nLab
momentum

Context

Physics

, ,

Surveys, textbooks and lecture notes

  • ,


,

, ,

    • , , , ,

      • ,

      • ,

      • ,

        • ,

      • and
    • Axiomatizations

          • ,
        • -theorem

    • Tools

      • ,

        • ,

        • ,
    • Structural phenomena

    • Types of quantum field thories

        • ,

        • , ,

        • examples

          • ,
          • ,
        • , , , ,

        • , ,

this entry needs attention

Momentum is, according to Newton, the ‘amount of motion’ in an object. For particles it is proportional to the object's mass and to its velocity, and thus measured in units of mass times velocity. For photons it is proportional to the frequency of the photon.

In Lagrangian mechanics (including relativistic and quantum versions), to every (generalized) coordinate q iq_i one associates a (generalized) momentum

p i=Lq˙ i p_i = \frac{\partial L}{\partial \dot{q}^i}

where LL is the Lagrangian of the system. More generally, we can say that p ip_i is the coordinate of dq i\mathrm{d}q^i in the action form.

If LL takes the form 12 im iq˙ i 2U(q 1,,q n)\frac{1}{2} \sum_i m_i \dot{q}_i^2 - U(q_1,\ldots,q_n), then we have p i=m iq˙ ip_i = m_i \dot{q}_i. Thus, the linear momentum of a point particle in Newtonian mechanics is traditionally defined as the intrinsic mass mm times the velocity q˙\vec{\dot{q}}. However, this breaks down in some situations:

In Hamiltonian mechanics, the momentum coordinates are half of the basic coordinates on phase space. If phase space is derived as the cotangent bundle of a configuration space, then the momentum coordinates are the new coordinates on the cotangent bundle, dual to the original coordinates on configuration space; we have the symplectic form ip idq i\sum_i p_i \mathrm{d}q^i. However, Hamiltonian mechanics cares only about phase space as a symplectic manifold with a Hamiltonian; there is no inherent distinction between position and momentum coordinates, and you can even make a canonical transformation that swaps them (up to a minus sign).

In quantum mechanics, the canonical quantization? process formally replaces momentum by ix,-i \hbar \frac{\partial}{\partial x}, where xx is position. (Equivalently, in momentum space?, canonical quantization replaces xx by ip.)i \hbar \frac{\partial}{\partial p}.) In this situation, position and momentum fail to commute.

There is a generalization of momentum in symplectic geometry, so called moment map.

on

p,1 ψ k x exp(ik μx μ) (x,x 0) exp(ikx+ik 0x 0) (x,ct) exp(ikxiωt) \array{ \mathbb{R}^{p,1} &\overset{\psi_k}{\longrightarrow}& \mathbb{C} \\ x &\mapsto& \exp\left( \, i k_\mu x^\mu \, \right) \\ (\vec x, x^0) &\mapsto& \exp\left( \, i \vec k \cdot \vec x + i k_0 x^0 \, \right) \\ (\vec x, c t) &\mapsto& \exp\left( \, i \vec k \cdot \vec x - i \omega t \, \right) }
symbolname
cc
\hbar
\,\,
mm
mc\frac{\hbar}{m c}
\,\,
kk, k\vec k
λ=2π/|k|\lambda = 2\pi/{\vert \vec k \vert}
|k|=2π/λ{\vert \vec k \vert} = 2\pi/\lambda
ωk 0c=k 0c=2πν\omega \coloneqq k^0 c = -k_0 c = 2\pi \nu
ν=ω/2π\nu = \omega / 2 \pi
p=kp = \hbar k, p=k\vec p = \hbar \vec k
E=ωE = \hbar \omega
ω(k)=ck 2+(mc) 2\omega(\vec k) = c \sqrt{ \vec k^2 + \left(\frac{m c}{\hbar}\right)^2 }
E(p)=c 2p 2+(mc 2) 2E(\vec p) = \sqrt{ c^2 \vec p^2 + (m c^2)^2 }