Doppler Effect or Not?

[Andrew Troup]

Originally Posted by Paul Elliott

It does seem to me that relative wind over sea should be the critical factor in wave height, regardless of the current. (Ground wind + current = sea wind.) I've been told that this is not the case, and I suppose I believe it, but I have no idea why. I've not heard a satisfactory explanation.

I suspect it has to do partly with a changing velocity gradient with depth. A bit like wind shear.

I reason this way: a (say) west-flowing tidal current in a particular zone is due to the surface water level being higher to the east than to the west.

Say the difference is 100mm. Looking at it simplistically: This is really only a problem for the uppermost slice of water: below 100mm depth, there is no discrepancy to be put right: the underlying body of water is all at the same level.

The deeper water will be dragged along by frictional interaction with the layers above (and also because I'm probably oversimplifying the cause), but most of the flow will happen near the surface. Once you get past a few metres down, it doesn't seem there would be much tidal flow.

So the energy distribution of a wave coming into the zone from the west is interfered with in a non-linear way, as a function of depth.

When a wave "feels the bottom" coming into a shallow beach, the energy distribution is interfered with at the bottom, and in crude terms it seems to make sense that the wave would steepen as a result of the friction with the seafloor.

However I don't actually think that explanation stands up to scrutiny: it's more the case that the energy distribution gets lopsided up into the crest region - it's misleading, I think, to think in terms of friction, because the water is not travelling along with the wave - at least, not until it breaks.

And at this point my explanation for current effects also starts to wobble: I can't yet make a plausible case for the waves steepening as a result of the current shear. (For simplicity, I'm assuming there's no wind, and that we're dealing with swells generated far away)

I could actually mount a vaguely plausible explanation for the opposite, but it's a bit like the faulty 'friction' explanation above, and of course we know that in practice, the waves do steepen dramatically.

My explanation also fails to explain why ocean currents are also notorious for causing waves to steepen. It's entirely possible that the reverse gradient could apply in the case of an ocean as distinct from a tidal current: increasing speed with depth.

I disagree with Dockhead that the frequency will change: I don't see how that is possible. If x000 waves enter a particular zone from the west every hour, I don't see how any number other than x000 could exit to the east in the same period of time. Where else would they go to, or come from?

[2Wind]

Originally Posted by Albro359

wrong !!!

It has NOTHING TO DO WITH THE MEDIUM

It has EVERYTHING to do with the movement of the source and/or receiver.

IT IS ONLY the relative velocities of the source and receiver that cause the Dopper effect.

This is BASIC physics.

Gee! Wrong again! Yes I was.....

So after due consideration:

1. For the purpose of the analogy, we assume speed of sound is constant within the medium (although it can vary as a consequence of factors that influence air density, such as temperature, humidity, altitude...)
2. As the medium is moving the sound does arrive faster than expected at the observer when down wind.
3. The relevant physical relationship is C=fλ
i.e. The velocity of sound in the medium = frequency x wavelength
4. To the observer the sound velocity appears to have increased by the additional velocity of the medium because the sound has actually arrived faster than expected
5. So as the velocity increases, it is wavelength that changes (lengthening) NOT frequency
6. So the pitch is not altered at the observers ears.

Its all clear to me now, thank you for making me think a lot harder.

[jongleur]

Forgive me for posting on this thread yet again, but
serendipitously, I came across this link:

Supersonic Stereo

If you haven't read xkcd, you've missed some good
stuff.

[Dockhead]

Originally Posted by Andrew Troup

I suspect it has to do partly with a changing velocity gradient with depth. A bit like wind shear.

I reason this way: a (say) west-flowing tidal current in a particular zone is due to the surface water level being higher to the east than to the west.

Say the difference is 100mm. Looking at it simplistically: This is really only a problem for the uppermost slice of water: below 100mm depth, there is no discrepancy to be put right: the underlying body of water is all at the same level.

The deeper water will be dragged along by frictional interaction with the layers above (and also because I'm probably oversimplifying the cause), but most of the flow will happen near the surface. Once you get past a few metres down, it doesn't seem there would be much tidal flow.

