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?
4098
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.
64718
I suspect an error in my “Classical Dynamics” lecture notes but my lecturer doesn’t agree with me so I need your help!  We assume that in the $n$-body problem, the force between particle $i$ and $j$ can be derived from a potential i.e  $$\vec{F_{ij}}=-\vec{\nabla_i}V_{ij}$$  where $\vec{\nabla_i}$ is the gradient with respect to position vector $\vec{r_i}$   $V$ would usually only depend on $\Vert\vec{r_i}-\vec{r_j}\Vert$  My lecturer then defines the total energy of the $n$-body system as  $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{ij}V_{ij}$$  However, I am convinced that this is wrong since we’d be double counting potentials. My lecturer says we should indeed be double counting potential. In my opinion, the expression should read   $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{1\le{i}\lt{j}\le{n}}V_{ij}$$  For the second equation we also get:  $$\dot{E}=\sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{1\le{i}\lt{j}\le{n}}\vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}}  + \vec{\nabla_j}V_{ij} \bullet \vec{\dot{r_j}} = \sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{i}\sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} = \sum_i \vec{\dot{r_i}} \bullet \Bigl(m_i \vec{\ddot{r_i}} + \sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \Bigr) = \sum_i \vec{\dot{r_i}} \bullet \Bigl( m_i \vec{\ddot{r_i}} - \vec{F_{ij}^{tot}} \Bigr) = 0 $$  So this seems reasonable to me. Is this actually correct?  We also have an example of the two-body problem in the notes where we only use $V_{12}$ as the potential, not $V_{12}+V_{21}$
37516
I am looking for help on how to find the solution of the following differential equation,  $$x\ddot{x}-\dot{x}^2=1,$$  which comes from solving the Euler-Lagrange equations for the lagrangian $L=x\sqrt{\dot{x}^2+1}$. It is from an exercise in my field theory course. My professor provided me with the solution $x(t)=c_1\cosh{\left(\frac{t-c_2}{c_1}\right)}$, which is a solution when plugged into the equation, but does not give any derivation.   Is there a way to solve this DE other than an educated guess? If so how?
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.
43513
,I got one confusion when reading Goldstein's Classical Mechanics (page 20, third edition). After getting the equation $$ \sum \left\{\left[\frac{\mathrm{d}}{\mathrm{d}t}{\left(\frac{\partial T}{\partial \dot{q_{j}}} \right)} - \frac{\partial T}{\partial q_{j} } \right] - Q_{j} \right\} \ \delta q_{j} = 0 $$ then it says that     Note that in a system of Cartesian coordinates the partial derivative of $T$ with respect to $q_{j}$ vanishes. Thus, speaking in the language of differential geometry, this term arises from the curvature of the coordinates $q_{j}$. In polar coordinates, e.g., it is in the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears.   My question is:  Is the above statement general, i.e., that the kinetic energy $T$ does not depend on the position. I wonder why velocity can't depend on the particle's position vector. I mean, why couldn't we have cases where $\vec{v} = \vec{v}(\vec r ,t)$, so that the kinetic energy depends on $q_{j}$ or $\vec{r}$?    I will use the example of the second answer in this post   the second answer,   suppose we have a velocity distribution of gas whose function is $\vec {v}=\vec {v}(x,y,z,t)$  then the partial derivative with respect to $x$ is  $$\frac{\partial T} {\partial x}=mv\frac{\partial v} {\partial x}$$  since $v$ depends on $x$, I think in this case. $\frac{\partial T}{\partial q_{j} }$ does not vanish.
103685
I solved the Lagrangian of a simple pendulum (with help from online examples as this concept is new to me) and ended with the following: $$\ddot{\theta}+\omega^2\theta=0$$ But in the example I was following, the next line written was this: $$\theta(t) = A\cdot\sin(\omega t + \Phi)$$ And I don't know what steps they took to get there?
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!
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?
105556
From definition of Lagrangian: $L = T - U$. As I understand for free particle ($U = 0$) one should write $L = T$.  In special relativity we want Lorentz-invariant action thus we define free-particle Lagrangian as follows:  $$L = -\frac{ m c^2}{\gamma} - U$$  At the same point, we have that definition of 4-momentum implies the kinetic energy is: $$T = (\gamma - 1) m c^2.$$  As you might guess, 1) question is how to relate all these formulas?  2) I do not understand why there is no $1/\gamma$ near $U$ in relativistic Lagrangian?  3) What is meaning of the first term in $L$ for relativistic case?
