Transformation Riemann tensor to an accelerating frame

[wnvl2]

Two reference frames moving relative to each other at a constant velocity are related by a Lorentz transformation. As a result, the invariant properties of the Riemann curvature tensor will remain the same in both reference frames. This means that all inertial observers will measure the same local curvature of spacetime at a given point in spacetime. When considering non-inertial reference frames, will inertial and non-inertial observers measure different local curvatures at the same point in spacetime? How does the Riemann tensor transform in such case?

 

[Ibix]

If a tensor is zero in one coordinate system it's zero in every one. Generally, to transform an upper index tensor $T$ from coordinate system $x^i$ to coordinate system $y^i$ you contract $T^j$ with $\partial y^i/\partial x^j$, and to transform a lower index you contract $T_j$ with $\partial x^j/\partial y^i$. So the Riemann $R^i{}_{jkl}$ expressed in coordinates $x^i$ transforms to $R^p{}_{qrs}$ in coordinates $y^i$ like$$R^p{}_{qrs}=
\frac{\partial y^p}{\partial x^i}
\frac{\partial x^j}{\partial x^q}
\frac{\partial x^k}{\partial x^r}
\frac{\partial x^l}{\partial x^s}R^i{}_{jkl}$$Note that since every term in the implied sum on the right contains a component of the Riemann which is zero in this case, the result is obviously zero independent of the functional form of the partial derivatives.

 

[wnvl2]

I assume the note (last sentence) applies to flat space time?

The curvature scalar remains the same, also in non flat space time?

 

[Ibix]

In curved spacetime at least some of the Riemann components are non-zero, so the result of transforming it depends on the relationship between the coordinates and at least some components will be non-zero in any system.

The curvature scalars will be the same since they are scalars. This isn't always obvious when the coordinates are very different, but it is so.

 

[Orodruin]

This isn't always obvious when the coordinates are very different, but it is so.

I would say it is obvious as they are scalars …

The curvature properties of spacetime (or any other manifold with an affine connection) are ultimately geometrical properties of the spacetime itself, not the coordinates imposed on it.

Edit: Grammar

 

[Ibix]

I would say it is obvious as they are scalars …

Sure. But the expressions for the fields can look different in different coordinates. For a slightly silly example, consider a spherically symmetric spacetime expressed in spherical polars, which would have a Kretschman scalar $K(r)$. In cylindrical polars it would be $K(\rho,z)$ and would look different.

Obviously in this case you would find that $K(\rho,z)$ would actually be $K(\sqrt{\rho^2+z^2})$ and obviously the same. But if you have two different classes of spacetimes parameterised by different things that overlap in one special case, comparing the curvature scalars would be one way to spot the overlap - but it may not be obvious if the coordinate systems you've used are very different and you don't necessarily know the explicit transformation.

 

[Orodruin]

Sure. But the expressions for the fields can look different in different coordinates. For a slightly silly example, consider a spherically symmetric spacetime expressed in spherical polars, which would have a Kretschman scalar $K(r)$. In cylindrical polars it would be $K(\rho,z)$ and would look different.

Obviously in this case you would find that $K(\rho,z)$ would actually be $K(\sqrt{\rho^2+z^2})$ and obviously the same. But if you have two different classes of spacetimes parameterised by different things that overlap in one special case, comparing the curvature scalars would be one way to spot the overlap - but it may not be obvious if the coordinate systems you've used are very different and you don't necessarily know the explicit transformation.

Ah, but then we are talking a different type of obvious than what I take away from the OP. It is obvious that the invariants will not change in a coordinate transformation. It may not be as obvious that two given coordinate descriptions describe the same manifold.

 

[Ibix]

we are talking a different type of obvious

Obviously!

 

[wnvl2]

I was a little bit in doubt after the reply from Gandalf on this topic here on stackexchange.
Question was:
"Since tensors are invariant under coordinate transformations, and a moving observer is just in another coordinate system, he should measure the same Riemann curvature tensor (with different components).

Is that right for all observers, regardless of how they move, whether or not they are in an inertial reference frame?"

Could what is highlighted in the red rectangles be incorrect?

ps I don't have enough reputation to comment on StackExchange.

 

[Orodruin]

Could what is highlighted in the red rectangles be incorrect?

It is incorrect.

 

[Ibix]

Yep, it's wrong.

The simplest tensor is a vector. Imagine I'm standing pointing in some direction. An observer might say I'm pointing in the +x direction, or the -y direction, or the $(\frac 1{\sqrt{2}},\frac 1{\sqrt{2}})$ direction or whatever - depends which direction they're facing. And an observer on a turntable might say the direction I'm pointing is rotating. The components are different in each case, even time varying, but the vector is the same one. And invariants, such as the length of my arm or the angle it makes with the floorboards, are the same for all observers. In particular, if I cross my arms and close my fists so I'm not pointing in any direction (the vector is (0,0)), everyone will agree on that whatever they're doing.

The same applies to more general tensors. You can describe stresses in materials with a tensor. Just imagine if you saw a different tensor from me becase you were accelerating - you might find that I crushed the glass I was holding while I know I'm holding it lightly. One of us would be surprised at the actual outcome!

I would guess Gandalf61 has heard "acceleration is the same as gravity" (which isn't true) and jumped to the (incorrect) conclusion that an accelerating observer must see a non-zero Riemann tensor. That's just a guess, though.