all the particles in a volume

V

, and the particles are treated as a continuum, then Newton's second law gives

$\frac{d \boldsymbol{P}_\text{mech}}{d t} = \int_V \left(\rho\boldsymbol{E} + \boldsymbol{J}\times\boldsymbol{B}\right) dV\,.$

The electromagnetic momentum is

$\boldsymbol{P}_\text{field} = \frac{1}{\mu_0c^2} \int_V \boldsymbol{E}\times\boldsymbol{B} dV\,,$

and the equation for conservation of energy for each component

i

of the momentum is

$\frac{d}{d t}\left(\boldsymbol{P}_\text{mech}+ \boldsymbol{P}_\text{field} \right) = \oint_S \sum_j T_{ij} n_j dS\,.$

The term on the right is an integral over the surface

S

representing momentum flow into and out of the volume, and

n j

is a component of the surface normal of

S

. The quantity

T i j

is called the Maxwell stress tensor, defined as

$T_{i j} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,.$ ^[32]

Media

The above results are for the microscopic Maxwell equations, applicable to electromagnetic forces in a vacuum (or on a very small scale in media). It is more difficult to define momentum density in media because the division into electromagnetic and mechanical is arbitrary. The definition of electromagnetic momentum density is modified to

$\boldsymbol{g} = \frac{1}{c^2}\boldsymbol{E}\times\boldsymbol{H} = \frac{\boldsymbol{S}}{c^2}\,,$

where the H-field

H

is related to the B-field and the magnetization

M

$\boldsymbol{B} = \mu_0 \left(\boldsymbol{H} + \boldsymbol{M}\right)\,.$

The electromagnetic stress tensor depends on the properties of the media.^[32]

Particle in field

If a charged particle

q

moves in an electromagnetic field, its kinematic momentum

m v

is not conserved. However, it has a canonical momentum that is conserved.

Lagrangian and Hamiltonian formulation

The kinetic momentum p is different to the canonical momentum P (synonymous with the generalized momentum) conjugate to the ordinary position coordinates r, because P includes a contribution from the electric potential φ(r, t) and vector potential A(r, t):^[21]

	Classical mechanics	Relativistic mechanics
Lagrangian	$\mathcal{L}=\frac{m}{2}\boldsymbol{\dot{r}}\cdot\boldsymbol{\dot{r}}+e\boldsymbol{A}\cdot\boldsymbol{\dot{r}}-e\phi\,\!$	$\mathcal{L} = -m\sqrt{1-\left(\frac{\dot{\boldsymbol{r}}}{c}\right)^2} + e \boldsymbol{A}\cdot\dot{\boldsymbol{r}} - e \phi \,\!$
Canonical momentum $\bold{P} = \frac{\partial L}{\partial \dot{\boldsymbol{r}} }$	$\boldsymbol{P} = m\boldsymbol{v} + e\boldsymbol{A}$	$\boldsymbol{P} = \frac{m\dot{\boldsymbol{r}}}{\sqrt{1-\left(\frac{\dot{\boldsymbol{r}}}{c}\right)^2}} + e\boldsymbol{A} \,\!$
Kinetic momentum $\bold{p} = m\boldsymbol{\dot{r}}$	$m\boldsymbol{v} = \boldsymbol{P} - e\boldsymbol{A}$	$\boldsymbol{P} - e\boldsymbol{A} = \frac{m\dot{\boldsymbol{r}}}{\sqrt{1-\left(\frac{\dot{\boldsymbol{r}}}{c}\right)^2}} \,\!$
Hamiltonian	$\begin{align} \mathcal{H} & = T + V \\ & = \frac{\boldsymbol{p}^2}{2m} + V \\ & = \frac{(\boldsymbol{P}-e\boldsymbol{A})^2}{2m} + e\phi \end{align}$	$\begin{align} \mathcal{H} & = \boldsymbol{P}\cdot\dot{\boldsymbol{r}} - L \\ & = {m\over \sqrt{1-\left(\frac{\dot{\boldsymbol{r}}}{c}\right)^2}} + e \phi \\ & = \sqrt{(\boldsymbol{P} -e\boldsymbol{A})^2 + (mc^2)^2} + e \phi \end{align}$

where ṙ = v is the velocity (see time derivative) and e is the electric charge of the particle. See also Electromagnetism (momentum). If neither φ nor A depends on position, P is conserved.^[6]
The classical Hamiltonian $\mathcal{H}$ for a particle in any field equals the total energy of the system - the kinetic energy T = p²/2m (where p² = p·p, see dot product) plus the potential energy V. For a particle in an electromagnetic field, the potential energy is V = eφ, and since the kinetic energy T always corresponds to the kinetic momentum p, replacing the kinetic momentum by the above equation (p = P − e''A) leads to the Hamiltonian in the table.
These Lagrangian and Hamiltonian expressons can derive the Lorentz force.

Canonical commutation relations

The kinetic momentum (p above) satisfies the commutation relation:^[21]

$\left [ p_j , p_k \right ] = \frac{i\hbar e}{c} \epsilon_{jk\ell } B_\ell$

where: j, k, ℓ are indices labelling vector components, B_ℓ is a component of the magnetic field, and ε_kjℓ is the Levi-Civita symbol, here in 3-dimensions.

Quantum mechanics

Further information: Momentum operator

In quantum mechanics, momentum is defined as an operator on the wave function. The Heisenberg uncertainty principle defines limits on how accurately the momentum and position of a single observable system can be known at once. In quantum mechanics, position and momentum are conjugate variables.
For a single particle described in the position basis the momentum operator can be written as

$\boldsymbol{p}={\hbar\over i}\nabla=-i\hbar\nabla\,,$

where ∇ is the gradient operator, ħ is the reduced Planck constant, and i is the imaginary unit. This is a commonly encountered form of the momentum operator, though the momentum operator in other bases can take other forms. For example, in the momentum basis the momentum operator is represented as

$\boldsymbol{p}\psi(p) = p\psi(p)\,,$

where the operator p acting on a wave function ψ(p) yields that wave function multiplied by the value p, in an analogous fashion to the way that the position operator acting on a wave function ψ(x) yields that wave function multiplied by the value x.
For both massive and massless objects, relativistic momentum is related to the de Broglie wavelength

λ

$p = h/\lambda\,,$

Electromagnetic radiation (including visible light, ultraviolet light, and radio waves) is carried by photons. Even though photons (the particle aspect of light) have no mass, they still carry momentum. This leads to applications such as the solar sail. The calculation of the momentum of light within dielectric media is somewhat controversial (see Abraham–Minkowski controversy).^[34]

Bismillah

Jumat, 01 November 2013

lanjutan

Relativistic mechanics

Lorentz invariance

Four-vector formulation

Classical electromagnetism

Vacuum

Media

Particle in field

Lagrangian and Hamiltonian formulation

Canonical commutation relations

Quantum mechanics

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