Say we construct the Lagrangian for a system and minimise the action, what happens if this is not unique? In other words the action is minimised by two distinct (not infinitesimally separated) paths. Is there something else that governs the evolution of the system, or is there always a unique solution to the Euler-Lagrange equations?
62682
A few months ago I was telling high school students about Fermat's principle.   You can use it to show that light reflects off a surface at equal angles. To set it up, you put in boundary conditions, like "the light starts at A and ends at B". But these conditions by themselves are insufficient to determine what the path is, because there's an extra irrelevant stationary time path, which is the light going directly from A to B without ever bouncing off the surface. We get rid of this by adding in another boundary condition, i.e. that we only care about paths that actually do bounce. Then the solution is unique.  Of course the second I finished saying this one of the students asked "what if you're inside an elliptical mirror, and A and B are the two foci?" In this case, you can impose the condition "we only care about paths that hit the mirror", but this doesn't nail down the path at all because any path that consists of a straight line from A to the mirror, followed by a straight line to B, will take equal time! So in this case the principle tells us nothing at all.  The fact that we can get no information whatsoever from an action principle feels disturbing. I thought the standard model was based on one of those!  My questions are   Is this anything more than a mathematical curiosity? Does this come up as a problem/obstacle in higher physics? Is there a nicer, mathematically natural way to state the "only count bouncing paths" condition? Also, is there a "nice" condition that specifies a path in the ellipse case? What should I have told that student?
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.
43754
Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined? For example, given the  for a  $$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$ with $(+,-,-,-)$ Minkowski sign convention, is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian? As another example: consider the Lagrangian $$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$ which can also be cast in the form $$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$ I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a  equation for each component of the field: $$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$ but why does this prevent me from considering $A^\mu$ a spin-1 massive field?  [1]: From Matthew D. Schwartz's , p.114:  A natural guess for the Lagrangian for a massive spin-1 field is $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,$$ which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.
79959
I'm reading a proof about Lagrangian => Hamiltonian and one part of it just doesn't make sense to me.  The Lagrangian is written $L(q, \dot q, t)$, and is convex in $\dot q$, and then the Hamiltonian is defined via the Legendre transform  $$H(p,q,t) = \max_{\dot q} [p \cdot \dot q - L(q, \dot q, t)]$$  Under the right conditions there exists a function $\dot Q (p,q,t)$ such that  $$H(p,q,t) = p \cdot \dot Q(p,q,t) - L(q, \dot Q(p,q,t), t)$$  i.e. when some $\dot Q(p,q,t)$ satisfies $p = \frac{\partial L}{\partial \dot q}\rvert_{(q, \dot Q(p,q,t), t)} = \frac{\partial L}{\partial \dot q}(q, \dot Q(p,q,t), t)$  (Finding this function is usually called "inverting p")  By taking partials in the $p$ variable and using the relationship, we can obtain the relationship  $$\dot Q = \frac{\partial H}{\partial p}$$  Because of the notation I chose, I get the strong urge to say $\dot q = \frac{\partial H}{\partial p}$  , and in fact this is what the textbook does. But have we proved this?    In other words, how can we deduce that  $$q'(t) = \frac{\partial H}{\partial p}(p(t), q'(t), t)$$ for any differentiable vector valued function $q$? (or maybe there are more conditions we need on $q$? Here  $$p(t) = \frac{\partial L}{\partial \dot q}(q(t), q'(t), t)$$ according to Lagrange's equations.
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?
94603
It is often stated that the Lagrangian formalism and the Hamiltonian formalism are equivalent.   We often hear people talk about eigenvalues of Hamiltonians but I have never ever heard a word about eigenvalues of Lagrangians.   Why is this so? Is it not useful? is it not possible to do it?
88519
From the Lagrangian I've got the following equations of motion for the  in 2D. (The masses are different but the lengths of the two pendula are equal.) Let $m_2$ be the lowest-hanging mass.     $$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ -2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-(m_1+m_2)g/l\sin(\theta_1)$$   and     $$m_2\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ 2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-m_2g/l\sin(\theta_1)$$   In the small angle approximation these become, respectively     $$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2= -2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1(m_1+m_2)g/l$$   and      $$m_2\ddot{\theta_1}+2m_2\ddot\theta_2= 2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1m_2g/l$$.   Most sources don't have the terms of order $\dot\theta$. This is because they apply the small angle approximation to the Lagrangian before taking the derivatives, thereby ignoring terms of order $\theta.$ What justification do we have for getting rid of these terms?
13459
So I was reading this:  and while the answers for the first question are good, nobody gave much attention to the second one. In fact, people only said that it can be proved without giving any proof or any.  So, if I have a Lagrangian and ADD an arbitrary function of $\dot{q}$, $q$ and $t$ in such a way that the equations of motion are the same, does this extra function MUST be a total time derivative?   EDIT Ok, I changed my question a little bit:  Question: If I have a function that obeys the Euler-Lagrange equation off-shell, this implies that my function is a time derivative? (This was used in Qmechanic's answer of this other question: , equation 14.)  Also, why people only talk about things that change the Lagrangian only by a total derivative? If this is not always the case that keeps the equation of motion the same, so why is it so important? And why in the two questions I posted about the same statement on Landau&amp;Lifshitz's mechanics book only consider this kind of change in the Lagrangian?
