Refraction of Ocean waves: Wide Headland vs Narrow

[Timtam]

When a wave encounters the shallow water of a headland the shallow section slows while the deeper section continues traveling at a faster speed. This causes refraction of the wave ray towards the shallower headland section

For wave rays encountering perpendicular a shallow headland does the width of the shallow section have any impact on the amount the wave ray bends ?

Attempt at a solution
Assuming a depth difference equating to shallow water speed of half the deep- I drew out a few scenarios with varying widths of Shallow and intermediate sections,
Wide Headland

Narrow headland

Narrow Intermediate Zone

Wide Intermediate Zone

The width of the narrow section did not seem to impact the angles of refraction only the width of the intermediate section changes the angle of the wave .

I don't think this is correct as taken to its limit (no width to the intermediate section) then there would be no refraction and the wave rays would separate ? Also Thinking of Huygens principle wouldn't the wavelets (which are traveling on oblique angles) pass thru the shallow section of a narrower headland quicker allowing them return to their normal speed ?

Would this not impact the refraction of the wave ray ?

 

[sophiecentaur]

Your four diagrams would probably be better if there is an angle between the wave front and the depth contours. In those diagrams, as they stand , there will be no refraction.

 

[Timtam]

Your four diagrams would probably be better if there is an angle between the wave front and the depth contours. In those diagrams, as they stand , there will be no refraction.

I don’t understand this is the classical refraction example ? Are you saying because the wave Ray is perpendicular to the depth change/ shallow water that there will be no refraction ? How will the the two propagation speeds reconcile ?

 

[sophiecentaur]

I don’t understand this is the classical refraction example ? Are you saying because the wave Ray is perpendicular to the depth change/ shallow water that there will be no refraction ? How will the the two propagation speeds reconcile ?

If you shine light normal to a glass surface, is its direction changed? All that changes is the wavelength.
Edit: Google Snell's law and you will see what I mean.

 

[Timtam]

Yup but Is light the best example ? I’m worried it might broaden the thread or get it off topic.
What do you think is causing the convergence in this picture ? I’ve only ever read that it’s caused by refraction there has never been described as requiring an angle only a velocity differential.

 

[sophiecentaur]

I’ve only ever read that it’s caused by refraction there has never been described as requiring an angle only a velocity differential.

Refraction changes the angle (when it's not normal). I just edited my last post to suggest looking up refraction / Snell's Law.
Edit: What's wrong with light as an example? a wave is a wave is a wave.

 

[Timtam]

Edit: What's wrong with light as an example?

Because every prior time I have deviated a thread from an OP to some other example it ends up off topic and not helping

a wave is a wave is a wave.

I'm not sure and don't want to get into it in this thread, regardless If that is the case then we can stick to surface waves as in the OP

What do you think is causing the convergence in this picture ?

There is something seriously wrong with a lot of websites if this is not refraction

 

[jbriggs444]

Your four diagrams would probably be better if there is an angle between the wave front and the depth contours. In those diagrams, as they stand , there will be no refraction.

As I read those diagrams, the waves are incoming from the top/North and proceeding downward/Southward. The depth profile has a uniformly shallow strip in the middle and a uniformly deep strip to either side. There is a ramping/shoaling region to the right and left of the shallows where depth increases uniformly.

The waves already arrive at right angles to the depth gradient. There will be refraction, as indicated in the diagrams.

 

[jbriggs444]

I don't think this is correct as taken to its limit (no width to the intermediate section) then there would be no refraction and the wave rays would separate ?

I think that this is correct.

Consider that as the shoaling region gets narrower and narrower, the wave power delivered to the North end of that region gets smaller and smaller.

The diagonal wave fronts in that region are stretched in the North/South direction. Accordingly, one would expect the wave amplitude to be reduced.

In the limit of a zero width shoaling region, the wave front connecting the deep water waves to the right and left with the shallow water waves in the center would be long and have a zero amplitude. It might as well not exist.

However, @sophiecentaur will likely school us both on how diffraction comes into play at such discontinuities so that the result will not have a sharp square wave cut-off at the ends of the shallow and deep wave fronts.

 

[sophiecentaur]

Your diagrams are making more sense to me now.

