将sympy矩阵转换为双形式（sympy.diffgeom）

Is there any way to convert a symmetric SymPy matrix to two-form ?

有没有办法将对称SymPy矩阵转换为双形式？

When I try to use

当我尝试使用时

sym.diffgeom.metric_to_Christoffel_1st(expr)

for the following matrix

对于以下矩阵

Matrix([[a**2/(cos(theta) - cosh(eta))**2, 0, 0], 
        [0, a**2/(cos(theta) - cosh(eta))**2, 0], 
        [0, 0, a**2*sinh(eta)**2/(cos(theta) - cosh(eta))**2]])

I get the error

我收到了错误

ValueError: The input expression is not a two-form.

The matrix expression is the metric for the transformation between Cartesian and Toroidal coordinates. I get the same error when I try to use the metric on the other sympy.diffgeom.metric_to_* functions.

矩阵表达式是笛卡尔坐标和环形坐标之间转换的度量。当我尝试在其他sympy.diffgeom.metric_to_ *函数上使用度量时，我得到相同的错误。

1 个解决方案

#1

Supposing your metric matrix is stored in variable m, you can do:

假设您的度量矩阵存储在变量m中，您可以执行以下操作：

diff_forms = your_coord_system.base_oneforms()

metric_diff_form = sum([TensorProduct(di, dj)*m[i, j] for i, di in enumerate(diff_forms) for j, dj in enumerate(diff_forms)])

Beware that for one-forms, TensorProduct(dx, dy) = -TensorProduct(dy, dx), this means that if your matrix is not diagonal, element m[i, j] will be summed to element m[i, j] for i != j.

请注意，对于单一形式，TensorProduct（dx，dy）= -TensorProduct（dy，dx），这意味着如果您的矩阵不是对角线，则元素m [i，j]将被求和为元素m [i，j]对于我！= j。

If you have a symmetric matrix, and you wish to add the off-diagonal elements only once, then add a 1/2 term for i != j.

如果你有一个对称矩阵，并且你希望只添加一次非对角线元素，那么为i！= j添加一个1/2项。

#1