From Mechanics; Landau and Lifshitz, it's stated on page 5:     Since space is isotropic, the Lagrangian must also be indpendent of the direction of $ \mathbf{v}$, and is therfore a function only of its magnitudie, i.e. of $ \mathbf{v} \cdot \mathbf{v}=v^2$:   $L = L(v^2)$   How is he able to exlude $L = L(| \mathbf{v}|)$?
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I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading &quot;Mechanics&quot; by Landau and Lifshitz recently and hit a few roadblocks right away.  Proving that a free particle moves with a constant velocity in an inertial frame of reference ($\S$3. Galileo's relativity principle). The proof begins with explaining that the Lagrangian must only depend on the speed of the particle ($v^2={\bf v}^2$): $$L=L(v^2).$$ Hence the Lagrance's equations will be $$\frac{d}{dt}\left(\frac{\partial L}{\partial {\bf v}}\right)=0,$$ so $$\frac{\partial L}{\partial {\bf v}}=\text{constant}.$$ And this is where the authors say   Since $\partial L/\partial \bf v$ is a function of the velocity only, it follows that $${\bf v}=\text{constant}.$$  Why so? I can put $L=\|{\bf v}\|=\sqrt{v^2_x+v^2_y+v^2_z}$. Then $$\frac{\partial L}{\partial {\bf v}}=\frac{2}{\sqrt{v^2_x+v^2_y+v^2_z}}\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix},$$ which will remain a constant vector $\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$ as the particle moves with an arbitrary non-constant positive $v_x$ and $v_y=v_z=0$. Where am I wrong here? If I am, how does one prove the quoted statement?  Proving that $L=\frac{m v^2}2$ ($\S$4. The Lagrangian for a free particle). The authors consider an inertial frame of reference $K$ moving with a velocity ${\bf\epsilon}$ relative to another frame of reference $K'$, so ${\bf v'=v+\epsilon}$. Here is what troubles me:   Since the equations of motion must have same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of coordinates and time (see the end of $\S$2).  First of all, what does same form mean? I think the equations should be the same, but if I'm right, why wouldn't the authors write so? Second, it was shown in $\S$2 that adding a total derivative will not change the equations. There was nothing about total derivatives of time and coordinates being the only functions, adding which does not change the equations (or their form, whatever it means). Where am I wrong now? If I'm not, how does one prove the quoted statement and why haven't the authors done it?
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This is a special case of , which, I think, might have drawn little attention because it was too general. In this question, I would like to consider a specific case.  Take a classical system given by the action $$ S[x] = \int dt \left[ \frac{\alpha}{2} \ddot{x}^2 + \frac{\beta}{2} \dot{x}^2 + \frac{\gamma}{2} x^2 \right]. $$  This action is Lagrangian, but it is not what we are usually dealing with in physics, because the Lagrangian contains second time derivatives.  The equations of motion are:  $$ \alpha \ddddot{x} - \beta \ddot{x} + \gamma x = 0. $$  This can be solved as usual by employing Fourier transform:  $$ x(t) = a_1 e^{i \omega_1 t} + a_1^{*} e^{-i \omega_1 t} + a_2 e^{i \omega_2 t} + a_2^{*} e^{-i \omega_2 t}, $$  where $\omega_1^2$ and $\omega_2^2$ are the two roots of $\alpha \omega^4 + \beta \omega^2 + \gamma$, given by  $$ \omega_{1,2}^2 = \frac{- \beta \pm \sqrt{\beta^2 - 4 \alpha \gamma}}{2 \alpha} .$$  Now by definition the phase space is the space of solutions of equations of motion. In this case it is parametrized by two complex co-ordinates $a_{1,2}$, meaning that it is 4-dimensional.  I would like to know if there's a natural way to associate the symplectic structure (Poisson bracket) on this phase space. I.e.,  $$ \{a_1, a_1^{*}\} = ? $$ $$ \{a_1, a_2^{*}\} = ? $$ $$ \{a_2, a_1^{*}\} = ? $$ $$ \{a_2, a_2^{*}\} = ? $$ $$ \{a_1, a_2\} = ? $$ $$ \{a_1^{*}, a_2^{*}\} = ? $$
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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?
