Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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{equation} \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} \end{equation}

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

share|improve this question
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 Z Oct 21 '12 at 16:46
add comment

1 Answer

up vote 4 down vote accepted

The problem is actually easy to spot if you compare to the entry for Lorentz transformation. Your primes (') should actually be on the other side. Otherwise it looks like you are trying to do an inverse transform with the wrong sign convention. Paying attention to signs might at first seem trivial, but is important when one realizes that one could inadvertently add a negative sign to the the determinant, which indicates a reflection in spacetime (see Diagrammatica page 2). Since a reflection for large objects would be akin to taking your mirror image (changing you from left to right handed), this is generally forbidden for classical objects.

share|improve this answer
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
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.