A moving wave surface in special relativity may be regarded as a hypersurface (a 3D subspace) in spacetime, formed by all the events passed by the wave surface. A wavetrain (denoted by some variable ) can be regarded as a one-parameter family of such hypersurfaces in spacetime. This variable is a scalar function of position in spacetime. The derivative of this scalar is a vector that characterizes the wave, the four-wavevector. The four-wavevector is a wave
four-vector that is defined, in
Minkowski coordinates, as: :K^\mu = \left(\frac{\omega}{c}, \vec{k}\right) = \left(\frac{\omega}{c}, \frac{\omega}{v_p}\hat{n}\right) = \left(\frac{2 \pi}{cT}, \frac{2 \pi \hat{n}}{\lambda}\right) \, where the angular frequency \tfrac{\omega}{c} is the temporal component, and the wavenumber vector \vec{k} is the spatial component. Alternately, the wavenumber can be written as the angular frequency divided by the
phase-velocity , or in terms of inverse period and inverse wavelength . When written out explicitly its
contravariant and
covariant forms are: :\begin{align} K^\mu &= \left(\frac{\omega}{c}, k_x, k_y, k_z \right)\, \\[4pt] K_\mu &= \left(\frac{\omega}{c}, -k_x, -k_y, -k_z \right) \end{align} In general, the
Lorentz scalar magnitude of the wave four-vector is: :K^\mu K_\mu = \left(\frac{\omega}{c}\right)^2 - k_x^2 - k_y^2 - k_z^2 = \left(\frac{\omega_o}{c}\right)^2 = \left(\frac{m_o c}{\hbar}\right)^2 The four-wavevector is
null for
massless (photonic) particles, where the rest mass m_o = 0 An example of a null four-wavevector would be a beam of coherent,
monochromatic light, which has phase-velocity v_p = c :K^\mu = \left(\frac{\omega}{c}, \vec{k}\right) = \left(\frac{\omega}{c}, \frac{\omega}{c}\hat{n}\right) = \frac{\omega}{c}\left(1, \hat{n}\right) \, {for light-like/null} which would have the following relation between the frequency and the magnitude of the spatial part of the four-wavevector: :K^\mu K_\mu = \left(\frac{\omega}{c}\right)^2 - k_x^2 - k_y^2 - k_z^2 = 0 {for light-like/null} The four-wavevector is related to the
four-momentum as follows: :P^\mu = \left(\frac{E}{c}, \vec{p}\right) = \hbar K^\mu = \hbar\left(\frac{\omega}{c}, \vec{k}\right) The four-wavevector is related to the
four-frequency as follows: :K^\mu = \left(\frac{\omega}{c}, \vec{k}\right) = \left(\frac{2 \pi}{c}\right)N^\mu = \left(\frac{2 \pi}{c}\right)\left(\nu, \nu \vec{n}\right) The four-wavevector is related to the
four-velocity as follows: :K^\mu = \left(\frac{\omega}{c}, \vec{k}\right) = \left(\frac{\omega_o}{c^2}\right)U^\mu = \left(\frac{\omega_o}{c^2}\right) \gamma \left(c, \vec{u}\right)
Lorentz transformation Taking the
Lorentz transformation of the four-wavevector is one way to derive the
relativistic Doppler effect. The Lorentz matrix is defined as :\Lambda = \begin{pmatrix} \gamma & -\beta \gamma & \ 0 \ & \ 0 \ \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} In the situation where light is being emitted by a fast moving source and one would like to know the frequency of light detected in an earth (lab) frame, we would apply the Lorentz transformation as follows. Note that the source is in a frame and earth is in the observing frame, . Applying the Lorentz transformation to the wave vector :k^{\mu}_s = \Lambda^\mu_\nu k^\nu_{\mathrm{obs}} and choosing just to look at the \mu = 0 component results in :\begin{align} k^{0}_s &= \Lambda^0_0 k^0_{\mathrm{obs}} + \Lambda^0_1 k^1_{\mathrm{obs}} + \Lambda^0_2 k^2_{\mathrm{obs}} + \Lambda^0_3 k^3_{\mathrm{obs}} \\[3pt] \frac{\omega_s}{c} &= \gamma \frac{\omega_{\mathrm{obs}}}{c} - \beta \gamma k^1_{\mathrm{obs}} \\ &= \gamma \frac{\omega_{\mathrm{obs}}}{c} - \beta \gamma \frac{\omega_{\mathrm{obs}}}{c} \cos \theta. \end{align} where \cos \theta is the direction cosine of k^1 with respect to k^0, k^1 = k^0 \cos \theta. So :
Source moving away (redshift) As an example, to apply this to a situation where the source is moving directly away from the observer (\theta=\pi), this becomes: :\frac{\omega_{\mathrm{obs}}}{\omega_s} = \frac{1}{\gamma (1 + \beta)} = \frac{\sqrt{1-\beta^2}}{1+\beta} = \frac{\sqrt{(1+\beta)(1-\beta)}}{1+\beta} = \frac{\sqrt{1-\beta}}{\sqrt{1+\beta}}
Source moving towards (blueshift) To apply this to a situation where the source is moving straight towards the observer (), this becomes: :\frac{\omega_{\mathrm{obs}}}{\omega_s} = \frac{1}{\gamma (1 - \beta)} = \frac{\sqrt{1-\beta^2}}{1-\beta} = \frac{\sqrt{(1+\beta)(1-\beta)}}{1-\beta} = \frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}
Source moving tangentially (transverse Doppler effect) To apply this to a situation where the source is moving transversely with respect to the observer (), this becomes: :\frac{\omega_{\mathrm{obs}}}{\omega_s} = \frac{1}{\gamma (1 - 0)} = \frac{1}{\gamma} ==See also==