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I have a problem deriving the conservation of energy from time translation invariance. The invariance of the Lagrangian under infinitesimal time displacements $t \rightarrow t' = t + \epsilon$ can be written as     \begin{equation} \delta L = L\left( q(t),\frac{dq(t)}{dt},t\right) -  L\left( q(t+ \epsilon),\frac{dq(t+ \epsilon)}{dt},t+\epsilon \right) = 0. \end{equation} Using Taylor series, keeping only first order terms this gives \begin{equation}\rightarrow \delta L =- \frac{\partial L }{\partial q} \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon -  \frac{\partial L }{\partial t}  \epsilon = 0. \end{equation} Using the Euler-Lagrange equation and assuming that the Lagrangian does not depend explicitly on time we get \begin{equation}\rightarrow \delta L =- \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \right) \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon  =0. \end{equation} Which we can write as  \begin{equation}\rightarrow \delta L  = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}}  \frac{\partial q}{\partial t} \right) \epsilon  = - \frac{d}{d t} \left(p  \frac{\partial q}{\partial t} \right) \epsilon = 0. \end{equation} But unfortunatly this  is not the Hamiltonian. This computation should yield  \begin{equation}  \rightarrow \frac{d}{dt} \left(    p \dot{q}  - L \right) = 0. \end{equation} But I can't find no reason why and how the the extra $-L$ should emerge. I can see that this term can be written at the place where it is written because we have $\delta L = - \frac{d L}{dt } \epsilon$ and therefore  \begin{equation} \rightarrow \delta L  = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}}  \frac{\partial q}{\partial t} \right) \epsilon  =  -  \frac{d L}{dt }  \epsilon. \end{equation} And then the desired equation would only say $0-0=0$. Any idea where i did a mistake would be much appreciated.
75939
A mass of point performs an effectively 1-dimensional motion in the radial coordinate. If we use the conservation of angular momentum, the centrifugal potential should be added to the original one.     The equation of motion can be obtained also from the Lagrangian. if we substitute, however, the conserved angular momentum herein then the centrifugal potential arises with the opposite sign. So if we naively apply the Euler-Lagrange equation then the centrifugal force appears with the wrong sign in the equations of motion.   I don't know how to resolve this "paradox".
91735
In considering the (special) relativistic EM field, I understand that assuming a Lagrangian density of the form  $$\mathcal{L} =-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}  + \frac{1}{c}j_\mu A^\mu$$  and following the Euler-Lagrange equations recovers Maxwell's equations.   Does there exist a first-principles derivation of this Lagrangian? A reference or explanation would be greatly appreciated!
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This arises from the question , which I'm afraid I answered just by looking it up in . However Schutz (as he frequently does) glosses over details he thinks are irrelevant or too simple to be worth explaining, and I have realised I don't understand an assumption he makes.  Schutz states without proof that if we have an equatorial orbit in a Schwartzschild metric then:     Independence of the metric of the angle $\phi$ about the axis implies that $p_\phi$ is constant.   In the non-relativistic world I assume this corresponds to angular momentum being constant in a central potential. So far so good. But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? The component $p^\phi$ is presumably not constant since (in this case) $p^\phi = p_\phi/r^2$.  Bonus points for also explaining his similar claim that time independance means that $p_t$ is constant rather than $p^t$.  I fear that Schutz didn't explain because it's an insultingly simple question, but if someone can provide a nice intuitive explanation I would be very pleased to read it.
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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.
39829
I feel like this question has been asked before but I can't find it. would the Euler Lagrange equation for, say, the standard model Lagrangian be $$\frac{\partial L}{\partial \phi}=\partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$$ Where $\phi$ is whatever field is an question and $\mu$ is (I believe) being summed from 0 to 3. Or, ist the correct equation $$\frac{\partial L}{\partial \phi}=D_\mu \frac{\partial L}{\partial (D_\mu \phi)}$$ Where $D$ is the covariant derivative of the theory. My intuition tells me its the second eqn but i just wanted to be sure, and I think I once saw someone say that the two were equivalent.
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What does $L^2$ gradient flow mean? Here is the Ginzburg–Landau free energy: $$\mathcal{E}(\phi):=\int_{\Omega}(F(\phi)+\frac{\epsilon^2}{2}|\nabla\phi|^2)\text{d}\mathbf{x}$$ Some references say the Allen-Cahn equation  $$\frac{\partial \phi(\mathbf{x},t)}{\partial t} = {\epsilon^2}\Delta\phi(\mathbf{x},t)-F'(\phi(\mathbf{x},t))$$ $$\frac{\partial \phi(\mathbf{x},t)}{\partial \mathbf{n}}=0, \ \text{on} \ \partial \Omega \times [0,T]$$ is the $L^2$-gradient flow of the total free energy $\mathcal{E}(\phi)$.  How to derive this?
