When deriving the Euler-Lagrange equations from the Principle of Stationary Action we start by insisting that the actual path of a particle minimizes the action $S$, defined as $$S=\int_{t_1}^{t_2}\mathcal{L}(q,\dot{q},t),\quad\text{where}\quad\dot{q}=\frac{dq}{dx}.$$ And then we show that a necessary condition for $S$ to be minimized is that $\mathcal{L}$ obeys the Euler-Lagrange equations, $$ \frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{q}}\right)-\frac{\partial\mathcal{L}}{\partial q}=0. $$ My problem is that nowhere in the derivation do we make use of the fact that the Lagrangian is of the form $\mathcal{L}=T-U$, where $T$ is the kinetic energy, and $U$ is the potential energy. So my question is: Is there an explanation for this identification beyond the fact that this what makes the physics work? My best guess at the answer is that if we instead started out with the Principle of Virtual Work, we eventually get to $$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_j}\right)-\frac{\partial T}{\partial q_j}=Q_j. $$ Where $Q_j$ is the generalized force, defined as $$ Q_j\equiv \sum_i\vec{F}_i\cdot\frac{\partial\vec{r}_i}{q_j}, $$ and if we assume that the applied forces on the system are conservative, so that $$ \vec{F}_i=-\vec{\nabla}U(\vec{r}_1,\vec{r}_2,...,\vec{r}_N), $$ and this potential, $U$, doesn't depend on the generalized velocities, we get $$ \frac{d}{dt}\left[\frac{\partial}{\partial \dot{q}_j}(T-U)\right]-\frac{\partial}{\partial q_j}(T-U)=0. $$ Which is exactly the E-L equation, with $\mathcal{L}=T-U$, so we can claim that both principles should give us the same physics, and therefore the Lagrangian we arrived at from the Principle of Stationary Action should also be $\mathcal{L}=T-U$. Maybe it's just me, but this seems unsatisfactory. Have we simply defined the action so that we have a more compact formulation of the physics we already knew, or can it stand by itself?
Is there a proof from the first principle that for the Lagrangian $L$,    $$L = T\text{(kinetic energy)} - V\text{(potential energy)}$$  in classical mechanics? Assume that Cartesian coordinates are used. Among the combinations, $L = T - nV$, only $n=1$ works. Is there a fundamental reason for it?  On the other hand, the variational principle used in deriving the equations of motion, Euler-Lagrange equation, is general enough (can be used to to find the optimum of any parametrized integral) and does not specify the form of Lagrangian. I appreciate for anyone who gives the  answer, and if possible, the primary source (who published the answer first in the literature).      Notes added on Sept 22: - Both answers are correct as far as I can find. Both answerers were not sure about what I meant by the term I used: 'first principle'. I like to elaborate what I was thinking, not meant to be condescending or anything near to that. Please have a little understanding if the words I use are not well-thought of. - We do science by collecting facts, forming empirical laws, building a theory which generalizes the laws, then we go back to the lab and find if the generalization part can stand up to the verification. Newton's laws are close to the end of empirical laws, meaning that they are easily verified in the lab. These laws are not limited to gravity, but are used mostly under the condition of gravity. When we generalize and express them in Lagrangian or Hamiltonian, they can be used where Newton's laws cannot, for example, on electromagnetism, or any other forces unknown to us. Lagrangian or Hamiltonian and the derived equations of motion are generalizations and more on the theory side, relatively speaking; at least those are a little more theoretical than Newton's laws. We still go to lab to verify these generalizations, but it's somewhat harder to do so, like we have to use Large Hadron Collider. - But here is a new problem, as @Jerry Schirmer pointed out in his comment and I agreed. Lagrangian is great tool if we know its expression. If we don't, then we are at lost. Lagrangian is almost as useless as Newton's laws for a new mysterious force. It's almost as useless but not quite, because we can try and error. We have much better luck to try and error on Lagrangian than on equations of motion. - Oh, variational principle is a 'first principle' in my mind and is used to derive Euler-Lagrange equation. But variational principle does not give a clue about the explicit expression of Lagrangian. This is the point I'm driving at. This is why I'm looking for help, say, in Physics SE. If someone knew the reason why n=1 in L=T-nV, then we could use this reasoning to find out about a mysterious force. It looks like that someone is in the future.
