Waves are the *raison d'etre* of physics, and maths.

Make of it what you will. If we are to judge by the accepted formulas of wave theory, then it was always self-evident that a wave will supply its *own* measure of space and time in the form of its wavelength and period. Space and time represents the *class* of all these possibilities. For let us take a simple travelling sine wave of the form

f(x,t) = Acos[wt - kx].

Further, let us assume that this represents the solution to any wave equation whatsoever. The quantity in brackets (I'll call it the wave argument) represents the number of cycles passed per the time t and length x, multiplied by 2pi. If we denote these by n and m respectively we obtain

f(x,t) = Acos[2pi(n - m)].

From elementary calculus we obtain (I will use d as a *partial* derivative)

df(x,t)/dt = (df(x,t)/dn).(dn/dt),
df(x,t)/dx = (df(x,t)/dm).(dn/dx).

But the terms (dn/dt) and (dm/dx) here are just the frequency and wavenumber (inverse wavelength). Denoting these respectively by v and s (supposed to be nu and sigma) we obtain

df(x,t)/dt = (df(x,t)/dn).v,
df(x,t)/dx = (df(x,t)/dm).s.

In fact, it's very easy to prove that the N'th derivative is

d^Nf(x,t)/dt^N = (d^Nf(x,t)/dn^N).(v^N),
d^Nf(x,t)/dx^N = (d^Nf(x,t)/dm^N).(s^N).

Similarly, we can associate these with the differential operators

d/dt = (d/dn).v
d/dx = (d/dm).s.

Conclusion 1). The variables x and t have disappeared altogether. Given any combination of sine wave, each and every one is equally qualified to represent 'space and time'.
Conclusion 2). It is false that these represent the 'solution' to differential wave equations, (or by implication, that the principles of mechanics precedes those of wave theory). If for example we assume that f(x, t) above represents a solution to the classical wave equation,

d^2f(x,t)/dt^2 = c^2.[d^2f(x,t)/dx^2]

where c = w/k then making the substitutions gives

(d^2f(x,t)/dn^2).(v^2) = c^2.[(d^2f(x,t)/dm^2).(s^2).

However, if we follow the wave front then n = m and we simple obtain

v^2 = c^2.s^2 or v/s = c

which we knew already. This is a tautology.

Here is a stark reminder that no one has or could ever observe a differential equation. Only the 'solutions' of them are observables. And this is precisely what nature has told us in the discovery that light and matter, in addition to sound, propagate through 'space and time' as waves.

Rick

On Tue, Jun 1, 2010 at 3:31 AM, rick <rick_ballan@yahoo.com.au> wrote:
>
> Make of it what you will. If we are to judge by the accepted formulas of wave theory, then it was always self-evident that a wave will supply its *own* measure of space and time in the form of its wavelength and period. Space and time represents the *class* of all these possibilities.

What do you mean by "wave theory?" What do you mean the "class" of all
these possibilities?

> For let us take a simple travelling sine wave of the form
>
> f(x,t) = Acos[wt - kx].
>
> Further, let us assume that this represents the solution to any wave equation whatsoever. The quantity in brackets (I'll call it the wave argument) represents the number of cycles passed per the time t and length x, multiplied by 2pi.

No it doesn't. The w is in units radians per unit time, and the t is
some unit of time. So the wt reduces to radians. The same applies to
kx. The k is in units radians per unit length, and the x is some unit
of length. So [wt - kx] has unit radians, which are really
dimensionless units. Or, if you'd like to convert to whole cycles and
multiply by 2pi cycles/rad, the whole thing ends up having units of
cycles, which are still dimensionsless units.

> If we denote these by n and m respectively we obtain
>
> f(x,t) = Acos[2pi(n - m)].
>
> From elementary calculus we obtain (I will use d as a *partial* derivative)
>
> df(x,t)/dt = (df(x,t)/dn).(dn/dt),
> df(x,t)/dx = (df(x,t)/dm).(dn/dx).
>
> But the terms (dn/dt) and (dm/dx) here are just the frequency and wavenumber (inverse wavelength). Denoting these respectively by v and s (supposed to be nu and sigma) we obtain
>
> df(x,t)/dt = (df(x,t)/dn).v,
> df(x,t)/dx = (df(x,t)/dm).s.
>
> In fact, it's very easy to prove that the N'th derivative is
>
> d^Nf(x,t)/dt^N = (d^Nf(x,t)/dn^N).(v^N),
> d^Nf(x,t)/dx^N = (d^Nf(x,t)/dm^N).(s^N).
>
> Similarly, we can associate these with the differential operators
>
> d/dt = (d/dn).v
> d/dx = (d/dm).s.

Makes sense to me.

> Conclusion 1). The variables x and t have disappeared altogether. Given any combination of sine wave, each and every one is equally qualified to represent 'space and time'.

Uh? No they haven't. They're right there in v and s. And you have
still the whole equation defined as f(x,t).

> Conclusion 2). It is false that these represent the 'solution' to differential wave equations, (or by implication, that the principles of mechanics precedes those of wave theory).

I'm still waiting for you to define "wave theory." Is this just a term
you've coined?

> If for example we assume that f(x, t) above represents a solution to the classical wave equation,
>
> d^2f(x,t)/dt^2 = c^2.[d^2f(x,t)/dx^2]
>
> where c = w/k then making the substitutions gives
>
> (d^2f(x,t)/dn^2).(v^2) = c^2.[(d^2f(x,t)/dm^2).(s^2).
>
> However, if we follow the wave front then n = m and we simple obtain
>
> v^2 = c^2.s^2 or v/s = c
>
> which we knew already. This is a tautology.

That is ridiculous math. What do you mean "if we follow the wave front
then n=m"? n will equal m if wt equals kx. And, from your initial
equation, that would make wt - kx equal to 0. This would also make v
and s equal by definition, so you get 1 = 1 = 1 = 1 all over the
place.

> Here is a stark reminder that no one has or could ever observe a differential equation. Only the 'solutions' of them are observables. And this is precisely what nature has told us in the discovery that light and matter, in addition to sound, propagate through 'space and time' as waves.

What?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> On Tue, Jun 1, 2010 at 3:31 AM, rick <rick_ballan@...> wrote:
> >
> > Make of it what you will. If we are to judge by the accepted formulas of wave theory, then it was always self-evident that a wave will supply its *own* measure of space and time in the form of its wavelength and period. Space and time represents the *class* of all these possibilities.
>
> What do you mean by "wave theory?" What do you mean the "class" of all
> these possibilities?

I'll get to that.
>
> > For let us take a simple travelling sine wave of the form
> >
> > f(x,t) = Acos[wt - kx].
> >
> > Further, let us assume that this represents the solution to any wave equation whatsoever. The quantity in brackets (I'll call it the wave argument) represents the number of cycles passed per the time t and length x, multiplied by 2pi.
>
> No it doesn't.

Yes it does. I said "multiplied by 2pi" at the end.

The w is in units radians per unit time, and the t is
> some unit of time. So the wt reduces to radians. The same applies to
> kx. The k is in units radians per unit length, and the x is some unit
> of length. So [wt - kx] has unit radians, which are really
> dimensionless units. Or, if you'd like to convert to whole cycles and
> multiply by 2pi cycles/rad, the whole thing ends up having units of
> cycles, which are still dimensionsless units.

Yes, dimensionless numbers (see below).
>
> > If we denote these by n and m respectively we obtain
> >
> > f(x,t) = Acos[2pi(n - m)].
> >
> > From elementary calculus we obtain (I will use d as a *partial* derivative)
> >
> > df(x,t)/dt = (df(x,t)/dn).(dn/dt),
> > df(x,t)/dx = (df(x,t)/dm).(dn/dx).
> >
> > But the terms (dn/dt) and (dm/dx) here are just the frequency and wavenumber (inverse wavelength). Denoting these respectively by v and s (supposed to be nu and sigma) we obtain
> >
> > df(x,t)/dt = (df(x,t)/dn).v,
> > df(x,t)/dx = (df(x,t)/dm).s.
> >
> > In fact, it's very easy to prove that the N'th derivative is
> >
> > d^Nf(x,t)/dt^N = (d^Nf(x,t)/dn^N).(v^N),
> > d^Nf(x,t)/dx^N = (d^Nf(x,t)/dm^N).(s^N).
> >
> > Similarly, we can associate these with the differential operators
> >
> > d/dt = (d/dn).v
> > d/dx = (d/dm).s.
>
> Makes sense to me.
>
> > Conclusion 1). The variables x and t have disappeared altogether. Given any combination of sine wave, each and every one is equally qualified to represent 'space and time'.
>
> Uh? No they haven't. They're right there in v and s. And you have
> still the whole equation defined as f(x,t).

