I've seen these two forms from multiple sources for solutions to Maxwell's equations: $$ \cos(kz - \omega t) $$ and $$ \cos(\omega t - \vec{k} \cdot \vec{r}) $$

The first one shows a wave travelling in the positive $z$ direction. What does the second one show? What's the significance of having swapping the signs on the two terms? i.e .$$\cos(-kz + \omega t)$$

  • $\begingroup$ What is $\cos(-x)$ equal to? $\endgroup$ – David Aug 13 at 8:30
  • $\begingroup$ Yes, missed that thanks. What about the main part of the question can you assist? $\endgroup$ – Natalie Johnson Aug 13 at 8:33
  • $\begingroup$ My comment should address your question about swapping the signs no? Or is it the $k \cdot r$ vs $kz$ that bothers you? $\endgroup$ – David Aug 13 at 8:36
  • $\begingroup$ Yes unclear. Also, if time is increasing are k and z both increasing and positive? $\endgroup$ – Natalie Johnson Aug 13 at 9:04
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/questions/351975/… $\endgroup$ – Gulce Kardes Aug 13 at 9:26

$k = \dfrac {2\pi}{\lambda}$ where $\lambda$ is the wavelength of the wave.

$\cos(kz - \omega t)$

is the equation of a (plane wave) travelling in the positive $z$ direction.
You can think of it as $\cos(\vec k \cdot \vec r - \omega t)$ where $\vec k= k \,\hat z$ and $\vec r= z \,\hat z$.

The direction of the k-vector gives the direction of travel of the wave or the opposite direction if the form of the equation is $\cos(kz + \omega t)$.

$\cos(\omega t - \vec{k} \cdot \vec{r})$

is the equation of a wave in three dimensions travelling in the k-vector direction with $\vec k \cdot \vec r = k_{\rm x} \, x + k_{\rm y} \, y + k_{\rm z} \, z$

This two-dimensional plane wave visualisation may help you understand what is going on.

$\cos(kz - \omega t)$ and $\cos(-kz + \omega t)= \cos (-[kz - \omega t])$

describe the same wave travelling in the positive z-direction because $\cos(-kz + \omega t)= \cos (-[kz - \omega t]) = \cos(kz - \omega t)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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