TITLE: Covering Space of Circle
QUESTION [4 upvotes]: It is easy to see that
$p:\mathbb{R} \rightarrow S^1: p(t)=(\cos t, \sin t)$ is a covering map of $S^1$
(Indeed take a point $x_0$ in $S^1$, take $U=S^1\setminus \{-x_0\}$ as an open neighborhood and then $p^{-1}(U)=\cup_{n \in Z} J_n$ where $J_n=\{t\in\mathbb{R} : t_0-n-1/2<t<t+0+n+1/2\}$. $p$ then maps $J_n$ homeomorphically onto $U$.)
I can't see why we can't take a similar $p:(a,b) \to S^1$. For example, Massey states that if we take the open interval $(0,10)$ onto the circle, then some points in $S^1$ fail to have an elementary neighborhood.

REPLY [4 votes]: Look at a preimage of a small neighborhood $U$ of $p(10)$. It will have one component which doesn't map homeomorphically onto $U$.


TITLE: Find a side of smaller triangle
QUESTION [2 upvotes]: $\triangle ABC$ has side $\overline{AB}=160 cm$, $\overline{BC}=190 cm$, and $\overline{CA}=190 cm$.
Point $D$ is along side $\overline{AB}$ and $\overline{AD} = 100 cm$. Point $E$ is along side $\overline{CA}$.

Determine the length of $\overline{AE}$ if the area of $\triangle ADE$ is three-fifths the area of $\triangle ABC$.

Is my solution correct? Thanks in advance!

REPLY [2 votes]: Suppose a triangle $T_1$ has two sides $m,n$, and that the angle between those two sides is $\theta$. If we create a second triangle $T_2$ with two sides $\alpha \cdot m, \beta \cdot n$ and keep the angle between them as $\theta$, then
Area $(T_2) \;$ = $\; \alpha \beta \;\cdot \;$ Area $(T_1)$
This can easily be seen, by first keeping one of the sides (say $n$) fixed, and multiplying the other (in this case, $\alpha \cdot m$). This intermediate triangle would have the same height as the original, but the base would get multiplied by $\alpha$, so the area would also be multiplied by $\alpha$. Then the above result follows by multiplying the other side by its factor.
In your question, side AB gets multiplied by $\frac{100}{160}=\frac{5}{8}$, whereas the area gets multiplied by $\frac{3}{5}$. Thus the second multiplier (call it $x$) must satisfy
$$\frac{5}{8} x = \frac{3}{5}$$
which yields $x=\frac{24}{25}$.
Thus the length of AE will be
$$\frac{24}{25} (190) = \boxed{182.4}$$


TITLE: Is it possible to have a non mathematical explanation of the dependence of Pair Production cross section with energy?
QUESTION [1 upvotes]: Cross section of pair production with interaction of photon increases with energy. But why does that happen? I want a non mathematical answer on this.

REPLY [5 votes]: A photon, no matter how much energy it has, will not turn into a pair of particles because of conservation of momentum  and energy. The photon does not have a center of mass because it has no rest frame . Any pair of particle-antiparticle will have a rest frame, so there is a reductio ad absurdium.
In order for a pair to appear there must exist a  target nucleus which will interact with the gamma and together with the pair obey energy and momentum conservation, the nucleus taking up momentum and energy.
So any cross section will depend on the nucleus, i.e. the target, that the gamma hits. There are tables on this, depending on the nucleus.
This is a simple Feynman diagram that also shows pictorially  what is necessary to get a cross section to first order:

Feynman Diagram of electron-positron pair production. One can calculate multiple diagrams to get the cross section

Thus the cross section will depend on the field of the particular nucleus. For example this plot (fig 2.2) has the cross sections calculated for pair production in carbon, for the specific needs of an experiment:

The cross section rises with energy because  the higher the energy of the gamma, the more it  penetrates the nucleus, where the electric fields are stronger,(1/r^2). Higher frequencies probe smaller distances, and the probability of interacting with the electric field larger. There is a saturation when the gamma energy is high enough, it will  scatter with the quarks which exist in the nuclei, and create jets. The specific pair production crossection reaches a plateau.  Also as the energy rises, more pair production channels will open .
This is about non mathematical as I can get.