4
$\begingroup$

Consider a particle constrained to a manifold $\mathbb Q$ embedded within standard Euclidean 3-dimensional space that experiences no forces other than constraint forces keeping it within $\mathbb Q$. The Lagrangian in such a case is simply the kinetic energy $T$, so therefore the action $\mathcal S$ minimized by the subsequent motion is as follows, where we take the integration to be along the path taken by the particle.

$$\mathcal S= \int dt\ T$$

The resulting motion of a particle will both be (1) a path of stationary length and (2) constant in $T$. I'm am having trouble visualizing why both of these two properties follow from the action integral stated above.

Even in the simplest case, where $\mathbb Q$ is the one-dimensional $x$-axis of the Cartesian coordinate system, the motion will of course be described by $x(t) = x_0 + v_0 t$, but the action integral reads

$$\mathcal S = \frac{m}{2} \int dt\ \dot x^2.$$

It is not clear to me that minimizing $\int dt\ \dot x^2$ implies that $x(t)$ is linear. Heck, I can't even seem to evaluate that antiderivative in terms of $x$, $\dot x$, $\ddot x$, etc. (nor can WolframAlpha, so I'm not sure it can be evaluated with this level of generality.)

Any suggestions on how to make sense of this both in this special case as well as in general? I feel like there must be something simple here I'm missing given how simplified of a case this is. Yes, I could whip out the Euler-Lagrange equations and easily find $m\ddot x = 0$ in this specific case, but I'm looking for as direct and conceptual of a sense of why this is so as possible.

$\endgroup$
1
$\begingroup$

The simple dependence of the action allows us to explicitily show that $\dot x = v_0$ minimizes it. First I claim that the solution $\dot x = v_0$ (consistent with the initial conditions) minimizes the action $$ S[\dot x] = \int_1^2 \frac 1 2 m\dot{x}^2 \, dt, $$ with $1$ and $2$ representing the starting and end times respectively.

Now I vary the path by adding a general function of $t$ - let's call it $\delta x$, $$ x = x_0 + v_0 t \to x' = (x_0 + v_0t) + \delta x(t), $$ with $\delta x$ such that $\delta x(1) = \delta x (2) = 0$ in order to respect the boundary conditions. The action changes as follow $$ S[v_0] \to S[v_0 + \delta \dot{x}] = \frac 1 2 m \int_1^2 (v_0 + \delta \dot{x})^2 \, dt = S[v_0] + mv_0 \int_1^2 \delta \dot{x} \, dt + \frac 1 2 m \int_1^2 (\delta \dot{x})^2 \, dt. $$

We can easily show that the integral of $\delta \dot{x}$ vanishes because of the boundary conditions imposed on $\delta x$: $$ \int_1^2 \delta \dot{x} \, dt = \int_1^2 \frac{d}{dt} \delta x \, dt = \delta x(2) - \delta x(1) = 0. $$ The new action is thus $$ S[v_0 + \delta \dot{x}] = S[v_0] + \frac 1 2 m \int_1^2 (\delta \dot{x})^2 \, dt. $$ Since the integrand is always $\geq 0$, the action only increases when we vary the path. Therefore it is a minimum. You can get to the same conclusion in 3 dimensions by replacing $x$ and $\delta x$ by vectors $\mathbf{r}$ and $\delta\mathbf{r}$. This procedure works just because the action has a really simple depence on $x$ and $\dot{x}$ (I think it also works when there is a potential like $V(x) = kx$), but for more general lagrangians you'll not be able to show in this way that the specific solution minimizes the action.

$\endgroup$
0
$\begingroup$
  1. According to D'Alembert's principle the constraint forces produce no virtual work, so they can be ignored. Since there are no other forces in the otherwise free case, the Lagrangian $L$ for the constrained particle is just given by the pullback of the kinetic term in the ambient space $\mathbb{R}^3$ to the constrained submanifold $\mathbb{Q}$.

  2. The solution trajectories are affinely parametrized geodesics on $\mathbb{Q}$, cf. e.g. this Phys.SE post.

$\endgroup$
  • 1
    $\begingroup$ I agree with what you have said, but I'm not sure how that answers the question at hand. $\endgroup$ – Trevor Kafka Sep 24 '18 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.