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6

In this answer we formally show that a (quasi)symmetry of an action implies a corresponding symmetry of its EOM$^{\dagger}$. The answer does not discuss form covariance of EOM. For further relations between symmetries of action, EOM, and solutions of EOM, see e.g. this Phys.SE post. Let us first recall the definition of a quasi-symmetry of the action ...


6

The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...


5

Usually there's a great deal of overlap between the definitions of the momenta you've listed, so your confusion is understandable, but nonetheless there are cases (that I know of at least) where the distinction is more clearly enunciated: Momentum as known in Newtonian mechanics: The momentum is a vector quantity (its vectorial superposition for many ...


5

The Lagrangian provided is given by, $$L=\sqrt{t+\dot{y}^2t}$$ Clearly, $\partial L/\partial y = 0$. We compute the second term in the Euler-Lagrange equations: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = \frac{d}{dt} \left( \frac{t\dot{y}}{\sqrt{t+\dot{y}^2 t}} \right) = \frac{t(\dot{y} + \dot{y}^3 + 2t\ddot{y})}{2[t(1+\dot{y}^2)]^{3/2}} = 0$$ ...


4

Problem: Given Newton's second law $$\tag{1} m\ddot{q}^j~=~-\beta\dot{q}^j-\frac{\partial V(q,t)}{\partial q^j}, \qquad j~\in~\{1,\ldots, n\}, $$ for a non-relativistic point particle in $n$ dimensions, subjected to a friction force, and also subjected to various forces that have a total potential $V(q,t)$, which may depend explicitly on time. I) ...


4

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2 $$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ...


3

Look more closely at what you write. I'm not going to write it with all the partial derivative and Lagrangians, because the confusions are not dependent on that: The requirement for spatial infinity arises because $\int_\mathcal{M} \mathrm{d}\omega \neq \omega$, but $\int_\mathcal{M} \mathrm{d}\omega = \int_{\partial\mathcal{M}}\omega $ (Stokes' theorem). ...


3

It's Lie theory. When we have got a continuous symmetry, the means the symmetry group is a Lie group $G$. Saying that the Lagrangian $\mathcal{L}$ is invariant under $G$ means that $\rho(g)(\mathcal{L}) = \mathcal{L}\;\forall\; g \in G$, where $\rho$ is the representation the Lagrangian transforms in. The "expansion to first order" physicists so often do ...


3

Let us start from Minkowski spacetime $M$ and construct the trivial bundle $\Phi=\mathbb R \times M \to M$ whose sections $\phi : M \ni p \mapsto (p,\phi(p))$ are the scalar fields you want to discuss their dynamics. Since you correctly wish to see the partial derivatives of $\phi$ as variables independent from $\phi$ itself (this is your second raised ...


3

In optics, you can take the example of a concave mirror : the optical path chosen by the light to join two fixed points A and B is a maximum.


3

Example: Consider an action functional $$\tag{1} S[q]~=~\int_{t_i}^{t_f} \! dt ~L, \qquad L~=~\frac{1}{2}m\dot{q}^2-V(q),$$ with Dirichlet boundary conditions (BC) $$\tag{2} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f,$$ where the potential $V$ has a repulsive pole $$\tag{3} V(q_0)~=~+\infty$$ at $q=q_0$. Then it is possible to choose a ...


3

(This is a largely a response prompted by your comment.) You can get the answer by just remembering the commutation/anti-commutation properties of the $\gamma$ matrices, and the fact that ${\bar \psi} = \psi^{\dagger} \gamma^0$. To see the following, you would have to expand the exponential factor, up to linear order $e^{M} = I + M + \ldots$. (I'm not ...


3

If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.


2

The canonical momentum $p$ is just a conjugate variable of position in classical mechanics, for we have the relation $p=\frac{\partial L}{\partial \dot{r}}$. When making the transition to quantum mechanics: we need substitute $p$ by an operator $-ih\nabla$ in the Hamiltonian; similarly, we need substitute $r$ by $i\hbar \nabla_p$ in momentum representation. ...


2

If we solve equations of motion for a particle with mass $m=1$ in some potential, e.g. $U=x^4-4x^3+4.5x^2$, fixing two points like $x(0)=0$ and $x(1)=2.651$, we'll get infinite number of solutions, here're some of them: They differ by initial velocity. Now each of them satisfies equations of motion, but only one makes the action minimal, here's how $S$ ...


2

As said in this answer, velocity and position are not varied independently. Indeed, when deriving Euler-Lagrange equations, we explicitly use the fact that $\delta v=\frac d{dt}\delta x$. So, when I add the constraint $v_i=\frac{x_{i+1}-x_i}{\Delta t}$, specifying $x_1$ and $x_n$ remains the only additional thing to converge to the solution. For example, ...


2

The first term vanishes because you have assumed that the potential is independent of $\dot{q}=v$ (as you have said, $V(q)$). The second term vanishes because speed and possition are independent.


2

An intuitive way to see why a term proportional to $F_{\mu\nu}F^{\mu\nu}$ is natural is recognizing the fact that this term introduces dynamics for the gauge field. If you make the dependence on the gauge field explicit, you will see that it contains a properly normalized kinetic term, allowing us to treat it as a field propagating in spacetime. This, as you ...


2

Recall that magnetic energies add to the Lagrangian while electric energies subtract from the Lagrangian--this is easily proven looking at the Lorentz force in the Lagrangian formalism. That is to say, the Lagrangian should be defined as $$ \mathcal L=T-V+W_b-W_e $$ where $W_n$ are the magnetic ($b$) and electric ($e$) energies. Thus, your Lagrangian is $$ ...


1

The version with the 'i' is correct. It is used in all references I checked, including printed books from pre-web days. The term without the 'i' may have come from some author using uncommon conventions or definitions, perhaps defining $\gamma^5 = \gamma^0 \gamma^1 \gamma^2 \gamma^3$ instead of the widely used standard definition $\gamma^5 = i\gamma^0 ...


1

When you write this: $$D_{t_2} S=-x_2^2\csc^2t_2.\tag6$$ You obviously mean total time derivative by $D_{t2}$. Which means: $D_{t2}S=dS/dt_2=\frac{\partial S}{\partial q}\frac{\partial q}{\partial t_2}+\frac{\partial S}{\partial t_2}\frac{\partial t_2}{\partial t_2}$ In our case: $q=x_2$ and you're missing this part: $\frac{\partial S}{\partial ...


1

Comments to the question (v1): The main point is that Ref. 1 is considering the (Dirichlet) on-shell action function $$ \tag{1} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl}], $$ not the (off-shell) action functional $$\tag{2} I[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ L(q(t),v(t),t)\right|_{v(t)=\dot{q}(t)}. $$ See e.g. this Phys.SE post. The Lagrangian ...


1

Hint to the question (v2): For a velocity-dependent force ${\bf F}$ (such as e.g. the Lorentz force), the relationship between force ${\bf F}$ and potential $U$ is $$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}. $$ See e.g. Goldstein, Classical Mechanics, Chapter 1. See also e.g. this and this Phys.SE ...



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