Showing the annihilation of an electron-positron pair How to show that the annihilation of an electron-positron pair under emission of one real photon (mass zero) is forbidden by energy-momentum conservation, the emission of two photons is allowed?
Why can't a single photon produce an electron-positron pair? In reading through old course material, I found the assignment (my translation):     Show that a single photon cannot produce an electron-positron pair, but needs additional matter or light quanta.   My idea was to calculate the wavelength required to contain the required energy ($1.02$ MeV), which turned out to be $1.2\times 10^{-3}$ nm, but I don't know about any minimum wavelength of electromagnetic waves. I can't motivate it with the conservation laws for momentum or energy either.  How to solve this task?
Some issues with timezones in PHP have been in the back of my mind for a while now, and I was wondering if there are better ways to handle it than what I'm currently doing.  All of the issues revolve around reformating database stored dates:  When dealing with a site that has to support multiple timezones (for users), to normalize the timezone offest of stored timestamps I always store it with the server timezone using the CURRENT_TIMESTAMP attribute or the NOW() function.   This way I don't have to consider what timezone was set for PHP when the timestamp was entered (since PHP time functions are timezone aware). For each user, according to his preference I set the timezone somewhere in my bootstrap file using:  date_default_timezone_set($timezone);   When I'm looking to format dates with the php date() function, some form of conversion has to take place since MySQL currently stores timestamp in the format Y-m-d H:i:s. With no regard to timezone, you could simply run:  $date = date($format,strtotime($dbTimestamp));   The problem with this is that date() and strtotime() are both timezone aware functions, meaning that if the PHP timezone is set differently from the server timezone, the timezone offset will apply twice (instead of once as we would like).  To deal with this, I usually retrieve MySQL timestamps using the UNIX_TIMESTAMP() function which is not timezone aware, allowing my to apply date() directly on it - thereby applying the timezone offset only once.  I don't really like this 'hack' as I can no longer retrieve those columns as I normally would, or use * to fetch all columns (sometimes it simplifies queries greatly). Also, sometimes it's simply not an option to use UNIX_TIMESTAMP() (especially when using with open-source packages without much abstraction for query composition).  Another issue is when storing the timestamp, when usage of CURRENT_TIMESTAMP or NOW() is not an option - storing a PHP generated timestamp will store it with the timezone offset which I would like to avoid.  I'm probably missing something really basic here, but so far I haven't been able to come up with a generic solution to handle those issues so I'm forced to treat them case-by-case. Your thoughts are very welcome Say we have a rod with a fixed axis of rotation. If we give it a push at some point other than the axis, it will start to rotate. Say our force is $F$, then we might write: $\Delta P=F\Delta t$. I’m wondering how to incorporate angular momentum here. We know that for a single particle it holds that $L=r\times p$. And for an extended object with a fixed axis of rotation, we generally might write $L=I\omega$. However, can I write $L=r\times p$ too for the extended object? I would think that I would need to integrate. So we have $p_\alpha=m_\alpha v_\alpha=m_\alpha r_\alpha\omega$, where we consider an infinitesimal mass $\alpha$. So should I write $L=\int r^2\omega\,dm$? I’m confused whether to work with this integral, or just use $L=r\Delta P$ at once.  The reason I'm asking, is because I'm trying to calculate the sweet spot of a baseball bat. So this is my attempt at solving the exercise. Quoted vs unquoted string expansion Movie/TV series, which bullets carried DNA of the shooter Every partial order can be extended to a linear ordering How do I show that every partial order can be extended to a linear ordering?      I think that I manage to prove that claim for finite set, how can I prove it for infinite set?  Thank you. Is Checksum to file back decoding possible? So, I was having this weird thought. If the checksums(MD5 or SHA...) are unique to any file and are modified if any part of the file is edited, why cant the checksums be decoded back to get the original file? That is if I want to send a large(say 700MB) video, instead I could send just the hash and the file be decoded locally in my friend's computer. I know the thought is stupid, but is it possible by any chance? Or am I missing some thing here? And forgive me if I'm wrong, I ain't a professional. How to run scripts on start up? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? How can one differentiate between "who" and "whom"? Error message "Strict standards: Only variables should be passed by reference" What is Ubuntu's status on the Meltdown and Spectre vulnerabilities? The most obvious answer everyone over the Internet is giving to this question is "Because anime is produced much faster compared to manga, they need to put in fillers in order to slow down the anime series". However, that shouldn't be the only reason.  