For what I know, the Action should be Lorentz-invariant in accordance to the axioms of SR. For example for a free particle the Action would be the integral of the proper time, which of course is invariant. Now my doubt is the following: does that mean that the Lagrangian should be Lorentz-invariant as well? In some texts I've read it is often stated that the Lagrangian should be a Lorentz scalar, but how does this assures the Action to be Lorentz-invariant too? For example, once one gets the Lorentz-invariant Lagrangian for the EM field $\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$, why should its action:$$\int -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} d^4x$$ be Lorentz invariant too? Aren't we integrating a scalar over a specific set of  coordinates which are not Lorentz-invariant?
This is regards to the lorentz invariance of a classical scalar field theory. We assume that the action which is $S= \int d^4 x \mathcal{L}$, is invariant under a Lorentz transformation. How do you prove that the integration measure $d^4 x$ is Lorentz invariant.
what are all the open subgroups of $(\mathbb{R},+)$ Can I use a HDD on my PS3 if I have already used another on the same console? I'm thinking of changing the Hard Disk of my PS3 (Slim 120GB), but I am unsure of what will happen to the data on the one I previously had if I do this:   PS3 has some data in the HDD. I turn off PS3, change HDD, install OS on new HDD. I download some data on new HDD from PS Plus.   Let's imagine I forgot to back something up...   I turn PS3 off, change HDD back.   Can I still use the old hard drive without needing to format it again? I understand that in a lot of big-lab fields it is common for the principal investigator to append their name to a paper even if they did not write the paper, design the experiment, or collect data since they spend energy securing funding, and managing the whole lab. What about for small labs?  What are the requirements for a supervisor to be included as an author on a paper, as opposed to just appearing in the acknowledgements? If you are working on your own projects independently of your supervisor, but using funding provided by your supervisor (how does this change when the funding provides resources versus just your salary), are you suppose to add them as authors or just acknowledge the source of funding? I got a brand new Windows 7 machine, installed the operating system, created one account and forgot its password. What can I do?  There is no external CD, the operating system is loaded from somewhere inside the machine.  I already tried to remember passwords and tried all candidates with all possible combinations of caps lock, num lock etc. How can author deal with late reviews? Max Serial Cable Length? What is the maximum length of serial cable I can use, to connection work properly? Is length of serial cable in connection with baud rate? I am using about 5 meter cable at speed 38400 bauds and RS232 communication, is it too long? If parsing the output of ls is dangerous because it can break on some funky characters (spaces, \n, ... ), what's the best way to know the number of files in a directory?  I usualy rely on find to avoid this parsing, but similarly, find mydir | wc -l will break for the same reasons.  I'm working on Solaris right now, but I'm looking for a answer as portable across different unices and different shells as possible. Specific system of differential equations I have the following system of equations:  \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray}  I want to prove that if a solutions starts (at time $t = 0$) in the region   $$ G = \{(x, y) \in \mathbb{R}^2 : x &gt; 0, y &gt; 0 \text{ and } 1 &lt; x^2 + y^2  &lt; 4\} $$  that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 &lt; x &lt; 2$ or the $y$-axis where $0 &lt; y &lt; 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x &gt; 0, y &gt; 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it.  In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} &lt; 0$ in $G$, so $x$ is decreasing.)  I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} &gt; 0$, so after crossing the circle, the solution would go back up to $G$.  Is the circle itself an orbit?  EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} &gt; 0 \Leftrightarrow 0 &lt; r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 &lt; r \leq 1$ satisfies $4r - r^3 &gt; 0$, and thus $\frac{dr}{dt} &gt; 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive.  Can someone tell me if this is correct? Can Blindsight "See" Through Walls, Floors, And Ceilings? I am currently play-testing an Awakened Mystic character in D&amp;D 5e, using the beta rules . As such, I have an ability that lets me choose between having Blindsight or Tremorsense.  BLINDSIGHT: ...[B]lindsight can perceive its surroundings without relying on sight, within a specific radius.... (Monster Manual p.8) TREMORSENSE: ...