- #1
gerald V
- 67
- 3
<Moderator's note: Moved from a mathematical forum.>
When one differentiates the determinant of a matrix, the adjugate of the matrix comes into play. The formula holds irrespective of whether or not the determinant vanishes.
I tried this for the 4x4 electromagnetic field tensor ##F##. But actually the result would hold for any skew-symmetric 4x4 matrix, since one can regard the components of ##E## and ##B## just as symbols for any numbers, I think.
As is known, the determinant of the electromagnetic field is ##(E \cdot B)^2## up to a numerical factor. When I calculated the adjugate, it turned out that the resulting (again skew-symmetric) matrix is proportional to ##E \cdot B##. This means that if the determinant of the electromagnetic field is zero, so is the entire adjugate matrix. However, the variation of ##\sqrt{\det F}## would be sensible even if the determinant is zero - very remarkable.
Is this correct or did I miscalculate?
Thank you very much in Advance.
When one differentiates the determinant of a matrix, the adjugate of the matrix comes into play. The formula holds irrespective of whether or not the determinant vanishes.
I tried this for the 4x4 electromagnetic field tensor ##F##. But actually the result would hold for any skew-symmetric 4x4 matrix, since one can regard the components of ##E## and ##B## just as symbols for any numbers, I think.
As is known, the determinant of the electromagnetic field is ##(E \cdot B)^2## up to a numerical factor. When I calculated the adjugate, it turned out that the resulting (again skew-symmetric) matrix is proportional to ##E \cdot B##. This means that if the determinant of the electromagnetic field is zero, so is the entire adjugate matrix. However, the variation of ##\sqrt{\det F}## would be sensible even if the determinant is zero - very remarkable.
Is this correct or did I miscalculate?
Thank you very much in Advance.
Last edited by a moderator: