Talk:Kundt's tube

Physical explanation
Now, the text states that the particles end up in the displacement nodes, because it's still and quiet there. To begin with, albeit not very relevant, is that the use of the word "quiet" in an article about acoustics perhaps shouldn't be used about the place in the system where the pressure amplitude has it maximum (the displacement node is a pressure antinode...). Furthermore, that's no explanation whatsoever, in my eyes. Whence the force? Any calculations done to first order would show that there is no time-averaged force on a particle, right? I like to put it up for discussion that the driving force might be the acoustic primary radiation force as given in the below references. In start contrast to the article as it stands, it would actually explain what's happening... Oh... and don't the particles end up in the pressure nodes and not the displacement nodes? They tend to in most experiments carried out, I think.

References: L. P. Gor’kov, ''Sov. Phys. Dokl.'', 1962, 6, 773–75. A. A. Doinikov, ''J. Fluid. Mech.'', 1994, 267, 1-21. Doing it in 3D on living cells: O. Manneberg, ''Appl. Phys. Lett.'', 2008, 93, 063901 (3pp). DOI:10.1063/1.2971030

(and yes, so no-one feels the need to point it out to me, I shamelessly picked a paper of my own. The two first ones are classic, though) --Osquar F (talk) 16:17, 4 February 2009 (UTC)
 * I wrote the parts in question. I don't have particular expertise in acoustics, and it sounds like you do, so maybe you are right.  Perhaps the problem is terminology.  The references I read, such as 1 and the one cited, said that the particles ended up at the points of no (or minimum) displacement, 'displacement nodes',  which would also be pressure maximums.  Naively that makes sense to me: the particles move due to viscous force from the surrounding air, and those are the only points where the air is not moving.  The pressure minimums, or 'pressure nodes', would be displacement maxima where the particles would feel maximum force. -- Chetvorno TALK 12:13, 5 February 2009 (UTC)
 * Your point about zero time averaged force on the particles is persuasive. Why do they move?  Perhaps it is that the particles have a higher density than the air, so their inertia causes their motion to become out of phase with the air motion.   But I would think that since a standing wave in a resonator is composed of two equal waves moving in opposite directions, any force in the direction of wave propagation would be cancelled out. -- Chetvorno TALK 12:13, 5 February 2009 (UTC)


 * Nice discussion! First, then, I have some knowledge of ultrasound acoustics; I'm on the last 6 months of a PhD doing, well, manipulation of microparticles using resonant ultrasonic fields (a quite small version of this in microfluidic chips). I "know for a fact" (as far as a physicist knows something, i.e. I've seen the theory predicting the existence of the force, and seen and done a fair share of experiments on it). It would seem like an easy way of determining where the particles end up in relation to the pressure/displacement nodes/antinodes (if we're going to not simply look at the equations, which I agree is boring) would be to try to find a good picture of a tube in action, such as this one. Note that the "particles" aggregate at a distance of about $$\lambda/4$$ from the top reflector, which to a good approximation in this case looks like a hard wall. The problem in explaining the effect is that there doesn't seem to be an "easy" explanation - the effect doesn't show up unless you do some pretty heavy math. Except for the two references above, I can recommend the chapter on acoustofluidics in the book "Theoretical Microfluidics" by Henrik Bruus. Very hand-wavingly, what you need to do is consider the time-averaged force on a particle, expanding the stress tensor $$\sigma_{ik}$$ to the second order in the amplitude of the sound field. The force on an area element in direction i will be $$dF_i=\sigma_{ik}n_kds$$. Thus, the time-averaged force on the particle will be
 * $$ \langle F_i \rangle_T = \langle \int_{s(t)}^{}\sigma_{ik}n_kds \rangle_T $$
 * where s(t) is the position of the particle surface (assume it to be a sphere, otherwise I'd say there's no chance in getting away with the rest of the math) at time t. From here it gets, at least to me, a bit messy. The main problem is that I can't think of a way to give a good explanation that's correct and actually gives some kind of intuituve explanation to why there is a force, and why it's directed as it is. I really wish I had one. Assuming a "perfect" 1-D standing wave, the force at least gets an expression that's not too horrible (in more complex sound fields, the force potential doesn't necessarily have it's minima at pressure minima); a common expression for the force on a small ($$ r<<\lambda $$) particle, taken from Gor'kov (see above ref.) is
 * $$ \mathbf{F}=-\nabla V \left(\frac{f_1 }{2\rho c^2}\langle p^2(\mathbf{r},t) \rangle_T-\frac{3}{4}\rho f_2 \langle\mathbf{v}^2(\mathbf{r},t)\rangle_T \right)$$
 * where
 * $$f_1=1-\frac{\rho c^2}{\rho_p c_p^2}$$ and  $$f_2=\frac{2\rho_p-\rho}{2(\rho_p+\rho)}$$.
 * In the above expressions, V is the volume of the particle, $$\langle\cdot\rangle_T$$ is the time average, $$\rho$$ is density and c is the speed of sound; index p pertains to the particle and indexless quantities to the medium in which the particle is immersed. For a 1-D standing wave in the x direction, this can fairly easily be shown to become (assuming a $$\cos(kx)$$ dependent pressure field, an sine-dependence gives the same but with a minus sign)
 * $$F=\frac{V p_0^2 k}{4 \rho c^2} \left( f_1+\frac{3f_2}{2} \right) \sin(2kx)$$
 * where p0 is the pressure amplitude.
 * Perhaps we should include these expressions, with a reference to Gor'kov's paper and a short derivation to the last equation? --Osquar F (talk) 09:02, 9 February 2009 (UTC)


