Wave vector

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A wave vector is a vector that specifies the wavenumber and direction of propagation for a wave. The magnitude of the wave vector indicates the wavenumber. The orientation of the wave vector indicates the direction of wave propagation.

For example consider a plane wave. A common representation of the oscillation at time (t) and a single point in space (z) along the direction of propagation is:

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

where A is the amplitude, φ is the starting phase of the wave, k is the angular wavenumber, and ω is the angular frequency. We can easily extend the formula by substituting the dot product of the wave vector k and the position vector r for the scalar product of the wavenumber k and the variable z as follows:

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

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 covariant and contravariant 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,

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

[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

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

$\frac{\omega_s}{c} \,$	$= \gamma \frac{\omega_{\mathrm{obs}}}{c} - \beta \gamma k^1_{\mathrm{obs}} \,$
	$\quad = \gamma \frac{\omega_{\mathrm{obs}}}{c} - \beta \gamma \frac{\omega_{\mathrm{obs}}}{c} \cos \theta. \,$

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

[edit] Source moving away

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

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