Elastic longitudinal waves on the ether

In a past article of mine, I mentioned the fact that Tesla had discovered how to produce longitudinal electromagnetic waves. These are waves that behave like a sound wave, a mechanic wave that propagates in an elastic mean. In the case of electromagnetic waves, the mean is the ether. Ether, as you know, behaves like a solid, because it is able to transmit also transversal waves. This requires a shear modulus, which is the characteristic of a solid.

Said this, we can try to obtain some more information about the ether. So, we have to study how a sound wave propagates on a solid. This is a somewhat difficult article, with some additional formula. So, you can skip directly to the conclusion, if you prefer.

Two starting hypothesis

To start, let’s try to do some further consideration on the speed of a longitudinal wave on a solid. To do this we can consider a volume of ether with the form of a prism. So, we can make the following two hypotheses:

  1. The elastic longitudinal wave appears like a train of longitudinal vibration parallel to the direction of propagation, the x-direction of the prism. Deformations are longitudinal but the material is stressed also in the perpendicular directions y and z. To explain this, think to an eraser. If you pull it, the eraser will lengthen but it also will shrink its section.
  2. The behavior of the material is elastic. It has a normal elastic modulus or Young modulus E, that describes how much the material is stiff. It also shows a shear modulus or Poisson coefficient ν, that describes the ratio between the elastic transversal deformation and the longitudinal deformation. In order to better understand, please, do not forget the eraser that stretches but also shrinks its section.

Let’s now consider some etheron contained in a very small cubic volume dV = dx*dy*dz, inside our prism. dV means that the volume is an infinitesimal part of the whole prism. The first face that the wave encounters has coordinate x, the second x+dx.

When the prism is run by a longitudinal elastic wave the said volume, dv, will have to cope with the two forces on the two faces perpendicular to the x-axis of propagation:

σx*(dy*dz) on the face with x coordinate and   (σx+dσx )*(dy*dz)on the face with coordinate x+dx.

σ is the pressure generated by the wave on the face. By multiplying it for the area of the face (dy*dz) we will obtain the force acting on it.

The writing of some formula

By considering the direction and, thus, the sign of these two forces, as shown in the picture, we can write the following equation:

Then we can rewrite it like this:

This is the resulting force acting of the volume dV. This force is equilibrated by the inertia forces.

Due to the elastic wave hitting the volume dV, the elongation of the etherons that are near to the face x is a little different from the one of the etherons near to the face x+dx. We can suppose however that dx is near to zero and thus the etherons vibrate the same way crossing the thickness of the little cube.

A generic equation

Let’s formulate now the equilibrium equation of the little volume. We want to arrive to formulate an equation. It should describe the position of a generic etheron depending on its position at rest (when the wave is absent) and in a lapse of time t.

We can start with the generic equation:

which is the differential equation for the Newton equation F=m*a.

We can apply it to our volume, for which we have calculated the resulting force acting on it:

where ρ(dxdydz) is the mass of the volume, and ρ is the density of our ether.

We have now:

Then we have to search for a relation between the tension σx and the position of the single etheron s.

The inverse of a speed

By making a few general remarks on deformation and its equations, we are now able to write the equation of the wave:

With a dimensional analysis, we can understand that the term is the inverse of a speed.

We can write that the speed of the longitudinal wave is:

Some conclusion

Much interesting is the consideration that a longitudinal wave is always faster than a transversal wave. If the light that is a transversal wave runs at 300000km/sec, a longitudinal wave can travel faster in the ether, denying thus the Einstein’s statement that the speed of light can’t be outmatched.

Moreover, considering the last formula I have just expressed, you can understand some more characteristic of the ether: to obtain a very high speed of the longitudinal wave the ether has to be very rigid with a very high elasticity module but also with a very low-density ρ.

Leave a Reply