Error-Handling in Swift-Language How to fix apt: Signature by key uses weak digest algorithm (SHA1)? I started setting up by adding repositories and then went to run a sudo apt-get update again before I started installing other software, and I get the Signature key lines and it stops. So it essentially won't let me update any packages now.  d@EliteBook:~/Downloads$ sudo apt-get update Ign:1 http://dl.google.com/linux/chrome/deb stable InRelease Hit:2 http://dl.google.com/linux/chrome/deb stable Release                      Hit:4 http://security.ubuntu.com/ubuntu xenial-security InRelease               Get:5 http://ca.archive.ubuntu.com/ubuntu xenial InRelease [247 kB] Hit:6 http://ca.archive.ubuntu.com/ubuntu xenial-updates InRelease Hit:7 http://ca.archive.ubuntu.com/ubuntu xenial-backports InRelease Fetched 247 kB in 0s (256 kB/s)                    Reading package lists... Done W: http://dl.google.com/linux/chrome/deb/dists/stable/Release.gpg: Signature by  key 4CCA1EAF950CEE4AB83976DCA040830F7FAC5991 uses weak digest algorithm (SHA1) d@EliteBook:~/Downloads$   I've never seen this before whenever I setup and start installing things in Ubuntu. Is there something else I can do? It is easy to find the minimum of $\|Ax-b\|_2$, when $A$ has full column rank. But how is the case when we add the constraint $\|x\|_2=1$? Or, to be explicit,  $$\min_{\|x\|_2=1}\|Ax-b\|_2=?$$  My idea is to construct the corresponding Lagrange function  $$L(x,\lambda) = (Ax-b)^T(Ax-b) + \lambda(x^T x - 1)$$  Differentiate the function with respect to $x,\lambda$, we have  $$A^T A x - A^T b + \lambda x = 0$$  $$x^T x=1$$  However, I find it hard to move forward to get the answer. Could anyone help me? Thanks a lot! Let $f(x)$ be a monotone non-increasing function such that $\int_{0}^{\infty}f(x) dx$ converges. Prove: $\lim\limits_{x \to \infty} xf(x)=0$. My question is, why can't I simply contradict any other possibility by using the Integral Limit Comparison Test with $1\over x$? I am, after all, to show that $f(x)=o(x)$ as $x\to \infty$. I really don't understand why monotony is crucial here. I could use some help. Process Unicode files with BOM correctly with POSIX tools Trying to use grep today, I ran into the  of the  in a Unicode file (UTF-8, in this case). Specifically, I was trying to find a file beginning with XYZ with the pattern grep '^XYZ', but of course grep treated the BOM as three separate characters and did not match the first line of the file if the first line started with XYZ. I even tried to update the regular expression to ignore spaces ('^[[:space:]]*XYZ'), but to no avail.  Other questions have dealt with  or , but I want to know if POSIX tools have a general option to handle Unicode files correctly. If grep handled the Unicode file correctly, it would consider the file contents to start after the BOM and match XYZ on the first line just like any other line. Do I have to code each case of this Grid full of plots separately? How to find the build / version of Windows 10 During the beta of Windows 10 it was hard know what version you were running unless it had it posted to the desktop. Once that wasn't there - how do you tell what version / build you are running?   This will become more of an issue as Microsoft starts releasing more builds with the new update mechanism. Landscape topography from grayscale image Is there some sort of classification of all minimal non-cyclic groups? Does there exist some sort of classification of all minimal non-cyclic groups (non-cyclic groups, such that all their proper subgroups are cyclic)  I know the following classes of such groups:  1) $C_p × C_p$, where $p$ is a prime  2) $Q_8$  3) $\langle a,b | a^p = b^{q^m} = 1, b^{−1}ab = a^{r}\rangle$, where $p$ and $q$ are distinct primes and $r ≡ 1 \pmod q$, $r^q ≡1 \pmod p$.  (These three classes completely cover the case, when our group is finite: )  4)$C_{p^{\infty}}$, where $p$ is a prime  5)$(\{ \frac{n}{p^m}| m, n \in \mathbb{Z} \}, +)$, where $p$ is a prime  (These two classes completely cover the case, when our group is infinite abelian: )  6)Infinite non-abelian groups, such that all their nontrivial proper subgroups are isomorphic to $C_{p}$ for a fixed prime $p$ (Tarski monster groups)  7)Infinite non-abelian groups, such that all their nontrivial proper subgroups are isomorphic to $C_{\infty}$ ().  