Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Waves at the Beach

  1. Apr 22, 2015 #1

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Go to any sand beach anywhere in the world. No matter which direction the beach faces, no matter the direction of wind and tide; waves always come in the direction offshore to onshore.

    Even on the lee side of an island, waves still come ashore and break on the beach. Waves may be bigger on some sides of the island, but I speak of direction, not magnitude.

    All around the perimiter of a convex island, or all arond the interior perimeter of a bay, waves come ashore approximately parallel to the shore.

    There must be a physical reason for this that escapes me at the moment.
     
  2. jcsd
  3. Apr 22, 2015 #2
    I think it's just the fact that the receding of the water isn't very spectacular, all you notice is the crashing of the waves onto the sand/cliff.
    So, just think of the water as sloshing back and forth, but that only the slosh towards the solid gets your attention.
     
  4. Apr 22, 2015 #3

    jbriggs444

    User Avatar
    Science Advisor

    If a wave originated at the beach, where would the energy come from?

    Edit: Misunderstood the thrust of your post. How about refraction as the explanation for coming in parallel?
     
  5. Apr 22, 2015 #4

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    A very good point. But couldn''t they originate elsewhere and pass perpendicular to the beach?
     
  6. Apr 22, 2015 #5
    Waves arriving on beach are not always perpendicular to the shore line..
    you often see a wave which breaks at first to your left then as it progreses it breaks to your right..
    The speed of the break going from left to right could actually exceed the speed of light without breaking any rules.
     
  7. Apr 22, 2015 #6

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Sorry for sloppy language. I mean waves nearly parallel to the shore that travel perpendicular to the shore, or approximately so, then they break in shallow water.

    One never sees wave fronts nearly perpendicular to the shore traveling approximately parallel to the shore, so that they never break.
     
  8. Apr 22, 2015 #7
    My best guess why that is would be that if in fact a wave came in sideways, because the water towards the shore is shallower, there's less water to displace. So, the incoming wave goes the path of least resistance and gets deflected towards the shore.

    Total hunch though.
     
  9. Apr 23, 2015 #8

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Thank you jbriggs444. Your "misunderstanding" made me think about it more, and I think you are correct, energy is the key.

    Consider a confused sea, with little wavelets moving and colliding in all directions. The net kinetic energy is zero, but there's lots of potential energy. Now consider a nearby linear beach. The beach only dissipates energy, never originates it. Like water that runs downhill, some of that offshore potential energy must move nearly monotonically toward the beach. There is an energy density gradient that drives it.


    Now, water waves move more slowly in shallow water. The rear end of waves travel faster than the fronts. As the rear catches up with the front, the wave front become more cohesive and grows in height. Components of motion parallel to the wave front are roughly balanced left-right, so they cancel each other. This makes the waves even more coherent, the wave fronts more parallel to the beach, and wave velocities more normal to the beach.

    Come to think of it, the analogous 3D action would be like debris in random orbits around a planet, that spontaneously form a highly coherent ring system, like Saturn's. So, like Saturn's rings are always normal to the vector toward the planet's center, so are sea waves approaching any beach orientation always traveling approximately normal to the shore line. Even though rings are make of discrete particle in free space (not liquids at a surface), I now believe that the basic physics of ring systems and beach waves are analogous. Am I right?

    It all seems so logical expressed in terms of energy, but it sounds nightmarish to express in water wave propagation equations. For example, there is nothing in the avove reasoning that hints what the period of the waves hitting the beach should be.

    upload_2015-4-23_9-1-37.png
     
  10. Apr 23, 2015 #9
    I believe that the volume on waves from the Berkley course (in 5 volumes) treats this very well. It's volume 3, by Crawford.
    If you don't have the book, try this:

    http://www.coastal.udel.edu/ngs/waves.html
     
  11. Apr 23, 2015 #10

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I think that's the answer. A shelving beach will always turn the waves towards the shore because the wave speed reduces as the water is shallower.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Waves at the Beach
  1. Waves ? (Replies: 4)

Loading...