Show $\frac{d^3\underline{k}}{2\omega_{k}(2\pi)^3}$ is Lorentz   invariant.      Hint: try to evaluate $\int dk_0\delta(k_0^2 - M^2)\theta(k_0)$ where $M^2 = \underline{k} + m^2$   My attempt is as follows,  $$ I = \int d^4k \delta(k_0^2 - M^2)\theta(k_0) = \int dk_0 \int d^3\vec{k}\ \delta(k_0^2 - (|k|^2 + m^2))\theta(k_0)\\ = \int dk_0 \int d^3\vec{k} \delta(k_0^2 - (|k|^2 + m^2))\theta(k_0) $$  then using $M^2 = m^2 + |k|^2 = \omega_k^2$ and $\delta(a^2-b^2) = \frac{1}{2|\omega_k|}\delta(a-b) + \frac{1}{2|\omega_k|}\delta(a+b)$ we get  $$ I = \int dk_0 \int d^3\vec{k} \Big( \delta(k_0 - \omega_k) + \delta(k_0 + \omega_k) \Big) \frac{\theta(k_0)}{2\omega_k} $$ using the property of the theta function, we get, $$ I = \int_{0}^{\infty} dk_0 \int d^3\vec{k} \Big( \delta(k_0 - \omega_k) + \delta(k_0 + \omega_k) \Big) \frac{1}{2\omega_k} $$ integrating over $dk_0$ and using the fact that $\theta(\omega_k) + \theta(-\omega_k) = 1$ $$ I = \int \frac{d^3\vec{k}}{2\omega_k(2\pi)^3} \Big( \theta(\omega_k) + \theta(-\omega_k) \Big) = \int \frac{d^3\vec{k}}{2\omega_k(2\pi)^3} $$ however, I still have an integral for $\int d^3\vec{k}$ I don't know how to loose this integral without loosing the $d^3\vec{k}$!
When we canonically quantize the scalar field in QFT, we use a Lorentz invariant integration measure given by $$\widetilde{dk} \equiv \frac{d^3k}{(2\pi)^3 2\omega(\textbf{k})}.$$  How can I show that it is Lorentz invariant?
How to verify $H\otimes K \cong \bigoplus\limits_{i\in I}H$ Let $H,~K$ be the Hilbert space. if $\{v_{j}\}_{j\in J}\subset H$ and $\{w_{i}\}_{i\in I}\subset K$ are the orthonormal bases, then how to construct the isomorphic mapping: $H\otimes K \rightarrow \bigoplus\limits_{i\in I}H$. Here, the $H \otimes K$ is the tensor product of two Hilbert space. If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity This is a reworking of a previous question  which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.  I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:     If $u$ is harmonic and bounded in $0 &lt; |z| &lt; \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|&lt;\rho$ when $u(0)$ is properly defined.   My Solution So Far:   Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected.   If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 &lt; |z| &lt; \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.  One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$  in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 &lt; |z| &lt; \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?  Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant... When I want to move a file to the Trash, Nautilus give me an message saying this:     You can't move example.file to the trash can. Do you want to remove it immediately?   I can give you this photo but it's in Spanish What are the advantages/disadvantages of the new encryption feature in ubuntu 13.10 I'm in the middle of installing 13.10 and the Ubuntu page I was following on how to install from a USB stick says nothing about these two choices (on the Installation type screen):     1) Encrypt the new Ubuntu installation for security (you will choose a security key in the next step).      2) Use LVM with the new Ubuntu installation (this will set up Logical Volume Management. It allows taking snapshots and easier partition resizing)   I'm a writer using the computer for mostly writing ebooks but I also do all the things most people do: buy stuff online, online banking etc. I want my information to be safe but I don't want to overburden my system with security stuff either. And I've read a lot of the similarly themed questions here on the forum but they've all been about stuff more technical than I'm concerned about.   Can somebody please just tell me in real layman's terms about the implications of these two choices? Thank you. You can also contact me to customize your template, or create a new unique one.   Do I need to put a comma between the new and unique adjectives in this sentence?  In my native language (if I am not mistaken) we don't put a comma if the adjectives are of a different type. Not sure about English though, and if it's the case here anyway. Any help would be appreciated! Are airlines in the EU obligated to provide water for passengers? Add/Delete table rows dynamically using JavaScript I was trying to remove all the lines of a file except the last line but the following command did not work, although file.txt is not empty.  $cat file.txt |tail -1 &gt; file.txt  $cat file.txt   Why is it so? Understanding minipages - aligning at top I would like to be able to align two minipage at the top, without worrying about what is in the minipage.  I am aware of the optional [t] parameter, but still have some confusion as to how this all works.  I tend to think of a minipage as a regular page in that whatever is placed on it starts at the top and fills the page going across and then down the page, but this does not appear to be the case.  In the MWE example below, I have color coded the sections for reference purposes.   The first case (black) works just as I would expect. The two minipages are aligned exactly as I want with the [t] specifier.  No question here.  The next case (red) only works if the inner minipage also has the [t] specifier. Why is that necessary?  I would have thought that any content would be placed at the top of this page, and since both the outer minipage had a [t] things should have aligned at the top. I would like to place a graphic in the second mini page, and still have the two aligned at the top.   The simple case (blue) of using \includegraphics directly does not result in the content being aligned at the top.  So, then I attempt to put this in a \parbox (green) and this too does not yield the desired results.  I also try to place the graphic inside an inner minipage (brown) and specify the [t] alignment, and even though this worked in the red section, it does not here.  The only difference between the red and brown being that the red section had text, whereas the brown section has a graphic.  I am aware of question regarding  which suggests to place \vspace{0pt} at the top of the minipage, and that does provide the desired alignment (cyan), but adds vertical space at the top.  