I was wondering how was the formula for Doppler effect in light i.e. $f=\sqrt{\frac{(1-v/c)}{(1+v/c)}}f$ was derived. I understood how it is derived for sound but I am unable to understand how it is derived for light . I read a few books but they were difficult for me to comprehend.A little hint of derivation would also do.
This is about a step in a derivation of the expression for the relativistic Doppler effect.  Consider a source receding from an observer at a velocity $v$ along the line joining the two. Light is emitted at frequency $f_s$ and wavelength $\lambda_s$. The frequency $f_0$ and wavelength $\lambda_0$ received by the observer will be different.  Some textbooks now argue that the wavelength received by the observer is given by $\lambda_0=(c+u)T_0$, where $T_0$ is the time period of the wave in the observers frame. The argument given is that the successive "crests" will be an extra distance $uT_0$ apart due to movement of the source. Upon relativistically transforming the time period in the source frame to the observer frame, the correct result $f_0=f_s\sqrt{\frac{c-u}{c+u}}$ is obtained.  But the argument for the wavelength seems reminiscent of the classical Doppler effect for sound. Is it really applicable here through this argument? And is there a way to show the same result for $\lambda_0$ mathematically?
Notation for infinite product in reverse order A function for which one-sided limits are zero and infinity gpgkeys: key C927EBE00F1B0520 not found on keyserver Possible Duplicate:    I am using 10.04 LTS - Lucid Lynx. I shutdown the computer while a software update was running. Then after I restarted my computer and while the update manager was running, I got this error:  Could not download all repository indexes The repository may no longer be available or could not be contacted because of network problems. If available an older version of the failed index will be used. Otherwise the repository will be ignored. Check your network connection and ensure the repository address in the preferences is correct. GPG error: http://toolbelt.herokuapp.com ./ Release: The following signatures couldn't be verified because the public key is not available: NO_PUBKEY C927EBE00F1B0520 Failed to fetch  http://ppa.launchpad.net/rabbitvcs/ppa/ubuntu/dists/**DISTRIBUTION**/main/binary-i386/Packages.gz 404  Not Found   I went through the answers to  and I tried this $ sudo launchpad-getkeys  Please wait... launchpad-getkeys is running an update so it can detect the missing GPG keys Trying to import all the missing keys gpg: requesting key 0F1B0520 from hkp server keyserver.ubuntu.com gpgkeys: key C927EBE00F1B0520 not found on keyserver gpg: no valid OpenPGP data found. gpg: Total number processed: 0 launchpad-getkeys has finished importing all missing GPG keys.  Try running sudo apt-get update - you shouldn't see any key  errors anymore  But when I run sudo apt-get update again, it again shows that the key was not found on the keyserver. How to solve this problem? I left my cast iron pot on the gas stove, on high flame, and then forgot about it for 3 hours. It was empty and I was drying it before I oiled it to store it. After I discovered it and let it cool, I oiled it with a paper towel and all this rusty stuff came off and continues to come off. Is my pan ruined? What should I do? Unfortunately hard disk of my system is damaged, now I recovered my database. But it cannot be attached again to SQL management studio 2005. The error is:  myfile.mdf is not a primary database file. (SQL server , error 5171) How to replace the_post_thumbnail template tag and show the first inside the post image instead Inner product of two continuous maps is continuous Let f,g be continuous maps of S in $\mathbb{R}^n$, to $\mathbb{R}^m$. Show the inner product h(x) = $\langle f(x), g(x)\rangle$ is continuous.  My attempt: Since f,g are continuous, their product is a continuous mapping, so for $\epsilon \gt 0\: t.e\:  \delta \gt 0$ such that $\parallel x - a\parallel \lt \delta $ implies $\parallel \langle f(x),g(x)\rangle - \langle f(a)g(a)\rangle \parallel \le\parallel f(x)g(x) - f(a)g(a)\parallel \lt \epsilon$ by schwartz inequality. Limiting double to 3 decimal places Comparison of backup tools Possible Duplicate:          My reputation score has gone down, but the reputation tab in my profile does not indicate why. Is this a bug? What else could have caused it? Limit with multiple variables I need to evaluate the limit $$\lim_{n\rightarrow \infty}(n-\sqrt{n+\alpha}\sqrt{n+\beta})$$  My work: $$(n-\sqrt{n+\alpha}\sqrt{n+\beta})=n-\sqrt{n^2+\alpha n+\beta n+\alpha\beta}$$ I understand that both $\alpha$ and $\beta$ are just constants. But, now how should I factor this? I understand the calculus behind limits, but the algebra troubles me. Normal subgroup of a normal subgroup Let $F,G,H$ be groups such that $F\trianglelefteq G \trianglelefteq H$.  I am asked whether we necessarily have $F\trianglelefteq H$. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true? Let $n \in ℕ$. Show that the ring $ℤ/nℤ$ is a field if and only if $n$   is prime.   Let $n$ prime. I need to show that if $\bar{a} \neq 0$ then $∃\bar b: \bar{a} \cdot \bar{b} = \bar{1}$. Any hints for this ?  Suppose $ℤ/nℤ$ is a field. Therefore: for every $\bar{a} \neq 0$ $∃\bar b: \bar{a}\cdot \bar{b}=1$. How can I show that $n$ must be prime ? Looking for story title - younger gen waters down age-slowing drink of older gen, crammed into small apartment I remember a short story in which multiple generations of people being crammed into a small apartment in a city. The magical drink that either reduced or completely stopped aging was something like a soda that came in a six pack. The story focused on visual aspects of the drink at some point. The younger family members complained that their elders were living too long and apparently decided to water down the soda, I think that conclusion was ultimately left up to the reader. Many thanks for any help tracking down this story, might be one of Kurt Vonnegut's perhaps? Show that the set of all $x$ such that $x \in P$ and $x \notin Q$ exists I have to show that set of all $x$ such that $x \in P$ and $x \notin Q$ exists. So, I think I have to assume that sets $P$ and $Q$ already exist. So, with that, I can use  . This axiom is $$ \forall w_1,\cdots, w_n \, \forall A \,\exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in A \wedge \varphi(x,w_1, \cdots , w_n, A)] \right) $$ So, I will let $\varphi(x, Q, P) = x \in P \text{ and } x \notin Q $. Also, I will let $A = P$ and  $w_1 = Q $. So using universal instantiation, I get the following $$ \exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge \varphi(x, Q, P)] \right) $$ And using, existential instantiation, there exists a set $B$ such that $$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge \varphi(x, Q, P)] \right) $$ Using the definition of $\varphi(x, Q, P)$, this can be simplified as following $$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge x \in P \wedge x \notin Q] \right) $$ $$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge x \notin Q] \right)$$ $$ \forall x \left( x \in B \Longleftrightarrow x \in \{ x \,| x \in P \text{ and } x \notin Q \} \right)$$ Now, using , it will follow that $$ B = \{ x \,| x \in P \text{ and } x \notin Q \}  $$ And, since $B$ exists, this means that $\{ x \,| x \in P \text{ and } x \notin Q \} $ also exists. Is the proof good ? Is it acceptable to say "fed up OF" - surely it should be "fed up WITH" or "bored WITH something" rather than "bored OF something"? I hear people using of when I have always thought it was with in the circumstances explained above.  It seems to be getting more common. Isekai manga where the mc is reincarnated through his own magic The MC reincarnated himself through his own magic after his party defeated the demon king. Before he used his magic to do so, he took the daughter from the demon king and traveled with her over the world. After he used his reincarnation magic he was born in a noble family where the daughter of the demon king is now working as a maid in this family. He also has an older/younger (not sure) brother. Through his knowledge he starts his training of body, swordsmanship and magic as a baby. Years later his brother wants to play with him but he denies him every time. But one day he accepts it to play tag. His brother can't catch him. After that, he requests from the mc to teach him his magic. He denied again, but made an offer. Only when he can catch him, would he teach him his magic. I hope I'm not mixing up different mangas. mailto link with HTML body Escaping quotes in zsh alias Following on from  about stripping newlines out of text, I want to turn this into a zsh alias as follows:  alias striplines=' awk " /^$/ {print \"\n\"; } /./ {printf( \" %s \",$0);}"'   I've tried escaping the quotes inside the awk script, but I'm getting this error:  awk: (FILENAME=bspsrobustness FNR=1) fatal: division by zero attempted   (The file is called bspsrobustness)  Is there a way to do what I want? I suppose I could turn this into an awk script rather than a zsh alias, is that my best option? Isomorphism equivalence implies fundamental group is abelian