I'm very confuse with the method using Hamiltonian to derive the equation of the movement. In example I have, it's easier to derive the equation of the movement using classical method (namely 2nd law of Newton). Can someone give me an example where using Hamiltonian is much easier than classical method? Also, is there a reason so that the Hamiltonian is the Legendre transform of the Lagrangian $T-V$ (where $T$ is the kinetic energy and $V$ the potential energy)? By the way, why is this the Lagrangian? All this looks a bit magic for me.
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?
62067
The least-action principle is a statement in classical physics saying that all bodies in a system follow a trajectory that minimize the following functional (ignoring explicit time dependence for now): $$ S[L] = \int dt L(x(t), \dot{x}(t)) \qquad\rightarrow\qquad  \frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{x}}\Big) - \frac{\partial L}{\partial x} = 0. $$ The Hamiltonian of the same system can be constructed using the Legendre transform: $$ H(p, x) = \dot{x}\cdot p - L(x, \dot{x})\;;\quad p \equiv \frac{\partial L}{\partial\dot{x}}. $$ Pedagogically, one first studies Lagrangian Mechanics starting from the least-action principle, and eventually constructs an equivalent Hamiltonian Mechanics framework. But suppose I was a weird physicist who wanted to teach Hamiltonian mechanics first, and then later construct Lagrangian Mechanics. What would be the best way to do this?
109997
I recently started studying simple harmonic motion and I came across two equations for the displacement of a particle, as given in my textbook:  \begin{align} y &amp;= a\sin(\omega t +\phi)\\ x &amp;= a\cos(\omega t+ \phi) \, . \end{align}  I really have no clue when to use which one. It would help me a lot if someone could give an explanation on this.
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Undergraduate classical mechanics introduces both Lagrangians and Hamiltonians, while undergrad quantum mechanics seems to only use the Hamiltonian. But particle physics, and more generally quantum field theory seem to only use the Lagrangian, e.g. you hear about the Klein-Gordon Lagrangian, Dirac Lagrangian, Standard Model Lagrangian and so on. Why is there a mismatch here? Why does it seem like only Hamiltonians are used in undergraduate quantum mechanics, but only Lagrangians are used in quantum field theory?
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?
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If there is a system, described by an Lagrangian $\mathcal{L}$ of the form   $$\mathcal{L} = T-V = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) + \frac{k}{r},\tag{1}$$   where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system   $$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$  If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as  $$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$  where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.    In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as  $$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$  where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake?
61143
Is there a way of explaining why no differential equations in physics exceed order two without delving into Lagrangian and Hamiltonian mechanics - i.e. from Newtonian mechanics? Moreover, is there such an alternative proof, which focuses more on physical reasoning than abstract mathematics?
7056
I've always used the method of Lagrange multipliers with blind confidence that it will give the correct results when optimizing problems with constraints. But I would like to know if anyone can provide or recommend a derivation of the method at physics undergraduate level that can highlight its limitations, if any.
19173
Since the equation of mechanics are of second order in time, we know that for $N$ degrees of freedom we have to specify $2N$ initial conditions. One of them is the initial time $t_0$ and the rest of them, $2N-1$ are initial positions and velocity. Any function of these initial condition is a constant of motion, by definition. Also, there should be exactly $2N-1$ algebraically independent constants of motion.  On the other hand, Noether's procedure gives us integrals of motion as a result of variational symmetries of the action. These integrals of motion are also conserved but they are not always $2N-1$ in number. In consequence, we classify the system by their integrability.  So, what is the difference between the  and ? Why do non- have less integrals of motion when they should always have $2N-1$ constants of motion?
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?
46159
I am a graduate student in mathematics who would like to learn some classical mechanics. However, there is one caveat: I am not interested in the standard coordinate approach. I can't help but think of the fields that arise in physics as sections of vector bundles (or maybe principal bundles) and would love an approach to classical mechanics or what have you that took advantage of this.  Now for the questions:   Is there a text book you would recommend that phrases the constructions in classical mechanics via bundles without an appeal to transition functions?   What are the drawbacks to this approach other than the fact that it makes computations less doable? (if it does that)   Are there benefits to thinking about things this way, ie would it be of benefit to someone attempting to learn this material to do it this way?
