Do polar coordinates define an inertial frame or not?
Everywhere in GR, the authors of all the books talk about bring the metric to diag(-1, 1,1,1) which would show that a Local Inertial Frame exists at each point on the manifold. And the coordinate in this local frame would be $(t, x, y, z)$.
But can't Polar coordinates define an inertial frame. Can't the metric be brought to \begin{equation} diag(-1, 1, r^2, r^2 \sin^2(\theta))\end{equation} at every point. And the frame attached to the point is still inertial but just the coordinates used will be spherical instead of cartesian. Is this wrong.
In fact books begin constricting inertial frames using perpendicular rods and this cartesian coordinates. Can't we construct inertial frame using polar coordinates.
Edit- after an answer
Lorentz Transformations are transformations between different frames. Will a transformation from cartesian coordinate to polar coordinates be called a Lorentz transformation or called just a coordinate transformation. Now I have read that Lorentz transformations are linear. Transformation from cartesian to polar or from $(r, \theta, \phi) -> (r', \theta', \phi') $ would be Non Linear.
So will they be Lorentz Transformation.
Best Answer
Yes, polar coordinates can be used as coordinates for an inertial frame. For example, when we solve the gravitational two-body problem in Newtonian mechanics we typically use polar coordinates to write $\mathbf F=m\mathbf a$. The equations then take a nicer form than in Cartesian components. Simply using polar coordinates in a non-rotating frame does not introduce any non-inertial forces, as does happen when one considers a rotating frame.
Courses in Special Relativity typically restrict their discussion of Lorentz transformations to Cartesian coordinates for simplicity; one is mainly interested in what happens in the direction of the relative motion between the two frames, and perpendicular to it, so Cartesian coordinates are natural, especially when one takes one of the Cartesian axes to be along the relative velocity. But the geometry of Minkowski spacetime is the same regardless of whether its metric is written in Cartesian coordinates, spherical polar coordinates, or any other coordinates describing a flat spacetime.
For details of the Lorentz transformation in cylindrical and spherical coordinates see this paper.