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0

My answer is the same as Chris', but formulated in a different way (it's essentially the same as this wiki article): In polar coordinates, the position vector is $$ \mathbf{r} = r\,(\cos\varphi,\sin\varphi) = r\,\mathbf{\hat{r}}, $$ with $\mathbf{\hat{r}}$ the radial unit vector. The velocity is then $$ \mathbf{v} = \dot{r}\,(\cos\varphi,\sin\varphi) + ...


0

The canonical (total) momentum is the sum of the kinetic (mechanical) momentum and the potential momentum. Potential momentum occurs only if the potential energy depends explicitly on velocity.


2

The "associated scalar equation" is just the formula for the time evolution of the scalar magnitude of the displacement, $r$, rather than all its vector components. It really only makes sense to write such an equation if the right-hand side can be expressed in terms of $r$ only, and not $\mathbf{r}$. Then you can use it to analyze the evolution of $r$ in ...


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First expand the product in $\mathcal{L}_0$: $$\mathcal{L}_0=-\frac{1}{2}\left(\partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu}-\partial_{\nu}A_{\mu}\partial^{\mu}A^{\nu}\right)-\frac{1}{2}m^2A_{\mu}A^\mu.$$ Now in the term $\partial_\mu A_\nu\partial^\mu A^\nu$, the vector fields $A$ have the same index, hence the metric tensor in the first term of $D_{\rho ...


6

I) OP essentially asked (v1): If two Lagrangian densities ${\cal L}$ and $\tilde{\cal L}$ have the same eqs. of motions, must they necessarily differ by a total divergence? Answer: No, one e.g. can always multiply a Lagrangian density ${\cal L}$ with a constant factor $\tilde{\cal L}=\lambda {\cal L}$ different from one $\lambda\neq 1$ without altering ...


0

The real reason is in following. Let's assume Majorana field: $$ \Psi_{M} = \Psi_{L} + \hat{C}\bar{\Psi}^{T}_{L}, \quad \hat{C} = i\gamma_{2}\gamma_{0}, \quad \Psi_{L} = \begin{pmatrix} \psi_{L} \\ 0 \end{pmatrix}. $$ By using this notation it's not hard to see that kinetic term is equal to $$ \bar{\Psi}_{M}\gamma^{\mu}\partial_{\mu}\Psi_{M} = ...


0

The short asnwer to your question is that the overall factor $\frac{1}{2}$ from the Lagrangian of a Majorana field (in the 4-component notation) $$\mathcal{L}=\frac{1}{2}(\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi -m\bar{\psi}\psi)$$ compared to the general Dirac Lagrangian is usual for self-conjugate fields and it is introduced to ensure a consistent ...


0

Yes lagrangians and hamiltonians are indeed used by engineers. To be precise, used by some types of engineers like aeronautical engineers, aerodynamics etc.. For example: http://www.osti.gov/eprints/topicpages/documents/starturl/47/566.html As far as i know electrical engineers dont use the lagrangian nor hamiltonian forms of mechanics nor ...


1

I'm a electrical engineer, and have never used either one in over 30 years of designing circuits. I vaguely remember that we went over them briefly in school, but since I haven't used them (knowingly) since, I can't tell you what the physical meaning of either is, which of course perpetuates the fact that I'm not going to use them.


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In civil engineering they use it for structures, and strength of materials in the elastic realm. It goes by the name of the enegy method. Google books might give an indication. Some authors are Beer and the mechanical engineer Stephen Timoshenko. This is for some what "static" indeterminant structures. So, there is no time element. But, I am sure it ...


0

I don't know where you're getting those $m$s from, or what substitution you're making. The appropriate substitution to perform is $$ p = \frac{\partial L}{\partial \dot{q}} = \omega \dot{q}. $$ If you do this, then the hamiltonian becomes $$ H = p\dot{q} - L = \frac{p^2}{\omega} - \frac{p^2}{2\omega} + \frac{1}{2} \omega q^2 = \frac{1}{2\omega}\left(p^2 + ...


4

The principle of Least (Stationary) Action (aka Hamilton's Principle) is derived from Newton's axioms plus D'Alembert's principle of virtual displacements. Because D'Alembert's principle allows to account for the (reactions of the) bonds between the components of a system in a transparent way, the Lagrangian and Hamiltonian formulations are possible. ...


0

Calculating a Hamiltonian from a Lagrangian leads us to a new quantity, which is a function of coordinates and momenta. If you simply substitute velocity as a function of momentum into the Lagrangian, then the "Hamiltonian" will depend on velocity implicitly. This is not what we want. Let me show you an example to explain why Hamiltonian (the usual one) ...


2

The key point in all of this is that general relativity is a gauge theory, and, as the saying goes, "the gauge always hits twice" (apparently attributed to Claudio Teitelboim). What this means is that (1) you have an arbitrary freedom in defining your evolution, corresponding to the ability to make gauge transformations, and (2) some of the evolution ...


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It is interesting to look at a linearized version of gravity, with $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ If you choose the Lorentz gauge : $$\partial^\mu \bar h_{\mu\nu}=0 \quad\quad \bar h_{\mu\nu} = h_{\mu\nu} - \frac{1}{2} h^i_i \,\eta_{\mu\nu} \tag{0}$$ the equations of movement in the vaccuum are simply : $$\square \bar h_{\mu\nu}=0 \tag{1}$$ ...