[T]remorsense can detect and pinpoint the origin of vibrations within a specific radius, provided that the [observer] and the source of the vibrations are in contact with the same ground or substance. Tremorsense can't be used to detect flying or incorporeal creatures.... (Monster Manual p.9)  I have been wondering what the difference is between these two, aside from the fact that Tremorsense requires contact with the ground and cannot detect flying or incorporeal creatures. Specifically, I want to know whether Blindsight cannot detect creatures through solid matter such as walls, floors, and ceilings, whereas Tremorsense can? Can Blindsight detect creatures on the other side of a dungeon wall, or on another floor of said dungeon? Are there monsters in 5th Edition that possess both Blindsight and Tremorsense, and if so, when and why would they use one as opposed to the other? I am aware of and appreciate  regarding Blingsight's capability to detect invisible creatures, but it doesn't exactly address my issue of whether solid matter effectively &quot;blocks&quot; Blindsight. I will gladly consider either rules-as-written or rules-as-interpreted answers. If there is a published game mechanic or a posted developer ruling I have overlooked, please include and cite them in your answer. Are layer files compatible between ArcGIS 10 and 9.3? If I save layer files in ArcGIS 10, can they be opened and modified in ArcGIS 9.3? Run true multiple process instances of gnome-terminal How does the /proc//exe symlink differ from ordinary symlinks? If I start a process and then delete the binary of it, I can still recover it from /proc/&lt;pid&gt;/exe:  $ cp `which sleep` . $ ./sleep 10m &amp; [1] 13728 $ rm sleep $ readlink /proc/13728/exe                            /tmp/sleep (deleted) $ cp /proc/13728/exe ./sleep-copy $ diff sleep-copy `which sleep` &amp;&amp; echo not different not different $ stat /proc/13728/exe    File: ‘/proc/13728/exe’ -&gt; ‘/tmp/sleep (deleted)’   Size: 0           Blocks: 0          IO Block: 1024   symbolic link   On the other hand, if I make a symbolic link myself, delete the target and attempt to copy:  cp: cannot stat ‘sleep’: No such file or directory   /proc is an interface to the kernel. So does this symbolic link actually point to the copy loaded in memory, but with a more useful name? How does the exe link work, exactly? An entire function whose real part is bounded must be constant. Greets  This is exercise 15.d chapter 3 of Stein &amp; Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer:  Define $g(z)=f(1/z)$ for $z\neq{0}$,then by the hypothesis we must have that for any $\epsilon&gt;0$ $g(D_{\epsilon&gt;0}(0)-\{0\})$ is not dense in $\mathbb{C}$, then the singularity at $0$ of $g$ is not essential, this implies $f$ must be a polynomial, but if $f$ is a non-constant polynomial, it is easy to see that its real part must be unbounded, so $f$ must be constant.  I would like to know an answer with the the maximum modulus principle.  Thanks The Follow Questions and Answers feature is now live across the Network We are happy to announce that the  follow question and answer feature is now live across the Network, including Stack Overflow, all Stack Exchange sites, and all Meta sites. (International Stack Overflow sites will have it turned on in a day or two once we have translations all set.) You can follow any question or answer (that you did not author) by clicking on the [follow] button that is shown in the menu immediately below the post (alongside the [share] button):    After you have followed a post, you will get inbox notifications for all new answers (in the case where you followed a question), comments, edits, and notices. You will not receive notifications for any action that you performed. As was mentioned in the earlier post, we are not making changes at this point to the notifications received by a post owner, or due to @mentions. The initial release had been planned to only include question following. Thanks to many folks who chipped in over the past few weeks, we are happy to release both question and answer following at the same time. These are available in both the regular and mobile views. We are still planning on two more related releases:  Follows profile tab and question listing filter: A tab for follows will be added to your user profile activity page. Each user will be able to view this tab that will provide a listing of followed questions and answers, with standard sorts, and the ability to unfollow from the listing. Users (with the exception of moderators and authorized staff) will not be able to see this information about other users (nor will it be made public in the API, SEDE, or data dumps). We will be renaming the Favorites feature to Bookmarks. As we mentioned in the preview post, the feature will be the same as Favorites, with the name and icon updated to more accurately represent user expectations and usage. And you will be able to both &quot;bookmark&quot; and &quot;follow&quot; a question.  