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.
53416
I am studying from Goldstein's Classical Mechanics, 3rd edition. In section 2.4, he discussed Hamiltion's principle with semiholonomic constraints. The constraints can be written in the form $f_\alpha(q_1,...,q_n;\dot{q_1},...,\dot{q_n};t)=0$ where $\alpha=1,...,m$. Using variational priciple, we get  $$\delta\int_{t_1}^{t_2}\left(L+\sum_{\alpha=1}^m \mu_\alpha f_\alpha\right)\mathrm dt=0 \tag{2.26}$$  where $\mu_\alpha=\mu_\alpha(t)$.  But how can he get the formula   $$\frac{d}{dt}\frac{\partial L}{\partial\dot{q_k}}-\frac{\partial L}{\partial q_k}=-\sum_{\alpha=1}^m \mu_\alpha \frac{\partial f_\alpha}{\partial\dot{q_k}} \tag{2.27}$$   for $k=1,...,n$ from the previous formula?  When I go through the steps as in section 2.3, I get $$\frac{dI}{d\beta}=\int_{t_1}^{t_2}\sum_{k=1}^n\left(\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial\dot{q_k}}+\sum_{\alpha=1}^m \mu_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial\dot{q_k}}\right)\right)\frac{\partial q_k}{\partial\beta}\mathrm dt$$ where $\beta$ denotes the parameter of small change of path: \begin{align} q_1(t,\beta)&amp;=q_1(t,0)+\beta\eta_1(t)\\ q_2(t,\beta)&amp;=q_2(t,0)+\beta\eta_2(t)\\ &amp;\ \,\,\vdots \end{align} Using the same argument as in the part of holonomic constraint in section 2.4, I get $$\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial\dot{q_k}}+\sum_{\alpha=1}^m \mu_\alpha \left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial\dot{q_k}}\right)=0$$ for $k=1,...,n$.  What am I missing?
98758
In the book of Goldstein, at page 337, while deriving the Hamilton's equations (canonical equations), he argues that  The canonical momentum was defined in Eq. (2.44) as $p_i = \partial L / \partial \dot q_i$; substituting this into the Lagrange equation (8.1), we obtain $$ \dot p_i= \frac{\partial L}{\partial q_i} \tag{8.14}$$ so Eq. (8.13) can be written as $$ dL = \dot p_i dq_i + p_id \dot q_i + \frac{\partial L}{\partial t}dt \tag{8.13′}$$ The Hamiltonian $H(q,p,t)$ is generated by the Legendre transformation $$ H(q,p,t) = \dot q_i p_i - L(q, \dot q, t), \tag{8.15}$$ which has the differential $$ dH = \dot q_i d p_i - \dot p_i d q_i - \frac {\partial L}{\partial t}, \tag{8.16} $$ where the term $p_i d \dot q_i$ is removed by the Legendre transformation. Since $dH$ can also be written as $$ dH = \frac{\partial H}{\partial q_i}d q_i + \frac{\partial H}{\partial p_i}d p_i + \frac{\partial H}{\partial t}d t, \tag{8.17} $$  However, if $H$ is defined to be a function of $q,p,t$, then how can we define $H(q,p,t) = \dot q *p - L(q,\dot q,t)$, i.e $\dot q$ is not an argument of $H$ whereas it is in its definition. Moreover, when he is taking the differential of $H$, he argues that $pd\dot q$ is removed, but he does not say why. I mean mathematically speaking this whole argument is plan wrong, as far as I can see, so assuming that it is not the case, what am I missing in here ?
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?
36921
The other day, I was listening to  on the Lagrangian for a charged particle in an electromagnetic field, and at one point in the video, the lecturer mentions that we can add any total time derivative of a function $f(q, t)$ to the Lagrangian without altering its equations of motion.   This is nothing new to me, and I understand it fully, but shortly afterwards (approximately two minutes after the linked starting point), he goes on to say that you can, in fact, add a total time derivative of a function $f(q, \dot{q}, t)$, given certain conditions. This definitely surprised me, and I would love to know more about it, but the lecturer quickly moves on, so my question is as follows: under what conditions can one add the total time derivative of a function which depends on the particle's generalized velocities in addition to its generalized coordinates and time without affecting the particle's equations of motion?
