Can every physically sound differential equation, that is covariant, deterministic etc. be derived by extremising a suitable action using a suitable lagrangian, that may be arbitary. Is this a mathematical theorem?
9610
We see variational principles coming into play in different places such as Classical Mechanics (Hamilton's principle which gives rise to the Euler-Lagrange equations), Optics (in the form of Fermat's principle) and even General Relativity (we get Einstein's equation from the Einstein-Hilbert action). However, how do we explain this very principle, i.e., more mathematically, I want to ask the following:      If I am given a set of generalized positions and velocities, say, $\{q_{i}, \dot{q}_{i}\}$, which describes a classical system with known dynamics (equations of motion), then, how do I rigorously show that there always exists an action functional $A$, where    $$A ~=~ \int L(q_{i}, \dot{q}_{i})dt,$$    such that $\delta A = 0$ gives the correct equations of motion and trajectory of the system?    I presume historically, the motivation came from Optics: i.e., light rays travel along a path where $S = \int_{A}^{B} n ds$ is minimized (or at least stationary). (Here, $ds$ is the differential element along the path). I don't mind some symplectic geometry talk if that is needed at all.
11074
While deriving Hamiltonian from Lagrangian density, we use the formula $$\mathcal{H} ~=~ \pi \dot{\phi} - \mathcal{L}.$$ But since we are considering space and time as parameters, why the formula  $$\mathcal{H} ~=~ \pi_{\mu}\partial^{\mu} \phi - \mathcal{L}$$ is not used? Is there any particular book/lecture notes dealing with these kind of issues in theoretical physics, I would love to know them?
72578
Is the Schrodinger equation covariant under Galilean transformations? I am only asking this question so that I can write an answer myself with the content found here:  and here:  I learned about these pages in a comment to  by Ron Maimon. I think Ron Maimon is the original writer of this content. This is creative commons, so it's ok to copy it here. It is not on any textbook on non-relativistic quantum mechanics that I know of, and I thought it would be more accessible (if to no one else, at least for myself) and safe here. I hope this type of question is not in disagreement with site policy.
101891
When proving the Galilean invariance of Newton's laws is it tacitly assumed that all equations are covariant, i.e. that they are form invariant?   For example, it is fairly trivial to show that the right-hand side of Newton's 2nd law is invariant under Galilean transformations. That is $$m\mathbf{a}=m\mathbf{a}'$$ however, does one simply assume that the relationship $\mathbf{F}=m\mathbf{a}$ holds in both inertial frames? It seems reasonable to assume that a given force acting on a body shouldn't change depending on the frame of reference, i.e. that $\mathbf{F}=\mathbf{F}'$, since physical laws should be observer independent and hence the equations describing them should also be observer independent.  As for other quantities, such as momentum and kinetic energy, these are clearly not Galilean invariant quantities, yet their equations are covariant, i.e. $\mathbf{p}=m\mathbf{v}\rightarrow\mathbf{p}'=m\mathbf{v}'$ and $E_{k}=\frac{1}{2}m\mathbf{v}^{2}\rightarrow E'_{k}=\frac{1}{2}m\mathbf{v}'^{2}$, but in each case $\mathbf{p}'\neq\mathbf{p}$ and $E_{k}'=E_{k}$. In this case is it simply that momentum and kinetic energy are simply defined by the equations $\mathbf{p}=m\mathbf{v}$ and $E_{k}=\frac{1}{2}m\mathbf{v}^{2}$, and so trivially $\mathbf{p}'=m\mathbf{v}'$ and $E'_{k}=\frac{1}{2}m\mathbf{v}'^{2}$?
88735
We have two inertial coordinate systems, $K'$ and $K$. $K$ is moving with infinitesimal velocity ${\epsilon}$ relative to $K'$. Using Galilean relativity we can transform this into $v'=v+{\epsilon}$. Here Lagrangian is only a function of speed $L=L(|v|^2)$. $L'$ may differ from $L$ only by a total derivative of function of coordinates and time, i.e. $L'=L+\frac{d}{dt}f(q,t)$  $"$We have $L'=L(|v|^2+2{v}·{\epsilon}+{\epsilon}^2)$. [1. ] Expanding this expression in powers of ${\epsilon}$ and neglecting terms above the first order, we obtain $L(|v'|^2)=L(|v|^2)+\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ [2.] The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $v$ $"$  I understand nothing in the quoted part. How do we get [2.]? Why does $\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ being a linear function of velocity imply that it's a total time derivative?
