- #1

Jianbing_Shao

- 102

- 2

- TL;DR Summary
- zero curvature

In relativity, a flat space is always regraded to be endowed with an invariant metric field ##g_{\mu\nu}(x)= \eta_{\mu\nu}##, So in a flat space the corresponding connection ##\Gamma_{\mu\nu}^\rho(x)=0## It means that if we parallel transport a vector ##v^\nu(x_0)## in the space. Then it obviously will obey rule ##\partial_\mu v^\nu(x)=0 ##. So if we parallel transport a vector ##v^\nu(x_0)## along an arbitrary loop ##C##, then we will get an invariant vector field ## v^\nu(x)= v^\nu(x_0)##.

From the definition of the curvature of a space, we know that if the curvature ##R_{\mu\nu\rho}^\lambda(x)=0##, then when we parallel transport a vector ##v^\nu(x_0)## from point ##x_0## along an arbitrary loop ##C## in the space, then when it returns back to point ##x_0##. We will find that ##\delta v^\nu(x_0)=0##. Here when vector moves along the loop, there is no need to restrict the vector must be invariant when it moves. In fact, if there exists a vector field ## v^\mu(x) =G^\mu_\nu(x)v^\nu(x_0), G^\mu_\nu(x_0)=I##, then it obviously fulfill the relation:

$$\partial_\mu v^\nu(x)= \partial_\mu G^\nu_\alpha(x) (G^{-1})^\alpha_{\rho } v^\rho(x)$$.

So if we define ##\Gamma_{\mu\rho}^\nu(x)= -\partial_\mu G^\nu_\alpha(x) (G^{-1})^\alpha_{\rho }(x) ##. So the vector field confined on the loop ## v^\mu(x) ## can be regarded to be generated by parallel transporting a vector ##v^\mu(x_0)## along loop ##C## in the space with a connection ##\Gamma_{\mu\rho}^\nu(x)##. Directly from the definition of curvature we can conclude that the curvature ##R^\rho_{\mu\nu\lambda}(x)=0##. And after a simple calculation we can find the conclusion is right. So we can draw such a conclusion the zero curvature space is not flat, but a flat space is obviously a zero curvature space. And the geodesics in the two different kinds of space will be different. How do we distinguish between the two in general relativity.

From the definition of the curvature of a space, we know that if the curvature ##R_{\mu\nu\rho}^\lambda(x)=0##, then when we parallel transport a vector ##v^\nu(x_0)## from point ##x_0## along an arbitrary loop ##C## in the space, then when it returns back to point ##x_0##. We will find that ##\delta v^\nu(x_0)=0##. Here when vector moves along the loop, there is no need to restrict the vector must be invariant when it moves. In fact, if there exists a vector field ## v^\mu(x) =G^\mu_\nu(x)v^\nu(x_0), G^\mu_\nu(x_0)=I##, then it obviously fulfill the relation:

$$\partial_\mu v^\nu(x)= \partial_\mu G^\nu_\alpha(x) (G^{-1})^\alpha_{\rho } v^\rho(x)$$.

So if we define ##\Gamma_{\mu\rho}^\nu(x)= -\partial_\mu G^\nu_\alpha(x) (G^{-1})^\alpha_{\rho }(x) ##. So the vector field confined on the loop ## v^\mu(x) ## can be regarded to be generated by parallel transporting a vector ##v^\mu(x_0)## along loop ##C## in the space with a connection ##\Gamma_{\mu\rho}^\nu(x)##. Directly from the definition of curvature we can conclude that the curvature ##R^\rho_{\mu\nu\lambda}(x)=0##. And after a simple calculation we can find the conclusion is right. So we can draw such a conclusion the zero curvature space is not flat, but a flat space is obviously a zero curvature space. And the geodesics in the two different kinds of space will be different. How do we distinguish between the two in general relativity.