弹性张量的种群

huangapple
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英文:

population of elasticity tensor

问题

我试图创建一个用于填充张量的代码，该张量应该具有36个独立的分量和对称性 C[i,j,k,l] = C[j,i,k,l] = C[i,j,l,k]。

我正在进行以下操作：

  1. C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
  2. var = sym.symbols('C1:%d'%82)
  4. ct = 0
  5. for i in range(3):
  6.     for j in range(3):
  7.         for k in range(3):
  8.             for l in range(3):
  9.                 if C[i,j,k,l] == 0:
  10.                     C[i,j,k,l] = var[ct]
  11.                     C[j,i,k,l] = var[ct]
  12.                     C[i,j,l,k] = var[ct]
  13.                     ct += 1

这给我提供了45个不同的条目。我做错了什么？

英文:

I am trying to create a code to populate a tensor with the symmetries of an elasticity tensor which should have 36 independent components and the symmetries C[i,j,k,l] = C[j,i,k,l] = C[i,j,l,k].

I am doing:

  1. C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
  2. var = sym.symbols('C1:%d'%82)
  4. ct = 0
  5. for i in range(3):
  6.     for j in range(3):
  7.         for k in range(3):
  8.             for l in range(3):
  9.                 if C[i,j,k,l] == 0:
  10.                     C[i,j,k,l] = var[ct]
  11.                     C[j,i,k,l] = var[ct]
  12.                     C[i,j,l,k] = var[ct]
  13.                     ct += 1

This is giving me 45 different entries. What am I doing wrong?

答案1

得分: 1

你似乎缺少对称性。如果你提到的两个对称性成立：

C[i,j,k,l] = C[j,i,k,l]（对称性1）
C[i,j,k,l] = C[i,j,l,k]（对称性2）

这意味着对每个元素都成立：

C[a,b,c,d] = C[b,a,c,d]（对称性1）
C[b,a,c,d] = C[b,a,d,c]（对称性2）

因此：

C[a,b,c,d] = C[b,a,d,c]或
C[i,j,k,l] = C[j,i,l,k]（隐含的对称性）

如果你在你的代码中包含这部分，你将得到：

  1. C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
  2. var = sym.symbols('C1:%d' % 82)
  4. ct = 0
  5. for i in range(3):
  6.     for j in range(3):
  7.         for k in range(3):
  8.             for l in range(3):
  9.                 if C[i,j,k,l] == 0:
  10.                     C[i,j,k,l] = var[ct]
  11.                     C[j,i,k,l] = var[ct]
  12.                     C[i,j,l,k] = var[ct]
  13.                     C[j,i,l,k] = var[ct]
  14.                     ct += 1

这将给你36个不同的条目。

英文:

You seem to be missing a symmetry. If the 2 symmetries you mentioned:

C[i,j,k,l] = C[j,i,k,l] (symmetry 1)
C[i,j,k,l] = C[i,j,l,k] (symmetry 2)

hold for every element that means that:

C[a,b,c,d] = C[b,a,c,d] (symmetry 1)
C[b,a,c,d] = C[b,a,d,c] (symmetry 2)

and therefore:

C[a,b,c,d] = C[b,a,d,c] or
C[i,j,k,l] = C[j,i,l,k] (implied symmetry)

If you include that in your code you'll get:

  1. C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
  2. var = sym.symbols('C1:%d'%82)
  4. ct = 0
  5. for i in range(3):
  6.     for j in range(3):
  7.         for k in range(3):
  8.             for l in range(3):
  9.                 if C[i,j,k,l] == 0:
  10.                     C[i,j,k,l] = var[ct]
  11.                     C[j,i,k,l] = var[ct]
  12.                     C[i,j,l,k] = var[ct]
  13.                     C[j,i,l,k] = var[ct]
  14.                     ct += 1

And that should give you 36 different entries.

huangapple
  • 本文由 发表于 2023年8月4日 23:03:12
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  • python
  • sympy
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