16688
states that, for every continuous symmetry of an action, there exists a conserved quantity, e.g. energy conservation for time invariance, charge conservation for $U(1)$. Is there any similar statement for discrete symmetries?
119901
If two Lagrangian (densities) $\mathcal{L}$ give the same equations of motion, are they equivalent?
75408
Why is the  formulated in terms of virtual displacements rather than real displacements in time?  EDIT In other words, which step of the derivation of D'Alembert's principle (or Largrange's equation) will not work if one uses real displacements and why?
20303
The following is exercise 8.2 in 3rd edition (and exercise 8.19 in 2nd edition) of Goldstein's Classical Mechanics.   Adding the total time derivative of a function of $q_i$ and t to the Lagrangian will not change the the Euler-Lagrangian equation. So if we make the following change to Lagrangian,   $$L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q_1,q_2,...,q_n,t)}{dt}$$ we can get  $$ \frac{d}{dt}\frac{\partial{L'}}{\partial{\dot{q_i}}} - \frac{\partial{L'}}{\partial{q_i}} = 0 $$ from  $$ \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_i}}} - \frac{\partial{L}}{\partial{q_i}} = 0 $$  How can we get the corresponding Hamiltonian equation part? This is to prove  $$ \dot{p'_i} = \frac{\partial{H'}}{\partial q_i} $$ $$ -\dot{q_i} = \frac{\partial{H'}}{\partial p'_i}  $$  from   $$ \dot{p_i} = \frac{\partial{H}}{\partial q_i} $$ $$ -\dot{q_i} = \frac{\partial{H}}{\partial p_i}  $$ where $p'_i = \frac{\partial L'}{\partial \dot q_i}$.  Edit  The corresponding $H'$ is $$ H' = \sum_k{p'_k \dot{q_k}} - L' $$ where $p'_k = \frac{\partial L'}{\partial \dot q_k}$.
57592
In the second chapter of Peskin and Schroeder, An Introduction to Quantum Field Theory, it is said that the action is invariant if the Lagrangian density changes by a four-divergence. But if we calculate any change in Lagrangian density we observe that under the conditions of equation of motion being satisfied, it only changes by a four-divergence term.  If ${\cal L}(x) $ changes to $ {\cal L}(x) + \alpha \partial_\mu J^{\mu} (x) $ then action is invariant. But isn't this only in the case of extremization of action to obtain Euler-Lagrange equations.   Comparing this to $ \delta {\cal L}$  $$ \alpha \delta {\cal L} = \frac{\partial {\cal L}}{\partial \phi} (\alpha \delta \phi) + \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \partial_{\mu}(\alpha \delta \phi) $$  $$= \alpha \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi \right) + \alpha \left[ \frac{\partial {\cal L}}{\partial \phi} - \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \right) \right] \delta \phi.    $$   Getting the second term to zero assuming application of equations of motion. Doesn't this imply that the noether's current itself is zero, rather than its derivative? That is:  $$J^{\mu} (x) =  \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi .$$  I add that my doubt is why changing ${\cal L}$ by a four divergence term lead to invariance of action globally when that idea itself was derived while extremizing the action which I assume is a local extremization and not a global one.
46448
It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others.  Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs".   Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian?  I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism.
22381
From Landau &amp; Lifshitz The Classical Theory Of Fields it is said:     To determine the action integral for a free material particle (a particle not under the influence of any external force), we note that this integral must not depend on our choice of reference system, that is, it must be invariant under Lorentz transformations.   This seems understandable. But in comments on  Ján Lalinský says that "there is no good physical reason why action should be invariant". Further he suggests another Lagrangian than that given in L&amp;L, namely, if we denote L&amp;L Lagrangian as $L_0$, then the example could be $L_0+Cv_x$, which is clearly anisotropic. Clearly, the equations of motion must not change with this Lagrangian, because $Cv_x$ is a total time derivative (of $Cx$).  On the other hand, in  Luboš Motl says that "the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. [snip] If $S$ depended on the inertial system, so would the terms in the equations $\delta S=0$, and these laws of motion couldn't be Lorentz-covariant".  How could I connect L&amp;L and Luboš's arguments with Ján's example? Can both sides be simultaneously right? They seem to contradict each other.
