Potentially the balance of a lightsaber is very different from that of a sword. With a sword, the weight and momentum of the blade significantly affect the depth of the cut.  When the lightsaber is active, does the weight increase?  Or is swinging it just like swinging the handle?  Perhaps like swinging a Wii controller.
In-universe, a lightsaber is basically a controlled, fixed-length plasma stream that can be used as a sword. As such, the blade wouldn't have any mass (or infinitesimal mass if you want to get technical about E=mc2).  Anyone familiar with fencing or swordplay knows that the weight of the handle and blade, and thus the overall balance of the weapon, is critically important while fighting. A sword that is too blade-heavy is unwieldy; it is difficult to start, stop, and change the path of the blade as it's swung, which is vital to outmaneuvering your opponent in a duel. Conversely, a sword with too light a blade (equivalently, too heavy a hilt) is too easy to swing around. it's hard to "feel" where the blade is, and thus to make the minute changes needed to block a blow or land one. It can also be just as exhausting, in a different way, to wield such a sword; because most of the mass is behind your hand, it takes muscle power to point the blade forward instead of upward, while a balanced weapon would point directly forward from the user's hands. You also get no mechanical advantage while swinging it.  Obviously, out-of-universe the props were full "swords" that were then rotoscoped to paint the beam blade over the white wooden (later plastic) dowel that protruded from the saber. This gives the actual prop used by the actors a more realistic weight and balance, which can be seen in the duels (Darth Maul has to heave a bit to get that double-saber staff moving, and you can see the effort behind a full baseball swing of the sword by Luke or Vader in the original trilogy).  Was there any attempt in-universe to explain the weight and balance that an active lightsaber seemed to possess?
In Ubuntu ISO MBR, why does a GPT partition overlap the boot partition, and make MBR uneditable? Are token creatures considerd as a creature spell? How can I diagnose and repair missing drive space? I have a small ext3 / partition on a 2T drive which appears to have a discrepancy in its free space. How can I determine what is causing it and how it can be fixed.   After trying everything I could think of, I forced a fsck on boot which didn't have any affect.  I have other systems configured identically which don't exhibit this condition.  # df -h / Filesystem            Size  Used Avail Use% Mounted on /dev/sda1             9.7G  8.9G  318M  97% /  # du -shx --max-depth 1 / 5.2G    /  # lsof | awk 'BEGIN {t=0} /(deleted)/ {t+=$7} END {print t}' 0  # tune4fs -l /dev/sda1 | grep -E 'state|Free|Reserve|size|Inode' Filesystem features:      has_journal ext_attr resize_inode dir_index filetype needs_recovery sparse_super large_file Filesystem state:         clean Inode count:              2621440 Reserved block count:     131029 Free blocks:              212317 Free inodes:              2487281 Block size:               4096 Fragment size:            4096 Reserved GDT blocks:      639 Inodes per group:         32768 Inode blocks per group:   1024 Reserved blocks uid:      0 (user root) Reserved blocks gid:      0 (group root) Inode size:               128   Edit:  fsck says the partition is 2% non-contiguous.  For comparison to du above (sparseness):  # du -hxs --apparent-size / 4.9G    / Prove any orthogonal 2-by-2 can be written in one of these forms.... I'd like to prove that any orthogonal $2 \times 2$ matrix can be written   $$\bigg(\begin{matrix}    \cos x &amp; -\sin x  \\     \sin x &amp; \cos x   \end{matrix}\bigg) \hspace{1.0em} \text{or} \hspace{1.0em} \bigg(\begin{matrix}    \cos x &amp; \sin x \\    \sin x &amp; -\cos x \end{matrix}\bigg)$$  for $0 \le x &lt; 2\pi$.  I have no idea how to do this. I feel like this is something I've missed and is really straightforward but I can't find it in my book... I understand what the game already tells you in the general sense... it changes enemy damage, resistances, and chance of legendary monsters appearing.  But, specifically, how does the difficulty level affect the game? Is going from Normal to Hard a 5% increase in damage? 50%? 200%? Are there any additional things changed that aren't mentioned. Does Survival mode add any extra mechanics, or is it just even bigger numbers than Very Hard? Different results for integration using Mathematica and MATLAB Non-principal ideal in $K[x,y]$? Let $K$ be a field. Then $K[x,y]$ is not a PID because $x$ is irreducible but the quotient $K[x,y]/(x)$ is isomorphic to $K[y]$, which is not a field.  So there must be an ideal in there which is not principal. What is an example of such an ideal? What's the difference between the snap folder in /home/UserName and the snap folder in root (/)? How do I transition to SSL without affecting PageRank? Difference between staticmethod and classmethod What is the difference between a function decorated with  and one decorated with ? Mac renames itself automatically - host name is incorrect in terminal I've got a handleful of macs (all with unique names) and on occasion it prompt a notice about saying another computer on the network as this name and adds a (2) or (3) to my mac.    