So, my second question is why is the \vspace{0pt} necessary when a graphic is included, but not when it is text.  This appears to be a hack and wondering if there is a replacement for minipage that works better without having to worry about exactly what is in the minipage?  Also, why did wrapping the figure in a parbox (green) or another minipage (brown) not solve this problem?   Here is the MWE:  \documentclass{article} \usepackage[demo]{graphicx} \usepackage{lipsum} \usepackage{xcolor}  \begin{document} \section{This Works - Aligned at top as desired} \begin{minipage}[t]{0.6\linewidth}     \lipsum[1] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \lipsum[2] \end{minipage}  \color{red} \section{Why does inner mini page required the [t]?} \label{Minipage-in-minipage} \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \begin{minipage}[t]{\linewidth}% Why is [t] required here         \lipsum[4]     \end{minipage} \end{minipage}  \color{blue} \section{Why are These not aligned at top?} \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \includegraphics[width=\linewidth]{foo} \end{minipage}  \color{green} \section{Why are These not aligned at top?} \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \parbox{\linewidth}{\includegraphics[width=\linewidth]{foo}} \end{minipage}  \newpage \color{brown} \section{Why does this not work with the [t] as it did in red section?} This worked fine in Section \ref{Minipage-in-minipage}  \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \begin{minipage}[t]{\linewidth}% Even with [t] here, this does not work         \parbox{\linewidth}{\includegraphics[width=\linewidth]{foo}}     \end{minipage} \end{minipage}   \color{cyan} \section{Adding a \textbackslash{}vspace\{0pt\} works} \begin{minipage}[t]{0.6\linewidth}     \vspace{0pt}% Hack!! But does do the job     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \vspace{0pt}% Hack!! But does do the job     \begin{minipage}[t]{\linewidth}% Why is [t] required here         \parbox{\linewidth}{\includegraphics[width=\linewidth]{foo}}     \end{minipage} \end{minipage} \end{document} I have a qualifying exam coming up in a couple days and I am just trying to understand some pathological examples I have in my notes. I will list a similar problem which I know the solution to and then the question.  True or False:   let $A,B \in \mathbb{Q}^{n \times n }$  Suppose $xI-A$ and $xI-B$ are equivalent then $\det(xI-A) = \det(xI-B)$.     I wrote in my notes this is true because any two matrices are equivalent in $\mathbb{Q[x]}^{n \times n}$ if and only if they have the same invariant factors.  Since the characteristic polynomial is a product of the invariant factors it follows that $\det(xI-A) = \det(xI-B)$.        True or False:   let $A,B \in \mathbb{Q}^{n \times n }$. If $\det(xI-A) = \det(xI-B)$ (in $\mathbb{Q}[x]$) then $xI-A$ and $xI-B$ are equivalent  in $\mathbb{Q}[x]^{n \times n}$   I think this is false because of the fact that equality of determinants is not strong enough to guarantee all invariant factors are equal but I do not have an example. JavaFX accordion with multiple open panes When working with an image in Photoshop, I'd like to have a simple keyboard shortcut, like in , that automatically does:   Export > Export As or Export > Save For Web  Resize to biggest dimension = 600px (either width or height) Add suffix _600.jpg to filename   How to do that?  I'm not sure how to edit a recorded macro to do that (add the suffix and choose 600px for the biggest dimension).  Example: you open myphoto.png, you press F6 and then you automatically get myphoto_600.jpg. What is the best technique for long-exposure photography in good light? Name for the center amplitude of a sine wave I am trying to find the name for the value at the center of a sine wave $$ y(t) = A\sin (kx \pm \omega t+\varphi) + D $$ i.e. the offset $D$ of the wave in the $y$-direction.  calls it Center Amplitude, but further googling this name is difficult and does not provide any further sources. I also found the expression , but I'm not sure that it refers to the same thing. Is there a commonly used name for the $y$-direction-offset of a sine wave? Where is the .Trash folder? While watching  I noticed a very characteristic camera shot. At a scene where the main character, Wendy, sits in her room at the computer (watching the supposedly foretelling fotos), her sister talks to her from the outside (asking if she can have the camera/fotos). There is a certain camera shot, where on the left side a frontal close-up of Wendy's face is shown, while on the right side you see her sister standing in full height inside the open door.    This shot immediately struck me as being somehow characteristic and out of place and infused feelings to me of some kind of older Hitchcock or Brian De Palma thriller or something the like. I also think to remember of having seen this kind of shot or something similar used in , where Jessica Lange is shown in front of the dresser talking to Nick Nolte who stands in the door or father away.  I would like to know if this technique has a certain name or theory behind it and where its origins lie, or if there is any particular director or movie who made this famous. It may be even that Final Destination 3 uses this especially in reference to some director or movie. Is there any difference between "offense" and "offence"? "Offense" vs. "offence", which is more correct? If both are correct, are there any differences in shades of meaning and/or usage? How to check if installing a package will require a reboot? The number of nonnegative integer triples with sum equal to $12$ How many triples $x,y,z$ satisfy  $x+y+z=12$, where  $x,y,z\ge 0$ are integers? Note that these are not ordered: $ (12,0,0)$ and $(0,12,0)$ are treated as same. Progress I have tried the formula ${13\choose 2}$ to select values of $x$ and $y$ from $0$ to $12$ as  $z$ would keep changing according to $x$ and $y$. (That is, $z=12-x-y$.) The problem is that that certain combinations of $x$ and $y$ would give value of $x+y$ greater than $12$. I can't seem to get around this. Is there a way to set transparency of windows in GNOME?