 * Yeah, there seem to be no pictures of Kundt's tube showing the node structure in relation to the ends except the one you found. As you said, it seems to show the 'particles' end up at the pressure nodes, although the styrofoam chips are much larger than the powder traditionally used, and might violate the assumption of being small compared to the wavelength.  Also, I now understand there are more forces on the particles than simple viscous force, which I assume account for the complexity of the math you mentioned; the Bernoulli force due to differential velocities, and also maybe interactions between nearby particles - it says  here that particles separated in the x direction repel, while particles separated in the y (lateral) direction attract each other.  Maybe these effects cause the particles to gather at the pressure nodes, in opposition to my common sense, although I'd like to see more sources on that before the article is changed.
 * I'm afraid I can't follow tensor mathematics. But your last equation, if I understand it, seems to show that the force on the particles does go to zero at equally spaced x-intervals, but at intervals of λ/4 - that is, at the displacement nodes  and  the pressure nodes (displacement antinodes).  Is this right? -- Chetvorno TALK 03:13, 11 February 2009 (UTC)


 * Sorry about the indices, the same equation on vector form would look something like
 * $$ \langle \mathbf{F} \rangle_T = \langle \int_{s(t)}^{} \overleftrightarrow{\mathbf{\sigma}} \cdot \mathbf{n} ds \rangle_T $$,
 * if that helps. And yes, there are a number of forces, but the primary radiation force is the one responsible for the gathering of the particles in the pressure antinodes in the picture. The direction-dependent force you mention sounds like the secondary radiation force (or secondary Bjerknes force, or just Bjerknes force, depending on who writes and if they work mainly with bubbles or particles), which is directed on a line connecting two particles and has a magnitude of
 * $$ F = 4 \pi a^6 \left( \frac{(\rho_p-\rho)^2 (3 \cos^2 \theta-1)}{6 \rho d^4} v^2 - \frac{\omega^2 \rho (\tfrac{1}{\rho_p c_p^2} - \tfrac{1}{\rho c^2})^2 } {9 d^2} p^2 \right)$$,
 * where d is the center-to-center distance, a the particle radius and $$\theta$$ is the angle between the line between the particles and the "propagation" direction of the wave. This force becomes important when the particles comes close together, and is in part responsible for creating the closely-packed "discs" in the pressure nodal planes (where the second term becomes negligible). Note that the magnitude of this force is such that that primary radiation force as given above will be orders of magnitude larger unless the particles ar very close, and that this force is an interparticle force, and not responsible for where in the sound field the particles end up. Two particles coming together from this force will simply act like a larger particle and go to the pressure node faster. That this is done by the primary radiation force, I think, is well underbuilt enough from looking at the results of Gor'kov and Doinikov (if you want to include thermal effects or other effects, check out the rest of Doinikovs papers. He's been productive...). In most cases of (ultra)sound manipulation, the Gor'kov equation gives good enough results to be the one used due to its relative simplicity.
 * I'm afraid I can't agree with your last (implied) conclusion, though (assuming you mean that the particles should aggregate at all those locations the force is zero). The function is indeed zero where you say it is, but the force being zero is a necessary but not sufficient condition for a stable equlibrium. We must also demand the potential to have a (local) minimum, i.e. the force must strive to return a particle as soon as it moves out of place. This will be true in the pressure nodes, but in the pressure antinodes (displacement/velocity nodes), the potential has a local maximum: The force is zero in exactly that point, but pushes the particle away as soon as it is disturbed.--Osquar F (talk) 14:29, 11 February 2009 (UTC)
 * Oh, and about the particles not being much smaller than the wavelength: They seem to be roughly 1/50 wavelength in radius, which for these purposes is just fine. The point here is not how exact the equations would be (they're flat and porous, to begin with), but the origin of the force and the fact that they end up in the pressure nodes. --Osquar F (talk) 14:43, 11 February 2009 (UTC)