Consider two different theories of string vibration determined by whether the string stretches as it moves.
1) If the string elongates as it bends, then the force acting on a point along the string is determined by the displacement of the point from its center of motion. The force acting upon the string is therefore greatest at the midpoint where the displacement is greatest and force decreases across the string to zero at the endpoints where no displacement can occur. This is the non-uniform theory of string motion. (See the Wikipedia page "String Vibration")
2) But if the string bends but does not elongate, the force acting on the string must be uniform across the string. This is because if the string does not elongate the curvature is constant, and if the curvature is constant then so is the field strength across the string is uniform. This is the uniform theory of string motion.
There are several reasons to believe the string orbit is uniform: 1) Gauss’s theorem says surfaces bend without elongation so curvature is constant. The string is a surface. 2) In the Hamiltonian formalism there is a tangent-cotangent vector field defined at a point on the string that results from the string as a bilinear form H: R2n x R → R. Since the tangent is perpendicular to the string, the motion of a point cannot be along the string axis. 3) The shape of the string must seek the lowest energy level and by Newtonian determinacy the shape must be a function of the initial state of the string. Therefore, even if tension and length somehow can vary, it must still be true the equations of motion are determined by length and tension at rest. 4) The use of partial differential equations based on the nonuniform theory leads to sine wave functions which have no normal vector and are defined in a plane. There is no way that sine wave functions can make a minimum surface of revolution for the string manifold.
This is an important question, I think, because if the nonuniform theory of string vibration is not correct then an entire field of mathematics and physics is also not correct. I say non-uniformity is nonsense. I do not see any mention in the literature that string curvature is constant. But how can it be understood in any other way?
Sorry, maybe I am missing the point. If you send a progressive sine wave of large amplitude along an ideal, stretched, originally at rest, straight string, then how can the wave travel along the string, if the string does not stretch? Therefore the straight string must stretch to allow the wave to travel.
Thank you for your answer. If the string has a time-independent Hamiltonian dH/dt = 0 so the string is a standing wave then there is no movement along the axis of the string. That is because the boundary condition on the string is dx/dt = 0. This is newton's law because it says that a point on the string cannot move unless acted upon by an external force. Points on the string move in a disc-shaped orbit that is strictly perpendicular to the string.
If you send sine waves along the string, the points move closer and further apart because the length of the sin wave curve is not constant. Higher modes on the string last like 0.1 sec. But if the string bends but does not stretch then the string must assume a minimum surface of revolution. Then the shape of the string is cosh x not sin x.
Clearly the string seeks the lowest energy level, which is the first mode.
The curvature of the string must be constant. The motion must be uniform. The velocity at a point must be constant. Then when the string reaches the outer limit of its orbit, it does not stop but continues with constant velocity since the potential energy of tension does not change.
Sine waves cannot make a closed manifold. The modes of the string clearly do not converge on a fixed-point limit so how can Morse Critical Point theory be understood.
You comment is greatly appreciated.
Hello! I am sorry but I am still confused. Say you have a straight 1-D string at rest, stretched and fixed between two poles. Then it is a straight line and therefore the shortest path between two poles. If there is a stationary wave, then the string will bend. If it bends, then it is no longer straight. Thus it's no longer the shortest path. Thus, it is elongated. So then what is the issue?
But according to Gauss's theorem when a surface bends it does not elongate. Its has a Latin name Theorema Egregium that means something like the amazing theorem. "The curvature of the surface does not change if one bends the surface without stretching it."This is like viewing the string as a rigid rod. So there are two models: the string elongates or not.
There is a model of string vibration based on Hook's law which says the string is like little masses that are connected by springs so points on the string move closer together then further apart. But what forces is this that moves the string in this way along the string if the tension is constant.
This model precedes Lagrange's point mass system.
Do you think waves travel down the string, reflect around the end points, and meet to make a standing wave. Why have two waves when only one is required. The problem seems to lie with d'Alembert's priniciple which gives the d'Alembert-Lagrange model when it should be the Jacobi Euler-Lagrange equation which gives a 1-periodic solution.
How can the string motion not be uniform? How can the string have high and low energy waves at the same time and not violate least action?
Thanks again for you interest. I am trying to write a question for physics stack exchange. Would you be interested in looking at it?
Yes, and you will hear that the pitch rises if the amplitudes becomes large enough And flattens as the vibration dies out. Particularly easy to hear if the tension is rather low.
The vibrating string is a «kink» travelling back and forth the string length, so there is no real «curvature» of the string. What we see is the «kink trace» doring the travel, which is very fast. 440 times a second for an A. There is some bedning of the string at the «kink», and it is dependant on the start conditions, or the bowing parameters, if we talk about bowed strings. This shape is reflected in the sound from the strings, usually filtered through a radiating body, like a guitar or a violin.
