How to install Windows 7 in Dual Boot with Ubuntu (Ubuntu pre-installed)
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Intuitive understanding of the elements in the stress-energy tensor There is an image in the Wikipedia about the :    I have a rough understanding of the stress tensor: you imagine cutting out a tiny cube from the fluid and form a matrix out of the forces on each side of it: it's not hard to see that the forces that push the faces outwards so the x direction of the force on the x side, the y direction of the force on the y side, etc. is the pressure, while other forces directions are the shear stress (these can be non-zero is some jelly-like substance).  Now the first question: Why is the lower half of the spatial part is the momentum flux, while only the upper one is the shear stress on the image? Is the image wrong? Isn't the spatial part describe the classical stress tensor?   The second question is: what's the intuition behind the temporal part?  So if I have a Minkowski spacetime and cut out a cube of it, then it's quite hard to see why "pressure" on the temporal face means density, also why "shear stresses" on the temporal face translate to momentum density, not speaking of the "temporal force components" the spatial faces...  Also stresses are measured in Pascals, mass density is measured in $kg/m^3$ while momentum density is in the units of $Ns/m^3$. These units doesn't seem to be compatible, but still they are in the same matrix, why?  EDIT: It has been pointed out that momentum flux refer to the entire blue part, it indeed have a lighter frame I didn't notice. Then yet another subquestion: what's the difference between the momentum density and momentum flux? Why isn't the gamma function defined so that $\Gamma(n) = n! $? As a physics student, I have occasionally run across the gamma function  $$\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)!$$  when we want to generalize the concept of a factorial.  Why not define the gamma function so that   $$\Gamma(n) = n!$$  instead?  I realize either definition is equally good, but if someone were going to ask me to choose one, I would choose the second option.  Are there some areas of mathematics where the accepted definition looks more natural?  Are there some formulas that work out more cleanly with the accepted definition? Multiplicative inverse in $\operatorname{GF}(2^8)$? I know how to do multiplication over ${\rm GF}(2^8)$:  uint8_t  gmul(uint8_t  a, uint8_t  b) {     uint8_t p=0;     uint8_t carry;     int i;     for(i=0;i&lt;8;i++)     {         if(b &amp; 1)             p ^=a;         carry = a &amp; 0x80;         a = a&lt;&lt;1;         if(carry)             a^=0x1b;         b = b&gt;&gt;1;     }     return p; }   So, I tried to create a ${\rm GF}(2^8)$ multiplication table using this code. I've given below the values in the 3rd row of the table, but I don't think they're correct:      0  1  2  3  4  5  6  7   8  9  A  B  C  D  E  F 0 1 2   0  2  4  6  8  A  C  E  10 12 14 16 18 1A 1C 1E  3 . . E F   I don't know what went wrong. I built the table by multiplying the values in the first row with those in the first column.  E.g. in the third row, I multiplied 2 &times; 0, 2 &times; 1, &hellip;, 2 &times; E, 2 &times; F.  How can I create a multiplication table for arithmetic in ${\rm GF}(2^8)$?    Also, how can I find the multiplicative inverse of a number in ${\rm GF}(2^8)$?  For example, how can I determine that the inverse of 95 is 8A?  I tried to do this using the multiplication table above, but when I took 9th row and the 5th column in the multiplication table I got 2D, not 8A. Resizing an image in a post? Ubuntu installation does not recognize drive partitioning I have a 1TB drive and installed Windows 7 on a 128GB partition.  When I now try to install Ubuntu 11.04 it does not recognize the Windows partition but offers the complete 1TB drive to install Ubuntu on instead. It displays:    However, in the Ubuntu Disk Utility the Windows partitions are recognized.    What do I need to do in order for Ubuntu to recognize the Windows 7 partition and install Ubuntu as a dual boot?   Response to comments  The following commands were executed and the results are shown below:  fdisk -l  WARNING: GPT (GUID Partition Table) detected on '/dev/sda'! The util fdisk  doesn't support GPT. Use GNU Parted.  Disk /dev/sda: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x34a38165     Device Boot      Start         End      Blocks   Id  System /dev/sda1   *           1          13      102400    7  HPFS/NTFS Partition 1 does not end on cylinder boundary. /dev/sda2              13       16318   130969600    7  HPFS/NTFS  Disk /dev/sdb: 500.1 GB, 500107862016 bytes 255 heads, 63 sectors/track, 60801 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x14a714a6     Device Boot      Start         End      Blocks   Id  System /dev/sdb1               1       60801   488384001   83  Linux   parted -l  Warning: Unable to open /dev/sr0 read-write (Read-only file system).  /dev/sr0 has been opened read-only. Error: /dev/sr0: unrecognised disk label Combine the output of two commands in bash Is it possible to combine output from these two commands?  