Tag Info

New answers tagged

0

It is actually clear that $T$ being NOT homogeneous quadratic in the velocities of the generalized coordinates does not imply that energy is not conserved. E.g. one can consider a boost transformation for a free particle in 1D and then $T=\frac{1}{2}m\dot{x}^2$ becomes $T=\frac{1}{2}m(\dot{x}-c)^2$. Clearly, $E$ is conserved. Unfortunately, I wasn't ...


2

If only the cyclic coordinate $q(t)$ varies with time (if it doesn't, $q$ is superfluous), the Lagrangian, or the essential physical situation, doesn't vary. Hence the initial value of $q$ doesn't determine the path, which is only possible if the path is closed.


0

You should check out Marsden and Hughes's "Mathematical Foundations of Elasticity". It is a book that requires a lot of work (I am going through it myself and it is not easy!), using the very general framework of tensors on manifolds + functional analysis, but it is an invaluable resource. 2 chapters in that book (5. Hamiltonian And Variational ...


3

I) OP's question (v2) seems to be essentially a question of mathematical precision vs. the way physicists express themselves concisely when speaking of "infinitesimals" without becoming too technical by introducing epsilons and deltas and what not. See also this related Phys.SE post. Perhaps the easiest and most elementary way to make sense of the ...


0

As you state in the comments, $$ \frac{dF}{dt}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t} $$ So popping this into the Lagrangian, $$ L'=L+\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t} $$ The Hamiltonian $H=p\dot q-L$ implies $$ H'=p'\dot{q}-L'=p\dot q+something\tag{1} $$ where $something$ is for you to work out. ...


3

The Hamiltonian is so useful because it is actually the operator providing translation in time (in autonomous systems). We know that any physical quantity on the phase space in the Hamiltonian formalism is evolved like $$\frac{df}{dt} = \{f,H \}$$ Where $\{\}$ is the Poisson bracket. It is thus natural to say $$\frac{d}{dt} = \{\cdot,H\}$$ and a small ...


6

I agree with Qmechanic but just to put a different perspective. While one may write down formulae for the Lagrangian, like $$ L = \frac{p^2}{2m} - U(x) $$ which only differs from the Hamiltonian by the minus sign, and while it's possible to simply put hats above all the operators, unlike the Hamiltonian, the Lagrangian isn't a natural operator in any sense. ...


2

Comment to the question (v1): Unlike the Hamiltonian $H$ (which is a constant of motion if there is no explicit time dependence), the Lagrangian $L$, as an observable, is typically not conserved in time. Think e.g. of a harmonic oscillator.


4

The Brownian motion $x(t)$ is non-differentiable, so a particular trajectory $x(t)$ can't extremize an action $S$ which would be a functional of $x(t)$ and its derivative, $\dot x(t)$, because the derivative isn't even well-defined and any expression of the type $\int [\dot x(t)]^2 dt$, the usual kinetic term in the action, diverges. (See e.g. middle of page ...


0

I can't write comments right now So here is what i wanted to write Coupling constant's value is an indication of how strong or weak the intraction and also for ascribing correct dimensions to lagarangian( as mentioned by @winther) is i.e an interaction lagarangian of two particles with higher value of coupling constant will result in stronger interactions ...


1

There is a distinction between points and vectors. Points are positions in space, and vectors are directions. One can easily mix up the two, because in Euklidean space they look rather similar. $\theta$ in this case is a coordinate, i.e. part of the description of a point. The vector associated to that coordinate could be called $\hat{e}_\theta$, and point ...


1

John R. Taylor's Classical Mechanics has a couple of chapters on Lagrange and Hamilton that I found very helpful.


0

You probably know that while the Cauchy stresses are objective, its stress rate (material derivative) is not. If $Q(t)$ is an orthogonal tensor representing a change of frame, the stress is the new frame is $$ T^* = Q T Q^T $$ However, if you take material derivatives on both sides, you have, $$ \dot{T^*} = \dot{Q} T Q^T + Q \dot{T} Q^T + Q T \dot{Q}^T ...


0

$ r^2 \dot{\theta} $ is known as the specific angular momentum. Also, the correct formula for the 2 body Lagrangian is actually: $$ \mathcal{L} = \frac{\mu}{2} (\dot{r}^2+r^2 \dot{\theta}^2) + \frac{GMm}{r} = \frac{Mm}{M+m} \frac{\dot{r}^2+r^2 \dot{\theta}^2}{2} + \frac{GMm}{r} $$ where $ \mu = \frac{Mm}{M+m} $ is the reduced mass.


3

To derive the field equations more quickly, consider that all the terms in the Lagrangian are "squares": a tensor contracted with itself on all indices. The variation of a square is $\delta(F_{ab} F^{ab}) = 2 (\delta F_{ab}) F^{ab}$ in analogy with $d/dx\; f^2 = 2ff'$. Using this, we have for the variation of the Lagrangian $$\delta \mathcal L = - (\delta ...


0

For reference sake, here is the computation of (G) mentioned above in @Qmechanic's answer: The Lagrangian is given by: $$ ...


4

Comment to the question (v2): P&S is using the notation of a 'same-spacetime' functional derivative. To illustrate this notation, let us for simplicity stay within first variations, and leave it to the reader to generalize to higher-order variations. I) First of all, functional/variational derivatives should not be confused with partial derivatives. In ...


1

It's obvious that the kinetic energy must be compute with respect to the inertial frame. Using Lagrangian method for your problem is very similar to applying this method in order to derive the equations of motion of a gyroscope that can be found anywhere (for example here). The only difference between your case and a gyroscope is that in your case, the ...


4

The first thing you should realize is the fact that while $\phi$ has an equation of motion with second time derivatives, it is not the wave function, and therefore there is no problem with QM. The field is just an operator (more or less), not a state. Acting with the fields on the vacuum state you generate the other states which do evolve with an hamiltonian ...


1

Hint: If the transformation equations $$\mathbf{r}_i~=~\mathbf{r}_i(q_1,\ldots, q_n ,t) \tag{1.38} $$ do not contain the time explicitly, then the explicit time derivative $\frac{\partial \mathbf{r}_i}{\partial t}=0$ is zero. Note that the position $\mathbf{r}_i$ of the $i$'th point particle also depends implicitly on time $t$ through the generalized ...


3

I) Since total divergence terms do not contribute to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post, one could just integrate the Faddeev-Popov $\bar{c}c$ term by part so that there are no more than first derivatives present and the standard form of the EL equations applies. II) Alternatively, in the presence of higher derivatives, the EL ...



Top 50 recent answers are included