Exercise on gauge invariance and global and local symmetry
Exercise on local and global symmetry for a given Lagrangian density
The man page for GNU find states:    -exec command ;     [...] The  string `{}'  is  replaced  by the current      file name being processed everywhere it occurs in the      arguments to the command, not just in arguments where      it is alone, as in some  versions  of  find.     Both  of  these  constructions might need to be escaped      (with a `\') or quoted to protect them from expansion      by the shell.     That's from the man to find (GNU findutils) 4.4.2.   Now I tested this with bash and dash, and both don't need to have the {} being masked. Here is a simple test:  find /etc -name "hosts" -exec md5sum {} \;    Is there a shell, for which I really need to mask the braces? Note, that it doesn't depend upon whether the file found contains a blank (invoked from bash):   find ~ -maxdepth 1 -type d -name "U*" -exec ls -d {} \;  /home/stefan/Ubuntu One   This changes if the found file is passed to a subshell:  find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d {}' \;  ls: cannot access /home/stefan/Ubuntu: No such file or directory ls: cannot access One: No such file or directory   which can be solved by:  find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d "$0"' {} \;   in contrast to:   find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d "{}"' \;  /home/stefan/Ubuntu One   but that's not what the man page is talking about, is it? So which shell treats {} in a different way? I tried a couple of substitutions but have so far gotten nowhere. Can anyone guide me in the right direction? Thanks for your time. Limit point compactness implies compactness. I have a question about the following counterexample, why  $U_n = \{n\} \times Y$ has not finite collection covering $X$?   Is it because if you take a finite set $S\subset U_n$ then it's impossible to have $X\subset \bigcup S$ since X is infinite?  Counterexample:  Let $Y$ consist of two points; give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point. It is not compact, for the covering of $X$ by the open set $U_n = \{n\} \times Y$ has not finite collection covering $X.$ Does Java have buffer overflows? Does Java have buffer overflows? If yes can you give me scenarios? Question about mosfet gate resistor I have n mosfet gate connected to  logic and Id is about 100mA. Both 4043 and mosfet has +5v. I have plan to use  mosfet  Questions are: How large gate resistor i need between 4043 and mosfet? Logic output is sometimes put on rapidly. How fast? Motherboard hdd led controls that. Do i need to place pull down resistor from logic to -0v, between 4043 and mosfet? Why I cannot ping 192.168.0.108 (my computer IP?) Ipconfig says that my IP is 192.168.0.108 C:\Users\vbcbu&gt;Reply from 192.168.0.108: Destination host unreachable.ip 'Reply' is not recognized as an internal or external command, operable program or batch file.  C:\Users\vbcbu&gt;ipconfig  Windows IP Configuration   Ethernet adapter Ethernet:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Local Area Connection* 12:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Local Area Connection* 13:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Wi-Fi 2:     Connection-specific DNS Suffix  . : www.tendawifi.com    Link-local IPv6 Address . . . . . : fe80::3804:2d04:a591:7142%6    IPv4 Address. . . . . . . . . . . : 192.168.0.108    Subnet Mask . . . . . . . . . . . : 255.255.255.0    Default Gateway . . . . . . . . . : 192.168.0.1  Ethernet adapter Bluetooth Network Connection:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  C:\Users\vbcbu&gt;  I can ping router at 192.168.0.1 I can't ping 192.168.0.108 C:\Users\vbcbu&gt;ping 192.168.0.1  Pinging 192.168.0.1 with 32 bytes of data: Reply from 192.168.0.1: bytes=32 time=2ms TTL=64 Reply from 192.168.0.1: bytes=32 time=24ms TTL=64  Ping statistics for 192.168.0.1:     Packets: Sent = 2, Received = 2, Lost = 0 (0% loss), Approximate round trip times in milli-seconds:     Minimum = 2ms, Maximum = 24ms, Average = 13ms Control-C ^C C:\Users\vbcbu&gt;ping 192.168.0.8  Pinging 192.168.0.8 with 32 bytes of data: Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable.  Ping statistics for 192.168.0.8:     Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),  C:\Users\vbcbu&gt;ping 192.168.0.8  Pinging 192.168.0.8 with 32 bytes of data: Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. When to use "If I was" vs. "If I were"? In indirect speech we generally backshift the tense. For ex - Direct speech : "I am the most powerful person." Indirect speech : "He said that he was the most powerful person."  What if someone reports like -  He denied the charges claiming that he is/was the most powerful person. So here which is correct- is or was ?  Thanks in advance. Real Time LaTeX? Possible Duplicate:          I'm looking for an application in which I can write pure LaTeX in one half of the screen and on the other half to see the compiled document in real time. This seems really like an obvious way to do LaTeX authoring but it doesn't look like any software is able to do it. LyX is not what I'm looking for. I want to edit the actual LaTeX and see it rendered in real time. Why isn't this already being done somewhere? Every finitely generated algebra over a field is a Jacobson ring Knowing that the polynomial ring in $n$ variables over a field $k$ is a Jacobson ring, how can we prove from it that every finitely generated $k$-algebra is a Jacobson ring?   EDIT: We define a ring to be Jacobson if every prime ideal is an intersection of maximal ideals. Is Cyclops’ mutant power just an organic version of Iron Man's repulsor tech? They occupy the same universe so we have to assume the same physics apply. Iron Man's repulsor tech is described as an inertia-less one-way kinetic blast. At higher energy levels, it can can even disintegrate matter in front of it. Cyclops’ eye beams are described exactly the same way. Is there any place in comics or the animation where the two beams are shown to have different powers? In newer Google Maps I can make area available offline.  Do other applications that rely on Google Maps (utilize Google Maps Android API) use offline maps? Excel function that evaluates a string as if it were a formula? Why is it impossible to override a getter-only property and add a setter? Why is the following C# code not allowed:  public abstract class BaseClass {     public abstract int Bar { get;} }  public class ConcreteClass : BaseClass {     public override int Bar     {         get { return 0; }         set {}     } }      CS0546 'ConcreteClass.Bar.set': cannot override because 'BaseClass.Bar' does not have an overridable set accessor I have just downloaded "ubuntu-14.04.1-desktop-amd64.iso" and "ubuntu-14.04.1-server-amd64.iso" through "".   Please can someone confirm the maximum RAM that is supported.  According to many sites, Maximum RAM supported by the OS  ubuntu-14.04.1-desktop : " 2^64 bytes " -- Kindly confirm  ubuntu-14.04.1-server : " 2^64 bytes " -- Kindly confirm Norms Induced by Inner Products and the Parallelogram Law Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.  It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality $$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 $$ holds for all pairs $u, v \in V$.  I'm having difficulty with the converse.  Assuming the parallelogram identity, I'm able to convince myself that the inner product should be $$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} $$  I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$.  How would one go about proving this? Splitting string into pair of characters in Ruby I am a Canadian citizen, and I intend to travel to the US for a student clinical elective at a university in Texas for the months of September and October. The university will help me obtain an F-1 status. However, I want to attend a few conferences and educational seminars which are being held in November/December for which I will be using a B1/B2 visa.  Will I have to leave the US and come back in November using a B1/B2 visa, or is there another way to change my visa status, or can I apply for both visas upon entering the US? Blown out blue/red light making photos look out of focus Group of order $p^2$ is abelian. Yes, I know that there are tons of solutions of this up here, but I, essentially, wanted to try it a different way and ah, well.  Let $|G| = p^2$ for some prime $p$.  Consider $x \in G$. So, $|x| = p, $ or $ p^2$. If it is the latter, then, we are done.  So, if $|x|= p$, then, let, $H = [h|  h=x^i, i \in Z] \Rightarrow |H|= p$.  As $p$ is the smallest prime dividing $|G|$, $H$ is a normal subgroup.  Now, consider some $y \in G$, but not in $H \Rightarrow y^p= e$. Then, its easy to show that $G = \langle x,y \rangle$.  Now, consider a homomorphism $f:G \rightarrow \operatorname{Aut}(H), g \to c_g $, wherein $c_g$ represents conjugacy by $g$. So, for example, $c_g(x) = gx g^{-1}$.  Then, it is clear that, $H \subset \operatorname{Ker}(f) $, as it is cyclic and therefore abelian. So, we need only look at $G/H$.  Basically, I am asking for a hint as to why $c_y(x) = yxy^{-1} = x = c_e(x)$, where $x$ and $y$ are the generators of $H$ and $G/H$, respectively?  Because, if I establish it for the generators, then it follows that it applies for for all elements, i.e, that $\operatorname{Im}(f)$ is trivial and therefore that $G$ is abelian.  Thank You!