The wave is $$\bar{E} = E_{0} \ \mathrm{sin}\left(\frac{2\pi z}{\lambda} + \omega t\right) \bar{i} + E_{0} \ \mathrm{cos} \left(\frac{2 \pi z}{\lambda}+\omega t\right) \bar{j}.$$

Let's simplify with $z = 1$. Now the $xy$-axis is defined by the parametrization $$\left(\mathrm{sin}\left(\frac{2\pi }{\lambda}+\omega t \right), \mathrm{cos}\left(\frac{2\pi }{\lambda} + \omega t\right)\right),$$ where $t$ stands for time and $\lambda$ is the wavelength. This parametrization satisfies the equation $1^2=x^{2}+y^{2}$, a circle.

Now, let's variate the value of $z$. We know now that it cannot move into $x$ or $y$ coordinates, or do we? Not really, the latter simplification is naive — $x-y$ parametrization depends on the dimension $z$ — but can we see something from it? If so, how to proceed now?

The solution is that the wave moves along the $z$-axis to the negative direction as $t$ increases, a thing I cannot see.

The way I am trying to solve this kind of problems is:

  1. Parametrize the equation
  2. suppose other things constant and change one dimension, observe
  3. check other variable

However, I find it hard to parametrize the $z$, so I'm a bit lost. How can I visualize the wave with pen-and-paper?

  • 1
    $\begingroup$ "Let's simplify with $z=1$." Almost certainly not what you want to do, as this is probably meant to be $E = E(z,t) = \dots$. You don't simplify out one of the independent variables. $\endgroup$ May 19, 2011 at 17:35
  • $\begingroup$ @dmckee: sure I won't, it tried a see what the wave looks in different situation but it did not help. I am trying to do it that way because I am unable to parametrize the equation, more here. $\endgroup$
    – hhh
    May 19, 2011 at 17:37

1 Answer 1


Would you agree that $\vec{E}$ depends only on $\frac{2 \pi z}{\lambda} + \omega t$ (taking $E_0$ to be a constant)?

If so, we can imagine picking some spot it space and time, taking note of the value of $\vec{E}$ at that point and looking to see how we have to move to keep the value constant in time

$$ \frac{2 \pi z}{\lambda} + \omega t = C $$

where C is determined entirely by our initial choice of space--time location. So:

$$ z = z(t) = \frac{ \lambda }{2 \pi} \left( C - \omega t \right) $$

represents a locus of $z$-positions as a function of time where $\vec{E}$ continues to have the same value it had at our starting point. And those positions move in the negative $z$ direction as time increases.

Question for the studuent: how fast do they move?

You should be able to answer by inspection.


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