Discrete Lagrangian Homework: Minimize S, Find EoM's & Discrete Trajectory

[Christoffelsymbol100]

Homework Statement

In this exercise, we are given a discrete Lagrangian which looks like this: http://imgur.com/TL0P61r. We have to minimize the discrete S with fixed point $$r_i$$ and $$r_f$$ and find the the discrete equations of motions.
In the second part we should derive a discrete trajectory for U(r) = 0 and U(r) = mgz.

Homework Equations

$$j = t_i + j \Delta t$$ ; $$\Delta t = (t_f - t_i)/N$$
$$r_j = r(t_i + j \Delta t)$$
$$v_j = \frac{r_j - r_{j-1}}{\Delta t}$$[/B]
j = 0, ..., N
U(r) is the potential energy
And the discrete action:
$$S(r_j) = \sum_{j=1}^N \Delta t \left(\frac{1}{2}m\left(\frac{\boldsymbol{r_j} - \boldsymbol{r_{j-1}}}{\Delta t}\right)^2 - \frac{1}{2}\left(U(\boldsymbol{r_j})+ U(\boldsymbol{r_{j-1}})\right)\right)$$

The Attempt at a Solution

Since S is a function, I thought I could take the gradient and set that to 0 to derive the discrete equations of motions: $$0 = \frac{\partial}{\partial r_k} S = \sum_{j=1}^N \Delta t \left(\frac{1}{2}m\frac{\partial}{\partial r_k}\left(\frac{\boldsymbol{r_j} - \boldsymbol{r_{j-1}}}{\Delta t}\right)^2 - \frac{1}{2}\left(\frac{\partial}{\partial r_k} U(\boldsymbol{r_j})+ \frac{\partial}{\partial r_k}U(\boldsymbol{r_{j-1}})\right)\right) \\ \rightarrow \sum_{j=1}^N \Delta t \left(m \left(\frac{\boldsymbol{r_j} - \boldsymbol{r_{j-1}}}{\Delta t}\right)\left(\frac{\delta_{jk}-\delta_{j-1,k}}{\Delta t}\right) -\frac{1}{2} \left(\frac{\partial U(\boldsymbol{r_j})}{\partial r_k}\delta_{jk} + \frac{\partial U(\boldsymbol{r_{j-1}})}{\partial r_k}\delta_{j-1,k}\right)\right) \\ \rightarrow 0 = m \left(\frac{-\boldsymbol{r_{k+1}}-\boldsymbol{r_{k-1}}+2\boldsymbol{r_k}}{\Delta t^2}\right) - \frac{1}{2} \left(\frac{\partial U(r_k)}{\partial r_k} + \frac{\partial U(r_{k+1})}{\partial r_k}\right)$$

The first term seems to be right, it looks like F = ma at least. I am not sure about the second part though because of that factor of 1/2 and I am not sure if the derivatives of the potential energy are correct.
For part b, I think I would have to insert the potential energy in the discrete equations of motions and integrate two times with respect to the time. But I am not sure how an integral would work here because of those indices.
Anyways, I would be really grateful for any advice! Thank you

 

[TSny]

Since S is a function, I thought I could take the gradient and set that to 0 to derive the discrete equations of motions: $$0 = \frac{\partial}{\partial r_k} S = \sum_{j=1}^N \Delta t \left(\frac{1}{2}m\frac{\partial}{\partial r_k}\left(\frac{\boldsymbol{r_j} - \boldsymbol{r_{j-1}}}{\Delta t}\right)^2 - \frac{1}{2}\left(\frac{\partial}{\partial r_k} U(\boldsymbol{r_j})+ \frac{\partial}{\partial r_k}U(\boldsymbol{r_{j-1}})\right)\right) \\ \rightarrow \sum_{j=1}^N \Delta t \left(m \left(\frac{\boldsymbol{r_j} - \boldsymbol{r_{j-1}}}{\Delta t}\right)\left(\frac{\delta_{jk}-\delta_{j-1,k}}{\Delta t}\right) -\frac{1}{2} \left(\frac{\partial U(\boldsymbol{r_j})}{\partial r_k}\delta_{jk} + \frac{\partial U(\boldsymbol{r_{j-1}})}{\partial r_k}\delta_{j-1,k}\right)\right)$$

I think everything is good so far.

$$\rightarrow 0 = m \left(\frac{-\boldsymbol{r_{k+1}}-\boldsymbol{r_{k-1}}+2\boldsymbol{r_k}}{\Delta t^2}\right) - \frac{1}{2} \left(\frac{\partial U(r_k)}{\partial r_k} + \frac{\partial U(r_{k+1})}{\partial r_k}\right)$$

I don't think the very last term on the right is correct. You might reconsider what the expression $\sum_{j=1}^N \frac{\partial U(\boldsymbol{r_{j-1}})}{\partial r_k}\delta_{j-1,k}$ reduces to.

 

[Christoffelsymbol100]

Wouldn't it reduce to $$\frac{\partial U(r_{k})}{\partial r_k}$$ $$j = r_{k+1}$$ because only for j=k+1 the kroenecker delta would give me a 1? And when you would have $$\frac{\partial U(r_{k+1-1})}{\partial r_k} = \frac{\partial U(r_{k})}{\partial r_k}$$ and then I would have two $$\frac{\partial U(r_{k})}{\partial r_k}$$ and I could cancle the 1/2 out?

 

[TSny]

Yes.

 

[Christoffelsymbol100]

Hey thank you very much! Now the discrete equation looks like Newtons 2nd law just as was asked :')