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：
C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
var = sym.symbols('C1:%d'%82)
ct = 0
for i in range(3):
for j in range(3):
for k in range(3):
for l in range(3):
if C[i,j,k,l] == 0:
C[i,j,k,l] = var[ct]
C[j,i,k,l] = var[ct]
C[i,j,l,k] = var[ct]
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:
C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
var = sym.symbols('C1:%d'%82)
ct = 0
for i in range(3):
for j in range(3):
for k in range(3):
for l in range(3):
if C[i,j,k,l] == 0:
C[i,j,k,l] = var[ct]
C[j,i,k,l] = var[ct]
C[i,j,l,k] = var[ct]
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：
C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
var = sym.symbols('C1:%d' % 82)
ct = 0
for i in range(3):
for j in range(3):
for k in range(3):
for l in range(3):
if C[i,j,k,l] == 0:
C[i,j,k,l] = var[ct]
C[j,i,k,l] = var[ct]
C[i,j,l,k] = var[ct]
C[j,i,l,k] = var[ct]
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:
C = sym.MutableDenseNDimArray.zeros(3,3,3,3)
var = sym.symbols('C1:%d'%82)
ct = 0
for i in range(3):
for j in range(3):
for k in range(3):
for l in range(3):
if C[i,j,k,l] == 0:
C[i,j,k,l] = var[ct]
C[j,i,k,l] = var[ct]
C[i,j,l,k] = var[ct]
C[j,i,l,k] = var[ct]
ct += 1
And that should give you 36 different entries.
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