2

I'll assume you want to know the equilibrium points. The Lagrangian tells you everything you need to know about the system. Because variation of generalized momentum is: $$ \frac{dp_k}{dt} = Q_k + \frac{\partial T}{\partial q_k} = -\frac{\partial V}{\partial q_k}+ \frac{\partial T}{\partial q_k} = \frac{\partial L}{\partial q_k} $$ Then: $$ ...


0

You don't need to express $t$ in terms of $z$ and $p$, explicit time dependence is permissible in the Hamiltonian. (It would not even be possible without restoring to $\dot z$.) In the other question it was not mentioned because it was not needed, the Lagrangian was time-symmetric and consequently so was the Hamiltonian. This is very often the case so many ...


1

To see that this is the simplest possible non-relativistic quantum field theory for fermions, it's useful to derive the dynamics. The canonical momentum for $\psi(x,y,z)$ is the Lagrangian's derivative with respect to $\partial_\tau \psi(x,y,z)$ – and it is $\psi(x,y,z)^\dagger$ (up to signs and $\pm i$ which depend on conventions). At any rate, the ...


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When you introduce $\phi(x) \phi(y)$ for $x \ne y$, you postulate an action at a distance, whichever the interval between said events is: time-like, null, or what. In other words, you admit some essence that isn’t a field, but propagates through the spacetime directly, in a point-to-point fashion. I am not sure you can’t maintain causality is such theory, ...


1

Not the exact title, but stripping some non-essential bits of the search to crash course lagrangian dynamics pdf, I found a 12-page document titled Crash Course in Discrete Lagrangian Dynamics (found here). This is likely not what you are looking for, but it could be useful nonetheless.


2

Classical lagrangians of fermions are always constructed out of Grassmann numbers. No exception. In both of OP's cases, the mass term is nonvanishing: In the first case the mass term is proportional to $\psi_1^*\psi_2^*-\psi_2^*\psi_1^* = 2\psi_1^*\psi_2^* \neq 0.$ In the second case, I write in the chiral basis: $$\gamma^0=\begin{pmatrix}& \sigma^0 ...


1

Tong is alluding to the standard trick in the derivation of Noether's theorem by promoting the (infinitesimal) $x$-independent parameter $\epsilon$ to become $x$-dependent, see e.g. this Phys.SE post.


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Let $Q$ denote the set of all possible configurations of the system (the configuration manifold). Consider a point $q_0\in Q$. For the sake of conceptual clarity, and to make contact with physics notation, let's work in some local coordinate patch around $q_0$. Suppose that $q_0$ represents the position of the system under consideration at time $t_0$. ...


0

As I understand, it must be a displacement in generalized coordinates. If they are orthogonal space coordinates they are not virtual.


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In short: virtual displacement is "pretend you are moving, but don't really move". In other words - you move by such a small amount that you don't change the state of the system - but it gives you insight (through work done etc) in what would happen if you did move. In other words - if the system is really moving, you can look at an interval $dt$ to see how ...


3

Let's assume a typical fermionic mass-term (interacting leptons and quarks are spin 1/2-particles): $$ \tag 1 \bar{\Psi}\Psi = \bar{\Psi}\left(\frac{1 + \gamma_{5}}{2} + \frac{1 - \gamma_{5}}{2}\right)\Psi = \left| \bar{\Psi}\left( 1 \pm \gamma_{5} \right) = \left( (1 \mp \gamma_{5})\Psi\right)^{\dagger}\gamma_{0} \right| = $$ $$ =\bar{\Psi}_{L}\Psi_{R} + ...


1

First of all, two tiny mistakes: your $\theta$ equation is missing an $\ell$, and the sign of $g \sin \theta$ is flipped. That is, it should be $$ 2 \dot{\ell} \dot{\theta} + \ell \ddot{\theta} = -g \sin \theta $$ Now, I have not seen Bessel equation emerge for pendulum dynamics for arbitrary length variability. It only arises, as far as I have seen, for ...


1

The correct equation of motion for $\theta$ is \begin{equation} l\ddot \theta + 2 \dot l \dot \theta = -g \sin\theta. \end{equation} Now if you assume $l(t)=l_{0}+vt$ then you will get a Bessel Differential Equations. \begin{equation} \ddot \theta+\frac{2}{l}\dot \theta+\frac{g}{v^{2}l}\theta=0 \end{equation} The solution for which is the Bessel Function. ...


3

OP's system is two coupled harmonic oscillators $$\tag{1} L~=~\frac{1}{2}(m\dot{x}^2 - k x^2) + \frac{1}{2}(M\dot{y}^2 - K y^2) - \kappa xy. $$ It seems a steep price to pay to create a non-local formulation by integration out one variable by brute force as OP does. Here we instead find the normal modes of the system of two coupled harmonic oscillators. ...


2

1 : I don't think so 2 : Note that $L_{eff}$, may be written, thanks to an integration by parts $(\partial_t E)^2 = \partial_t(E\partial_t E) - E \partial_t^2E$, and neglecting the surface term due to the the total derivative : $$ L_{eff}=E\quad (\frac{1}{2g}(-\partial_t^2 -\omega _{LC}^2)+\frac{e^2}{2m}(\partial_t^2+\omega _{0}^2)^{-1})\quad E ...



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