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Discrete Lagrangian

  • #1

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 [tex]r_i[/tex] and [tex]r_f[/tex] 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


[tex] j = t_i + j \Delta t[/tex] ; [tex] \Delta t = (t_f - t_i)/N [/tex]
[tex]r_j = r(t_i + j \Delta t)[/tex]
[tex]v_j = \frac{r_j - r_{j-1}}{\Delta t}[/tex][/B]
j = 0, ....., N
U(r) is the potential energy
And the discrete action:
[tex]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)[/tex]

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: [tex]
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)
[/tex]

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
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,408
2,841
Since S is a function, I thought I could take the gradient and set that to 0 to derive the discrete equations of motions: [tex]
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)[/tex]
I think everything is good so far.
[tex]\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)[/tex]
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.
 
  • #3
Wouldn't it reduce to [tex] \frac{\partial U(r_{k})}{\partial r_k} [/tex] [tex]j = r_{k+1}[/tex] because only for j=k+1 the kroenecker delta would give me a 1? And when you would have [tex] \frac{\partial U(r_{k+1-1})}{\partial r_k} = \frac{\partial U(r_{k})}{\partial r_k} [/tex] and then I would have two [tex]\frac{\partial U(r_{k})}{\partial r_k}[/tex] and I could cancle the 1/2 out?
 
  • #4
TSny
Homework Helper
Gold Member
12,408
2,841
Yes.
 
  • #5
Hey thank you very much! Now the discrete equation looks like newtons 2nd law just as was asked :')
 

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