- #36

Qwet

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vanhees71 said:I don't know, how you want to even formulate Gauss's theorem without Hodge duality and how to define the Levi-Civita tensor without a metric.

Well, I would do it like this.

Let's write 3-dimensional Gauss's theorem similarly to the form, which Ostrogradskiy got.

## \int_V [\frac{∂A^1}{∂x^1}+\frac{∂A^2}{∂x^2}+\frac{∂A^3}{∂x^3}]dx^1dx^2dx^3= \oint_{∂V} [A^1sgn(n_1)|dx^2dx^3|+A^2sgn(n_2)|dx^1dx^3| + A^3sgn(n_3)|dx^1dx^2|]##

If ##A^i## is vector density, then ##\frac {∂A^i}{∂x^i}## (summation over paired indices) appears to be scalar density, so the volume integral on the left side is a scalar, and the theorem gets covariant meaning, i.e independent of change of variables. At the same time, the surface integral on the right side appears to be a scalar, of course.

Where is any need for metric here?