nLab plane wave

Contents

Contents

Idea

Let n∈ℕn \in \mathbb{N} and write ℝ n\mathbb{R}^n for the Cartesian space of dimension nn. Then a plane wave on ℝ n\mathbb{R}^n is a function

ℝ n⟶ℂ \mathbb{R}^n \longrightarrow \mathbb{C}

given by the exponential function with complex argument of the form

x→↦Ae 2πik→⋅x→. \vec x \;\mapsto\; A e^{2 \pi i \vec k \cdot \vec x} \,.

Here k→∈ℝ n\vec k \in \mathbb{R}^n is the wave vector of this plane wave (and A∈ℂA \in \mathbb{C} is its amplitude).

In Fourier analysis over Cartesian space, the Fourier transform expresses every function with rapidly decreasing partial derivatives as a superposition of plane waves.

If here ℝ n≃ℝ p,1\mathbb{R}^n \simeq \mathbb{R}^{p,1} is identified with Minkowski spacetime with canonical coordinates labeled (x 0,x 1,⋯x p)(x^0, x^1, \cdots x^p), then the 0-component of the wave vector

ν≔k 0 \nu \coloneqq k_0

is called the frequency of the wave (in this chosen coordinate system). If in this situation the wave vector satisfies k μk μ=0k_\mu k^\mu = 0, then the plane wave is a solution to the wave equation on Minkowski spacetime. If more generally it satisfies k μk μ+m 2=0k_\mu k^\mu + m^2 = 0 for some m 2∈ℝm^2 \in \mathbb{R}, the it is a solution to the Klein-Gordon equation on Minkowski spacetime.

plane waves on Minkowski spacetime

ℝ p,1 ⟶ψ k ℂ x ↦ exp(ik μx μ) (x→,x 0) ↦ exp(ik→⋅x→+ik 0x 0) (x→,ct) ↦ exp(ik→⋅x→−iωt) \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
ccspeed of light
ℏ\hbarPlanck's constant
\,\,
mmmass
ℏmc\frac{\hbar}{m c}Compton wavelength
\,\,
kk, k→\vec kwave vector
λ=2π/|k→|\lambda = 2\pi/{\vert \vec k \vert}wave length
|k→|=2π/λ{\vert \vec k \vert} = 2\pi/\lambdawave number
ω≔k 0c=−k 0c=2πν\omega \coloneqq k^0 c = -k_0 c = 2\pi \nuangular frequency
ν=ω/2π\nu = \omega / 2 \pifrequency
p=ℏkp = \hbar k, p→=ℏk→\vec p = \hbar \vec kmomentum
E=ℏωE = \hbar \omegaenergy
ω(k→)=ck→ 2+(mcℏ) 2\omega(\vec k) = c \sqrt{ \vec k^2 + \left(\frac{m c}{\hbar}\right)^2 }Klein-Gordon dispersion relation
E(p→)=c 2p→ 2+(mc 2) 2E(\vec p) = \sqrt{ c^2 \vec p^2 + (m c^2)^2 }energy-momentum relation

References

See also

Last revised on March 31, 2026 at 11:21:56. See the history of this page for a list of all contributions to it.