38928
Suppose I have an inertial frame with coordinate $\{q\}$. Now I define another reference frame with coordinate $\{q'(q,\dot q,t)\}$. I obtain the equation of motion in $\{q'\}$ in two different ways:   First obtain the equation of motion in $\{q\}$ by the Euler Lagrange equation $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0$$ and then rewrite the equation in terms of $\{q'\}$. First transform $L(q,t)$ to $L'(q',t)=L(q(q',t),t)$ and then obtain the equation of motion $$\frac{d}{dt}\left(\frac{\partial L'}{\partial \dot q'}\right)-\frac{\partial L'}{\partial q'}=0.$$   Are the two answers just the same?
59992
Suppose a particle with mass $m_1$ and speed $v_{1i}$ undergoes an elastic collision with stationary particle of mass $m_2$. After the collision, particle of mass $m_1$ moves with speed $v_{1f}$ in a direction of angle $\theta$ above the line it was moving previously. Particle with mass $m_2$ moves with speed $v_{2f}$ in a direction of angle $\phi$ below the line which particle with mass $m_1$ was moving previously. Using equations for conservation of momentum and kinetic energy, how can we prove these two equations  $\frac{v_{1f}}{v_{1i}}=\frac{m_1}{m_1+m_2}[\cos \theta \pm  \sqrt{\cos^2 \theta - \frac{m_1^2-m_2^2}{m_1^2}}]$  and  $\frac{\tan(\theta +\phi)}{\tan(\phi)}=\frac{m_1+m_2}{m_1-m_2}$ ?  EDIT. Here is what I've done:  For the first one, set the $xy$ coordinate system so that the positive direction of the $x$ axis points toward the original path of the particle with mass $m_1$. So we have three equations:  $m_1v_{1i}=m_1v_{1f}\cos \theta + m_2v_{2f} \cos \phi$  $0=m_1v_{1f}\sin \theta - m_2v_{2f}\sin \phi$  $m_1v_{1i}^2=m_1v_{1f}^2+m_2v_{2f}^2$.  From the second one, we get:  $v_{2f}=\frac{m_1v_{1f}\sin \theta}{m_2 \sin \phi}$  Plotting this into third equation, we get  $v_{1i}^2=v_{1f}^2(1+\frac{m_1 \sin^2 \theta}{m_2 \sin^2 \phi})$ (1)  From the first equation, we have  $\cos \phi =\frac{m_1(v_{1i}-v_{1f}\cos \theta)}{m_2v_{2f}}$  which after applying the equation we have for $v_2f$ becomes  $\sin^2 \phi = \frac{1}{1+\frac{(v_{1i}-v_{1f}\cos \theta)^2}{\sin^2 \theta \times v_1f^2}}$  Plotting this into equation (1), gives us an equation in terms of $m_1$, $m_2$, $v_{1f}$, $v_{1i}$ and $\theta$, but it is too far from what I expected.  For the second one, assigning the $xy$ coordinate in a way that the positive direction of the $x$ axis points toward the final path of the particle $m_2$, will give us three equations (two for conservation of linear momentum and one for conservation of kinetic energy), but I don't know what to do next.
29133
In mechanics problems, especially one-dimensional ones, we talk about how a particle goes in a direction to minimize potential energy. This is easy to see when we use cartesian coordinates: For example, $-\frac{dU}{dx}=F$ (or in the multidimensional case, the gradient), so the force will go in the direction of minimizing the potential energy.   However, it becomes less clear in other cases. For example, I read a problem that involved a ball attached to a pivot, so it could only rotate. It was then claimed that the ball would rotate towards minimal potential energy, however $-\frac{dU}{d\theta} \neq F$! I think in this case it might be equal to torque, which would make their reasoning correct, but it seems like regardless of the degrees of freedom of the problem, it is always assumed that the forces act in a way such that the potential energy is minimized. Could someone give a good explanation for why this is?  Edit: I should note that I typed this in google and found this  where it states that minimizing potential energy and increasing heat increases entropy. For one, this isn't really an explanation because it doesn't state why it increases entropy. Also, if possible, I would like an explanation that doesn't involve entropy. But if it is impossible to make a rigorous argument that doesn't involve entropy then using entropy is fine.  As a side note, how does this relate to Hamilton's Principle?