For example, consider One Piece. The show is going on for more than 500 episodes, and less than 10% are fillers. Comparing that to Naruto, the show has almost the same number of episodes (original and Shippuuden combined), but almost 50% of its episodes are fillers. Needless to say, the manga of Naruto is so far ahead that the anime won't be able to catch up even if they produce 100+ episodes without any fillers. How to align right within equation array Microsoft Edge missing in Start menu and pins How to create vector graphics In Photoshop? I know people prefer using CorelDraw or Adobe Illustrator, but I want to know if there is a technique I can follow to use Adobe Photoshop to make vector Images. A probability theory question about independent coin tosses by two players Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.    The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]&amp; = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}   How does one simplify this?      Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2. In Pillow Problems and a Tangled Tale Lewis Carroll presents the following problem:     A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?   One can easily reason that in the end, the bag must be in one of three states:   Black remains, new white taken New white remains, original white taken Original white remains, new white taken   In two of those states, a white counter remains to be drawn. Therefore, the chance is 2/3.  How can one arrive to this answer using Bayes' theorem? Foreword  This question was inspired by initial mistakes in . I wanted to explore the strange circle with $A&gt;\pi r^2$ and got lost into geometrical jungle.  A  is usually described by it's height $h$, radius of the base $a$ and radius of the sphere $r$. Besides we have a relation $a^2+(r-h)^2=r^2$ so the cap actually only has two degrees of freedom.  When considering a case of circle on a sphere (like the Earth), it's the surface of a spherical cap, but we can't see the height or other classic parameters of this spherical cap. Instead we encounter parameters like area $A$, perimeter $C$ and diameter $d$. I will try to write formulae for these: $$A=2\pi rh$$ $$C=2\pi a$$ $$d=2r\theta$$  Where $\theta$ is introduced with $\sin\theta=\frac{a}{r}$ or $\cos\theta=(1-\frac{h}{r})$.  As the cap only has two degrees of freedom, it should be a good problem with unique solution if only two values are known.  A-C Problem  The simple example: $A$ and $C$ are given. I show this only as example before the other tasks.  We can easily find $a=\frac{C}{2\pi}$ and then solve:  $$A=2\pi rh = 2\pi r (r-\sqrt{r^2-a^2})$$  to obtain  $$r=\frac{A}{2\sqrt{\pi(A-\pi a^2)}}$$  Parameters $a$ and $r$ are now known and it is straightforward to find $h$, $d$, $\theta$ and anything else.  C-d Problem  $C$ and $d$ are given.  First step once again is easy: $a=\frac{C}{2\pi}$. The knowing of $a$ suggests that we might prefer the $\theta$ definition with the sine. We obtain: $$d=2r\theta=2r\arcsin\frac{a}{r}$$  That's where I got stuck. Is there an easy way to solve this? Is there a hard way to solve this?  I tried solving numerically with Wolfram Mathematica and it went into complex numbers. When solving the geometrically equivalent $\sin\frac{d}{2r}=\frac{a}{r}$ I got some roots, but the search seemed very unstable and not finding the root closest to initial value. Besides it was finding a lot of different roots while I am pretty sure that there should only be one $r$ for each pair of $d$ and $a$. What would be the right formulation to solve this problem numerically?  A-d problem  $A$ and $d$ are given. This seems the hardest one. We obtain a system of equations:  $$A=2\pi r h$$ $$d=2r\arccos\left(1-\frac{h}{r}\right)$$  (The first equation suggested solving the problem in $h$ and $r$).  Is this system solvable analytically? Or numerically? My experiments with Wolfram Mathematica never converged.   Of course, I could express $h$ or $r$ from the first equation and leave only the second in one of these forms: $$d = 2r\arccos\left(1-\frac{A}{2\pi r^2}\right)$$ $$d = \frac{A}{\pi h}\arccos\left(1-\frac{2\pi h^2}{A}\right)$$  But I have no idea how to solve these either. Is it correct to write $\Bbb E[X]$ or $\Bbb E_{\theta}[X]$? Suppose that we observe the discrete random variable $ X = (X_1, \dotsc , X_n)$ with state space S, whose distribution we do not know but we are assuming that its joint p.m.f. belongs to a known family $\{f_θ : θ ∈ Θ\}$. We derive a sufficient statistic $T(X)$.  Is it correct to write $\Bbb E[X]$? If not, why? Would it be more correct to write $\Bbb E_{\theta}[X]$? I am from Nigeria and I'm doing a one year Internship program with The Marriott hotel in Wisconsin, I applied for the UK visitor visa two weeks ago to visit my fiance who lives and work in the UK, but I was refused the visa because the letter of my workplace that I provided was back dated of 2014 which they say they are not satisfy it accurately reflects my current employment circumstance. I also provided my pay slips, but they said they were not satisfied with it demonstrating the terms and conditions of my employment, and they also said I did not present any evidence to satisfy me having sufficiently strong ties and commitments in my country of habitual residence or elsewhere to motivate my timely departure from the UK. They said the documents my fiance presented were OK.  How soon can I re-apply for a UK visa after getting a refusal?