0
$\begingroup$

I have written code for calculating cristofell symbol using for loop with 3 iterators for each coordinate 

A = {
   {1, 0, 0},
   {0, r^2, 0},
   {0, 0, 1}
  } ;
Kor[n_] := \[Piecewise] {
   {(r), n == 1},
   {(\[Theta]), n == 2},
   {(z), n == 3}
  } 
ChristoffelSymbol1[i_, j_, k_] := 
 1/2 (-D[A[[i, j]], Korr[k]] + D[A[[j, k]], Korr[i]] + 
    D[A[[k, i]], Korr[j]])
For[i = 1, i < 4, i++,
    For[j = 1, j < 4, j++,
        For[k = 1, k < 4, k++,

            Print[Subscript[\[CapitalGamma], i, j, k], " = ", 
    ChristoffelSymbol1[i, j, k]]

              ]
        ]
    ]

I was wandering if I could use Map/Table to do the same without changing the functions too much. As far as I could tell I need to make the Piecewise function return symbols for i, j, k when they are entered instead of number so I dont get an error when I do ChristoffelSymbol1[1,j,k] and (this I don't know how) somehow define what happens when I try to access the matrix A with A[[1,j]] for example , so it doesn't give me "expression j cannot be used as a part specification" and leave it in symbolic notation so it can be used when j iterators start.

Improve this question
$\endgroup$
3
  • $\begingroup$ if you actually want to Print like that you would use Do. I dont see why you would have the part issue you are imagining -- try it. I wouldn't bother with that Piecewise by the way , just do Kor[1]=r;Kor[2]=theta;Kor[3]=z or do Kor={r,theta,z} and reference it as a list. $\endgroup$ – george2079 Sep 28 '17 at 18:22
  • $\begingroup$ You can use for example Tuples and then Map function over the obtained list. However, it is unclear from your question what exactly you want to do? $\endgroup$ – ercegovac Sep 28 '17 at 18:58
  • $\begingroup$ I want a table of {Symbol,value} pairs outputted in 1 cell,but for the symbol to keep the formatting from print. $\endgroup$ – Andrej Licanin Sep 28 '17 at 19:26
0
$\begingroup$ 
\[CapitalGamma]ddd[i_, j_, k_] := 
 1/2 (-D[A[[i, j]], Kor[[k]]] + D[A[[j, k]], Kor[[i]]] + 
    D[A[[k, i]], Kor[[j]]])

A = DiagonalMatrix[{1, r^2, 1}]
Kor = {r, \[Theta], z}

Table[{Subscript[\[CapitalGamma], i, j, k], \[CapitalGamma]ddd[i, j, 
    k]}, {i, 1, 3}, {j, 1, 3}, {k, 1, 3}] // TableForm
Improve this answer
$\endgroup$
0
$\begingroup$

Got it 

 A = {
  {1, 0, 0},
  {0, r^2, 0},
  {0, 0, 1}
 } ;  B = {
  {1, 0, 0},
  {0, 1/r^2, 0},
  {0, 0, 1}
 }; 
Kor[n_] := \[Piecewise] {
   {(r), n == 1},
   {(\[Theta]), n == 2},
   {(z), n == 3}
  } 
ChristoffelSymbol1[i_, j_, k_] := 
 1/2 (-D[A[[i, j]], Kor[k]] + D[A[[j, k]], Kor[i]] + 
    D[A[[k, i]], Kor[j]])
ChristoffelSymbol2[i_, j_, m_] := 
 Sum[B[[k, m]] ChristoffelSymbol1[i, j, k], {k, 1, 3}]
TableForm[
 Table[{Subscript[\[CapitalGamma], i, j, k], 
   ChristoffelSymbol1[i, j, k]}, {i, 1, 3}, {j, 1, 3}, {k, 1, 3}]]
TableForm[
 Table[{Subscript[Superscript[\[CapitalGamma], m], i, j], 
   ChristoffelSymbol2[i, j, m]}, {i, 1, 3}, {j, 1, 3}, {m, 1, 3}]]

When i tried to replace For with Table i just copied the iterators,they were going from 1-4 in the loop and the marix is 3x3 so it bugged out :p

Improve this answer
$\endgroup$

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.