Recent content by Arend

Would not be even possible to say that the electron (initial) position is inside of the plane wave wavelength? I guess that the Gaussian time dependency result is like a "random walk" for the electron.

So, the electron particle position uncertainty math model is not something that is fixed. I can not say, just as an example, that every electron, with low velocity, potentially free, has a standard deviation position of 5nm (or so) ?

Assuming the "static" state is pure theoretical, no trap procedure. The wave function can be only estimated by measurements? (can not be defined theoretically)

Does not position uncertainty exist inherently for an electron, even in a static state? So, there is a wave function Ψ associated, that results in a probability distribution |Ψ|^2, that can be represented by a Gaussian representation, no? Thanks

Found plenty of math expressions pointing to the equivalence of the wave function as a Gaussian wave packet, but always presenting the Gaussian parameters (~width, peak) as undefined. Is it possible to have a defined Gaussian for the electron position in this case? Thank you

You know PeroK, I have the naive or fooled idea that classical EM is just "messed" by the uncertainty, but still there ;) Thanks

The main point of the photoelectric effect as a particle phenomenon is the threshold frequency. Now, degrading our actual atom quantum model to the old Bohr model, the ("wrong") idea of electron orbital frequency is applicable. In this case, we can observe that there is a correlation, or...

Understood. tks!

For example, a not charged particle, such as a neutrino, even a very small mass would generate a huge gravitational field closer to such "point" mass. Changing the question, would the gravitational field have a limitation due to the quantum enviroment?

Sorry, incomplete question. For example, applying the classical gravitational force between two electrons at Planck distance results in a strong force (~1.35 x 10^20 N). I know, that the "size" of the electron is bigger, but could be the case for other scenario. The point is, the classical...

Hi. Following the classical mech. g=-GM/R^2, for quantum distances the gravity force would be strong. What are the quantum mechanisms that cause the effects of gravity to vanish? The position uncertainty or the probabilistic wave function can be a justification for suppressing strong...

Mean stress due to direct relativistic effect. For sure any mass under acceleration gets stressed due to the applied force. No stress or deformation forces due to the Lorentz contraction.

Now I can see, I should go deeper before going into paradoxical ideas. But got (wrong) perception due to some older papers (including Bell) that claimed the stress condition under relativistic acceleration. However, Minkowski pointed out more than a hundred years ago that relativistic length...

observer

Very interesting. So, for the case where it is applied a homogeneous acceleration to the object, the relativistic kinetic energy expenditure will be material dependent?