# Tag Info

0

FWIW, the simplest is probably to graph the mechanical energy $(y,\dot{y})\mapsto \frac{m}{2}\dot{y}^2 + mgy$ rather than the Lagrangian. Because of energy conservation the stationary paths would then be level sets.

2

Well, it is possible to work directly with the Lagrangian in Lagrange's equations, cf. e.g this Phys.SE post. However, if one wants to have a variational principle that leads to Euler-Lagrange (EL) equations, it is necessary to introduce the action functional $S=\int_{t_i}^{t_f}\! dt~L$. This leads to the principle of stationary action/Hamilton's principle. ...

1

The quantum equivalent to the optimal control of densities is a topic of interest in the areas of large population control, optimal mass transport and mean field games (try https://arxiv.org/abs/1810.06064) in the control theory. The idea in the related literature is to transport a density given the individual micro-dynamics. The question of the minimum time ...

0

There are various issues with problem 7.1: Firstly, when reading chapter 7, it becomes clear that Hamill confusingly is denoting a total spacetime derivative $d/d x^{\mu}$ as $\partial/\partial x^{\mu}$, cf. e.g. this Phys.SE post. Secondly, for dimensional reasons, the function $F^{\mu}$ in the change in the Lagrangian density $$\Delta{\cal L}~=~\sum_{\... 0 An inertial frame A calculates that the clocks in another one B, with a relative velocity v tick slowly. (A person in B is aging slowly for them). And that is symmetrical. The frame B also see A aging slowly. In an accelerated frame aF, a free fall frame that is momentarily at the same speed of aF will calculate the clocks in aF ticking ... 1 Since action has the units of angular momentum, the proportionality constant needs the units of energy by dimensional analysis. It must also be Lorentz-invariant, so is mc^2 times some real number. This number's modulus can be varied without changing the resulting equations of motion, but @Cryo's answer shows a modulus of 1 recovers the usual \int(T-V)... 2 Let us for simplicity consider Minkowski space although the generalization to curved spacetime is straightforward. Lorentz invariance suggests that the Lagrangian one-form for a massive point particle should be$$\mathbb{L}~=~ f(\dot{x}^2)\mathrm{d}\lambda, \qquad \dot{x}^2~:=~\eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}~>~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{...

4

The equation you mention is the action of a single point particle. $$S =-mc^2\int d\tau$$ The unit of action is energy multiplied by time, in the present case the rest energy which corresponds to the mass of the particle multiplied by the proper time of the particle. This equation refers to the action from the point of view of the reference frame of the ...

6

For a single particle, it does not matter what prefactor you use, the equations of motion and everything else stays the same. The factors only start to matter when you couple different systems to each other. For example, consider a charged particle in an electromagnetic field described by a vector potential $A_\mu$. The right action describing its movement ...

9

I prefer to think like this. Inertial observer ($X$) is at rest in his/hers reference frame. The world-line of $X$ is the longest possible route between any two events. This follows since, in its frame, $X$ is moving fully along the temporal axis, therefore $ds=cd\tau$ (displacement in time; $c$ is the speed of light), and $d\mathbf{r}=\mathbf{0}$, and one ...

3

Extremising the action just means finding the classical solution by the principle of stationary action, which eventually boils down to solving the Euler-Lagrange equations $$\partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} = \frac{\partial \mathcal{L}}{\partial \phi},$$ in this case $$\partial_\mu \partial^\mu \phi = V'(\phi) = m^2 \... 1 Note that the use of Hamilton's principle (a.k.a. the principle of stationary action) for systems with semi-holonomic constraints in Ref. 1 is inconsistent with Newton's laws, and has been retracted on the errata homepage for Ref. 1. See Ref. 2 for details. See also this & this related Phys.SE posts. For starters, Ref. 1 provides a wrong (or at best an ... 2 This is because you work in first order in \varepsilon. So you need to consider only the linear terms in \varepsilon:$$ \delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ \bar{g}^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ \frac{\partial \varepsilon^{\mu}}{\partial x_{\nu}}+\frac{\partial \varepsilon^{\nu}}{\partial x_{\mu}}...

0

This is the divergence of a vector. One can use Stokes theorem and say that this is equal to a boundary contribution at infinty which is what we can set to be equal to zero.

0

The principle of virtual work, as I understand it, says we can make any imaginary change to a parameter to find the amount by which dependent variables change as a result. While I am not sure, I think here lies the issue. You cannot make any imaginary change to a parameter, only changes that can be made while time is frozen. If time was frozen and the ...

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