Is "scalar wave" a legitimate term?

sophiecentaur said:

I am trying to deal with some Physics here

A big part of physics is the definition of the terms used. Pressure is a scalar, so a pressure wave is a scalar wave by the definitions of the terms used.

sophiecentaur said:

That wave equation is 'just maths' and ignores what the quantity 'u' is.

Yes, that is the point of mathematics: to abstract away the unnecessary details. Any quantity that can be represented as a scalar function on space and time is a valid ##u## for this purpose. Pressure is an example.

sophiecentaur said:

When energy is carried by a wave, there are always two physical quantities involved and how can you ignore displacement?

It is a red herring, that obviously I was not very good at ignoring yesterday. I will try to ignore the red herring today.

Neither the energy nor the displacement is relevant to the classification as a scalar wave. Pressure is a scalar, therefore by definition a pressure wave is a scalar wave. There is no requirement in the definition of the term scalar wave that depends on either energy or displacement.

sophiecentaur said:

The term Longitudinal Wave is accepted for describing a sound wave; you seem not to acknowledge this.

I have nothing against that. Multiple words often describe the same thing. The fact that sound waves can also be described as longitudinal waves is another red herring. Scalar waves and longitudinal waves are not mutually exclusive categories. It has no bearing on the fact that pressure is a scalar and therefore pressure waves are scalar waves.

Usually, I don't like formal "mathy" explanations, but I am having a hard time reconstructing the argument. It seems to be:

1. Given a scalar U that follows the wave equation,
2. [itex]\nabla U[/itex] is always a vector V
3. Therefore, there are no "true" scalar waves - there is always a vector V "behind the scenes".

Is that correct? I

Vanadium 50 said:

Therefore, there are no "true" scalar waves - there is always a vector V "behind the scenes".

And behind the vector there's a symmetric tensor ##\nabla_{i}\nabla_{j}U##, followed in turn by a rank-3 tensor, and so on ad infinitum. Where do we stop? What's the point?

sophiecentaur said:

I am trying to deal with some Physics here and that (I think) should involve dealing with cause and effect.

To do that you need to know the source of the waves, which appears nowhere in the wave equation we have been discussing. That wave equation describes how waves propagate after they have already left the source. A different equation describes how waves are generated at the source.

Again, you are making things a lot harder than they need to be by mixing up different aspects of the problem.

sophiecentaur said:

When energy is carried by a wave, there are always two physical quantities involved

What are the two physical quantities involved for a sound wave?

sophiecentaur said:

how can you ignore displacement?

I already addressed that in my previous post: you are mixing up different levels of description, which just makes a muddle.

sophiecentaur said:

it makes practical sense but using displacement is also valid

Please give a reference for how sound waves are described in terms of displacement. (Not water waves, sound waves.)

I am 90% sure there is something in Rossing about sound waves as displacement waves as a way to deal with boundary conditions. But I think we are better off dealing with the abstraction here. We don't want to say "Well, you proved/disproved it for X. But what about Y?"

There is nothing to prove. Pressure is a scalar so a pressure wave is a scalar wave.

That remains true even if red herrings are vectors and so red herring waves are vector waves. Red herrings and pressure are different.

Vanadium 50 said:

I am 90% sure there is something in Rossing about sound waves as displacement waves as a way to deal with boundary conditions.

Is the displacement a scalar or a vector? @sophiecentaur appears to be assuming it's a vector, but I don't think that's the case for a displacement treatment of sound waves (as opposed to, say, water waves).

For analyzing the behavior of, say, an old-fashioned microphone or an old-fashioned loudspeaker, a displacement treatment can be useful since the devices in question have an actual membrane whose amplitude of displacement from their equilibrium positions are the receiver or source of sound waves. But it's still a scalar wave model.

@sophiecentaur has been here for a while, so I am trying to get at the root of the misunderstanding rather than just shutting him down. "It;s a red herring" is isomorphic to "you just don't get it", which may well be true, but just because a statement is true doesn't mean it's helpful. And it's not like he's deliberately setting out to confuse.

Vanadium 50 said:

"It;s a red herring" is isomorphic to "you just don't get it",

Here “red herring” is just isomorphic to ”distraction”. I know that @sophiecentaur gets the idea that pressure is a scalar and therefore a pressure wave is a scalar wave.

He correctly pointed out that the molecular displacement is a distraction (red herring was his term and I understood it as meaning distraction). That made me realize that so were the macroscopic displacements and the pressure gradients.

Frabjous said:

Since the simplest acoustic approximation has P-P0=rho0 c0 u,I am going out on the limb and say every introductory textbook.

In that equation, ##u## is a scalar. So this model would still be a scalar wave model.

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