Recent content by anuttarasammyak

Deleted

@davLev Why we observe more electrons than positrons would become your question then.

The formula for the 2nd and the 3rd case is written as $$\frac{a^2+s^2-as(\sin\theta + \cos\theta)}{2 \sin\theta \cos\theta}$$ which is obviously symmetric between s and a, and also between sin and cos, i.e. symmetric wrt y=x axis.. Thus we observe that the 3rd case solution comes from the 2nd...

The answers for the first case and the 4th case are symmetric for exchange of s and a. May we expect the same for the 2nd and the 3rd cases?

The electromagnetic repulsive force is an obstacle that must be overcome to achieve nuclear fusion. This condition is satisfied in the interior of stars.

Yes. Ref. https://www.feynmanlectures.caltech.edu/II_01.html

@gen x Let us see the case of moving-spinning disk as sketched below. Orbit of O is a line. The disk is rotating around O with ##\Omega## along the orbit. Orbit of A is a helix. The disk is rotating around A with the same ##\Omega## along the orbit.

If I did it right, for 1 Bq radioactivity we should have ##5 \times 10^7 ## kg Tellurium sample, which is more or less weight of Titanic.

FYI Landau-Lifsitz's Mechanics explains in section 31 Then they investigate the case where O' which is distance a from O is taken. I recommed you to find a copy and read it.

A sample made of a lot of radioactive nuclei started to decay immediately after it was generated.

$$s=\sqrt{2}/2$$is the minimum value satisfying it though not larger any more. And also $$s=\frac{1}{2\sqrt{2}} $$is the maximum.

Suppose the object is rigid. Fixing the positions of any three points on the object is sufficient to construct a co-moving coordinate system. Any frame of reference that remains at rest with the system is also a co-moving coordinate system. Usually, one chooses a frame whose origin is at the...

It seems not necessary. For any angle configuration we get the same result.

The contraposition: "E[ X ] converges thus E[ |X| ] converges" seems easier to handle. $$ E[|x|]:=\int_{-\infty}^{+\infty} f(x)|x| dx = -\int_{-\infty}^{0} f(x)x dx + \int_{0}^{+\infty} f(x)x dx = E[x]-2 \int_{-\infty}^{0} f(x)x dx $$ As a finite value cannot be divided into two infinites...

Integral for even n diverges clearly to plus infinite. Integral for odd n is zero for integrand [-L,L]. However we do not have such a constraint thus the integral diverges plus or minus.