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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 .$$\cos(-kz + \omega t)$$

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  • $\begingroup$ 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
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$k = \dfrac {2\pi}{\lambda}$ 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 $\cos(-kz + \omega t)= \cos (-[kz - \omega t]) = \cos(kz - \omega t)$

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