Introduction to Differential Geometry with Tensor Applications. Группа авторов
of n linear equations.
Property 1.6.5. Considering the transformation zi = zi(yk) and yi = yi(xk), let N function zi(yk) be of independent N variables of yk so that
Here, N equation zi = zi(yk) is solvable for the z’s in terms terms of yi’s.
Similarly, yi = yi(xk) is a solution of yi in terms of xi’s so that
Now, we have by the chain rule of differentiation that
Taking the determinant, we get
(1.11)
Considering a particular case in which zi = xi, Equation (1.5) becomes
Or
This implies that the Jacobian of Direct Transformation is the reciprocal of the Jacobian of Inverse Transformation.
1.7 Differentiation of a Determinant
Consider the determinant
Then, the derivative of a with respect to x1 is given by
Therefore, in general, we can write
1.8 Examples
Example 1.8.1. Write the terms contained in S = aijxixj taking n = 3.
Solution: Since the index i (or j) occurs both in subscript and superscript, we first sum on i from 1 to 3, then on j from 1 to 3.
Example 1.8.2. Express the sum of
Solution: Here, the number of terms is 33 = 27.
Since the index i (or j or k) occurs both in subscript and superscript, we first sum on i from 1 to 3, then on each term of its 3 terms we sum j from 1 to 3. This results in 9 terms. Then, on each of the 9 terms we sum k from 1 to 3, which results in 27 terms. Like the last example, we sum
Example 1.8.3. If f is a function of n variables xi, write the differential of f.
Solution: Since f = f(x1, x2, … xn),
from calculus, we have
Example 1.8.4. (a) If apqxpxq = 0 for all values of the independent variables x1, x2, … xn and apq‘s are constant, show that aij + aji = 0.
(b) If apqrxpxqxr = 0 for all values of the independent variables x1, x2, … xn and apqr‘s are constant, show that akij + akji + aikj + ajki + aijk + ajik = 0.
Solution: Differentiating:
(1.12a)
with respect to xi
Differentiating (1.12b), with respect to xj, we get
(b) Differentiating
with respect to xi
Differentiating with respect to xj, we get
Differentiating in the same way, with respect to xk we get
Example 1.8.5. If
Solution: We have
Example 1.8.6. If
Solution: From above result
Example 1.8.7. If