node ~/projects/trunk/index.js  python ~/projects/trunk/run.py run   Neither command exits so I'm not sure how to do this. TikZ doesn't fill when node coordinates are used Can someone please explain me why the following code doesn't result in a fill?  And why, sometimes, I get such "very thin lines" artifact (that's visibile in my picture)?  \begin{tikzpicture} \node (A) at (0,0) {A}; \node (B) at (1,0) {B}; \node (C) at (1,1) {C}; \fill[red] (A) -- (B) -- (C) --cycle; \end{tikzpicture}   Here's what I get: If it is given that component of acceleration perpendicular to the velocity of a body has a constant, non-zero magnitude, how can we mathematically prove that the trajectory of the body will be circular?   I have basic knowledge of calculus, vectors and co-ordinate geometry. I can prove that the trajectory of a body in projectile motion is parabolic. I tried to follow a similar method to get an equation of circle but couldn't succeed. I tried forming parametric equations of x and y in terms of time t (the way we proceed when proving the parabolic path of a projectile) but could not reach the conclusion.  Edit: I have edited the question in order to improve it. However I still can not share my work on this question as I had asked the question almost 6 months ago and now I can not find where I tried solving it. Sincerely sorry for that. Why Smart's attack doesn't work on this ECDLP? Prove that $\left (\frac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ Prove that $\left (\dfrac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} &gt; a^a b^b c^c$ if $a$, $b$ and $c$ are distinct natural numbers.  Is it possible using induction? Optimization used in latex The exact phrase from the book of Swan (the context is choosing the words permit or allow) -      The passive structure with it is only possible with permit   Having said that...     It is not allowed to smoke in the kitchen - incorrect     It is not permitted to smoke in the kitchen - correct   I feel absolutely fine for the incorrect sentence.   Your inputs please. Make me understand this. Can internal forces do work? My Mechanics textbook claims that the sum of the work by internal forces is not generally zero.  translated to English the paragraph reads:     Notice about the work by internal forces: the work by the internal forces is in general not equal to zero although the sum of the internal forces is always equal to zero. Since the work also depends on the relative displacement of the two pointmasses in the material system, the work provided by these 2 internal forces will generally not be equal to zero.   Other sources claim that the work done by internal forces is always zero.  (answer by Mark Eichenlaub)     Then I didn't do any work on the system at all since, by definition, only external forces can do work.        the net work done by internal forces is zero since they do not cause motion of the object.   To me it would seem that these sources contradict eachother. Do they? If so, which one is correct? If not, please explain why they don't contradict eachother. Display the full URL of the home page? Is it possible to display the full, absolute URL of the home page in Drupal 7, like  www.example.com/The-best-example-possible ?  I'd like to display the full URL of the home page, not just the hostname. This copy of the Install OS X El Capitan application can't be verified. It may have been corrupted or tampered with during downloading Macbook Pro 13" 2015 with OS X Yosemite on board. I've been trying to install OS X El Capitan, but always get the following message:  "This copy of the Install OS X El Capitan application can't be verified. It may have been corrupted or tampered with during downloading."  By the way this is not a date time issue, date is correct on my computer.    How can I fix this? Who knows the solution, please help. Necessary and sufficient condition for the matrix $A = I - 2 x x^t$ to be orthogonal How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$ If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$? Imputation before or after splitting into train and test? Why are POSIX Find Parameters Different from Other Program styles? Why is a parameter in the POSIX find command added with a single hyphen for multi-character parameter names, while most other programs use single hyphen to indicate multiple single-character flags, and double-hyphens to indicate single parameter names with multiple characters?  This seems inconsistent, and I'm curious as to  A. why it's really not inconsistent, or  B. the history of that decision. Question about a bijection between $\mathbb R$ and $\mathbb R \cup \{\infty\}$ About a bijection between $\mathbb R$ and $\mathbb R \cup \{\infty\}$  If we'll take a line $f(x)=\begin{cases} ax &amp;, x\in\mathbb R\\ 0 &amp;, x=\infty  \end{cases}$ then we'll have a bijection.   But about this function: $g(x)=\begin{cases} x &amp;, x\in\mathbb R\setminus \{0\}\\ 0 &amp;, x=\infty  \end{cases}$ ? it appears to be a bijection and it's even simpler.