Do light and sound waves roll up and break like ocean waves?

In summary, when sea waves approach the shore they roll up and break due to different velocities of water layers formed due to the gradual change in water depth. The highest wave peaks move faster than all other layers and thus falls down. All other layers fall the same way but in a delay. This ends up with wave rolling up when approaching the shore.
  • #36
Jamison Lahman said:
OP's statement in the quoted post.
Which quoted post? You started off by quoting My Post, about shock waves being local but I was not the OP.
It would really help if you were a bit more fulsome with your writing. That last post still means very little to me (or many other readers, I fear).
Whilst you may know everything there is to know about this topic, if you want your ideas to be taken in by the readers of the thread, you need to help us a bit. We may not actually be thinking along your lines until you give more guidance.
 
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  • #37
sophiecentaur said:
Which quoted post? You started off by quoting My Post, about shock waves being local but I was not the OP.
It would really help if you were a bit more fulsome with your writing. That last post still means very little to me (or many other readers, I fear).
Whilst you may know everything there is to know about this topic, if you want your ideas to be taken in by the readers of the thread, you need to help us a bit. We may not actually be thinking along your lines until you give more guidance.
The post I quoted in my initial comment. Specifically, comment #15 in the thread.

The reason I am not being completely fulsome is because it is a digression from the topic and one OP may not even be interested in.
Exerts from Computational Physics, Koonin and Meredith for context:
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...
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  • #38
I really would appreciate you telling us how that apparently random cut and paste is relevant, either to the OP or to the topic of shock waves. Do you genuinely believe that it is a sequiter to any of the preceding posts?
Post #15 introduced Black Holes. Is that the link?
It's several months before April 1st.
 
  • #39
sophiecentaur said:
I really would appreciate you telling us how that apparently random cut and paste is relevant, either to the OP or to the topic of shock waves. Do you genuinely believe that it is a sequiter to any of the preceding posts?
Post #15 introduced Black Holes. Is that the link?
It's several months before April 1st.
It is directly relevant to the sentence quoted in my first post, as I have mentioned before. I genuinely can't clarify any more than that. I have also twice clarified it as a digression, yet you persisted I expand further. I genuinely don't know what more you want.
 
  • #40
Jamison Lahman said:
It is directly relevant to the sentence quoted in my first post,
Perhaps if you posted this as a new thread in the Mathematics Forum it may make more sense to more people.
 
  • #41
sophiecentaur said:
Perhaps if you posted this as a new thread in the Mathematics Forum it may make more sense to more people.
K.
 
  • #43
nikkkom said:

It took a bit of searching to find this section:

3.6.1.1.1 Free Surface Release Waves in Gases

Let us assume that material supporting the shock wave is a gas. Since the pressure change in a rarefaction wave is always continuous (no instantaneous pressure drops), the pressure at the leading edge of the escaping gas is zero (in keeping with the requirement of equal pressures at the gas/vacuum interface) and the process of converting internal energy into kinetic energy is complete. The front edge thus immediately accelerates to the final maximum velocity (escape velocity) when it reaches the free surface. Farther back in the release wave the pressure increases and the velocity decreases. At the release wave boundary (which moves backward at c_s) the pressure and velocities are the same as in the original shock wave.

The escape velocity of a gas is equal to:

Eq. 3.6.1.1.1-1

u_escape = (2*c_s)/(gamma - 1)If we use a frame of reference in which the unshocked gas was stationary we get:

Eq. 3.6.1.1.1-2

u_escape = (2*c_s)/(gamma - 1) + u_particleand the rear edge of the release wave travels backwards at

Eq. 3.6.1.1.1-3

v_release = u_particle - c_sThe release wave thus stretches itself out with time at a speed of:

Eq. 3.6.1.1.1-4

((2/(gamma - 1)) + 1)*c_swhich for a perfect monatomic gas is 4*c_s.

It contradicts the concept of sound waves crashing summed in one sentence, "Since the pressure change in a rarefaction wave is always continuous (no instantaneous pressure drops), the pressure at the leading edge of the escaping gas is zero (in keeping with the requirement of equal pressures at the gas/vacuum interface)"
 
  • #44
Rarefaction wave is not a shock wave, and can't become one.

You need section "3.4.3 Hydrodynamic Shock Waves"
 
  • #45
nikkkom said:
Which particular phrase / sentence / paragraph should I read?
If you have a point to make then why not help the reader with where it is? Do you expect a casual contributor to top through the whole of that link and discover where your point is being demonstrated? Full marks to jerremyjon for finding it but should he have had to go to all that trouble? PF normally tries to be helpful, rather than to set people examination questions.
But it seems to confirm that this sort of thing doesn't happen in a free wave and must involve significant differences in the motion of parts of the medium.
 