So the energy distribution of a wave coming into the zone from the west is interfered with in a non-linear way, as a function of depth.

When a wave "feels the bottom" coming into a shallow beach, the energy distribution is interfered with at the bottom, and in crude terms it seems to make sense that the wave would steepen as a result of the friction with the seafloor.

However I don't actually think that explanation stands up to scrutiny: it's more the case that the energy distribution gets lopsided up into the crest region - it's misleading, I think, to think in terms of friction, because the water is not travelling along with the wave - at least, not until it breaks.

And at this point my explanation for current effects also starts to wobble: I can't yet make a plausible case for the waves steepening as a result of the current shear. (For simplicity, I'm assuming there's no wind, and that we're dealing with swells generated far away)

I could actually mount a vaguely plausible explanation for the opposite, but it's a bit like the faulty 'friction' explanation above, and of course we know that in practice, the waves do steepen dramatically.

My explanation also fails to explain why ocean currents are also notorious for causing waves to steepen. It's entirely possible that the reverse gradient could apply in the case of an ocean as distinct from a tidal current: increasing speed with depth.

I disagree with Dockhead that the frequency will change: I don't see how that is possible. If x000 waves enter a particular zone from the west every hour, I don't see how any number other than x000 could exit to the east in the same period of time. Where else would they go to, or come from?

The apparent frequency changes, just like with the Doppler effect and sound. The structure of the waves is set up over one body of water, then when these waves run into another body of water which is moving in relation to the first body of water the structure is changed - stretched out if the second body is moving away, compressed if moving towards. Change of wavelength changes height and steepness of the waves, as the energy must go somewhere - conservation of energy in the wave train.

It has nothing to do with the bottom; cf infamous wind over tide effects in the Gulf Stream where there is no change of depth.

To answer your question - you are right - if the waves exit the moving stream and go back into a third body of water which is not moving in relation to the first one, then structure of waves (including wavelength, height, steepness) goes back to what it was.

[Paul Elliott]

So with the Gulf Stream (for example), the wave steepness is because the waves are generated in water with little current, then when these waves encounter an adverse current they are "compressed", thus becoming shorter in wavelength and therefore steeper? This makes sense, and is something I am quite familiar with at the entrance to the San Francisco bay. There, when the gentle ocean swells meet the quickly-flowing ebbing tidal current the waves can become quite nasty.

But this has nothing to do with "wind over current", which is usually how it's presented.

Notice that as with sound in moving air, the frequency of the waves is unchanged.

[Dockhead]

Originally Posted by Paul Elliott

So with the Gulf Stream (for example), the wave steepness is because the waves are generated in water with little current, then when these waves encounter an adverse current they are "compressed", thus becoming shorter in wavelength and therefore steeper? This makes sense, and is something I am quite familiar with at the entrance to the San Francisco bay. There, when the gentle ocean swells meet the quickly-flowing ebbing tidal current the waves can become quite nasty.

But this has nothing to do with "wind over current", which is usually how it's presented.

Notice that as with sound in moving air, the frequency of the waves is unchanged.

Yes, exactly. At least, that's the way it was explain to me by my little brother, who is a professor of physics.

Here's another explanation, only without resort to wave terminology: http://www.cruisersforum.com/forums/...tml#post356457

It says partially the same thing, only without resort to physics terminology. It also adds another aspect to the phenomenon -- that the two bodies of water running into each other adds energy to the wave train. Although my brother didn't say anything about this aspect of the phenomenon, this makes perfect sense to me.

In any case, wind over tide or wind over current (however you want to call it) results from the change in the form of the waves resulting from the transition between two bodies of water moving in relation to each other.

[Dockhead]

And here is the formula for what happens to the wavelength, height, etc. of waves when they hit a current (a body of water moving in relation to the body of water where they were formed):




Here's the textual explanation:

"The two curves called H (height) and L(length) show how these two wave characteristics are modified by a favourable current (to the right of the vertical axis), or more interestingly by an opposing current (to the left of the vertical axis.

