I've been trying to solve this problem but there are no resources that help. I've tried different approaches to solve this problem but every one of them leads to a dead end. I've found one approach that seems promising but I cannot solve the differential equation for $r^\sigma(t)$, which is the path the particle takes.

First of all, I have a vector field$$\Lambda^\alpha(x,y,z)$$ but when $x,y,z$ depends on $t$: $$\Lambda^\alpha(x(t),y(t),z(t))$$

Also I have a path that I am going to vary $$r^\sigma(t)$$ The components are $$r^1(t)=x(t)$$ $$r^2(t)=y(t)$$ $$r^3(t)=z(t)$$

The initial conditions are $$r^\sigma(0)=x^\sigma$$ and $$\frac{dr^\sigma(0)}{dt}=v^\sigma(t)$$

I want to find the path that the position vector and the vector field are pointing in the same direction. So I have to maximize the dot product between $r^\sigma(t)$ and $\Lambda^\alpha(t)$ for each tiny $dr^\sigma(t)$ along $\Lambda^\alpha(t)$, we can write this as: $$A=\int \Lambda_\alpha(t) \ dr^\alpha(t)$$ but we want to integrate over time for the path of stationary action so we get $$A=\int_0^t \Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}dt$$ Here we can see that: $$\mathcal{L}=\Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}$$ So using the euler lagrange equations: $$\frac{d}{dt}\frac{\partial \Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}}{\partial(\frac{dr^\sigma}{dt})}=\frac{\partial \Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}}{\partial r^\sigma}$$

I don't know how to solve these equations for $r^\sigma$ and I don't know how to implement the initial conditions into this equation. Because the paths should change completely as we implement them in, so it shouldn't be a linear term.

  • $\begingroup$ Why do you have to maximize the dot product? If you want $\Lambda$ to be parallel to $ r$ you should try and solve $\Lambda(\bf r) \times \dot {\bf r} =0$, where $\times $ is the cross product. $\endgroup$ – user617446 Aug 12 at 12:59
  • $\begingroup$ @user617446 The principle of stationary action will make it work in both cases, maximizing and minimizing the action. I think the dot product will just be easier to work with $\endgroup$ – hu huu Aug 12 at 15:06
  • $\begingroup$ Does your scheme take into account that $dr \over {dt}$ can be scaled by a scalar to be as large as you want? $\endgroup$ – user617446 Aug 13 at 5:37
  • $\begingroup$ @user617446 no but there will be a coefficient that scales r correctly, a bit like the gravitational constant. $\endgroup$ – hu huu Aug 15 at 15:36
  • $\begingroup$ Do you really want the position vector aligned with the vector field, or do you want the velocity vector aligned aligned with the vector field. If you want the position vector aligned with the vector field, the only solution is $\Lambda = f(r) r$. $\endgroup$ – Charles Hudgins Sep 4 at 21:12

We want to find the stationary points of $$ S[r(t)] = \int \Lambda_\alpha (r^\sigma (t)) \dot{r}^\alpha dt $$ So $$ S[r(t)] = \int L(r^\sigma(t) , \dot{r}^\sigma(t) ) dt $$ where $$ L(r^\sigma(t) , \dot{r}^\sigma(t) = \Lambda_\alpha (r^\sigma (t)) \dot{r}^\alpha $$ The Euler-Lagrange equations tell us that whenever $\delta S = 0$ we have $$ \frac{\partial L}{\partial r^\sigma} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r^\sigma}}\right) $$ For the LHS we have $$ \frac{\partial L}{\partial r^\sigma} = \frac{\partial \Lambda_\alpha}{\partial r^\sigma} \dot{r}^\alpha $$ For the RHS we have \begin{align} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{r}^\sigma}\right) &= \frac{d}{dt} \left(\Lambda_\alpha (r^\beta (t)) \delta^\alpha_\sigma \right) \\&= \frac{d}{dt} \left( \Lambda_\sigma( r^\beta (t))\right) \\&= \frac{\partial A_\alpha}{\partial r^\sigma} \dot{r}^\sigma \end{align} Putting this together, we have the rather uninspiring equation $$ \frac{\partial A_\alpha}{\partial r^\sigma} \dot{r}^\sigma = \frac{\partial A_\alpha}{\partial r^\sigma} \dot{r}^\sigma $$ What's the moral of the story? Sometimes the Euler-Lagrange equations are no help in solving a variational problem.

How could we have seen this coming? Note that $$ p_\sigma = \frac{\partial L}{\partial \dot{r}^\sigma} = \Lambda_\sigma $$ The canonical momenta are fully constrained. The system has no dynamics.

How else could we work this problem? Sometimes the simplest answer is the easiest. I'm going to assume you want the velocity vector of the particle aligned with the vector field. This is easily translated to a differential equation as $$ \Lambda^\alpha(r^\sigma(t)) = f(r^\sigma(t)) \dot{r}^\alpha $$ where $f(r^\sigma(t))$ is some scale factor that may depend on position.

  • $\begingroup$ If you want to learn more about what went on in this problem, I suggest you read up on singular Lagrangians. $\endgroup$ – Charles Hudgins Sep 4 at 23:01

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