How can I set the host name from the command line? Installing Broadcom Wireless Drivers Empty set does not belong to empty set Herbert in his book "Elements of set theory" on page no 3 says that      we can form the set $ \{ \emptyset \} $ whose only member is $\emptyset $. Note that $ \{ \emptyset \} \neq  \emptyset $, because $ \emptyset \in \{ \emptyset \} $ but $\underline{ \emptyset \notin \emptyset} $·   By the last argument $\emptyset \notin \emptyset$, is he saying that empty set is not a member of or does not belong to empty set  OR  it is a typo and he wanted it to be $ \{\emptyset \} \notin \emptyset $, that set containing empty set is not a member of empty set I've been given a proof that shows the following:  If $f:[a,b]\to \mathbb C$ is a continuous function and $f(t)=u(t)+iv(t)$ then $$\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$$ The proof begins by letting $\theta$ be the principle argument of the complex number $\int_a^b f(t)dt$ and there is one step in the proof I don't understand; can anyone explain to me why $$\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)$$  Would this not imply that the imaginary part of the left hand side is equal to $0$? I'm not sure why the LHS and RHS would be equal here. Solving $\cos x=x$ I would like to know how can we solve the equation $\cos x = x$, without graphing. I know that  there would only be one solution, that is obvious, that too in between $0$  and $\frac{\pi}{2}$. Is there any real expression in finite terms [perhaps we call that closed form, I am not sure] that could give $x$ or $\cos x$. Although I have not studied Taylor series, I know that it only gives an infinite series, which is not what I want. I suspect that it could not be done, but can anyone explain me why?  Just for completeness, Wolfram Alpha, gives the approximate answer $x = 0.7390851332151606416553120876738734040134$, but fails to give a exact solution. $|f'(x)| \le g(x)$ implies $|f(b) - f(a)| \le \int_a^b g(x) dx$, without assuming $f'$ to be integrable. Let $p$ be prime and $(\frac{-3}p)=1$. Prove that $p$ is of the form $p=a^2+3b^2$ Let $p$ be prime and $(\frac{-3}p)=1$, where $(\frac{-3}p)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.   My progress:  $(\frac{-3}p)=1 \Rightarrow$ $(\frac{-3}p)=(\frac{-1}p)(\frac{3}p)=(-1)^{\frac{p-1}2}(-1)^{\lfloor\frac{p+1}6\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\lfloor\frac{p+1}6\rfloor=2k$  I'm stuck here. This is probably not the way to prove that.  Also tried this way:  $(\frac{-3}p)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$ stuck here too.  Any help would be appreciated. I've recently bought a new Android smartphone. After setting it up and loading my contacts onto it I thought that would be it.  Now after some days, every time I unlock the phone it takes a few seconds before it opens the default browser and opens some spam website. I tried to see what was causing it. I removed any apps I didn't know I could trust, but the problem persisted. After trying to solve it for some time I gave up and reset the phone to factory default. Now after having it reset to the default it ran without issues for roughly a week before the problem reappeared.    I tried to uninstall some apps in order to see if they were to cause of it, but nothing changed after doing so. However I did notice that if I turn off the wifi it won't even try to open the browser (I haven't tried mobile data since I don't have a bundle). This makes me feel like it might be related to something on the network but that wouldn't explain why only that phone would have the issue when there are at least 6 android phones on that very same network.  I hope that there is someone who can help me find the cause of this issue and help me solve it.  TL;DR When unlocking my phone, it opens the browser and tries to open a spam website. But only does so while connected to the wifi.  Things I've tried so far:   Factory reset (only helped for a limited time) Clearing the browser cache and all related data  Removing any app I don't know to be reliable Attempting to find what triggers it (seems like some type of internet connection is required??)   The device is the Doogee Shoot 1. As for the browser, the default is set to be the Android browser, but if I change the default it would use Chrome just as well. It seems to really just use the browser that is set to default. How to run 32-bit app in Ubuntu 64-bit? How can I gzip standard in to a file and also print standard in to standard out? I want to execute a command, have the output of that command get gzip'd on the fly, and also echo/tee out the output of that command. i.e., something like: echo &quot;hey hey, we're the monkees&quot; | gzip --stdout &gt; my_log.gz  Except when the line executes, I want to see this on standard out: hey hey, we're the monkees