Wave vector

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A wave vector is a vector representation of a wave. The wave vector has magnitude indicating wavenumber (reciprocal of wavelength), and the direction of the vector indicates the direction of wave propagation.

The wave vector is most useful for generalizing the equation of a single wave into a description of a family of waves. As long as the family of waves all travel in the same direction and with the same wavelength, a single wave vector is valid for the entire family. The most common case of a family of waves that meets these requirements is the plane wave, in which the family of waves is also coherent, i.e. all the waves have the same phase.

For example, a common representation of a single wave at a single point in space is:

$\psi \left(t\right) = A \cos \left(\varphi + \omega t\right)$

where A is the amplitude, ω is the angular frequency, and φ is the starting phase of the wave (the independent variable t is time).

In order to generalize the equation to all points in the one-dimensional space of the direction of propagation, we add in an additional phase offset term:

$\psi \left(t , z\right) = A \cos \left(\varphi + k z + \omega t\right)$

where k is the wavenumber (2π/λ) and the new independent variable z is the distance along the wave.

Now, as long as we are dealing with a simple family of waves, with identical direction, wavelength, and phase (i.e. a plane wave), we can easily extend the formula by substituting the wave vector k for the wavenumber k, and the location in space vector r for the variable z:

$\psi \left(t , {\mathbf r} \right) = A \cos \left(\varphi + {\mathbf k} \cdot {\mathbf r} + \omega t\right)$

Wavevector is also closely related to wavelength by the following formula

$k = \frac{2\pi}{\lambda}$

1 In special relativity
- 1.1 Lorentz transform
  - 1.1.1 Source moving away
  - 1.1.2 Source moving towards
2 References

[edit] In special relativity

A wave packet of nearly monochromatic light can be characterized by the wave vector

$k^\mu = \left(\frac{\omega}{c}, \vec{k} \right) \,$

which, when written out explicitly in its contravariant and covariant forms is

$k^\mu = \left(\frac{\omega}{c}, k^1, k^2, k^3 \right)\,$ and

$k_\mu = \left(\frac{\omega}{c}, -k_1, -k_2, -k_3 \right) . \,$

The magnitude of this wave vector is then

$k^2 = k^\mu k_\mu = k^0 k_0 - k^1 k_1 - k^2 k_2 - k^3 k_3 \,$

$=\frac{\omega^2}{c^2} - \vec{k}^2 = 0. \,$

That last step where it equals zero, is a result of the fact that, for light in vacuum,

$k = \frac{\omega}{c}. \,$

[edit] Lorentz transform

Taking the Lorentz transform of the wave vector 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 transform as follows. Note that the source is in a frame S^s and earth is in the observing frame, S^obs. 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 $μ = 0$ component results in

$k^{0}_s = \Lambda^0_0 k^0_{\mathrm{obs}} + \Lambda^0_1 k^1_{\mathrm{obs}} + \Lambda^0_2 k^2_{\mathrm{obs}} + \Lambda^0_3 k^3_{\mathrm{obs}} \,$

$\frac{\omega_s}{c} \,$	$= \gamma \frac{\omega_{\mathrm{obs}}}{c} - \beta \gamma k^1_{\mathrm{obs}} \,$
	$\quad = \gamma \frac{\omega_{\mathrm{obs}}}{c} - \beta \gamma \frac{\omega_{\mathrm{obs}}}{c} \cos \theta. \,$ where $\cos \theta \,$ is the direction cosine of $k 1$ wrt $k 0, k 1 = k 0 cosθ.$

$\frac{\omega_{\mathrm{obs}}}{\omega_s} = \frac{1}{\gamma (1 - \beta \cos \theta)} \,$

[edit] Source moving away

As an example, to apply this to a situation where the source is moving directly away from the observer ( $θ = π$ ), this becomes:

$\frac{\omega_{\mathrm{obs}}}{\omega_s} = \frac{1}{\gamma (1 + \beta)} = \frac{\sqrt{1-\beta^2}}{1+\beta} = \frac{\sqrt{(1+\beta)(1-\beta)}}{1+\beta} = \frac{\sqrt{1-\beta}}{\sqrt{1+\beta}} \,$