I'm not sure and don't want to get into it in this thread, regardless If that is the case then we can stick to surface waves as in the OP

If you want to stick to surface waves then you have far fewer examples and even fewer readily understandable explanations to fall back on. If you can give a good reason for not using a generic wave to explain this phenomenon then why not use light? Can you think of a good reason, apart from the non linearity associated with breaking ocean waves?
However, the coastline is a very fine structure compared with the wavelengths involved and not typical of an optical lens system so perhaps a simple ray trace might not be adequate. Huygen's construction could give a better answer but I haven't ever used it personally. There is bound to be some software to do that for you.

Also Thinking of Huygens principle wouldn't the wavelets (which are traveling on oblique angles) pass thru the shallow section of a narrower headland quicker allowing them return to their normal speed ?

I think that the wavelets would have shorter wavelengths in the shallow water because of lower speed wave fronts. But you may not mean what I think. Do you mean the apparent speed along a normal to the shore?

 

[sophiecentaur]

However, @sophiecentaur will likely school us both on how diffraction comes into play at such discontinuities so that the result will not have a sharp square wave cut-off at the ends of the shallow and deep wave fronts.

I just read this. This is really what Huygens does (everything is diffraction) but it's not as straightforward as working out what comes out of a simple aperture. I found this link which is what I am talking about. On page 11 there is a diagram of how they take points on a wave front and calculate the shape of the next wave front by adding contributions of wavelets along the initial front (crude diffraction calculation) The method makes an assumption about the speed and there's a comment about the difficulty when the 'refractive index' (= 1/speed ?) is different over the width - and that's what you want. But the authors seems fairly convinced by it.
@Timtam I hope you don't want too much more than an arm waving discussion of this as it could get pretty heavy. Take a look at the link and say what you think.

 

[Timtam]

Consider that as the shoaling region gets narrower and narrower, the wave power delivered to the North end of that region gets smaller and smaller.
The diagonal wave fronts in that region are stretched in the North/South direction. Accordingly, one would expect the wave amplitude to be reduced.
In the limit of a zero width shoaling region, the wave front connecting the deep water waves to the right and left with the shallow water waves in the center would be long and have a zero amplitude. It might as well not exist.

I agree with this logic regarding the zero limit of the intermediate section presenting a wave a depth gradient at right angles to wave ray

My only concern (aside from the many real world surfing examples) is that if there is a discontinuity along the wave ray, we would have a situation where we have adjacent sections that are out of phase (as indicated in the below diagram)

If were to expand this to consider the spherical wavelets crossing the interface obliquely then these also would be out of phase and we would see deconstructitive interference along/across the shallow/deep interface.

I think that the wavelets would have shorter wavelengths in the shallow water because of lower speed wave fronts. But you may not mean what I think. Do you mean the apparent speed along a normal to the shore?

I was thinking that wavelets are traveling at oblique angles (say for example 45*) These are traveling at an angle to the interface

Could it be as simple as to say that the wavelets traveling from deep water travel faster so there is more of these wavelets crossing the interface (per unit time ) into the shallow region than is exiting to the deeper region ?

 

[jbriggs444]

Could it be as simple as to say that the wavelets traveling from deep water travel faster so there is more of these wavelets crossing the interface (per unit time ) into the shallow region than is exiting to the deeper region ?

I do not follow this. Surely frequency is the same everywhere. Wavelength and wave speed combine to make it so.

 

[Timtam]

I do not follow this. Surely frequency is the same everywhere. Wavelength and wave speed combine to make it so.

Uggh True True so not that simple

Ive been reading a few surfing posts about this issue in the interim and it seems that there is a bit of debate amongst surfers whether this phenomenon is Refraction or Diffraction
https://www.swellnet.com/forums/crystal-ball/331846

As a wave meets the obstruction the unaffected line of wavefront just moves past as it was with NO DIRECTION CHANGE. But some energy ,or wave leaks around the obstruction and tries to fill in. This is our diffraction and it is totally due to Huygens description of the spherical wavelets from the particles at the edge of the obstruction loosing some of the destructive interference on the obstructed side. You get a circular or spherical wave coming around the obstruction but no bending. So where does this leave us, it seems it's not refraction or diffraction with our ocean waves around headlands etc.