Yes they have. The variables (x, t) which we *started* with have now been substituted by the variables v and s at the *end*. However, the inverses of (x, t) DO have the same dimensions as v and s. Dimensional consistency at either side of the equality sign is a fundamental principle of physics. The fact that n, m are dimensionless, for eg, does not make them invalid. And of course timesing 2pi takes them into 'radian space'.
>
> > Conclusion 2). It is false that these represent the 'solution' to differential wave equations, (or by implication, that the principles of mechanics precedes those of wave theory).
>
> I'm still waiting for you to define "wave theory." Is this just a term you've coined?

Ok, fair enough Mike. If you follow through to the next step then I can explain it better.
>
> > If for example we assume that f(x, t) above represents a solution to the classical wave equation,
> >
> > d^2f(x,t)/dt^2 = c^2.[d^2f(x,t)/dx^2]
> >
> > where c = w/k then making the substitutions gives
> >
> > (d^2f(x,t)/dn^2).(v^2) = c^2.[(d^2f(x,t)/dm^2).(s^2).
> >
> > However, if we follow the wave front then n = m and we simple obtain
> >
> > v^2 = c^2.s^2 or v/s = c
> >
> > which we knew already. This is a tautology.
>
> That is ridiculous math. What do you mean "if we follow the wave front
> then n=m"? n will equal m if wt equals kx. And, from your initial
> equation, that would make wt - kx equal to 0. This would also make v
> and s equal by definition, so you get 1 = 1 = 1 = 1 all over the
> place.

No Mike, its a very standard condition that [wt - kx] = 0. For only then do we get wt = kx and w/k = v/s = x/t = c, called the phase velocity. If we set x = constant then the wave oscillates at a point in space. t = constant gives a 'snapshot' of the wavelengths stretching out in space. And setting both equal to a constant, here zero, follows the wavefront. But at no time does v = s by definition. It is vt = sx.
>
> > Here is a stark reminder that no one has or could ever observe a differential equation. Only the 'solutions' of them are observables. And this is precisely what nature has told us in the discovery that light and matter, in addition to sound, propagate through 'space and time' as waves.
>
> What?

It is generally regarded that a differential equation precedes its solution, as if it was the 'question' to which the sine wave like the one above is the 'answer'. It gives the impression that it came first and that wave theory grew out of mechanics (and is therefore a 'branch' of mechanics). More likely the differential equations were made to suit the observable conditions of waves and the process was inverted after the fact. (And I confess that my use of the word 'tautology' was a friendly dig at Gene for using it to me, just to point out that without observation, all maths reduces to tautology. I did after all keep saying "No, I don't mean ANY p/q but only the convergents". There IS a difference you know. ).

So getting back to your question, while I dislike definitions that try to reduce a whole lot of stuff into one sentence, I suppose that wave theory could be defined as the theory which says that the universe is 'wave-like' in character, as opposed to mechanical or 'particle-like'. This is because space and time are now *identified* with the wavelengths and periods of physical waves and not on the rods and clocks of our own making. In this model Fourier Analysis takes centre stage. The x-t axes can now in principle be inverted into wavenumber-frequency 'space'. I suppose it could also mean *that* part of each theory, acoustics, quantum mechanics, optics etc...which uses FA as its basis. It is the general conditions that they all have in common. But yes it is my own coinage, simply because I didn't know what else to call it.

I like arguing with you Mike.
>
> -Mike
>