Voltage is defined as:   Given the circuit:   Why would potential not drop, from the point differentially away from the positive terminal on the battery, to the point I have labeled. I understand that charges lose energy in resistors due to inter lattice collisions, where the acceleration generated due to thermal motion and voltage bias is translated from kinetic energy to heat energy. Before the resistor the resistance is neglible so the drift velocity will be high, and collsions will be minimal. Even so as the charges move from the positve terminal the Electric field will be strong and we will be losing the potential energy as we travel in the direction of the field. With this so why dosent potential drop as we move in the direction of the field? Is the energy associated with the charge essentially translated from potential to kinetic energy, where very little of this kinetic energy is translated to heat energy, and we assume the total potential is associated with the total energy the charge has at this point (KE + PE)?   Please address the question above. Further discussions will be encouraged though!  I am really looking for an answer to my question. It would be much appreciated if we address the core question I asking instead of side discussions!  Added picture:
Electric potential is related to Electric Field by negative integral of Field with distance.If charges move away from the positive terminal (Assuming conventional flow) why is there no drop in electric potential even though the charges are moving in the direction of electric field?  Some of the answers I read for the similar question said that the electrons in wires rearrange themselves in a way to make the electric field uniform in the wire.However,this doesn’t explain the absence of potential drop in the wires (due to electric field) since the field is still there.
The most prominent place I've noticed this is when SSH-ing at work, but I feel like I've observed this behaviour elsewhere too.  When I try to log into Linux servers from my Windows desktop at work, I've noticed that if I mis-type my password, it takes a good 5 seconds or so before I get "Access Denied" back. Then when I type my password correctly, the login (along with welcome messages etc) is virtually instant.  Is there any logical reason for this, or would it be down to some odd configuration that's specific to the machines here at work? Can I use Beetel 450TC1 as wireless router? Too many requests? I went to stackexchange.com and clicked on SO.  After a few minutes I got a message saying "We have detected too many requests from this IP.  If this is an error contact us."  That was the second SE page I tried to visit.  Second only to SE.com  I moved to another machine and it's working now.  The first machine usually works, I don't know if a got assigned a blacklisted IP or what.  Anyone know what's going on?  EDIT: I would really rather not post my IP on the web, but if a diamond or dev asks I'll email it.  EDIT 2: I looked at badp's dupe suggestion and can verify that I have no scripts frequently calling SO. A classic exponential inequality: $x^y+y^x>1$ Number of connected components of the complement of a compact Let $X$ be a connected, locally compact space and let $K$ be a compact subset of $X$. Is the number of connected components of the complement $X\setminus K$ finite? If it is not, are there some hypotheses over $X$ for finiteness of connected components? Newton's cradle "implements Runnable" vs "extends Thread" in Java From what time I've spent with threads in Java, I've found these two ways to write threads:  With implements Runnable:  public class MyRunnable implements Runnable {     public void run() {         //Code     } } //Started with a "new Thread(new MyRunnable()).start()" call   Or, with extends Thread:  public class MyThread extends Thread {     public MyThread() {         super("MyThread");     }     public void run() {         //Code     } } //Started with a "new MyThread().start()" call   Is there any significant difference in these two blocks of code ? I get this message while trying to install Ubuntu 12.04 as a guest in VirutalBox 4.1.18, on an Ubuntu 10.04 host.         This kernel requires the following feature not present on the CPU: pae   Some host specs:     The host's kernel is: Linux 2.6.32-41-generic-pae GNU/Linux   lscpu (host): Architecture: i686, CPU op-mode(s): 32-bit, 64-bit   grep --color=always -i PAE /proc/cpuinfo   does show pae in its output.   The 12.04 iso used is: ubuntu-12.04.0-desktop-i386.iso      As a comparison/check, I downloaded and installed Linux Mint 13 Cinnamon to the same host on the exact same VM (I just changed the .iso image). It worked fine. Its iso is: linuxmint-13-cinnamon-dvd-32bit.iso   It seems (to me) that I have pae.. what is going on here?   Update: I had assumed that Linux Mint also required pae (being Ubuntu based), but I've just run;   grep --color=always -i PAE /proc/cpuinfo   in the Mint VM.   It showed no output.   So it seems the issue may lie with VirtualBox.   If that is the case, how can I get Virtualbox into pae mode? Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ Let $a&lt;b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$    Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D&gt; 0$  Since the function is continuous at $x_0$, we have $$\forall \epsilon&gt;0, \exists \delta&gt;0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2&gt;0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f &gt;0$  Now since $\int_{[a,b]} f = 0$. we have  $$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f&lt;-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$  Is my proof correct? What is the reason behind "non-static method cannot be referenced from a static context"? What should I look for when buying a used bicycle? I'm moving to a city where I can ride a bicycle to the grocery store, coffee shop etc. I want to purchase a decent used bicycle. I'm leaning towards a bike that's in between a "10-speed" or road-bike and a mountain bike. What should I look for when choosing a particular bike?  I've never purchased a used bicycle. Other than classifieds (e.g., Craigslist), what types of places can I look? What other issues should I consider, such as size, visual appearance, leaking tires, etc.? How can I make sure it's not stolen? bash chained logical operator execution order, lazy evaluation Galactic Rotation Speeds - Ehrenfest Paradox, Gravitational time dilation, Dark Matter - all of the above? How did Anakin Skywalker survive Order Sixty Six? . When activated, troopers would immediately perceive Jedi as a threat and kill them.  Since Anakin was a Jedi until around the time of the Order, how did the troopers know not to kill Anakin? How can one Identify a Magic Item? I had a disagreement with one of my groups recently: They believe that the  spell is the only way to identify a magic item, while I believe a high enough Intelligence (Arcana) check is able to do the same.   What I came here to ask is: What are all the possible ways to identify a magic item? Determine if wire in switch box is safe to re-purpose? I need to pull a neutral wire to my light switch for a smart switch.  It's presently switching an outlet that I don't need switched, and the other side of the outlet already has unswitched power.  My plan is to use a smart switch, which requires neutral, to control a different device wirelessly.    I can't feasibly run a new wire, because the wires aren't run through conduit.  Instead, my plan is to repurpose the switched wire to get neutral from the outlet.    The switch, which controls an outlet has two wires, one black and one red.  The switched outlet has a white neutral wire, plus the red and black wires that I assume run directly to the switch.  I've completely disconnected the red and black wires from the outlet and the switch.  The black wire is measuring 120V AC when measured against ground using the box as ground, both at the outlet and at the switch. Using my DMM, the red wire is measuring 24V AC with respect to the box/ground at both sides.  I assume this is a  caused by induction from the charged black wire over to the floating red wire.    Our house is ~100 years old and this particular switch seems to be run using armored cable (note the paper wrap around each wire.)  There are literally only two wires in the switch box; no wire nuts.    If relevant, we do have some old knob and tube wiring active, but I'm pretty confident there's none on this circuit.  Is there some way I can determine with enough confidence that the red wire isn't connected elsewhere such that I can know it's safe to connect the red wire at the outlet to netural? Identity involving Euler's totient function: $\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$ Let $\varphi(n)$ be Euler's totient function, the number of positive integers less than or equal to $n$ and relatively prime to $n$.    Challenge: Prove  $$\sum_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}.$$   I have two proofs, one of which is partially combinatorial.    I'm posing this problem partly because I think some folks on this site would be interested in working on it and partly because I would like to see a purely combinatorial proof.  (But please post any proofs; I would be interested in noncombinatorial ones, too.  I've learned a lot on this site by reading alternative proofs of results I already know.)  I'll wait a few days to give others a chance to respond before posting my proofs.  EDIT: The two proofs in full are now given among the answers. How to create a Gant-Chart in LaTeX-Beamer I am trying to create a Gant-Chart in Beamer. I found the following example:   \documentclass{beamer}  \usepackage[utf8]{inputenc} \usepackage{pgfgantt}  \title{Sample title} \author{Anonymous} \institute{Overleaf} \date{2019}   \begin{document}  \frame{\titlepage} \begin{frame}{Gant-Chart}     \begin{ganttchart}[vgrid, hgrid,                         bar/.append style={fill=blue!80},                        milestone/.append style={fill=green}]{1}{15}         \gantttitle{2019}{15} \\         \gantttitlelist{1,...,15}{1} \\         \ganttbar{Task 1}{1}{2} \\         \ganttmilestone{Milestone}{7} \ganttnewline         \ganttbar{Final Task}{8}{12}     \end{ganttchart} \end{frame}    \end{document}   Now, I want to change it slightly. The first row shows the year. I want the second row to show months. And then the third row shows the number of weeks that I have planned for the task. Also, I don't want to see any arrows, but just have the rectangles. Is there a way to add a grid? Basically I want everything to just look as in an Excel-Table: Grids and then color the rectangles during which I am working.   Here is an example of what I would like to have:   Can somebody show me how to do this?   Here is another example that seems to get pretty close to what I want. Only in the week-row now it is always written "week 1", "week 2" and so on. I just want the number.   \begin{frame} \begin{figure}[h!bt]     \begin{center}         \begin{ganttchart}[             vgrid={*{6}{draw=none}, dotted},             x unit=.08cm,             y unit title=.6cm,             y unit chart=.6cm,             time slot format=isodate,             time slot format/start date=2019-02-01]{2019-02-01}{2019-07-31}             \ganttset{bar height=.6}             \gantttitlecalendar{year, month=name, week} \\             \ganttbar[bar/.append style={fill=blue}]{Task 1}{2019-02-11}{2019-02-17}\\         \end{ganttchart}     \end{center}     \caption{Time Plan} \end{figure} \end{frame} Why are 15 reputation points required to upvote? I think more reputation points should be required to add a new answer than to upvote an existing answer.  I don't continually watch for new questions, so usually when I find a question it has already been answered.  It seems bit inappropriate to answer an already-answered question in order to contribute to the conversation, but until you earn reputation, you can't upvote or comment; you can only add new answers, which prevents me from contributing at the moment as much as I would like to.  This seems backwards to me. If someone incorrectly upvotes a bad answer, the damage is minimal, as other users can down vote it and upvote the correct answers. On the other hand, an incorrect answer wastes the time of everyone who reads it.  So, IMHO, it should require more reputation points to add an answer than to upvote or comment on an existing answer.  I wouldn't say say that users should be prevented from answering unanswered questions until they have x reputation points, as that might prevent users who have answers from providing them, but perhaps it should be required if the question is already answered, to discourage users from reiterating previously given answers.  What do you think? Why aren't Christoffel symbols tensors? - asked from a fibre bundle perspective