Refraction Convergence and Amplitude change- Ocean waves

[Tom79Tom]

There are many explanations on the internet, of refraction and convergence of ocean waves entering shallow water around a headland

However they all go no deeper than this statement

"Where the water is shallow the wave rays converge wave energy is greater where the wave rays spread out the wave energy is less" Is this an example of positive interference or if not, what is the mechanism and there a formula relationship which explains the resultant amplitude change? Does it depend on the angle of refraction therefore wave speed therefore water depth ratio?

If it is interference ? What is the theoretical maximum this "energy concentration" ?
For example if two waves of 1m which are in phase interact it is additive and the maximum amplitude is 2m does it depend on the angle of refraction ?

 

[sophiecentaur]

Surface water waves are really the hardest version of waves to get started on despite the fact that they are the first ones we are aware of. The basic wave phenomena are far easier measured and calculated in the context of light, sound and radio, where the situations can be made more 'ideal'. The sums are the same throughout the World of Waves (i.e. the whole World!) and your diagrams of constructive and destructive interference apply everywhere. But it is Diffraction that you should be considering (an infinity of small wavelets interfering with each other) rather than just two wave sources in your ripple tank picture.
Refraction of light at a plane surface is the result of a step in change of wave speed and the resulting Diffraction pattern (although people tend not to acknowledge that it is diffraction) gives a nice change of direction for the whole ray.
On a coast, the depth changes as you approach the land ; you usually get an upwards slope which causes the waves to slow down. This 'pulls the direction' in towards the shore, in the same way that a light beam bends towards the normal. The diagram sort of shows that.
Your quotation:

"Where the water is shallow the wave rays converge wave energy is greater where the wave rays spread out the wave energy is less"

makes sense because the changes in wave speed tend to focus waves against a headland. The essential step that is missing is an initial diagram of waves incident on a long straight sloping shore at an angle, which will show the angle getting more and more Normal towards the beach. (Just like the optical example).

 

[sophiecentaur]

It just struck me that, if you have an ariel photograph of the coastline in question with visible waves, you could draw in the 'orthogonals', they would be the equivalent to the Power Flow Lines and you could compare the spacings between them at different locations. The wave power would be more or less proportional to 1/s where s is the spacing. That's on the basis that the original power density was uniform.

 

[tech99]

It just struck me that, if you have an ariel photograph of the coastline in question with visible waves, you could draw in the 'orthogonals', they would be the equivalent to the Power Flow Lines and you could compare the spacings between them at different locations. The wave power would be more or less proportional to 1/s where s is the spacing. That's on the basis that the original power density was uniform.

Where a wave enters a medium where propagation is slower, I thought the energy passing a point every second would be the same, corresponding to constant power. The energy per metre traveled would be more.

 

[jbriggs444]

Where a wave enters a medium where propagation is slower, I thought the energy passing a point every second would be the same, corresponding to constant power. The energy per metre traveled would be more.

The energy per meter in the forward and back direction is not relevant to the rate of received power along a stretch of beach. Yes, if the waves are slowing down, one has to pack the energy of each crest into the shortened wavelength within which that crest exists. Each crest has to become higher. But the rate at which crests reach the shore (or break on the shallows) is unchanged.

It is the power per meter in the side to side direction that dictates how much power arrives on a given stretch of beach.

 

[sophiecentaur]

Where a wave enters a medium where propagation is slower, I thought the energy passing a point every second would be the same, corresponding to constant power. The energy per metre traveled would be more.

The rate that the power arrives is the same over a linear wave (it can't build up anywhere) but what counts here is the density of the power arriving. It's the taper of those 'orthogonals' that tells you how the power is redistributed / focussed / concentrated. The mean power is the same as on the open sea (ignoring losses due to friction with the sea bed) but it is concentrated and spread out by the coastline profile.

 

[sophiecentaur]

The energy per metre traveled would be more.

This is one way of looking at it and it is 'not wrong' because there will be more wave crests per metre where the waves are traveling slower but how relevant is that - unless you could 'freeze' the wave and use the Potential energy of water at all the peaks in a particular area of water? What counts with waves is the Power passing through a given aperture and that is not dependent on the wavelength / speed.

 

[Tom79Tom]

Hi guys thanks for all the responses,I will try and explain my question a bit better

Using the Huygens-Fresnel principle which states that every point on a wave front is a source of wavelets I have constructed two scenarios -
1. A straight wave where the depth profile is uniform along the wave front and thus the wave speed is equal (same size wavelets across)

2. A refracted wave where the depth is profile is non uniform across the front and thus the wave speed varies with depth - Slower in the Center (smaller wavelets) and faster on the Outside (larger wavelets)

Examining the progression of wavelets perpendicular to the wave front explains quite clearly how the wave front gets refracted towards the shallower slower section - however it does not explain the convergence of wave energy towards the Center.

We know that there must be a net transfer of energy toward the Center so i examined the transverse wavelets (as the wave is advancing this ends up being the angled wavelets Purple Outside to Center and Blue wavelets Center to Outside

The straight example clearly demonstrates the net transfer of energy is Zero as these wavelets are merely exchanged .

The refracted example is confusing as the Outside to Center depth profile is the inverse of the Center to Outside so the average speeds are the same . However the Outside to Center wavelets have to cover less distance than the Center to Outside ones

At the same speed but greater distance Outside receives less from the inside than is received by the Inside section

This positive net exchange due to distance traveled is the only way i can intuitively explain why energy is concentrated in the shallower section

Is this correct ?

 

[sophiecentaur]

- however it does not explain the convergence of wave energy towards the Center.

I'm not sure what you are saying here. Applying Huygens principle to a flat wave front produces another flat wave front in a forward direction. If you are not getting that result with your method then your sums are wrong. You also have to consider contributions from the whole wave front (+-∞) or the resulting pattern will be (hardly surprising) the near field diffraction pattern of a finite aperture. You can use that as a check for your method and not the other way round.

We know that there must be a net transfer of energy toward the Center

Again, I have a problem with what you actually mean by this. There is only a focussing effect when there is a tilt in the wave phase and in this case it must be modeled by changing the radius of the wavelets as the slope decreases.

Frankly, IMO a graphical attempt at using Huygens principle is more trouble than it's worth and is something we are told early on and then we move on. i
If you are getting a 'wrong' answer from your method then perhaps you should approach it mathematically - which really does work. I don't know your technical level so I can't recommend an appropriate source but Google will give you plenty of sources to choose from.

the only way i can intuitively explain

Intuition can easily let you down when things get complicated.