# nLab momentum

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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 $q_i$ one associates a (generalized) momentum

$p_i = \frac{\partial L}{\partial \dot{q}^i}$

where $L$ is the Lagrangian of the system. More generally, we can say that $p_i$ is the coordinate of $\mathrm{d}q^i$ in the action form.

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

• Relativistically, the velocity factor is more complicated (tending to $\infty$ as velocity tends to the speed of light). However, it is possible to keep $p = m v$ using a velocity-dependent relativistic mass (especially since the mass may depend on other dynamical considerations).
• In the presence of velocity-dependent forces (notably those produced by a magnetic field), there is an additional position-dependent term.
• In nonlinear coordinates (such as for angular momentum), one not only needs to use a nonlinear velocity (such as angular velocity) but also replace mass with something more complicated (such as moment of inertia) which is often position-dependent.

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 $\sum_i p_i \mathrm{d}q^i$. 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 $-i \hbar \frac{\partial}{\partial x},$ where $x$ is position. (Equivalently, in momentum space?, canonical quantization replaces $x$ by $i \hbar \frac{\partial}{\partial p}.)$ In this situation, position and momentum fail to commute.

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

on

$\array{ \mathbb{R}^{p,1} &\overset{\psi_k}{\longrightarrow}& \mathbb{C} \\ x &\mapsto& \exp\left( \, i k_\mu x^\mu \, \right) \\ (\vec x, x^0) &\mapsto& \exp\left( \, i \vec k \cdot \vec x + i k_0 x^0 \, \right) \\ (\vec x, c t) &\mapsto& \exp\left( \, i \vec k \cdot \vec x - i \omega t \, \right) }$
symbolname
$c$
$\hbar$
$\,$$\,$
$m$
$\frac{\hbar}{m c}$
$\,$$\,$
$k$, $\vec k$
$\lambda = 2\pi/{\vert \vec k \vert}$
${\vert \vec k \vert} = 2\pi/\lambda$
$\omega \coloneqq k^0 c = -k_0 c = 2\pi \nu$
$\nu = \omega / 2 \pi$
$p = \hbar k$, $\vec p = \hbar \vec k$
$E = \hbar \omega$
$\omega(\vec k) = c \sqrt{ \vec k^2 + \left(\frac{m c}{\hbar}\right)^2 }$
$E(\vec p) = \sqrt{ c^2 \vec p^2 + (m c^2)^2 }$

Last revised on November 7, 2017 at 08:46:13. See the history of this page for a list of all contributions to it.