5422
What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations?    If someone on the street would flat out ask me that question, then I'd probably mumble something about physicists wanting to be able to use the Lagrangian approach. And to allow a positive rotation and translation invariant energy term, which allows for local propagation, you need something like $-\phi\Delta\phi$.  I assume the answer goes in this direction, but I can't really justify why more complex terms in the Lagrangian are not allowed or why higher orders are a physical problem. Even if these require more initial data, I don't see the a priori problem.  Furthermore you could come up with quantities in the spirit of $F\wedge F$ and $F \wedge *F$ and okay yes... maybe any made up scalar just doesn't describe physics or misses valuable symmetries. On there other hand in the whole renormalization business, they seem to be allowed to use lots and lots of terms in their Lagrangians. And if I understand correctly, supersymmetry theory is basically a method of introducing new Lagrangian densities too.   Do we know the limit for making up these objects? What is the fundamental justification for order two?
66220
In classical mechanics, the Hamiltonian is well defined by the Lagrangian. Whereas, energy is a very ambiguous term. We just say $E=T+U$, and usually it equals to Hamiltonian. Does there exist a way that, by just looking at the Lagrangian mathematically, we immediately know the relationship between the Hamiltonian and the energy of the system?  And if we have a system, the Hamiltonian of which does not equal to energy, what is the physical meaning of that difference?
41151
The Euler-Lagrange equation for particles is given by   $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q},\tag{1}$$  and for fields  it is   $$ \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac{\partial \mathcal{L}}{\partial \phi}.\tag{2} $$  Comparing the two equations, the first one has a total time derivative $\frac{d}{dt}$ but the other one appears to have partial derivatives $\partial_{\mu}$. These derivatives come from integration by parts in the derivation of the EL equation. I was wondering why the field version has partial derivatives and the particle version has total derivatives?   I have also seen for the specific example (in Quantum Field Theory for the gifted amateur) of 1 dimensional waves on a string, the corresponding Euler-Lagrange equation is  $$ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \left(\frac{d\phi}{dt}\right)} + \frac{d}{dx} \frac{\partial \mathcal{L}}{\partial \left(\frac{d\phi}{dx}\right)} = \frac{\partial \mathcal{L}}{\partial \phi}, \tag{3}$$  which uses total derivatives, so I am a bit confused.
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!
62771
It is often said that the classical charge $Q$ becomes the quantum generator $X$ after quantization. Indeed this is certainly the case for simple examples of energy and momentum. But why should this be the case mathematically?  For clarity let's assume we are doing canonical quantization, so that Poisson brackets become commutators. I assume that the reason has something to do with the relationship between classical Hamiltonian mechanics and Schrodinger's equation. Perhaps there's a simple formulation of Noether's theorem in the classical Hamiltonian setting which makes the quantum analogy precisely clear?  Any hints or references would be much appreciated!  Mathematical Background  In classical mechanics a continuous transformation of the Lagrangian which leaves the action invariant is called a symmetry. It yields a conserved charge $Q$ according to Noether's theorem. $Q$ remains unchanged throughout the motion of the system.  In quantum mechanics a continuous transformation is effected through a representation of a Lie group $G$ on a Hilbert space of states. We insist that this representation is unitary or antiunitary so that probabilities are conserved.   A continuous transformation which preserves solutions of Schrodinger's equation is called a symmetry. It is easy to prove that this is equivalent to $[U,H] = 0$ for all $U$ representing the transformation, where $H$ is the Hamiltonian operator.  We can equivalently view a continuous transformation as the conjugation action of a unitary operator on the space of Hermitian observables of the theory  $$A \mapsto UAU^\dagger = g.A$$  where $g \in G$. This immediately yields a representation of the Lie algebra on the space of observables  $$A \mapsto [X,A] = \delta A$$  $$\textrm{where}\ \    X \in \mathfrak{g}\ \ \textrm{and} \ \  e^{iX} = U \ \  \textrm{and} \ \  e^{i\delta A} = g.A$$   $X$ is typically called a generator. Clearly if $U$ describes a symmetry then $X$ will be a conserved quantity in the time evolution of the quantum system.  Edit  I've had a thought that maybe it's related to the 'Hamiltonian vector fields' for functions on a symplectic manifold. Presumably after quantization these can be associated to the Lie algebra generators, acting on wavefunctions on the manifold. Does this sound right to anyone?
9448
I thought the Hamiltonian was always equal to the total energy of a system but have read that this isn't always true. Is there an example of this and does the Hamiltonian have a physical interpretation in such a case?