120278
Why is the d'Alembert's Principle  $$\sum_{i} (  {F}_{i} - m_i \bf{a}_i )\cdot \delta \bf r_i = 0$$  stated in terms of "virtual" displacements instead of actual displacements? Why is it so necessary to "freeze" time in displacements? Also, what would $\sum_{i} (  {F}_{i} - m_i \bf{a}_i )\cdot d \bf r_i$ correspond to if anything at all? In other words, what will be the value of the expression with real displacements instead of virtual ones?
45649
Given the differential equation $$\dot x = -2x + u$$ determine the optimal control law $u = - kx$ that minimizes the performance index $$J = \int_0^{\infty} x^2 \, \mathrm d t$$  My approach was to find the state feedback $k$. But since the value of $R$ (positive semidefinite Hermitian) is not given, that means $R=0$. How do I determine the optimal control for this system where $R=0$?
56027
I have encountered a problem while re-reading the formalism of Hamiltonian mechanics, and it lies in a very simple remark.   Indeed, if I am not mistaken, when we want to do mechanics using the Hamiltonian instead of the Lagrangian, we perform a Legendre tranformation on the Lagrangian to get the Hamiltonian. This, in the case of a 1 dimensional problem, is written as follows: $$ H(p,q) = p\dot{q}-L(q,\dot{q}).$$ Notice that this transformation is such that $H=H(p,q)$, where $p = \frac{\partial L}{\partial \dot{q}}$. By variation of $H$, we can indeed verify it as function of $p$ and $q$, so that they are now considered the new independent variables of the problem.  So far so good. However, there is a problem, and it lies in the fact that for this construction to hold, we need the Lagrangian not to change convexity. Let me write what I know about the Legendre transformation, in a somewhat formal way:  Given a function $f: x\rightarrow f(x)$, we define the function $p: x\rightarrow p(x)$ by the relation $\frac{df}{dx} = p$. Supposing $\frac{df}{dx}$ is invertible, we can define the inverse of $p(x)$, which we call $g : p\rightarrow g(p)$. Then, the legendre transformation of $f$ is $f^* : p\rightarrow f^*(p)$ such that $\frac{df^*}{dp}=g(p)$. We can write, in a more familiar way, $g(p) = x(p)$ since it is the inverse of $p(x)$.  Anyway, we can prove with those assumption that $f^* = pg(p)-f(g(p))$ which is $f^* = px-f(x)$ written in the "familiar" way. All that is just to point out that, for all this construction to work, and hence for the existence of $f^*$, we need the condition that $f(x)$ be of constant convexity, otherwise $\frac{df}{dx}$ is not invertible and we cannot even define $g(p)$.  However, when we consider a general Lagrangian, I don't think that this is always the case. Taking simply $L = \dot{q}^3$ makes the Lagrangian not of constant convexity. And yet, we always use the Hamiltonian, without ever checking this convexity constraint. Why can we do this? Is it because we are interested in the local behavior of our Lagrangian? But even then, what would we do at an inflexion point?   Or is it because a general "physical" Lagrangian will always satisfy the condition of constant convexity?
17652
I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy.  Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations?  I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking.  My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that?
65341
I would like to recover the (timelike) geodesics equations via the variational principle of the following action:  $$ \mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}} $$  Using an arbitrary auxiliary parameter $\lambda$, then one is able to rewrite the action and obtains: $$ d\tau = \sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$ $$ -m \int d\tau = -m \int (d\tau / d\lambda) d\lambda =  -m\int\sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$  Now we can vary the paths $x^{\mu}(\tau) \rightarrow x^{\mu}(\tau) + \delta x^{\mu}(\tau)$.  I need the following equation to be true, so i get the geodesics equations, but I don't know how one can say the following: $$ \int d\tau \,\delta g_{\mu\nu}\frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} = \int d\tau\, g_{\mu\nu, \rho} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} \delta x^{\rho} $$ I know there is a simpler action, which leads to the same equation of motion, but there one has to do the same manipulation, which I sadly don't understand.  Thanks for helping me!