# Lorentz, waves, Einstein and bodies: transformations +/- gamma

1. Mar 25, 2014

### ANvH

I am using a wikipedia page, Derivation of the Lorentz transformations and a lot of historical papers. To follow through I came up with my own transformations that do not contain the gamma factor:

$x^{'}=x-\beta ct$​
$t^{'}=t-\beta \frac{x}{c}$​

When applying them to a waveform

$\omega t^{'}-kx^{'}=(1+\beta)\omega t-(1+\beta)kx$​

The speed of the wave is then
$u=\frac{\omega}{k}\frac{1+\beta}{1+\beta}=c$​

So for a waveform the above transformations suffice given the speed of the Doppler shifted wave is equal to the non Doppler shifted wave. However, using the wikipedia page: http://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations, the transformations need the gamma factor when I am following through with the above equations. For the waveform this will increase the Doppler shifts:

$\omega t^{'}-kx^{'}=\gamma (1+\beta)\omega t-\gamma(1+\beta)kx$​

The speed of the wave is then
$u=\frac{\omega}{k}\frac{\gamma(1+\beta)}{\gamma(1+\beta)}=c$​

I would conclude that the transformations without the gamma factor is sufficient to transform the waveform. The transformations with the gamma factor is apparently necessary when the waveform is not utilized to test the validity of the transformations, but comparing the above with the reasoning in Wikipedia seems to be confusing.

What am I thinking wrong?

2. Mar 25, 2014

### robphy

Note that the determinant of your transformation is not equal to 1.

3. Mar 26, 2014

### Bill_K

Lacking the gamma, your transformation includes a scale transformation. It preserves null vectors but does not preserve the length of spacelike or timelike four-vectors. Especially it does not preserve the rest mass of a particle.

4. Mar 26, 2014

### ANvH

Ok, the Lorentz transformation matrix

\begin{align}
A=\gamma\begin{pmatrix}1 & -\beta/c\\ -\beta c & 1\end{pmatrix}
\end{align}

gives a determinant

det$A=\gamma(1-\beta^{2})=\frac{(1-\beta)(1+\beta)}{\sqrt{(1-\beta)(1+\beta)}}=\sqrt{(1-\beta)(1+\beta)}$

5. Mar 26, 2014

### Bill_K

det$A=\gamma^2(1-\beta^{2})$

6. Mar 26, 2014

### ANvH

gosh..., thanks.