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  • #46
sophiecentaur said:
Which particular phrase / sentence / paragraph should I read?
If you have a point to make then why not help the reader with where it is? Do you expect a casual contributor to top through the whole of that link and discover where your point is being demonstrated?

You need section "3.4.3 Hydrodynamic Shock Waves"

But I also strongly recommend reading the entire FAQ. You will increase your knowledge of nuclear weapons physics x100.
 
  • #47
nikkkom said:
You need section "3.4.3 Hydrodynamic Shock Waves"
3.4.3 Hydrodynamic Shock Waves

Compression waves are fundamentally unstable. They naturally tend to steepen in time and eventually (if they propagate long enough) will become infinitely steep: a shock wave. On the other hand, rarefaction waves are stable and a rarefaction shock (a sudden pressure drop) is impossible.

from:
3.4.1 Acoustic Waves
Any local pressure disturbance in a gas that is not too strong will be transmitted outward at the speed of sound.
6E8D82C8-90B2-448E-B3C5-35F9E33F0BAB_w1023_r1_s.jpg
 
Last edited:
  • #48
I can't think of anything for light yet, but perhaps for sound waves: Make a long bar consisting of layers of material that conduct sound at different velocities. If in fact that is a characteristic of different materials. Eg wood, plastic, metal. Then strike the bar on the end and then find some way to (visually?) observe the sound in transit or arriving at the other end of the bar. Guess I should research sound/shock velocities in different materials.
DC
 
  • #49
nikkkom said:
You need section "3.4.3 Hydrodynamic Shock Waves"

But I also strongly recommend reading the entire FAQ. You will increase your knowledge of nuclear weapons physics x100.
Yes. I realize I am somewhat 'under-read' about a lot of topics - this is one of them. There must be a lot of people like me who have dipped into the thread and a bare reference to a long and difficult link is not going to encourage them to follow it. I always reckon that some personal input is worth while. A paragraph in one's own words, even when it's at a lower level than really necessary, is worth while, just to keep a thread going and to address a wider audience.
 
  • #50
jerromyjon said:
3.4.3 Hydrodynamic Shock Waves

Compression waves are fundamentally unstable. They naturally tend to steepen in time and eventually (if they propagate long enough) will become infinitely steep: a shock wave. On the other hand, rarefaction waves are stable and a rarefaction shock (a sudden pressure drop) is impossible.

from:
3.4.1 Acoustic Waves
Any local pressure disturbance in a gas that is not too strong will be transmitted outward at the speed of sound.
View attachment 210241
That picture made me realize that the surface wave that's set up when a boat is traveling at even a low speed, exhibits many of the characteristics of a breaking wave. The water at the top of the bow wave is traveling backwards, relative to the boat and the bow wave is continually 'breaking'. It makes me think that the equivalent must happen for other wave types but there is a very significant difference for water surface waves and that is they are very dispersive (speed is proportional to √λ - see link). You would need a very special medium for this with sound or EM, I think.
 
  • #51
If the boat is traveling slowly (i.e., less than the so-called "hull speed"), the bow wave can actually be fairly quite smooth (non-breaking), provided the boat has a reasonably fine entry and you aren't driving into much chop.

If you are getting close to hull speed then, yes, this starts to be reminiscent of a shock.
 
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  • #52
Referring to low frequency radio waves:
Conductivity of the surface affects the propagation of ground waves, with more conductive surfaces such as sea water providing better propagation.[8] Increasing the conductivity in a surface results in less dissipation.[9] The refractive indices are subject to spatial and temporal changes. Since the ground is not a perfect electrical conductor, ground waves are attenuated as they follow the earth’s surface. The wavefronts initially are vertical, but the ground, acting as a lossy dielectric, causes the wave to tilt forward as it travels. This directs some of the energy into the Earth where it is dissipated,[10] so that the signal decreases exponentially.

www.jpier.org/PIER/pier19/02.970718p.Ling.SU.pdf

So, yes apparently EM waves crash in the same place that water waves crash due to the change in conductivity between shore and ocean water.
 
  • #53
kbomeisl said:
So, yes apparently EM waves crash in the same place that water waves crash due to the change in conductivity between shore and ocean water.
Ground waves surprised early experimenters who expected much greater losses over the horizon for MF and LF waves. I can see how your analogy is attractive but ground waves are a linear phenomenon and happen the same for all signal levels. All that is happening is a tilted wave front so I don't think it really counts, unfortunately.
 

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