We hope that this new feature will allow you to have better access to the content that you care about and want to keep tabs on. Update (June 11, 2020): The user profile tab and the rename of Favorites to Bookmarks have been released. Mboxcox, interpreting difficult regressions Mboxcox in Stata suggests transforming my variables using a power of 0.1 for the independent variable, and 0.4 for the dependent variable.  I have run the model, and it fixes problems associated with the assumptions of OLS. But certainly, it complicates matters in terms of interpretations.  Please outline possible interpretations and solutions.  Dependent variable is in millions of dollars, and independent variables are number of years.  and a number of dummies which do not require transformation. Will the Lee 100mm filter holder work with a Samyang 14mm autofocus lens? Will this  work with a ? Which adapter ring would I need to buy for it? Thank you, and apologies if any of this comes across as confusing - I'm very new to filters and what-not. I posted on another account yesterday and received some very useful tips, but had to remove it as it was a duplicate account causing me login issues. Thanks, guys! What does a leading zero do for a php int? Understand PAM and NSS In the last days I have set up some Linux system with LDAP authentication and everything works fine, but there's still something I can't really understand regarding NSS and PAM, also after a lot of research. Citing:  NSS allows administrators to specify a list of sources where authentication files, host names and other information will be stored and searched for  and  PAM is a set of libraries that provide a configurable authentication platform for applications and the underlying operating system  What I don't understand is how PAM and NSS work and interact together. In  the architecture is explained pretty well: I configure PAM to use pam_ldap for LDAP accounts and pam_unix for local accounts, then I configure nsswitch.conf to fetch information from local files and LDAP. If I have understood correctly LDAP is used twice: first by pam_ldap and then by NSS which is itself called from pam_unix. Is that right? Is LDAP really used twice? But why do I need to configure both NSS and PAM? My explanation is that PAM performs different tasks than NSS and it is used by other programs. But, then, it should be possible to use only NSS or only PAM, as I have read in . So I experimented a bit and I have first tried to delete LDAP from the nsswitch.conf (and the authentication stopped to work as if only pam_ldap is not enough to do the job). Then I re-enabled LDAP in NSS and I deleted it from the PAM configuration (this time everything worked fine, as if pam_ldap is useless and NSS is enough to authenticate a user). Is there anyone who can help me to clarify this? Many thanks in advance. UPDATE I've just tried something now. I removed again all the pam_ldap entries in all pam configuration fields and I have also removed shadow: ldap from nsswitch.conf. As now in all the system there are only the lines: passwd: ldap files and group: ldap files in nsswitch.conf. Well... the login with LDAP users works perfectly, those two lines (plus /etc/ldap.conf) are enough to configure LDAP auth. From my knowledge PAM in independent from NSS, but my tests showed it's not. So I ask myself is it possible to completely disable NSS and use only PAM? I just started real analysis.  I don't have a background in proofs or logic, simply calculus.  So I'm trying to learn more about proofs--so forgive the basic question, please.  How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$.  This number is given by $x=b+(-a)$.  First part of the proof I understand--it's simple.  We simply use some axioms and do the following:  $a+b-a=b$, which of course is true.  This is just true from plugging in the $b+(-a)$ for $x$.  But how about the uniqueness issue?  In my text, it says:  $$(a+x)+(-a)=b+(-a) x=b+(-a)$$  How does this prove uniqueness?  Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?  How would I have known to set this up as part of the proof if I was asked to prove something is unique?  Sorry for the basic question--have to start somewhere. How can I get my usual interface when opening projects created by other people? I've bougth cup models from internet.   In the ZIP I get the blend file and pressing F12 I can render the project (Do you call it project?)  But I don't see anymore the "Layout" tab, that is what I use the most.   Why and how can I go back to my usual interface?    Like in the Opening Layout:    UPDATE 1  I could open the file with the default layout, however I'm not seeing the objects in there:    UPDATE 2  How to open Collection 1? It's grayed out.