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Background:

https://en.wikipedia.org/wiki/Longitudinal_wave

'Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or the opposite direction to, the direction of propagation of the wave. ... The other main type of wave is the transverse wave, in which the displacements of the medium are at right angles to the direction of propagation.'

Also see Derivation of Green's Function for Wave Equation where:

$ \delta(r-ct)/4\pi r$ is the time-space Green's function (or elementary solution) to the 3D wave equation.

EDITS -----------------------------------------------------------------------

My question is:

Is $ \delta(r-ct)/4\pi r$ a transverse or longitudinal wave, or is it either one? Please give references if possible. Also please explain how it can represent each type of wave, and (if possible) the differences in the required initial conditions needed by the wave equation for a transverse vs a longitudinal elementary solution.

Note that the terms "transverse" and "longitudinal" DO apply to scalar fields. They do not only apply to vector fields.

For example see:

https://www.physicsclassroom.com/class/waves/Lesson-1/Categories-of-Waves

https://www.khanacademy.org/science/ap-physics-1/ap-mechanical-waves-and-sound/introduction-to-transverse-and-longitudinal-waves-ap/a/transverse-and-longitudinal-waves-ap1

https://www.acs.psu.edu/drussell/demos/waves/wavemotion.html

http://230nsc1.phy-astr.gsu.edu/hbase/Sound/tralon.html

and many other references--search Google for 'transverse vs longitudinal waves'

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    $\begingroup$ Have you heard of the Hairy Ball Theorem? (e.g.: scienceblogs.com/builtonfacts/2011/08/12/… ) $\endgroup$
    – JEB
    Jul 30, 2018 at 17:29
  • $\begingroup$ In your "background" a displacement is discussed. This would imply a vector field, such as the gradient of the potential in question. $\endgroup$
    – my2cts
    Jul 30, 2018 at 21:39
  • $\begingroup$ Can that be related to a guitar string? There the displacement is orthogonal to the string and a vector field is not required. $\endgroup$
    – user45664
    Jul 30, 2018 at 21:47
  • $\begingroup$ Re. hyperphysics.phy-astr.gsu.edu/hbase/Sound/tralon.html $\endgroup$
    – user45664
    Jul 30, 2018 at 21:51

3 Answers 3

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The question doesn't make sense, since the terms "transverse" and "longitudinal" don't apply to scalar fields. They refer to the relationship between the polarization of a wave and its propagation direction, but a scalar wave has no polarization; its value at every point is just a number.

On the other hand, if you're using the scalar wave equation to describe a wave with polarization, you're simply using it as a simpler incomplete description; you cannot recover the polarization information.

Note that the terms "transverse" and "longitudinal" DO apply to scalar fields. They do not only apply to vector fields.

In all the examples you gave, we started with a physical phenomenon that we knew was either transverse or longitudinal already. Then, we chose to describe the configuration by taking a scalar value at each point. (This isn't even possible outside of the simplest of cases.) These scalar values satisfy the wave equation. You can clearly see this reasoning doesn't go backward; the scalar wave equation by itself doesn't contain all the information. You might as well ask, "is the scalar wave equation actually describing sound or light?"

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  • $\begingroup$ See my 'Background' paragraph. Referring to that it seems to me that it would apply to the scalar wave equation. Transverse/longitudinal refer to the direction of motion of the 'particles' in the medium vs the propagation direction. Eg. an acoustic wave. see quora.com/Is-a-sound-wave-a-vector-quantity-or-scalar-quantity $\endgroup$
    – user45664
    Jul 30, 2018 at 18:12
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    $\begingroup$ @user45664 The point is that if you just have the scalar wave equation, there's no such thing as transverse or longitudinal, while if you're using it to approximate/simplify something else, the answer is just indeterminate -- you've stripped out the necessary information. $\endgroup$
    – knzhou
    Jul 30, 2018 at 18:54
  • $\begingroup$ see also hyperphysics.phy-astr.gsu.edu/hbase/Sound/tralon.html . I think the term transverse/longitudinal do apply to scalar waves but I agree that maybe they are not distinguished by the scalar wave equation by itself. I think the waves do not have to be polarized. $\endgroup$
    – user45664
    Jul 31, 2018 at 18:18
  • $\begingroup$ please see my comment to GiorgioP . This is where I am confused now. $\endgroup$
    – user45664
    Dec 26, 2018 at 4:01
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    $\begingroup$ @user45664 Imagine you were doing a basic physics question about motion in one dimension, using $F = ma$. Given that $a = 3 \text{m/s^2}$, does that mean $a$ is really in the horizontal or vertical direction? Which direction is that one dimension? There simply isn't enough information there to say. $\endgroup$
    – knzhou
    Dec 26, 2018 at 9:43
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Usually transversality is discussed in the context of EM force fields. Here the question refers to a potential. For example for a Coulomb potential, the time like component of a four vector potential, the concept has no Lorentz or gauge invariant definition. Note that the Coulomb potential is not a scalar field.

Is the electric field of a point charge transversal? Almost, as $\vec \nabla \cdot \vec E = 0$ thus $\vec E$ is transversal everywhere except at the origin.

Here, the OP discusses a solution of the 3D wave equation, the source of which is dimensionless. The "force" associated with it is like the electric field of a point charge, transversal everywhere except in the origin.

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    $\begingroup$ A potential is also a field. It is a scalar field, as opposed to a vector field, but it is still a field. $\endgroup$
    – AccidentalFourierTransform
    Jul 30, 2018 at 20:21
  • $\begingroup$ @my2cts How does 'no Lorentz or gauge invariant definition' apply? $\endgroup$
    – user45664
    Jul 30, 2018 at 21:15
  • $\begingroup$ @my2cts For example a guitar string has transverse waves. $\endgroup$
    – user45664
    Jul 30, 2018 at 21:17
  • $\begingroup$ @AccidentalFourierTransform There are also vector potentials. But in general I do not think fields are required to discuss transverse/longitudinal waves. $\endgroup$
    – user45664
    Jul 30, 2018 at 21:24
  • $\begingroup$ @my2cts Sorry but i still don't understand. I am thinking in terms of the paragraph following 'Background:' in my question--and my comment there. $\endgroup$
    – user45664
    Jul 30, 2018 at 21:30
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Longitudinal and transverse 3D spherical waves obey the same wave equation. It is neither the equation nor the boundary conditions which can tell us whether we are dealing with a longitudinal or transverse wave. Once we have the direction of propagation of a wave, it is the physical interpretation of the model which allows to classify the wave as longitudinal or transversal.

This is true for all the solutions, including the elementary solutions (Green functions of the equation). Notice that even if one would be working with vector waves, it is not the function dependence on $r-ct$ which could tell anything about the transverse/longitudinal character.

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  • $\begingroup$ Re. "..nor the boundary conditions..", is that also true for the initial conditions? For the second initial condition giving initial 'velocity': in one case (longitudinal) the initial velocity would be in the propagation direction and in the other case (transverse) the initial velocity would be orthogonal to the propagation direction. $\endgroup$
    – user45664
    Dec 26, 2018 at 3:53
  • $\begingroup$ Sure. For a scalar hyperbolic equation like a wave equation the initial conditions correspond to provide function and its time derivative over the whole space at the same time. These are two independent functions and there is no place where to distinguish between transverse or longitudinal waves. For pure scalar waves, distinction enters only at level of interpretation of what the wave field is. But it is understood that if one models a phenomenon with a differential equation the meaning of the scalar field is known and consequently the transverse/longitudinal character of the wave. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Dec 26, 2018 at 6:55

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