# Lorentz transformations for sine wave in random direction

Hello i have been struggling to derive Lorentz transformations for a sine wave, which is traveling at random direction. I started by prooving that phase $\phi$ is invariant for relativity and that equation $\phi = \phi'$ holds.

By using the above equation i am now trying to derive Lorentz transformations for angular frequency $\omega$, and all three components of the wave vector $k$, which are $k_x$, $k_y$ and $k_z$.

This is my attempt:

\begin{align}\phi' &= \phi\\ \omega'\Delta t' + k' \Delta r' &= \omega \Delta t + k \Delta r\\ \omega'\Delta t' + [{k_x}' , {k_y}' , {k_z}'] [\Delta x' , \Delta y' , \Delta z'] &= \omega \Delta t + [k_x , k_y , k_z][\Delta x , \Delta y , \Delta z]\\ \omega' \Delta t' + {k_x}'\Delta x' + {k_y}' \Delta y' + {k_z}' \Delta z' &= \omega \Delta t + k_x \Delta x + k_y \Delta y + k_z \Delta z \end{align} Now with just the left side: \begin{gather} \omega' \gamma \left(\Delta t - \Delta x \frac{u}{c^2}\right) + {k_x}' \gamma \Bigl(\Delta x - u\Delta t \Bigl) + {k_y}' \Delta y + {k_z}' \Delta z\\ \gamma \left(\omega' \Delta t - \omega' \Delta x \frac{u}{c^2}\right) + \gamma\Bigl({k_x}' \Delta x - {k_x}' u \Delta t \Bigl) + {k_y}' \Delta y + {k_z}' \Delta z\\ \gamma \left(\omega' \Delta t - {k_x}'c\, \, c \Delta t \, \frac{u}{c^ 2}\right) + \gamma \Bigl({k_x}' \Delta x - \frac{\omega'}{c} u\frac{\Delta x}{c} \Bigl) + {k_y}' \Delta y + {k_z}' \Delta z\\ \Delta t \, \gamma \Bigl(\omega' - {k_x}' u \Bigl) + \Delta x \, \gamma \Bigl({k_x}' - \omega' \frac{u}{c^2} \Bigl) + {k_y}' \Delta y + {k_z}' \Delta z \end{gather}

From this I can write down the Lorentz transformations.

$$\begin{split} \gamma\Bigl(\omega' - {k_x}' u \Bigl) &= \omega\\ \gamma \Bigl({k_x}' - \omega' \frac{u}{c^2} \Bigl) &= k_x\\ {k_y}' &= k_y\\ {k_z}' &= k_z\\ \end{split}$$

My professor said that my signs are wrong, but what am i doing wrong?

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 I edited your equation to make it fit in the post. (For everyone's information, formatting problems like that aren't a valid reason to delete a question.) – David Zaslavsky♦ Oct 21 '12 at 16:46

 I don't think i understand. Should i put in the Lorenz transformations for $\Delta t$ and $\Delta x$ (instead of $\Delta t'$ and $\Delta x'$) in the 5th row? – 71GA Sep 24 '12 at 11:23 Your placing your boosts on the wrong side of your derivation, you are boosting x -> x' for instance, not the other way around. – Hal Swyers Sep 24 '12 at 12:04 p.s. It really is spelled Lorentz and not Lorenz, there actually are a lot of similarly named people that have different math and physics concepts named after them, and the naming convention needs to be followed to prevent confusion. – Hal Swyers Sep 24 '12 at 12:08 I think that if i place boosts on the right side instead odf the left side i would only get Lorentz tr. instead of reverse Lorentz transformation. How does this solve my sign issue? – 71GA Sep 24 '12 at 12:18 If you take all your primes(') on the LHS and put them on the RHS and change nothing else, you have solved the problem. – Hal Swyers Sep 24 '12 at 12:29