How Does the Metric Tensor Relate to a General Tensor B in Tensor Calculations?

[redstone]

I'm trying to understand what kind of relation the metric can have with a general tensor B.

$$d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}=d{{s}^{2}}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=1$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{ab}}}{d{{s}^{2}}}=\frac{1}{D}g_{a}^{a}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}g_{a}^{d}}{d{{s}^{2}}}=\frac{1}{D}g_{d}^{a}g_{a}^{d}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}g_{a}^{d}}{d{{s}^{2}}}A_{m}^{n}=<br /> \frac{1}{D}g_{d}^{a}g_{a}^{d}A_{m}^{n}$$
Define: $$B_{am}^{dn}=g_{a}^{d}A_{m}^{n}$$
substitute in
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}}{d{{s}^{2}}}B_{am}^{dn}=\frac{1}{D}g_{d}^{a}B_{am}^{dn}$$
$$\frac{d{{x}^{a}}d{{x}^{b}}{{g}_{db}}}{d{{s}^{2}}}B_{am}^{dn}=\frac{1}{D}B_{am}^{an}$$

It all looks Ok to me. Does all of the following look reasonable, or is there a problem somewhere?

 

[member 553083]

Yes, the above looks reasonable. It shows that the metric has a relationship with a general tensor B, such that the metric can be used to calculate the rate of change of the tensor B with respect to distance. Specifically, it shows that the rate of change is equal to the tensor divided by the determinant of the metric.