It's a well known fact that wind will cause very big waves if blowing against a strong current, for example in the gulf stream off the US coast, or in the bay of San Francisco with a very strong (5 knots+) tidal current.

Notably, a 30 knots wind against a 3 knots current will create way bigger waves than a 33 knots wind against no current at all, even-though the relative speeds are obviously the same.

What is the explanation of this phenomenon?

  • $\begingroup$ If it was a narrow current, faster in the middle and slower at the edges, there could be a focusing effect, since the wavefronts at the edges are moving faster in the upstream direction. $\endgroup$ Mar 11 '18 at 22:32
  • 1
    $\begingroup$ Interesting question. It would be great if you could provide some references documenting the phenomena. $\endgroup$ Mar 12 '18 at 3:47
  • 1
    $\begingroup$ Wide currents like the Gulf Stream seem to be made up of many narrower swirls. $\endgroup$ Mar 12 '18 at 5:14
  • 1
    $\begingroup$ You can also check the info in: earthscience.stackexchange.com/questions/2860/… $\endgroup$
    – arkaia
    Mar 12 '18 at 12:37
  • 2
    $\begingroup$ This paper has some mathematical theory and references. On p.4 they mention a channel width of 500 m. Their Fig. 1 is what I was imagining. $\endgroup$ Mar 12 '18 at 14:48

There are two physical processes at play here:

  • Wind speed relative to the ocean surface
  • Wave focusing and blocking in opposing current

Wind speed relative to the ocean surface

As you describe in your question, wind speed relative to the ocean surface varies depending on the current orientation and magnitude. The rate of wave growth is proportional to the wind speed $U$ in direction of the wave relative to the phase speed of the wave $c_p$, squared:

$$ S_{in} \sim \left( U - c_p \right)^2 $$

However, the wave phase speed is modulated by the current:

$$ c_p = c_{p0} + u $$ where $c_{p0}$ is the intrinsic phase speed (phase speed in absence of currents) and $u$ is in the direction of the wave. Negative value of $u$ is then an opposing current. Wave growth is then:

$$ S_{in} \sim \left( U - c_{p0} - u \right)^2 $$

Let's plug in some numbers here. Let's say $U = 10\ m/s$, $c_{p0} = 3\ m/s$, and $u = \pm 1\ m/s$:

  • In the case of following current, $u = 1\ m/s$, $S_{in} \sim 36\ m^2/s^2$ ;
  • In the case of opposing current, $u = -1\ m/s$, $S_{in} \sim 64\ m^2/s^2$ ;

Thus, by changing the orientation of a moderate current in characteristic wind and wave conditions, we increased the rate of wave growth by a factor of 1.78! This effect will be more powerful the smaller the relative wind speed $U - c_{p0}$ is.

Wave focusing and blocking in opposing currents

Wave energy is advected (transported) by the waves' own group speed $c_{g0}$ and also by currents. If for simplocity we drop all the other terms in the wave energy balance equation, we get:

$$ \dfrac{\partial E}{\partial t} = -\dfrac{\partial \left[\left(c_{g0} + u \right)E\right]}{\partial x} $$

The crux here is in the modulated group speed $c_{g0} + u$. If the waves propagate into an area of increasingly stronger opposing current, there will be a convergence of wave energy in that space, which will manifest itself in increase of high and steep waves. At the point where $c_{g0} = -u$, that is, the current is exactly countering the group speed of the waves, they can no longer propagate in that direction! Wave energy accumulates locally and vigorous wave breaking ensues.


Wave blocking is a different process from the wave growth dependence on relative wind speed described above, but, they often correlate because young windsea (short waves locally generated by wind) are almost always aligned with wind direction, so all you need is the currents to oppose wind for both of these effects to come into play at the same time.

Both are likely to occur in strong and steady western boundary currents such as the Gulf Stream, Kuroshio, Aghulas, and Brazil current, and also in any region with strong tidal currents.

More on this topic in the answer to this question.


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.