If I have an ideal wire and I connect a battery to it, then the potential drop is zero along the wire, right? So, all its points must be at the same potential. Well, that means charges ( which are initially assumed to be at rest) will not go from one point to another in the wire. There's one more question to it. I've read in books that if the terminals of a battery are connected by an ideal wire, then all the points on the wire have the same potential as the potential of the positive terminal of the battery. But why don't all the points of the wire have the potential same as that of the negative terminal of the battery instead by similar logic? (because all the points of the wire are also connected to the negative terminal with zero resistance, and hence, zero potential drop along the way).  Here's how I tried to explain it. Please tell me if I'm wrong: Even if all the wire has zero resistance, then still charges are transferred through it because it's points are not at the same potential even if the wire has zero resistance. Yes, there is no potential drop due to resistance but there is a potential drop due to the kinetic energy gained by the charges as they more along the wire. So, it's points are at different potentials even with no resistance, because, at the end of the day, what we're talking about is the usual electric field, which will push charges to points of lower potentials even if there is no resistance along the way.  But according to this explanation of mine, a potentiometer would never work, because if my explanation is true, then charges would still be transferred along the wire containing the galvanometer even if both the points of that wire are at the same potential. So, where am I wrong?  It's not a duplicate. I've asked one more question in the details.
According to , if there is a potential difference, $V$, across a resistor then there is a current, $I$, flowing through it.   Since we assume that points along the connecting wire are at the same potential, how can current, $I$ flow between points at the same potential, $V$?
When I type $\overset{\circ}{Z}^\lambda$ the lambda is placed slightly above and to the right of the circle. Is there a way to make it appear as a superscript on the Z? I tried $\overset{\circ}{Z^\lambda}$, but then the circle is too high and to the right.  As a bizarre aside: a friend uses the exact same code with no problem. However, when I type it into MathOverflow it looks the same as in my compiler (i.e. bad). Uploading an image from the web can leave paste broken in editor SEO regarding using separate  tags for the logo and website name Is is true &lt;h1&gt; tag should not have more than 1 on every page for SEO?   Every pages on the header, it includes and &lt;h1&gt; tag. Website Name which show an image from logo class, like this:  &lt;div id="header"&gt;   &lt;h1 class="logo"&gt;      &lt;a href="#"&gt;Website Name&lt;/a&gt;   &lt;/h1&gt; &lt;/div&gt;   and I also include &lt;h1&gt; on the contact page, help, etc and article page for example:  &lt;h1&gt;Contact Us&lt;/h1&gt; &lt;h1&gt;Name of the Article Title here&lt;/h1&gt;   On the homepage, it just has one &lt;h1&gt; for the logo Site Name... other pages have 2 &lt;h1&gt; tags Plotting complex numbers App keeps crashing & rendering error since update to 1.0.68 Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.  Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)  This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?    Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right. Is $G^{(X, Y)} = (G^X)^Y?$ ($A^B$ just means that $B$ is mapped to $A$) What are my options if I need to move a database from SQL Server 2012(32bit) to SQL Server 2005(64bit)?  I know I cannot:   restore a backup of the database on SQL Server 2005 detach &amp; attach   I know I can:   use import data wizard, and I tried it on one database, but it only moves data, and even that was troublesome as I needed to do a lot of work creating temp tables to maintain identity columns, recreating all FKs, indexes etc.    Is there an easier option? Story about lost colonized planet sent into orbit around "free" star what is the surface area of a cap on a hypersphere? According to , let the sphere have radius $R$, then the surface area a spherical cap of height $h$ and base radius $a$ is given by $$S=2\pi Rh=2\pi(a^2+h^2).$$  What is this value for an n-dimensional hypersphere? If it helps simplify the problem we can assume $R=1$ and $a=0.5$.   Many thanks. I have an unusual problem with Chrome when I try to open PDFs. If I have already downloaded a PDF and try to download it again, I can open it by tapping its name; in the screenshot below, I could tap "pdf.pdf" to open the file.   The issue is that tapping this link actually opens RollerCoaster Tycoon Classic! If I choose to download again, Chrome redownloads and opens the PDF correctly, so it's not a huge issue, but still an inconvenience.  