"The horizontal axis shows the ratio between current speed and wave period, the vertical axis shows: 1. Hc/Hs, ratio between wave height with current and wave height without current; 2. Lc/Ls, wave length with current and wave length without current.
"R" is a measure of wave steepness, the ratio between height and length. It is usually accepted that beyond a 1:7 steepness the wave breaks. Suppose one is sailing in a 1:10 steep wave, say 2m high, 20m long, and all of a sudden an opposing current brings us along the dashed line in the graph.

"Along line L one can find Lc (point A): it's 70% of Ls. The current shortens the wave, which is now only 14m long. Along line H one can find Hc: 1.4 times Hs, the opposing current brings wave height to 2.8m. The theoretical steepness of the resulting wave would be 1:5, in practice the wave would have broken already...

"Of course, real world waves are not of constant length nor height, but rather a group of relatively smaller ones, followed by a group of higher waves and so on... in practice the boat finds herself like periodically jumping in a hole in the water, usually with the loudest crash. When water depth is sufficiently low, the effect is magnified."

Brancaleone, dalla Bretagna in giro per l'Atlantico: ENG - Waves and tidal streams


So you see:

1. The wavelength is shortened if the current is moving against the direction of the waves. So the frequency is increased from a point of view tied to the ground.

2. The steepness of the waves are changed as a function of the wavelength.


Another Curious Fact about wind over tide or wind against tide:

The phenomenon exists even in a dead calm. Think about that for a minute.

[Paul Elliott]

Originally Posted by Dockhead

[very interesting stuff!]

1. The wavelength is shortened if the current is moving against the direction of the waves. So the frequency is increased from a point of view tied to the ground.

Let's beat this dead horse one more time.

I say that the frequency is not changed, even if observed from an island. As someone previously pointed out, some number of waves per minute enter the current-zone. These waves pass through the current-zone. While the wavelength of these waves is changed, no new waves are created, and none are lost. Therefore, the *frequency* of the waves is unchanged regardless of the observing reference, since we still see that same number of waves per minute.

[goboatingnow]

Originally Posted by Paul Elliott

Let's beat this dead horse one more time.

I say that the frequency is not changed, even if observed from an island. As someone previously pointed out, some number of waves per minute enter the current-zone. These waves pass through the current-zone. While the wavelength of these waves is changed, no new waves are created, and none are lost. Therefore, the *frequency* of the waves is unchanged regardless of the observing reference, since we still see that same number of waves per minute.

If the wavelength changes then so does the frequency. The wave train is a continuous process not a time quantum.

Dave

[Dockhead]

Originally Posted by Paul Elliott

Let's beat this dead horse one more time.

I say that the frequency is not changed, even if observed from an island. As someone previously pointed out, some number of waves per minute enter the current-zone. These waves pass through the current-zone. While the wavelength of these waves is changed, no new waves are created, and none are lost. Therefore, the *frequency* of the waves is unchanged regardless of the observing reference, since we still see that same number of waves per minute.

Well, then, the physicists are wrong, the formula is wrong, and the phenomenon of wind over tide remains a mysterious riddle . . .

Wavelength and frequency are reciprocals of one another, given a fixed observation point. So you can't have a change of wavelength without a change of frequency.

Waves pass through and then, let's say, out of the current zone. When they get back into a piece of water which is not moving in relation to the piece of water in which they were formed, then they resume their original wavelength and, naturally, their original frequency. So there is no mystery, in any case, about the number of waves which come in and then go out.

But -- all I know is what was explained to me by people who forgot yesterday more than I will ever know about physics. So I can't really prove anything. You can make your own choice, obviously, what to believe.

[Dennis.G]

Originally Posted by Dockhead

So I can't really prove anything. You can make your own choice, obviously, what to believe.

And I believe that Dockhead and goboating now have in correct, while Paul has it wrong.

Back to the OP's question, if wind velocity was the same at source of sound and at listening station, the sound frequency would be unchanged.

If however the wind velocity changed (between sound's sending and receiving points), a change in sound frequency should occur. Example: A ship off shore in a 20 kt onshore wind sounds his horn. Wind dies down to 5 kt when reaches shore. An listener on shore would hear the horn at a very slightly lower frequency (lower pitch) than would be heard at the ship. This would be analogous to seas flattening (frequency decreasing) when entering an estuary during a flood tide.

