Graduate Curvature tensor in all flat space coordinates

[mertcan]

hi, I am just curious about, and really wonder if there is a proof which demonstrates that curvature tensor is 0 in all flat space coordinates. Nevertheless, I have seen the proofs related to curvature tensor in Cartesian coordinates and polar coordinates, but have not been able to see that zero curvature tensor dominates all flat space coordinates ( besides the polar and cartesian coordinates ). Could you please explain or help me reach the proof ?

 

[ShayanJ]

hi, I am just curious about, and really wonder if there is a proof which demonstrates that curvature tensor is 0 in all flat space coordinates. Nevertheless, I have seen the proofs related to curvature tensor in Cartesian coordinates and polar coordinates, but have not been able to see that zero curvature tensor dominates all flat space coordinates ( besides the polar and cartesian coordinates ). Could you please explain or help me reach the proof ?

The Riemann tensor transforms under a coordinate transformation as $R'^\sigma_{ \ \rho \mu \nu}=(S^{-1})^\sigma_\alpha S^\beta_\rho S^\gamma_\mu S^\delta_\nu R^\alpha_{ \ \beta \gamma \delta}$ which means its components in the new coordinate system are linear combinations of its components in the old coordinate system. So if the components are all zero in the old coordinate system, they'll also be zero in the new coordinate system. That's why you only need to prove they're zero in only one coordinate system to prove that the space is flat.

 

[Orodruin]

To add to Shyan's reply, this is true for any tensor. If it is zero in one set of coordinates, it is zero in all coordinates.

 

[Markus Hanke]

As Shyan and Orodruin have already pointed out, the proof lies in the fact that Riemann is a tensor. As geometric objects, tensors are by their nature and definition independent of the choice of coordinate basis - if you perform a transformation into a different coordinate system, then the expressions for each individual component of the tensor may change, but the relationships between the various components will not, meaning the overall geometric object remains the same.

In a perhaps more tangible sense, a change in coordinates just signifies that you choose to label the same events in spacetime in a different way - and since labels are completely arbitrary, these events will still remain related in the same ways, so the overall geometry cannot change simply by your act of "re-labelling". Therefore, you need to calculate Riemann only once, in any suitable coordinate system of your choice, and the result ( flatness or not ) will be applicable for all other equivalent choices of coordinate basis as well.

That is the beauty of tensors.

 

[pervect]

I'd like to add a few more comments on the notion of a tensor. If one views a tensor in an abstract, coordinate-free manner, one says that tensors are "independent of the coordinates" - the tensor is regarded as an abstract entity that is independent of the specific coordinates used to express it.

If one views a tensor as as a multi-dimensional array of numbers (which are called the components of the tensor), then the numbers (or components) that make up the tensor do change when one changes coordinates. However, in order for a tensor to be a tensor, these numbers must change (transform) in a specific, standardized manner. One can find a brief and rather terse summary of how a tensor must transform in the above posts, or look up a detailed account elsewhere. A very short summary that gets the essence is that a tensor must transform as a multi-dimensional linear array. Given the transformation equations, one can prove (as several posters have already done) that a tensor that has all zero components in one coordinate system must also have all zero components in any other coordinate system. This result basically follows from the linearity of the transformation.

Not every quantity is a tensor - tensor quantities are very special in the simplicity of how they transform, a simplicity which makes them special and useful.