While not very well written this comment makes sense if we view the speed change as the obstruction and diffraction is causing "swell energy to diffract around it" in this case the swell energy is diffracting into the area in front of the obstructed shallower water wavefront , of course this is slowed as well as it enters the shallower water

or that the bending is due to plain old friction "The wave particles are tied together as a group and so the slower particles pull the faster particles around with them."

Both of these make some sense to me in explaining the real world physical phenomenon we see of waves bending from the normal due to a change in depth

 

[olivermsun]

View attachment 222992
What do you think is causing the convergence in this picture ? I’ve only ever read that it’s caused by refraction there has never been described as requiring an angle only a velocity differential.

I guess I am confused by some of the "tangents" in this thread.

Based on the picture, I think the convergence around the headland happens because the shape of the headland continues under water. The incoming waves encounter the depth contours obliquely and are refracted because the surface waves slow down as they shoal.

If you encounter a (wide) depth discontinuity, then you can try to model this by matching conditions at the discontinuity, just as you would for a light ray crossing from one medium to another, which results in Snell's Law.

Alternatively, you could get diffraction if you have waves impinging obliquely on a very narrow (referring to actual extent, not abruptness of the depth transition), in which the wave crests will "bend around" the obstruction, i.e., the usual demonstration of Huygens' Principle. You don't need depth change to explain this effect, however.

 

[Timtam]

The incoming waves encounter the depth contours obliquely and are refracted because the surface waves slow down as they shoal.

Yes this the point I feel @jbriggs444 made very well but the most common surfing examples of this happening Mavericks Nazare Jaws it’s actually abrupt depth changes across the wave ray (deep offshore channels or ridges) that cause the most significant Focussing ... where the 'intermediate' depth profile approaches the zero limit case the larger the convex focusing occurs.

The only thing that seems to be evident from real world examples is that the wave rays must stay ‘joined’ and you can see large portions of consistent deeper depth profile outside of the shallow headland that should progress at the same speed ‘dragged’ by the slower shallower sections to ensure that the wave Ray remains joined . I’m scared to suggest it but almost like a boundary layer

 

[olivermsun]

Yes this the point I feel @jbriggs444 made very well but the most common surfing examples of this happening Mavericks Nazare Jaws it’s actually abrupt depth changes across the wave ray (deep offshore channels or ridges) that cause the most significant Focussing ... where the 'intermediate' depth profile approaches the zero limit case the larger the convex focusing occurs.

I believe you are referring to the abrupt depth changes oblique to the wave crests, (rays are in the direction of wave phase propagation, which is normal to the crests). That's what tends to focus the incident waves into the characteristic 'vee' shape.

The only thing that seems to be evident from real world examples is that the wave rays must stay ‘joined’ and you can see large portions of consistent deeper depth profile outside of the shallow headland that should progress at the same speed ‘dragged’ by the slower shallower sections to ensure that the wave Ray remains joined . I’m scared to suggest it but almost like a boundary layer

If by "almost like a boundary layer" you mean that physical quantities such as a velocity and pressure in the case of the waves must match across the discontinuity, then the physics require it. Wave crests are "piles" of water, so if you suddenly "cut off" the wave crest on one side the piles will tend to slump off to that side. That's why jbriggs444 and sophiecentaur mentioned earlier in the thread that wave phase has to be continuous across the discontinuity, and that you should look at Snell's law for geometric optics or Huygen's principle for wave diffraction, etc. They can all help explain what is observed in these real world examples.

 

[sophiecentaur]

there is a bit of debate amongst surfers whether this phenomenon is Refraction or Diffraction

Diffraction is the general term for how a wave is influenced when passing an obstruction / aperture or some more subtle change in propagation speed. That includes lenses, mirrors (or the equivalent for surface waves. That debate is coming from a direction where 'classification' has been viewed as more important then 'understanding'. It doesn't take the argument anywhere useful.

you should look at Snell's law for geometric optics

It's a possible first stab but I think that ray tracing wouldn't work so well in the case of surface waves that are not much shorter than the dimensions of the coast and seabed. I think the bullet needs to be bitten as with the Huygens' approach in the paper I linked higher up.