[Paul Elliott]

Originally Posted by Dockhead

Well, then, the physicists are wrong, the formula is wrong, and the phenomenon of wind over tide remains a mysterious riddle . . .

Wavelength and frequency are reciprocals of one another, given a fixed observation point. So you can't have a change of wavelength without a change of frequency.

See below

Originally Posted by Dockhead

Waves pass through and then, let's say, out of the current zone. When they get back into a piece of water which is not moving in relation to the piece of water in which they were formed, then they resume their original wavelength and, naturally, their original frequency. So there is no mystery, in any case, about the number of waves which come in and then go out.

But -- all I know is what was explained to me by people who forgot yesterday more than I will ever know about physics. So I can't really prove anything. You can make your own choice, obviously, what to believe.

I'm sticking to my guns here.

The formula is Wavelength = Vprop / Frequency.

What you are not considering is the propagation velocity of the medium. Wavelength, frequency, and Vprop are *all* interrelated. I say the frequency remains constant in the cases we are considering. What changes is the Vprop, and the wavelength.

Recall my previous analogy of a radio signal traveling through free space, and through coaxial cable. Air has a Vprop of about 0.9999 c, while coax has a Vprop of about 0.6 c. The frequency remains unchanged, but the wavelength is definitely changed as the signal moves from one medium to the other. This isn't relativistic, it's simple algebra.

The same effect can be demonstrated with sound, with an old-fashioned "tin can telephone". Using string or, better yet, wire to connect the two cans, the pitch of your voice is unchanged even though the sound waves travel through the string or wire more rapidly than they do through the air.

In a manner similar to a more or less dense medium, the moving water or air also changes the Vprop, and the wavelength is affected in an identical manner. The frequency is unchanged.

Consider your example of waves moving through a current zone. All the waves can do is speed up or slow down, right? Every time a wave peak is launched into the current zone, It will travel at some speed through the current zone. There will be no more, nor fewer, waves leaving than entered the zone. In the middle of the zone there is still a one-for-one correspondence. The *only* change is the wavelength, and that is entirely due to Vprop.

[Dockhead]

Originally Posted by Paul Elliott

See below

I'm sticking to my guns here.

The formula is Wavelength = Vprop / Frequency.

What you are not considering is the propagation velocity of the medium. Wavelength, frequency, and Vprop are *all* interrelated. I say the frequency remains constant in the cases we are considering. What changes is the Vprop, and the wavelength.

Recall my previous analogy of a radio signal traveling through free space, and through coaxial cable. Air has a Vprop of about 0.9999 c, while coax has a Vprop of about 0.6 c. The frequency remains unchanged, but the wavelength is definitely changed as the signal moves from one medium to the other. This isn't relativistic, it's simple algebra.

The same effect can be demonstrated with sound, with an old-fashioned "tin can telephone". Using string or, better yet, wire to connect the two cans, the pitch of your voice is unchanged even though the sound waves travel through the string or wire more rapidly than they do through the air.

In a manner similar to a more or less dense medium, the moving water or air also changes the Vprop, and the wavelength is affected in an identical manner. The frequency is unchanged.

Consider your example of waves moving through a current zone. All the waves can do is speed up or slow down, right? Every time a wave peak is launched into the current zone, It will travel at some speed through the current zone. There will be no more, nor fewer, waves leaving than entered the zone. In the middle of the zone there is still a one-for-one correspondence. The *only* change is the wavelength, and that is entirely due to Vprop.

Yes, I think you're right, actually! Here's another explanation:


"I'm an ex-physicist. I've just tried dusting the cobwebs off an old textbook on waves to see if there were any diagrams I could still understand, but there was none that was appropriate to this.

"I suspect the main effect is pretty much the same as the Doppler Effect. The waves are generated at a certain frequency based upon wind speed (and of course fetch etc.) and the velocity of propogation is determined by that (could look up all the numbers for that I guess but haven't yet).

