Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Relativistic Doppler formula.

Posted by peeterjoot on July 6, 2009

[Click here for a PDF of this post with nicer formatting]

Transform of angular velocity four vector.

It was possible to derive the Lorentz boost matrix by requiring that the wave equation operator

\begin{aligned}\nabla^2 = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \boldsymbol{\nabla}^2 \end{aligned}

retain its form under linear transformation ([1]). Applying spatial Fourier transforms ([2]), one finds that solutions to the wave equation

\begin{aligned}\nabla^2 \psi(t,\mathbf{x}) = 0 \end{aligned}

Have the form

\begin{aligned}\psi(t, \mathbf{x}) = \int A(\mathbf{k}) e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} d^3 k \end{aligned}

Provided that \omega = \pm c {\left\lvert{\mathbf{k}}\right\rvert}. Wave equation solutions can therefore be thought of as continuously weighted superpositions of constrained fundamental solutions

\begin{aligned}\psi &= e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ c^2 \mathbf{k}^2 &= \omega^2 \end{aligned}

The constraint on frequency and wave number has the look of a Lorentz square

\begin{aligned}\omega^2 - c^2 \mathbf{k}^2 = 0 \end{aligned}

Which suggests that in additional to the spacetime vector

\begin{aligned}X = (ct, \mathbf{x}) = x^\mu \gamma_\mu \end{aligned}

evident in the wave equation fundamental solution, we also have a frequency-wavenumber four vector

\begin{aligned}K = (\omega/c, \mathbf{k}) = k^\mu \gamma_\mu \end{aligned}

The pair of four vectors above allow the fundamental solutions to be put explicitly into covariant form

\begin{aligned}K \cdot X = \omega t - \mathbf{k} \cdot \mathbf{x} = k_\mu x^\mu \end{aligned}

\begin{aligned}\psi = e^{-i K \cdot X} \end{aligned}

Let’s also examine the transformation properties of this fundamental solution, and see as a side effect that K
has transforms appropriately as a four vector.

\begin{aligned}0 &= \nabla^2 \psi(t,\mathbf{x}) \\ &= {\nabla'}^2 \psi(t',\mathbf{x}') \\ &= {\nabla'}^2 e^{i(\mathbf{x}' \cdot \mathbf{k}' - \omega' t')} \\ &= -\left(\frac{{\omega'}^2}{c^2} - {\mathbf{k}'}^2 \right) e^{i(\mathbf{x}' \cdot \mathbf{k}' - \omega' t')} \\  \end{aligned}

We therefore have the same form of frequency wave number constraint in the transformed frame (if we require that
the wave function for light is unchanged under transformation)

\begin{aligned}{\omega'}^2 = c^2 {\mathbf{k}'}^2  \end{aligned}

Writing this as

\begin{aligned}0 = {\omega}^2 - c^2 {\mathbf{k}}^2 = {\omega'}^2 - c^2 {\mathbf{k}'}^2  \end{aligned}

singles out the Lorentz invariant nature of the (\omega, \mathbf{k}) pairing, and we conclude that this pairing
does indeed transform as a four vector.

Application of one dimensional boost.

Having attempted to justify the four vector nature of the wave number vector K, now move on to application of a boost along the x-axis to this vector.

\begin{aligned}\begin{bmatrix}\omega' \\ c k' \\ \end{bmatrix}&=\gamma\begin{bmatrix}1 & -\beta \\ -\beta& 1 \\ \end{bmatrix}\begin{bmatrix}\omega \\ c k \\ \end{bmatrix} \\ &=\begin{bmatrix}\omega - v k \\ c k - \beta \omega\end{bmatrix}  \end{aligned}

We can take ratios of the frequencies if we make use of the dependency between \omega and k. Namely, \omega = \pm c k. We then have

\begin{aligned}\frac{\omega'}{\omega}&= \gamma(1 \mp \beta) \\ &= \frac{1 \mp \beta}{\sqrt{1 - \beta^2}} \\ &= \frac{1 \mp \beta}{\sqrt{1 - \beta}\sqrt{1 + \beta}} \\  \end{aligned}

For the positive angular frequency this is

\begin{aligned}\frac{\omega'}{\omega}&= \frac{\sqrt{1 - \beta}}{\sqrt{1 + \beta}} \\  \end{aligned}

and for the negative frequency the reciprocal.

Deriving this with a Lorentz boost is much simpler than the time dilation argument in wikipedia doppler article ([3]). EDIT: Later found exactly the above boost argument in the wiki k-vector article ([4]).

What’s missing here is putting this in a physical context properly with source and reciever frequencies spelled out. That would make this more than just math.


[1] Peeter Joot. Wave equation based Lorentz transformation derivation [online].

[2] Peeter Joot. Fourier transform solutions to the wave equation [online].

[3] Wikipedia. Relativistic doppler effect — wikipedia, the free encyclopedia [online]. 2009. [Online; accessed 26-June-2009].

[4] Wikipedia. Wave vector — wikipedia, the free encyclopedia [online]. 2009. [Online; accessed 30-June-2009].


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: