Using theoretical framework of the special relativity, we can show that the quantity that we classically regard as energy does have a property of inertia. And particularly, if the total energy of a box is $L$ then its inertia is $L/c^2$. Now suppose I have an infinitely large parallel plate capacitor which is at rest in an inertial frame $O$. Uniformly charged plates of the capacitors have such a charge density so that the uniform electric field between the plates is $E$.   Now in a frame of reference $O'$ which is moving uniformly along the area vector of the plate plane with respect to $O$, the electric field will be $E$ and magnetic field would still remain zero.  In $O$ frame, the total energy of capacitor is $L_1 = l_0 \int\frac{1}{2}\epsilon_0E^2 dA $ and thus its mass is $m_1 = L_1/c^2$.Where $l_0$ is the perpendicular separation between the plates and $dA$ is the infinitesimal small area element of one of the plates in $O$ frame.  In $O'$ frame, the total energy of the capacitor is $L_2 = l_0\sqrt{1-v^2/c^2} \int\frac{1}{2}\epsilon_0E^2 dA $ . The infinitesimal $dA$ and limits of integration remain exactly the same because no length contraction happen in the direction perpendicular to the relative motion. Here also the mass is thus $m_2 = L_2/c^2$.  So, $m_2$ = $m_1$ $\sqrt{1-v^2/c^2}$. But in general, in relativity we prove that $m = m_0/\sqrt{1-v^2/c^2}$. So why does this apparent contradiction is happening? Does it have to do anything with the fact that actually both the $m_1$ and $m_2$ are infinities? Because I doubt that it might be the case that we cannot directly write  $m_2$ = $m_1$ $\sqrt{1-v^2/c^2}$ because $m_2$ and $m_1$ are infinities. But on the other hand that relation seems to hold pretty well from their corresponding expressions in the integral form.
Suppose I look at a parallel plate capacitor in its rest frame and calculate the electrostatic energy, $E$.  Next, I look at the same capacitor in a primed frame boosted in the direction perpendicular to the plane of the plates. In this frame, the $E$-field is the same strength, there is no magnetic field, and the volume over which the $E$-field extends is less by a factor $1/\gamma$. This suggests $E' = \frac{1}{\gamma} E$, but relativity states that energy transforms as $E' = \gamma E$.   Where is the missing energy?
BAD signature when trying to verify SHA256SUMS for Bionic Beaver ISO I am attempting to verify my download of Bionic Beaver on Windows 7.  I feel like I am missing something very obvious.  I went to  and clicked on both SHA256SUMS and SHA256SUMS.gpg  It seems like this does not download the files, but opens the text in the files in new tabs.  I used Ctrl-A and then copied the text into Notepad, naming the files SHA256SUMS and SHA256SUMS.gpg.  I made sure that windows didn't add any weird extensions when saving.  I manually downloaded both keys I used from https://keyserver.ubuntu.com, one with fingerprint  C598 6B4F 1257 FFA8 6632 CBA7 4618 1433 FBB7 5451   and the other with fingerprint  8439 38DF 228D 22F7 B374 2BC0 D94A A3F0 EFE2 1092   After I downloaded both keys I certified them with my personal key using Kleopatra, after verifying the key fingerprint with those at .  When I ran   gpg --verify SHA256SUMS.gpg SHA256SUMS   I got   gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using DSA key 46181433FBB75451 gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key &lt;cdimage@ubuntu.c om&gt;" [full] gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using RSA key D94AA3F0EFE21092 gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key (2012) &lt;cdimage@u buntu.com&gt;" [full]   When I ran   gpg --verbose --verify SHA256SUMS.gpg SHA256SUMS   I got  gpg: armor header: Version: GnuPG v1 gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using DSA key 46181433FBB75451 gpg: using pgp trust model gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key &lt;cdimage@ubuntu.c om&gt;" [full] gpg: binary signature, digest algorithm SHA512, key algorithm dsa1024 gpg: Signature made 11/29/18 16:27:43 US Mountain Standard Time gpg:                using RSA key D94AA3F0EFE21092 gpg: BAD signature from "Ubuntu CD Image Automatic Signing Key (2012) &lt;cdimage@u buntu.com&gt;" [full] gpg: binary signature, digest algorithm SHA512, key algorithm rsa4096   I tried copying the data from SHA256SUMS and SHA256SUMS.gpg tabs over TOR instead, in case it was a local network problem, but no dice either.  What am I missing here? Is there some way to download the SHA256SUMS and SHA256SUMS.gpg files directly, as opposed to copying the data into a text editor and saving it?  Any help would be very appreciated. I am very confused and frustrated at this point. What do you call a person who is never content with anything that is given to him? I am looking for an adjective for a person who, no matter what you give him/her (materially or otherwise), is never content and always expects more. I've come across a small issue which occurs when a question has been closed but the question itself or one of its answers are being edited.  Steps to reproduce:   Visit a closed question. (even closed years ago)   In a different tab or browser, edit the question or one of its answers. "an edit has been made to this post; click to load" message will appear above the edited post in the first tab/browser, click it. The .question-status div becomes hidden, which includes the closed notice below a question and, for a duplicate-closed question, the list of duplicates above the question body. This gives the appearance that the question is no longer closed (until you refresh).   Some screenshots:                I know it isn't common/important but it did confuse me when looking at a particularly bad question - I thought it had been reopened.   Is this an intentional feature? If so, what was the thinking behind it?    Another example  Before    After Let $M$ be an A-module and $f \in \text{End}(M)$ an idempotent endomorphism. Show that $M = \ker(f) \oplus \ker (f-\text{id}_M)$. I don't know how to approach this exercise, using only properties of modules and direct sums, without projections. Can you help me? Thanks in advance. When viewing a question, when I hover over the star icon (under the voting module) the tooltip says I've made it a favorite when I haven't yet. Virtualbox Kernel driver not installed I tried to use my Virtual Box and this is the error I received - (I just updated to the new Ubuntu also) and I am VERY new at this Linux OS.     Kernel driver not installed (rc=-1908)      The VirtualBox Linux kernel driver (vboxdrv) is either not loaded or   there is a permission problem with /dev/vboxdrv. Please reinstall the   kernel module by executing      '/etc/init.d/vboxdrv setup'      as root. Users of Ubuntu, Fedora or Mandriva should install the DKMS   package first. This package keeps track of Linux kernel changes and   recompiles the vboxdrv kernel module if necessary. I have looked it up and the resources all talk about schengen visa, travelling with eu national family member etc. so I would like to get a more assuring answer about my situation:   I am a non-EU national with an Italian residence permit. My father is an Italian citizen. I will be travelling to Sofia and Bucharest next month with my girlfriend who is a British citizen. Do I need to get a visa to enter these two cities? Filling triangle doesn't seem to work with PGF/TikZ I was wondering if anyone had any clue on why my triangles are not filled in red within the following piece of code:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc,3d}  \usepackage[active, pdftex, tightpage]{preview} \PreviewEnvironment[{[]}]{tikzpicture} \newcounter{number}  \begin{document} \begin{tikzpicture}[x  = {(-0.5cm,-0.5cm)},                     y  = {(0.9659cm,-0.25882cm)},                     z  = {(0cm,1cm)}] \begin{scope}[canvas is yx plane at z=0] \foreach \x / \y / \label in {1 / 0.4 / 1, 0.9 / 0.4 / 0, 1 / 0.5 / 1, 0.7 / 0.3 / 0, 0.8 / 0.4 / 0, 0.9 / 0.5 / 0, 1 / 0.6 / 1, 0.7 / 0.4 / 0, 0.4 / 0.4 / 0, 0.6 / 0.4 / 0, 0.5 / 0.4 / 0, 0.8 / 0.5 / 0, 0.9 / 0.6 / 0, 1 / 0.7 / 1, 0.7 / 0.5 / 0, 0.5 / 0.5 / 0, 0.4 / 0.5 / 0, 0.6 / 0.5 / 0, 0.8 / 0.6 / 0, 0.9 / 0.7 / 0, 1 / 0.8 / 1, 0.7 / 0.6 / 0, 0.6 / 0.6 / 0, 0.5 / 0.6 / 0} { \addtocounter{number}{1} \pgfmathparse{\value{number}}\let\k\pgfmathresult \draw (8 * \x, 8 * \y) node[inner sep=0, outer sep = -0.1pt,color=black,fill,circle] (node-\k) {}; };  \foreach \x / \y / \z / \trash in { 4/5/8/0, 9/11/16/0, 9/16/17/0, 11/10/18/0, 11/18/16/0, 10/8/15/0, 10/15/18/0, 8/5/12/0, 8/12/15/0, 5/2/6/0, 5/6/12/0, 2/1/3/0, 2/3/6/0, 17/16/24/0, 16/18/23/0, 16/23/24/0, 18/15/22/0, 18/22/23/0, 15/12/19/0, 15/19/22/0, 12/6/13/0, 12/13/19/0, 6/3/7/0} { \draw[fill=red]  (node-\x) --  (node-\y) --  (node-\z) -- cycle; } \end{scope} \end{tikzpicture} \end{document}) I once heard the late John Ciardi (NPR's "A Word in Your Ear") try to explain that the 1920s idiom, "copacetic" (meaning completely satisfactory), entered into the African-American vocabulary in Harlem from the days when Jews and African Americans lived there together.  He argued that copacetic has the same meaning as the Israeli idiom "kol b'seder" which literally means "all is in order."  The problem with that, said my Harlem-raised father-in-law, is that the Jews in Harlem spoke Yiddish and kol b'seder was not used in Yiddish.  The dictionary I've got is not helpful. Can someone come up with a better explanation? Find the angle ADE of the given triangle. Given :  Angle ABE = 50, Angle DAC = 20, Angle BAD = 60, Angle ACB = 20 and Side CD = Side DA.Unit of angle values given is degree.  Find : Angle ADE.    I'm stuck and I need help here. Installation of Miktex How to start Skype 4 with Ubuntu 12.10? I've installed Skype and it's running all OK but I want it to start automatically when I log with my user.  How can I do that? Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$ What if I accidentally run command "chmod -R" on system directories (/, /etc, ...) I accidently ran  sudo chmod 755 -R /   instead of  sudo chmod 755 -R ./   I stopped it after few seconds, but now there is some problems such as  sudo: must be setuid root   How can I revert permissions back? Naming threads and thread-pools of ExecutorService I am very new to LaTeX and I am writing my thesis. In a chapter, I want to show my codes, which are in Mathematica.   I use the listings package as follows:  \documentclass{article}   \usepackage{listings}   \begin{document}   \lstset{language=Mathematica}    \begin{lstlisting}   Do[i^j,{i,1,10},{j,1,10}]   \end{lstlisting}    \end{document}   However, the output is much different to what the code looks like in the Mathematica IDE (even if I use other languages in \lstset{language=&lt;name&gt;}):    I read some questions on the site but, in some of them, the font or color of the codes were not the same. Could you please help me to display something like the picture below in LaTeX?  If it is important, I am using the BiDi package. This is a question that's been nagging me for some time. I always thought that C++ should have been designed so that the delete operator (without brackets) works even with the new[] operator.  In my opinion, writing this:  int* p = new int;   should be equivalent to allocating an array of 1 element:  int* p = new int[1];   If this was true, the delete operator could always be deleting arrays, and we wouldn't need the delete[] operator.  Is there any reason why the delete[] operator was introduced in C++?  The only reason I can think of is that allocating arrays has a small memory footprint (you have to store the array size somewhere), so that distinguishing delete vs delete[] was a small memory optimization. I extracted simple example:  require 'pp' x = 1..3 pp x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end pp x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} }  pp(x.map do |i| {:value =&gt; i, :double =&gt; (i*2)} end) pp(x.map { |i| {:value =&gt; i, :double =&gt; (i*2)} })   I am wondering why first pp produces:  [1, 2, 3]   While all the oders are giving:  [{:value=&gt;1, :double=&gt;2}, {:value=&gt;2, :double=&gt;4}, {:value=&gt;3, :double=&gt;6}]   I assume it has something to do with operator precedence. Where can I find good explanation? How does Azkaban's management work? Invisible space when trying to remove file (but exclude all files in a list)? I'm trying to accomplish a simple task using the excellent post : I would like to remove all files from a folder except those listed in a .txt file contained in the folder. The syntax works fine, but for some reason all files are removed - basically the script is saying that there are no correspondences between the filenames in the folder and those in the .txt file - but I can see that there are!   I thought maybe there might be some invisible white space after my file names that I cannot see (in the folder or the .txt file). But I don't think so...Is there another solution that I am not thinking of?