There are torison vaves travelling, as well as the kink, for the bowed and plucked string.
In a piano with impacted strings, there is also a longitudional component in the strings and thus in the sound. If this is removed, the sound will not be natural. In addition there are the major musical note vibrations of two or three strings in unison vibrating at very slightly different frequencies causing beats, slow beats in a tuned piano. The crosswise compoents live longer than the perpendicular, and the piano is reverberant in its nature partly because of the coupling of these two components. I think the guiitar are too.
The longitudinal component is a comression vave that tr
Thank you, Anders Buen, for your answer.
The problem with the idea that opposing forces move along the string is the resulting two wave solution violates least action. Why two waves when one will do? What is the force that bends the string into a higher energy state?
The Euler-Lagrange least action integral gives a wave equation that is defined in a plane. Least action is defined on displacement but if the string does not stretch, displacement is not proportional to force. So how do you get a smooth manifold from a bunch of waves that have no upper or lower boundary condition. you don't think the string minimizes action on the lowest mode.
Remember this is not current flux in a wire.
The least action needs to be based in the Jacobi least action integral not the Euler-Lagrange equation. The Euler -Lagrange is very general and very useful but it does not apply to the string. It is older than the Hamilton-Jacobi.
Do you really think a string is like a chain of masses separated by springs. That is not continuous and cannot be integrated. And again you have two elements when you only need a point mass.
A point on a string moves strictly in a disc that is orthogonal to the string. So if the string does not stretch by theorema egregium, then waves on the string are a a 1-period solution to the Jacobi-Hamilton equations, not sine waves. My forces are perpendicular to yours.
Terence B Allen The string length of a vibrsting string is longer than a string at rest, as the travelling kink gives the string a triangular form. The vibrating string has to be stretched!
I'm really lost here!
From my perspective, if I bend a string that I'm fretting, on my guitar fretboard, the note becomes higher. Because it's that bit tighter.
If I loosen the string, it becomes lower, as if I was doing dropped D tuning, where the bottom E is tuned to a D.
When you bend a guitar sting, the string elongates and tension increases. Pitch rises. The pitch is constant unless you apply force with your finger. This is Newton's first law: the string at rest or moving continues in that state unless acted upon by an external force.
During vibration, the string is rigid and does not elongate. Then the string tension is constant, and pitch is an inertial constant of motion.
In the modern theory of string vibration, the distance between two points on the string increases and decreases so the tension along the string is not constant but instead oscillates. But what internal force can push two points on the string closer together and then pull them a part?
The reason this is important to you as a guitarist is because you want to understand how to transform music in standard tuning into music in Drop D when you tune your bass string from E down to D. You need a way to understand what is happening, because it takes time to relearn the music in a new tuning. It can seem daunting, even impossible. It can also seem like all tunings are equally good; they are not.
For example, if you are in the Key of E standard you probably want to go to the Key of D Drop D. That is, instead of adding two frets to the low E string, you want to subtract two frets from all the strings except the low E. To see this, you have to write the tablature out, and when you do this, you will learn the grammar of the Dorp D tuning. It is all geometric but there is no way to visualize how tuning changes music. But for each note, it is easy to understand the math; it is just the big picture of Open G that is impossible to sus out. Open G is big, way bigger than you think.
My advice would be skipping Drop D and instead go directly to Open G or Open D tuning because these are far richer musically. Drop D is close to standard but Open D and Open G are completely different and very beautiful.
So, the problem is we have a number of languages on guitar, where each key in each tuning has a different language-level structure, and the modern theory of string vibration gives us no clue about a way to understand what makes guitar music in one tuning algorithm so richly expressive and weak or useless in another key or tuning.
But we know enough about algorithms to understand guitar, we just don't have anyone interested in both guitar music and language theory.
Changing tuning is difficult but changing the whole guitar pitch level is actually worse. It is not too hard to change from Open G to Open D and back but changing from Open D to Open E is more difficult. You may think that going up or down in pitch is not important because you don't have to relearn how to play, but you will find the problem is re-naming the note, and that the effect on the tonality of the music is profound. So if you think the recorded music is In Open D but it is actually in Open C (two steps down), you will find the music unrecognizable because the notes in Open C cannot be found in Open D. But in Open D, you can easily recognize music in Open E because the notes are all their, just two steps higher.
Mathematicians have completely overlooked the mathematics of guitar. They are the high priests of knowledge and do not countenance music as truly mathematical. The do not understand music theory and do not realize they do not understand string vibration either.
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