In Settings, it says that RollerCoaster Tycoon Classic has no defaults, and I am unable to change the "open supported links" setting.    How do I stop RollerCoaster Tycoon  trying to be a PDF reader? Run 32bit applications on Windows 10 64 Bit I've got a Windows 10 Computer (home edition), which has 4GB of Ram and an i5 processor. I need to run an old 32 bit accounting programme. The vendor said it won't release a 64 bit version (although a 64bit version exists, it's missing key features from the older version).  The programme installs fine, but when I click on the programme icon, it says the programme won't work on a 64 bit system.   What options do I have?    I've done a fair bit of research, but it seems to run up dead ends.   1) WoW64     Apparently, this is supposed to let you run 32 bits on 64 bit Windows. However, I am not sure how to install it. I think it's for web servers only. However,  says it can work on Windows 7, but doesn't explain how to set it up (or if it will work on Windows 10).  2) XP mode     A 32bit Windows XP emulator. However, it seems to be for Windows 7 only. Also according to this , you need Windows Professional.   3) Virtual Box Machine    Let's you emulate Windows. Seems overkill for my purposes. Plus I would need to buy a 32 bit Windows license. I think it would run slowly with 4GB of Ram.  4) Partition drive into 32bit and 64 bit versions Apparently, you can use the free tool  to split the drive into two and then install separate 32bit and 64 bit versions.   Again, seems overkill and it looks like I would need buy a 32 bit Windows 10 license.   Are these my only options?     Images  Installation    First run (From the error message I guess it's a 16-bit program)   Clicking on programme Can a non-EU citizen travel freely within the Schengen area without passport? My current situation is that I have residence permit of Germany (as a stamp in passport) and I wish to apply for UK visa in next few days from Germany and won't have my passport anymore during the processing time.  Instead of waiting for the process, I wish to travel to Scandinavian countries immediately on holidays, later receive my passport from visa center after I return back to Germany.   I am wondering if photocopy of passport and residence permit suffice as identity proof to border control (if I face them)? What else are valid documents that I can show as valid identity proof to border control? What are the exact problems I will face if I face the border control, and they are not satisfied with my travel documents? Frobenius Inequality Rank I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:  Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$ Is trace invariant under cyclic permutation with rectangular matrices? I'm working with trace of matrices. Trace is defined for square matrix and there are some useful rule, i.e. $\text{tr}(AB) = \text{tr}(BA)$, with $A$ and $B$ square, and more in general trace is invariant under cyclic permutation. I was wondering if the formula $\text{tr}(AB) = \text{tr}(BA)$ holds even if $A$ and $B$ are rectangular, namely $A$ is $n$-by-$m$ and $B$ is $m$-by-$n$. I figured out that if one completes the involved matrices to be square by adding zero entries in the right places, then the formula still works... but I want to be sure about this! I reviewed  in the First Posts queue.  Which said      i changed it to 'localsystem' and done. thanks much   So it obviously was a potential low quality post. But sometimes such answers can be spot on, so I checked it out, better safe than sorry. I turned out the post had been deleted. Great! So I clicked "No action needed". Then I got this:    You can imagine I felt "wrongfully accused".  I think that deleted posts should not be used for review audits, because the meaning of "No action needed" is not clear in this context. Polygamma function series: $\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2$ Applying the , I found: $$S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2$$ where $\Psi^{(1)}(k)$ is the polygamma function. Is it known any sharper bound for the sum $S$? Thanks. What is a word for someone that only does something to benefit them? Take this paragraph for example: I am a follower, not a leader. Hell, I'm not even that good of a follower either, just someone who will do your bidding for the right price. Even then, if it doesn't benefit me -- or put you in debt to me -- I won't do it. I just manually summoned the Eye of Cthulhu, going into battle with copper armour, tungsten broadsword, gold bow, jester arrows and plenty of lesser healing potions. I came rather close to death but I managed to stay alive. However, the Eye took off when I had his health down to ~1700. I'm assuming this means you only have a few minutes to finish it off after a certain point. Is there any equipment I'm missing to help kill the Eye faster and to help my survivability? If it's relevant, my current life is 120. How do I separate hydraulic disc brake pads after pulling the brake while the wheel is off?