"When the waves enter an area of moving water the wave is slowed down by the same amount as that water is moving (I'm assuming for simplicity that the moving water is moving in completely the opposite direction to the waves) yet the frequency remains the same. Therefore the wavelength will shorten.

"The above assumes constant depth of water. When you combine that with the effect of the shallowing of the water (which is often the case where you get waves encountering a strong ebb) thats when that bit of sea becomes famous and gets its picture in the pilot books."

Wind against tide - physical reasons why this is dangerous ? - Page 2


I assumed that a change of wavelength must needs means a change of frequency. Assume = "ass" out of "u" and "me" The mistake was my own stupid one, and not my brother's. Sorry for muddying the waters, and thank you for enlightening me and everyone else.

[Andrew Troup]

Hmmmmm ... verrra interrresting discussion !

Waves certainly slow down when they travel into shallow water - one of the earliest physics experiments which caught my imagination was a long tray, into which we dropped a wedge diagonally across one corner. The waves, generated by a small slotcar motor with an out of balance crankbob, stuck with plasticene to the bottom of the tray (anyone remember plasticene? .... sigh ... I fondly remember massaging it into my classmates' hair, presumably as a mark of esteem ... )

As I was saying before being temporarily overcome by nostalgia, the wave fronts, having travelled straight along the tray, slowed down first at the end which encountered the shallowing water. The wave front slewed or wheeled around, aligning itself perpendicular to the axis of the wedge, as the other end, still in deeper water, maintained the original speed.

I found this a fascinating explanation for (inter alia) the way surf waves always seem to come in more or less perpendicular to a long, gradually sloping beach.

I went quiet in this interchange some time ago because I was having trouble with the concept of waves slowing down when they met a contrary current. Something about that was hard for me to feel confident about. I was thinking that electromagnetic waves propagating through empty space would not know or care if space was moving (the speed of light being constant from all frames of reference) but that was a distraction: cosmic radiation, light and so on, I think, form a special case, as they doesn't (AFAIK) need a medium.

Another difficulty with my musings, I think, was because one line of mental enquiry took me into standing waves - and I think on reflection that standing waves in a tiderace must be a different beast from standing waves in a river: otherwise if you slowed a river current down, formerly standing waves would start to move upstream, which clearly they do not: they're a geographical phenomenon, it seems to me, reflecting the underwater topology in that locality.

Here's how I convinced myself that displacing the medium must affect the speed of propagation.
Imagine a long violin string which has a winch drum at each end: If you pluck one end, a wave will travel along the string. If you start both the winches in the opposite direction, maintaining the tension, does that affect the wave travel speed? This seems to me like a valid analogy for changing the current speed for an existing oceanic wave train.

By imagining, instead of the winches being activated, the whole apparatus mounted on a truck driving in the 'opposite direction' mentioned above, it's clear to me that the wave propagation speed would be affected in both cases in the same way: effectively cancelled when the speeds are equal, temporarily making a sort of 'standing wave'

(all bets are off when the wave reaches the far end of the string, of course, and turns into a standing wave in the frame of reference of the structure carrying the string....)


When I get some time I must have a go at analysing what happens to the orbital motion associated with ocean waves when a contrary current is encountered: do they become ellipses? If so, are the ellipses tilted?

It seems to me some posters are talking about a more complex question of wind against tide

The purer question, easier to analyse, is when a swell train encounters an adverse ocean current: no wind, and no current speed difference with depth. Which Paul Elliott's explanation seems to me to embrace nicely.

Thanks Paul for teasing it out, and kudos to Dockhead. A bit more of both these people's mature attitudes to discourse wouldn't go amiss in these pages, it seems to me...

I'm not sure what happened to those who stormed off when their (sometimes shouted) assertions failed to satisfy those questioners who sought convincing explanations rather than scriptural truths.

And I think those questioners are now vindicated.

[Dennis.G]

Originally Posted by Paul Elliott

See below

In a manner similar to a more or less dense medium, the moving water or air also changes the Vprop, and the wavelength is affected in an identical manner. The frequency is unchanged.

If one is in a boat and moves towards oncoming waves, the frequency at which one encounters the waves increases. Would not the same be the case if one remains stationary and the surrounding water instead moved towards you?