121167
Let’s say you have a 3000 lb vehicle moving very slow at 1 kilometer per hour. It seems obvious that you could simultaneously determine its position and momentum at any point along the way but does the uncertainty principle still apply even here?  You should be able to calculate this at a future point. If not then why not? If so then at what point do you revert back to uncertainty?
29432
In Sean Carroll's book "The Big Picture," he states ():      Classical mechanics, the system of equations studied by Newton and   Laplace, isn't perfectly deterministic. There are examples of cases where a unique outcome cannot be predicted from the current state of the system. This doesn't bother most people, since cases like this are extremely rare—they are essentially infinitely unlikely among the set of all possible things a system could be doing. They are artificial and fun to think about, but not of great import to what happens in the messy world around us.   What are some of these situations he is referring to?   My first guess was something like the conceptual equivalent of a ball sitting exactly atop a hill and asking which direction it will roll. But that doesn't seem right—this would be a case where the "true" solution is that the ball wouldn't roll at all, while in reality there will always be tiny disturbances that cause the ball to roll in some direction. I don't think this is what Carroll is referring to. He seems to suggest that the "true, exact mathematical outcome" is not uniquely determined for certain classical systems.  I've heard that uniqueness theorems for solutions to differential equations don't generally apply to non-linear equations, and wonder if this may have something to do with it. Elucidation and examples would be much appreciated.
98700
Why do we get the same differential equations from both principles? Surely there is a fundamental connection between them? When written out, the two seem to have nothing in common.   $$\sum _i ( \mathbf F _i - \dot{\mathbf p}_i) \cdot \delta \mathbf r _i = 0$$  $$S[q(t)] = \int ^{t_2} _{t_1} \mathcal L (q,\dot{q},t)dt$$  After playing with d'Alembert's principle  we find that we can rewrite the whole thing as $$\sum _i \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} \right) - \frac{\partial T}{\partial q_i}-Q_i\right] \delta q_i$$  This can further be rewritten under certain conditions so we get the exact form of the E-L equation.   It seems to me that both ways of arriving at the result are fundamentally different.   A function must obey the E-L equations in order to minimize the action over a path, but when we look at the virtual work, it appears that they come from the fact that (quoting Goldstein) "particles in the system will be in equilibrium under a force equal to the actual force plus a 'reversed effective force'."  I think I understand the principle of stationary action, I can see how it leads to the E-L equations, but d'Alembert's Principle seems so arbitrary, I can't see any motivation for it.
93991
What are the difference between:    the action $\int_{t_{1}}^{t_{2}}(L+H) dt$ that we use in the principle of least action, and  the action integral $\int_{t_{1}}^{t_{2}}L dt$ that we use in Hamilton's variational principle?   References:   H. Goldstein, Classical Mechanics, Section 8.6. H. Goldstein, Classical Mechanics, Chapter 2.
70364
What exactly are  (and )?   I want to self-study QM, and I've heard from most people that Hamiltonian mechanics is a prereq. So I wikipedia'd it and the entry only confused me more. I don't know any differential equations yet, so maybe that's why.    But what's the difference between Hamiltonian (&amp; Lagrangian mechanics) and Newtonian mechanics?  And why is Hamiltonian mechanics used for QM instead of Newtonian?  Also, what would the prereqs for studying Hamiltonian mechanics be?
112980
Is there a way to know if a system described by a known equation of motion admits a Hamiltonian function? Take for example $$ \dot \vartheta_i = \omega_i + J\sum_j \sin(\vartheta_j-\vartheta_i)$$ where $\omega_i$ are constants. How can I know if there exist suitable momenta and a Hamiltonian?
20123
I have always been puzzled by how do you arrive at Lagrangians?  That is, how do you know that the functional you need to get Newton's equations is  $$L  =  T-V(x)~?$$ Do you derive the Lagrangian first somehow or do you just guess the one which would satisfy the action to give equation's of motion? (because then you would need to know the equations of motion first.) Also, is the Lagrangian always equivalent to something resembling $T-V$, or there are other ways to determine it? More generally, is the action known before you know the equations of motion, or vice versa?  What I really wish to know is, was the Einstein Hilbert action discovered before field equations were? If so, how? My major doubt is how do you come across the Lagrangian if you don't know the equations of motion, and you can't guess $L$ then?
6543
The action  $$S=\int L \;\mathrm{d}t$$  is an important physical quantity. But can it be understood more intuitively? The Hamiltonian corresponds to the energy, whereas the action has dimension of energy × time, the same as angular momentum.  I've heard the action being described as a measure of change, although I don't know how this description can be justified.
108739
I can't find an answer for whether the Double Pendulum equations can be derived solely using Newtonian Mechanics (Newton's Laws of Motion). The reason I ask for this answer is because wherever I go, I only see the Lagrangian solution. I am also aware that there are certain problems in Classical Mechanics that can't be solved solely using Newtonian Mechanics, so I wanted to know if this is one of those cases... Any help on this is greatly appreciated!