ok, breaking surface tension with harmonics, i'm dumb, help?

Question

is it possible to use a well placed sonic vibration to hyper-accelerate air molecules, allowing easier displacement of them and less friction becuase of displacing them?
so if i beamed a sonic wave in front of a rocket, would it allow me to slip through air easier, since the median of said air is moving, causing it to be more likely to be compressed and even give that fraction versuses the air pocket thing.

Answer 1

kind of criss-crossed your question, and interlacing concepts
of physics with hydrodynamics,oscillations and measurement techniques,photons,gravity,thermodynamics ,to help in your
question,since mathemathics can solve everything probably you can rely in this formula for a start,however it can be the other way around it will only contribute to make you more confused;

....q satisfies p Delta q = cos . The resulting functions of the form Phi(cos( p; q) are called zonal. It is useful to expand Phi(p Delta q) in terms of the spherical harmonics Y j on S n (cf. 11, 12] This we can do by employing the famous Addition Theorem for spherical harmonics [11]: P (n 1; p Delta q) n d n ( dn ( X j=1 Y j (p) Y j (q) 3.2) Here, d n ( is the dimension of the space of n 1 dimensional harmonic polynomials homogeneous of degree and n is the volume of S n . This results in the expansion Phi(p Delta q) 1 X =0 dn ....
where = 1 ; Gamma1 ) by x 1 = cos 1 x 2 = sin 1 cos 2 x 3 = sin 1 sin 2 cos 3 : x Gamma1 = sin 1 sin 2 : sin Gamma2 cos Gamma1 x = sin 1 sin 2 : sin Gamma2 sin Gamma1 : The Laplace operator can be written in the form [14] (8:1) Deltap = p rr Gamma 1 r p r 1 r 2 Delta p where Delta is a second order elliptic operator in ; for = 3, and Delta p = 1 sin 2 (sin p ) 1 sin 2 2 p 2 : Consider a surface S ffl : r = 1 fflf ( with jf j C 2 1, and ....

....(and elliptic estimates) kr Delta k Fk L 2 (B) kFk H 2k 1 (B) K 1 kr Delta k Fk L 2 (B) Thus, it suffices to show that (8.14) holds with kFk H s (B) replaced by k Delta k Fk L 2 (B) and kr Delta k Fk L 2 (B) when s = 2k and s = 2k 1, respectively. On the other hand, since [14] Delta X m;n Fnm (r)Y nm ( X n;m ( 1 r Gamma1 r (r Gamma1 r Fnm ) Gamma n(n Gamma 2)Fnm )Y nm ; we have, using the othogonality properties of Ymn , 8:16) k Delta k Fk L 2 (B) X m;n Z 1 ffi r Gamma1 j[ 1 r Gamma1 r (r Gamma1 r ) Gamma n(n ....

Answer 2

It sounds like OK physics. The device beaming the force would need to be without recoil, as this would make it a rocket as well.
You would want your wave of force to 'make' air move quickly beneath the plane and slowly over it, to give lift like a wing, except here the the plane uses the air to pull itself forward. Sounds a bit like a propelor plane.

Answer 3

Droplets of a fluid can be formed by the tips of waves on the surface of the fluid breaking off. This droplet ejection occurs when the surface waves reach a certain critical amplitude. Hence, droplets can be produced through any mechanism that causes growing surface waves. One such mechanism is Faraday Excitation, where the bottom of the container of fluid is subjected to a forcing oscillation/vibration. I produced a short animation (in mpg and avi formats) illustrating a growing droplet ejecting wave caused by the Faraday Excitation.

Previous research has shown that the size of an ejected droplet is determined by the wavelength of the surface that ejects it.[1-3] There is in fact a universal constant factor that relates the diameter of a droplet to the wavelength of the surface wave that ejected that droplet. Currently, Professor Tom Donnelly’s particle sizing research group (HMC Physics) is generating droplets through the Faraday mechanism, and precisely measuring the size of these droplets through MIE scattering techniques. The goal of my research has been to provide the Donnelly research group with predictions for the wavelength of the surface waves generated by Faraday Excitation on any fluid with any set of values for the oscillation amplitude and oscillation frequency of the forcing vibration.

I used linear stability analysis to the Faraday Excitation model as my main method of analysis of this problem. My analysis is similar to that used by Kumar and Tuckerman in 1994.[4] In their work, Kumar and Tuckerman generated neutral stability curves for a system of two fully viscous fluids subjected to Faraday Excitation. These stability curves predict the regions of parameter space for which unstable surface waves, surface
waves that grow without bound, can occur. Since only growing fluid surface waves can eject droplets, this analysis predicts the conditions under which droplets will form for the two fluid problem.

If in the experimental setting, the amplitude of the oscillator is slowly increased until droplets are observed, this exclusively selects the minimum amplitude unstable wave mode. In other words, the waves that are producing droplet ejection must correspond to the point on the neutral stability curve that is a minimum in amplitude.

I generated the neutral stability curves for a generalized fluid over a 25 decade range of forcing frequency. Figure 1 is the plot of the neutral stability curves at low frequency, while Figure 2 is a plot of the neutral stability curves at high frequency. I also produced a short animation (in mpg and avi formats) that shows how these neutral stability curves change qualitatively as the non-dimensional forcing frequency is increased from 10^-10 to 10^5.

Answer 4

In short no.Why? have you heard of for every action there is an equal but opposite reaction. The energy used to create the sonic wave would reduce the energy of the rocket and actually end up with a loss of acceleration.

Answer 5

1st nope. Such "well placed sonic vibration" wouldn't exist in a places for a rocket to fly. 2nd, yes, but sonic wave would help but not enough net force to "slip" through air easier. It wil lbe easier, but not...like SLIP. Quite the logical theory. aerospace major?

Answer 6

it would reduce air friction, because moving particles of air do have less pressure, supposedly, however it is not likely that causing the vibrations would be efficient enough to become practical. in order to have that large of an effect you would need a wave with a fairly large amplitude...

also, once you broke the sound barrier it would become inneffective anyways