However, I do not know, whether there exists anything, that does not fall into these classes. I only know, that if such groups exist, they have to be infinite non-abelian. Why isn't temperature frame dependent? In (non-relativistic) classical physics, if the temperature of an object is proportional to the average kinetic energy ${1 \over 2} m\overline {v^{2}}$of its particles (or molecules), then shouldn't that temperature depend on the frame of reference - since $\overline {v^{2}}$ will be different in different frames?  (I.e. In the lab frame  $K_l = {1 \over 2} m\overline {v^{2}} $, but in a frame moving with velocity $u$ relative to the lab frame, $K_u =  {1 \over 2} m \overline {(v+u)^{2}}$). Prove that $\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$ Prove that $$\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$   I tried to look at $$ f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n $$  And maybe taking it's derivative but it didn't work out well.  Any ideas? In-Page Editor Line Distortion The SFDC In-page editor has always had distortion for me, and I'm just wondering if it's just a lousy browser plugin they're using or if someone has found a good browser / operating system combination that stops the obnoxious distortion.   Obviously Salesforce doesn't have much interest in fixing it: as it's been this way on multiple operating systems and browsers I've used since I've used Salesforce.  However, my frustration with their other developer tools has lead me to use it occasionally and it's always the same: What is the formula for armor/magic resist reduction? As far as I can tell, it's something like this:  damage reduction = armor / (armor + 100)   Is this accurate? My passport is expiring; can I still travel? Answer didn't load temporary How can I make Chrome open bookmarked pages in new tab by default? Current situation:  I am currently browsing a page (say, www.google.com). I want to open a bookmarked page (say, www.bing.com), but I want to keep the current page (Google) opened. I need to click on new tab, then click on the bookmark for Bing. Alternatively, I can right click and open Bing in new tab.  But I want none of those because why click twice when one is enough?    What I want:  I am currently on Google. I want to open Bing, so I click on the bookmark and Bing opens in a new tab. Google remains open.    I am aware of a duplicate question linked . However it didn't really provide a satisfactory answer. Changing every link to javascript breaks the icon and is a pain to do; middle clicking doesn't really work because I'm on a laptop and the touchpad does not support middle clicking. I hope there is an extension for Chrome that can redefine this behavior to the way I want. A hidden setting in Chrome would be even better but I doubt Chrome offers that. Let $p,q,r$ be integers such that the symmetric sum of fractions $\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}$ is an integer. Prove that each of the numbers: $\dfrac{pq}{r},\dfrac{qr}{p},\dfrac{pr}{q}$ is an integer. How to do this? Someone posted but deleted soon. Leave only two decimal places after the dot Almost universal class I so stuck with a problem of set theory. But first a recursive definition:   Define $R_0=\emptyset$ If $R_\alpha$ is defined, then $R_{\alpha+1}=\mathcal{P}(R_\alpha)$ (the power set). For a limit ordinal $\gamma$, if $R_\alpha$ is defined for all $\alpha&lt;\gamma$, then define $R_\gamma=\displaystyle\bigcup_{\alpha&lt;\gamma} R_\alpha$   We define $\text{BF}=\displaystyle\bigcup_{\alpha\in\text{OR}} R_\alpha$ (the class of well founded sets).  Next, my problem     Take $A\subseteq \text{BF}$ a proper transitive class such that $(A,\in)\models\text{ZF}$. Prove that $A$ is almost universal.    I think that the exercise is false because if it was true, then we could conclude that the strongly inaccessible cardinals doesn't exists. This because if $\kappa$ is strongly inaccessible then $R_\kappa$ satisfies the hypothesis but $R_\kappa$ isn't almost universal. I really appreciate any hint or/and suggestion.   Edit: my counterexample is wrong. But, then, how can I solve the exercise? Given Ohm's law, how can current increase if voltage increases, given fixed resistance? According to Ohm's law, $V=IR$ (voltage equals current times resistance).  So if the voltage increases, then the current increases provided that the resistance remains constant.  I know that Voltage or potential difference means work done per unit positive charge in bringing that charge from one point to another.  So according to Ohm's law, if the work done per unit charge increases then current will increase. How can this be true? Point out my mistakes.