This is a continuation of a few "off-topic" discussions, with no more convenient home. My apologies if you are looking for "joelcampbell1735-research." I will be back to that soon.
Waves
I have spent more time in recent years staring at waves. I live on a 'coast' and almost daily walk the beach and watch the end of the wave lifecycle as they peak, topple, and crash into the shore. Occasionally I sail on a tall ship and view the wash basin that is the normal sea. Waves of all heights, directions, and wavelengths. Sometimes waves appear organized...traveling in a group, all with the same direction and speed.
Wakes
One well known wave (or set of waves) is that which extends out from the bow of a ship and appears to travel at the speed of the ship. A similar wake is seen behind a duck, a kayak, or a swimmer. In fact, way back in August of 1887, William Thomson, announced that these wakes all had the same angle regardless of the speed or size of the moving object. Thomson, also known as Lord Kelvin, a 63-year-old world-renowned scientist in Glasgow, delivered a lecture at the Conversazione in the Science and Art Museum in Edinburgh, giving the angle as one made by a line from the bow of a ship (x=0, y=0) to a point of x/y = 81/2. [If you remember your trigonometry, y/x would be the tangent of this angle, so the angle is....
tan-1(8-1/2) = 19.47 degrees.]
Daderot, Public domain, via Wikimedia Commons
Getting to the Solution
How Lord Kelvin arrived at this elegant solution is not revealed in his 1887 address. [You can read the paper at Archive.org] It has been solved on-line and in textbooks, sometimes with a bit too much fancy mathematics and hand-waving for this author. In my opinion, there is a hierarchy of learning needed to arrive at an understanding of the solution. To understand the wake angle, you need to understand "group velocity." But to understand group velocity, you must understand "gravity waves." But to understand gravity waves you need to understand the "Law of Squares."
The Law of Squares
In nature, many relationships are linear. For example, if you double the speed of an object, then you double the distance the object travels. Other relationships are related by the Law of Squares. Gravity is one of those. As the distance between two objects increases the amount of mass needed to obtain the same gravitational force increases by its square. It can be written
g α d2 / m
where g is some gravitational constant which is proportional to the square of the distance separating two bodies.
Gravity Waves
Surface waves on a body of water are governed by gravity and are termed 'gravity waves.' It should come as no surprise that the Law of Squares is involved in the description of gravity waves. In fact, the speed of a gravity wave is given as
v = (gL/2π)1/2
where g the gravitational constant and L is the distance between wave crests.
You can see that water waves are much different than waves of light or sound. The speed of light or sound through a media is fixed, whereas water waves travel at many speeds with their wavelengths increasing by the square root of 2 when the velocity doubles.
We can also write the above equation in terms of a wave number (k=2π/L) or
v = (g/k)1/2
For all waves, frequency (F) is velocity multiplied by the wave number or
F = v k = (g/k)1/2k = (gk)1/2 Sometimes called the frequency dispersion relationship for gravity waves.
Once again, the Law of Squares determines the relationship between the frequency of a gravity wave and its wave number.
Group velocity
Group velocity is not an easy concept to digest. Fortunately, it can be readily observed. Just this weekend, I sat on my paddleboard and watched a 'group' of waves from a motorboat pass under me. When I focused on the lead wave in the group, I saw it disappear and a new wave appear at the rear of the group. The group had a slower velocity than an individual wave. The excellent graphic below was borrowed from the Wiki page on Group Velocity,
On my paddleboard, I was focusing on the red square. But the group is actually moving at the speed of the green dot.
What causes this? It is the result of the interference of waves of similar wavelengths.
The group velocity (vg) can be derived as the derivative of the frequency with respect to the wave number or
vg = dF / dk
From our equation above for gravity waves for F, the derivative is
vg = dF / dk = ½ (g/k)1/2 = ½ v
For gravity waves, the group velocity (the speed of the visible wave front) is one half the wave speed. It is a direct result of gravity waves having a Law of Squares relationship.
Wake Angle
The wake waves that trail the bow of a ship appear stationary with respect to the ship. They differ from waves generated from a point source, such as a pebble dropped into a pool, from which ripples proceed out in concentric circles.
A moving ship, on the other hand, is continually creating waves, some of which the ship is keeping up with, or from the perspective of a person on the ship, is 'dragging along.' Observationally, there is no doubt that a ship produces waves that appear stationary with respect to the ship and are contained within a wedge bounded by a straight line.
From our earlier discussion, that a wave group moves at one half the speed of the wave phase velocity, one might think that there may be an analogy here to the wedge produced by a supersonic object traveling at twice the speed of sound. In the Figure below, the supersonic object has moved from x=8 to x=0, while the sound created at x=8 has only moved to x=4 (the BLUE circle). Similarly, the sound made when the object was at x=4 has only moved to x=2 (the RED circle). The wedge of sound waves, moving at one half the speed of the object, is bounded by a straight line defined by sin a = SpeedOfSound/SpeedOfObject = ½, or a = 30°.