 

[olivermsun]

It's a possible first stab but I think that ray tracing wouldn't work so well in the case of surface waves that are not much shorter than the dimensions of the coast and seabed. I think the bullet needs to be bitten as with the Huygens' approach in the paper I linked higher up.

For the idealized cases that the OP showed, the coast and seabed seem to vary only along one dimension, so ray tracing may give a reasonable view of what happens in the regime before breaking becomes important. If the headland has a complicated structure that has very short dimensions relative to its depth (change), then yes, I agree that diffraction will need to be considered also. However, in many situations the waves have already shortened significantly (due to shoaling) as they approach the headland.

 

[Timtam]

In the limit of a zero width shoaling region, the wave front connecting the deep water waves to the right and left with the shallow water waves in the center would be long and have a zero amplitude. It might as well not exist.

Diffraction is the general term for how a wave is influenced when passing an obstruction / aperture or some more subtle change in propagation speed.

the headland has a complicated structure that has very short dimensions relative to its depth (change), then yes, I agree that diffraction will need to be considered also.

I managed to find a ripple tank photo that I believe replicates the 'wavefronts approaching an abrupt transverse depth change from the normal’ scenario we are discussing

Here is the link and the blurb
https://fphoto.photoshelter.com/image/I00005PtiOGFXRRU
REFRACTING WAVES IN RIPPLE TANK
Waves refracting off shoal in ripple tank
When vertical plane waves encounter shoal, the angle of the wave changes due to the altered depth of the water. The wave bends around the shoal due to resistance.

Some very interesting observations, there seems to be a lot going on that doesn't appear evident from Snells law

1. There seems to be an effect BEFORE the shoal (the deep water section starts wrapping)
2.The wavelengths on top of the shoal initially appear to shorten
3. The wave fronts next to the shoal appear to lengthen (or become more cnoidal) as they progress they also appear to exhibit an s shape in deep water adjacent to the shoal
2. There is constructive/destructive interference along oblique wavefronts on top of the shoal from coinciding/joining the longer deeper wavelengths but interference is not evident here.
5. The oblique interference on top of shoal appears more destructive than constructive as in the overall original wavefront appears to diminish as the interference grows and the stronger wavefronts appears to become the oblique wavefronts as the wave progresses

Compared to ocean waves ripple tanks have a very small amplitude to wavelengths, Scale might be an issue where we are seeing that magnitude of the bending along the interface (especially that S shape) ... but does seeing this image give anyone more ideas ?

IMO it points more to diffraction effects than refraction

 

[sophiecentaur]

there seems to be a lot going on that doesn't appear evident from Snells law

Snell's Law can't be expected to apply there because the corner of the shallow region means there is no identifiable 'Ray'. Snell applies well enough well away from that transition. But sin(0°) is 1 so it doesn't mean a lot.
Your points 1 to 5 are worth noting but they're no surprise because the standing waves due to an interface are to be expected.
Ripple tank pictures are good to look at but I wouldn't reckon that they are much use for serious quantitative study. EM waves behave much better because they are mostly in a linear well behaved medium.

 

[Timtam]

But sin(0°) is 1 so it doesn't mean a lot.

I agree that Snells law doesn't apply at 0° But we are seeing a bending of the wave in the deep water are we not ? I don't think its refraction on that basis.

EM waves behave much better because they are mostly in a linear well behaved medium.

If you shine light normal to a glass surface, is its direction changed? All that changes is the wavelength.Edit: Google Snell's law and you will see what I mean

but i wouldn't expect -if we shine a light at the normal at a glass box - to see any bending of wave outside or inside the glass but we see this here this here.

Snell applies well enough well away from that transition

What do you mean by that (in the transverse direction away from the Shallow/Deep interface to deep water ? or the away from the initial frontal transition in a linear direction as the wave progresses? I disagree with both as we see the above noted oblique effects where as Snells predicts only the wavelength change

 

[sophiecentaur]

What do you mean by that

I mean that i and r are near enough zero and the wave fronts are straighter the further you go from the discontinuous bit. That's where Snells Law for zero degrees is working. It demonstrates the point I made way back that Snells Law doesn't suit the surfing problem. The full diffraction treatment is needed.
Snell appears to work on long straight shorelines where you can often see the waves ending up with almost normal incidence right near the beach.