Tag Info

Hot answers tagged

4

The equation of motion for a scalar massless relativistic point particle is $$ \tag{A} \dot{x}_{\mu}\dot{x}^{\mu}~\approx ~0, $$ where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is $$ \tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, ...


4

Ok, let us start from scratch. A function $g: \mathbb R^n \to \mathbb R$ with $f \in C^2(\mathbb R^n)$ is said to be convex if its Hessian matrix (i.e. the one with coefficients $\partial^2 f/\partial x_i \partial x_j$) is everywhere (strictly) positive defined. Let $\Omega \subset \mathbb R \times \mathbb R^n$ be an open set, and focus on a jointly ...


4

I) Lagrangian formalism. Let us suppress position dependence $q^i$ and explicit time dependence $t$ in the following, and also assume that the Lagrangian $L=L(v)$ is a smooth function of the velocities $v^i$, where $i=1, \ldots, n$. Define functions $$\tag{1} g_i(v)~:=~\frac{\partial L(v)}{\partial v^i}, \qquad i=1, \ldots, n; $$ $$\tag{2} ...


2

We may approach the problem via differential forms, or ordinary tensor calculus: Differential Forms: The field strength tensor $F$ is a differential form given by the exterior derivative of the 1-form $A$, i.e. $F=\mathrm{d}A$ which in components is $\partial_{[\mu}A_{\nu]}$. To add a total derivative to the form $A$ is equivalent to adding the exterior ...


2

I think you should first express the kinetic energy of each block, using $\frac{1}{2} m v^2$, where $v$ is the velocity of the block. Then just sum these up. It looks like for two of the blocks, the velocity is $\dot{x}$, and for one of them the velocity is $\dot{x} - \dot{y}$. Be careful to remember that for one of the blocks the mass is $2m$.


2

You can derive the equations of motion (equations of geodesics) for a particle in curved spacetime by using the Lagrangian $$L = \frac{1}{2} \sum_{\mu,\nu} g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt},$$ so the answer is yes. You could regard the configuration manifold as the manifold, it need not be the physical spacetime. I would like to clarify that ...


2

On one hand, by including the Lautrup-Nakanishi field $B^a$, we have an off-shell BRST formulation, i.e. we can prove the nilpotency of the BRST transformation without using the (Euler-Lagrange) equations of motion. On the other hand, for some applications, a simpler on-shell BRST formulation (where the Lautrup-Nakanishi field $B^a$ has been integrated ...


2

The quote is taken from just above eq. (1.32) in Ref. 1: [...] If the internal forces are also conservative, then the mutual forces between the $i$th and $j$th particles, ${\bf F}_{ij}$ and ${\bf F}_{ji}$, can be obtained from a potential function $V_{ij}$. To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance ...


2

For a convex function you can do the following: For each point on the graph of the function, draw the line tangent to the function at that point. That point can now be identified by its original $x$ and $y=f(x)$ coordinates, or by specifying the slope of that tangent line and its corresponding y-intercept. Each point maps to one and only one line, and ...


1

The fields satisfy the wave equation. We can therefore write \begin{equation} \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + ...


1

This is an interesting question, and although I don't know a rigorous answer, we can discuss some typical cases. Usually, the inverse exists, but the cases where this inverse does not exist are not necessarily pathological (sound models can have the problem that the inverse does not exist). For standard field theories (say, $\phi^4$, O(N) models, classical ...


1

The easiest answer is "because that generates Maxwell's equations". The slightly more difficult answer is that the Lagrangian density has to be gauge invariant and a Lorentz scalar. The objects we have at hand are $F_{\alpha\beta}$, $A_\alpha$ and $J^\alpha$. Now, something like $A^\alpha A_\alpha$ is out, because it is not gauge invariant, $A^\alpha ...


1

The force is proportional to the extension: $$ F = kx $$ where we subsume all the various constants like Young's modulus and area into the constant $k$. We know $dW = Fdx$, so: $$ dW = k x dx $$ and integrating this gives: $$ W = \tfrac{1}{2}kx^2 + C $$ If we define the work to be zero when the extension is zero the constant $C$ is zero, and we get the ...


1

Alternatively, there exists an extended approach to the Legendre transformation between the Lagrangian and Hamiltonian formalism, cf. e.g. Ref. 1. Let us suppress explicit time dependence $t$ from the notation in the following. Consider the extended Lagrangian $$\tag{1} L_E(q,v,p)~:=~ p_i(\dot{q}^i-v^i)+L(q,v)~=~p_i\dot{q}^i-h(q,v,p), $$ $$\tag{2} ...


1

The dynamics of a large class of mechanical systems can be described as a geodesic motion in some ambient space. This is the essence of the Kaluza-Klein theory. The basic and most elementary example is the case of a charged particle in $3D$ coupled to a magnetic field which can be described as a neutral particle geodesically moving in a background metric ...


1

I) OP talks about minimizing curves (rather than higher dimensional objects) so let us concentrate on point mechanics (as opposed to field theory) with Lagrangian $L$ (rather than Lagrangian density ${\cal L}$). We conventionally call the curve parameter time $t$, although it doesn't have to correspond to any physical time variable. Let us for simplicity ...


1

You don't need to assume that the path of least action is the path taken. You can show it from Newton's laws. See http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf The path of least action is the path for which $F = ma$ holds at each point. This is the geodesic. This is the shortest path through space-time. You get this path from the Lagrangian. You ...


1

Say the density of the string is $\mu$ and the tension is $T$. It's clear that the kinetic energy of an infinitesimal piece of string is $$dT = \frac{1}{2} (\mu \, dx) u_t(x)^2$$ The length of the infinitesimal piece of string from $(x, u(x))$ to $(x + dx, u(x + dx))$ is \begin{align} d\ell &= \sqrt{dx^2 + (u(x+dx) - u(x))^2} \\ &= \sqrt{dx^2 + ...



Only top voted, non community-wiki answers of a minimum length are eligible