Even though we have shown the group velocity is one half the phase velocity, the sonic shock wave is not analogous. The ship is generating waves of many speeds and directions that are constantly constructively and destructively interacting with each other. More importantly, the group wave that is stationary with respect to the ship, is the one whose phase velocity matches that of the visible wake, not the group wave that is moving at one half the speed of the ship in direction 90° - a.
Let's look at the problem from a pure trigonometric perspective. We know from observation that there are some wake waves that are stationary and are bounded in a wedge defined by a straight line. We know from our knowledge of gravity waves, that what we are observing are the group waves produced by waves whose phase velocity is twice that of the group. In the Figure below, a ship has proceeded from O to A. The BLACK lines are wave fronts of the waves creating the visible group waves. The group wave associated with this front has only moved to where the RED line appears. The PURPLE line is perpendicular to the phase and group wavefronts, and represents the wave velocity.
Because waves at this particular (unknown) phase appear stationary, they must have a velocity of vp = V sin a, where V is the speed of the ship and a is the angle of the phase wavefront with the ship. Hence the visible group wave has a group velocity of
vg = ½ vp = ½ V sin a
This is the stationary wave condition that differs from the shock wave situation with sound waves.
As suggested by the trigonometry, the group wavefront orientation and the visible wakefront are not parallel. This is clearly seen in the duck image above. The wake angle, b, is actually created by echelons of short group wave segments.
From this simple sketch of fronts and angles, what can we say about 'a' (the wave group angle) and 'b' (the visible wake angle)?
We can use the rule of trigonometry that the sine of any angle of a triangle divided by its opposite side is a constant. Take the triangle formed by OPA. Its angles are b, π/2 - a, and π - b -(π/2 - a) = π/2 - (b - a).
We can then write
sin(b)/OP = sin(π/2 - (b - a))/AO Eq.1
where AO = Vt where V is the ship velocity and t is the time to move from O to A
and OP = vgt = ½ vpt = ½ Vt sin(a).
We can rewrite Eq.1 as
sin(b)/[½ Vt sin(a)] = sin(π/2 - (b - a))/Vt
Knowing that sin(π/2 - X) = cos (X)
we can write 2sin(b)/sin(a) = cos(b - a)
Interestingly, the relationship between these two angles is independent of the ship velocity, V, which is what we observe.
We can perform more trigonometry to solve this equation, which I will show. If this is of no interest, skip down to the result shown in Eq. 2.
Multiply both sides by sin(a)
2sin(b) = sin(a)cos(b - a)
But 2sinAcosB = sin(A+B) + sin(A-B)
So 2sin(b) = ½[sin(a+(b-a)) + sin(a-(b-a))]
4sin(b) = sin(b) + sin(2a-b)
3sin(b) = sin(2a-b) Eq.2
The right side of Eq. 2 must be ≤ to 1. Therefore
sin(b) ≤ 1/3
b ≤ sin-1(1/3)
b ≤ 19.47 °
So the visible wake angle must be 19.47 ° or less. This agrees with more robust derivations of the wake angle.
We can also solve for the angle of the group wave front, a. Once again, the right side of Eq. 2 must be ≤ to 1.
sin(2a - b) ≤ 1
then 2a-b ≤ π/2 or 90°
2a ≤ 90° + b
a ≤ (90° + b)/2
a ≤ 54.7°
Simple trigonometry has led to values for the Kelvin wake angle and the angle of the visible group waves. [Note that the Figure above is constructed to scale with a = 54.7° and b = 19.47°]
Wavelengths
One last simple observation can be made about the wake waves seen in the duck picture above. We know that the wavelength for gravity waves is given by v = (gL/2π)1/2 or L = 2πv2/g. If the speed of the ship is driving the waves of most interest, then L = 2πV2/g. Because the still picture cannot tell us how fast the duck is moving, perhaps we can guess by measuring the distance between the echelons in the duck wake. It looks like there are about 2 wavelengths per length of the duck. If the duck is about 12 inches long, the wavelength is 6 inches, L/2π = about 2.4 cm.
V = [9.8 m/s2 x .024 m]1/2
V = [3.1 x 0.16] m/s
V = 0.5 m/s or about 1 mph
My swim speed is about 2 mph, so the result seems reasonable.
The reader can find many derivations on the internet with much more scientific approaches to derive the wake waves and their patterns. But we have been able to confirm some of our observations of wakes with just a knowledge of gravity waves, group velocity, and trigonometry.
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