Recent content by eneacasucci

This explanation is super informative and accurate, I cannot thank you enough for this. My notes were wrong and I was so confused about it.

I've found this image online (ref: https://edurev.in/t/188018/Origin-of-Energy-Bands ), it should be the graphical representation of the potential binding energy between two nearest-neighbor atoms. I don't understand how it can be correlated to the Lennard-Jones pontential graph: , in which we...

I found this image on the wikipedia italian page for the Lennard-Jones potential and I think the derivative displayed are wrong: not only in that reange (below r_eq) the negative derivative of the force should be negative and vice versa, but also the physcal meaning is that F(r) = -...

thank you so much, that is exactly what i was asking

##2\pi \int_{-1}^1 d\mu' = 4\pi## we can't write this, in our case because we have a function ##\varphi## that depends on ##\mu## (or ##\theta##) so the only integral that we can directly solve is the one related to the azimuthal angle (that gives the ##2\pi##), because none of the functions...

Thank you so much! Could I ask you also about this mathematical part "the integral over Omega in spherical coordinates has a part that is shown with the polar angle theta and then there is another integral from 0 to 2\pi for the azimuthal angle. This last integral should have as an integrand 1...

I have a question about the neutron transport equation, my question is more about mathematics, from the book Duderstadt Hamilton I tried to make the calculations, it should be quite simple but still I don't understand where the 2\pi terms went... the integral over Omega in spherical...

So to make it discontinuous we should explicitly define a piecewise function like: f(x)=ln(x) for x>0 and f(x)=0 for x≤0? It is not enough to say "consider ln(x) over whole R" then, right?

it is correct to say that if we consider the whole of R as the domain, the function ln(x)is not continuous, whereas if we consider the domain of the function as the domain, then it is continuous?

oh yes right, I wasn't thinking about it but of course because as per definition: The radius of a circle is perpendicular to the tangent at every point on the circle

Thank you, i Think it is clearer now. I couldn't find on books proper information about those geometrical constructions. So it should be something like this

image construction with concave and convex mirrors I don't know exactly how to explain it but what I have noticed is that when I use the rules for image construction with concave and convex lenses the image is wrong if I don't use the extension of the ray to the tangent to the mirror (purple...

Thank you! But I think I made a mistake here because BCC metals, due to their more open lattice structure and lower packing efficiency, are generally more susceptible to thermal creep than FCC because thermal creep is mainly governed by dislocation climb (over dislocation glide) that is a...

@Astronuc I think I'll need your knowledge also to find an answer about this... of course everyone is more than welcome to join in! 😂

"Austenitic steels (fcc) irradiated and tested at high temperatures show no increase in either yield". I understand that irradiation hardening occurs because the movement of dislocations is hindered by the presence of defects. This phenomenon can be observed in both body-centered cubic (bcc) and...