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Question: As we know, (1) the macroscopic spatial dimension of our universe is 3 dimension, and (2) gravity attracts massive objects together and the gravitational force is isotropic without directional preferences. Why do we have the spiral 2D plane-like Galaxy(galaxies), instead of spherical or elliptic-like galaxies?   Input: Gravity is (at least, seems to be) isotropic from its force law (Newtonian gravity). It should show no directional preferences from the form of force vector $\vec{F}=\frac{GM(r_1)m(r_2)}{(\vec{r_1}-\vec{r_2})^2} \hat{r_{12}}$. The Einstein gravity also does not show directional dependence at least microscopically.  If the gravity attracts massive objects together isotropically, and the macroscopic space dimension is 3-dimensional, it seems to be natural to have a spherical shape of massive objects gather together. Such as the globular clusters, or GC, are roughly spherical groupings , as shown in the Wiki picture:     However, my impression is that, even if we have observed some more spherical or more-ball-like , it is more common to find more-planar  such as our ? (Is this statement correct? Let me know if I am wrong.)  Also, such have a look at this more-planar Spiral "galaxy" as this NGC 4414 galaxy:    Is there some physics or math theory explains why the  turns out to be planar-like (or spiral-like) instead of spherical-like?  See also a   p.s. Other than the classical stability of a 2D plane perpendicular to a classical angular momentum, is there an interpretation in terms of the quantum theory of vortices in a macroscopic manner (just my personal speculation)?  Thank you for your comments/answers!
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 $E'_{k}=\frac{1}{2}m\mathbf{v}'^{2}$?
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My text book reads      If a particle is acted upon by the forces which are conservative; that is, if the forces are derivable from a scalar potential energy function in manner $ F=-\nabla V $.   I was just wondering what may be the criteria for force to be expressed as negative gradient of scalar potential energy and HOW DO WE PROVE IT?
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I saw this Lagrangian in notes I have printed:  $$ L\left(x,\frac{dx}{dt}\right) = \frac{m^2}{12}\left(\frac{dx}{dt}\right)^4 + m\left(\frac{dx}{dt}\right)^2\times V(x) -V^2(x). $$ (It appears in the exercises in the first chapter of Goldstein.)  What is it? Is it even physical? It seems like it doesn't have the right units of energy.
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A lot of physics, such as classical mechanics, General Relativity, Quantum Mechanics etc can be expressed in terms of Lagrangian Mechanics and Hamiltonian Principles. But sometimes I just can't help wonder whether is it ever possible (in the future, maybe) to discover a physical law that can't be expressed in terms of Lagrangian Equations?  Or to put it in other words, can we list down for all the physical laws that can be expressed in terms of Lagrangian equations, what are the mathematical characteristics of them( such as it must not contain derivatives higher than 2, all the solutions must be linear etc)?
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It is well known that geodesics on some manifold $M$, covered by some coordinates ${x_\mu}$, say with a Riemannian metric can be obtained by an action principle . Let $C$ be curve $\mathbb{R} \to M$, $x^\mu(s)$ be an affine parametrization of $C$. (Using same symbol for coordinates and parametrization here, but it's standard). The action yielding geodesics is:  $$ S (C) =\int_CL ds $$ where  $$ L\equiv \sqrt{g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu} $$  Now the popular  makes the claim that variation of $$F \equiv \frac{L^2}{2}$$  Will yield the exact same action principle solutions. However, the reference above ONLY shows that:  $C$ solves euler Lagrange equation for $L$ $\implies$ $C$ solves euler Lagrange equation for $F$  And this can be shown through a direct brute force computation.   My question: Is the converse of the above true, and how does one go about proving it. In particular, how do we show that $F$ yields no extraneous solutions to the Euler Lagrange equations.
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Applying the divergence to Eq. $47$, we obtain $$ \mathbf{\nabla} \cdot \mathbf{B} = \frac{\mu_{0}}{4\pi} \int \nabla \cdot \left( \mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right) d\tau^{'}. \tag{48}$$ Invoking the product rule number $6$, $$\nabla \cdot \left(\mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right)=\frac{\hat{\mathbf{r}}}{r^2}\cdot(\nabla \times \mathbf{J})-\mathbf{J}\cdot\left(\mathbf{\nabla} \times \frac{\hat{\mathbf{r}}}{r^2}\right)\; .\tag{49} $$ But $\nabla \times \mathbf{J}=0$, because $\mathbf{J}$ doesn't depend on the unprimed variables, while $\mathbf{\nabla} \times \frac{\hat {\mathbf{r}}}{r^2}=0$, so $$\mathbf{\nabla} \cdot \mathbf{B}=0\;.\tag{50}$$  I have the Fourth Edition of Introduction to Electrodynamics by David J. Griffiths, published by Pearson. I have found this erratum in Chapter 5 Magnetostatics, page 234 . It's just a normal triple product manipulation, i.e. $$ \vec a \cdot \left(\vec b \times \vec c \right) = \vec b \cdot \left( \vec a \times \vec c \right) = \vec c \left(\vec b \times \vec a\right)$$ Therefore, I think that an &quot;equal to&quot; sign is missing between the expressions on the RHS of Eq. $(49)$. I searched for  but couldn't find it.