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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.

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Can't we construct inertial frame using polar coordinates.

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.

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  • $\begingroup$ Thanks I understood most of it. Just 2 small issues- 1) That means in GR we could equivalently say that if the metric at each point can be reduced to $diag(-1, 1,r^2, r^\sin^2\theta), then we would be having Local Inertial Frames at each point, just that their coordinates would be written in spherical coordinates. Would that be technically correct. Please let me know this, I have seen the Lorentz Transformations in spherical polar coordinates in the paper you linked and I think what I have said above should be correct. Kindly confirm. $\endgroup$ – Shashaank Dec 26 '20 at 6:01
  • $\begingroup$ I agree. However, even in a curved spacetime, there is a local inertial frame in a sufficiently small neighborhood of each point. $\endgroup$ – G. Smith Dec 26 '20 at 6:05
  • $\begingroup$ Secondly - 2a) 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. 2b) 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. Kindly let me know this as well. Added it in the Edit to the question as well for your reference. Thanks $\endgroup$ – Shashaank Dec 26 '20 at 6:06
  • $\begingroup$ yeh I know that, I just wanted to describe that coordinate in terms of polar coordinates rather than cartesian, like the books do. $\endgroup$ – Shashaank Dec 26 '20 at 6:08
  • $\begingroup$ added edit. Will polar to polar' won't be Non Linear and thus not qualify as Lorentz Transformation which need to be Linear. Also then the answer by Charlie is wrong, right. $\endgroup$ – Shashaank Dec 26 '20 at 6:11
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Polar coordinates are not inertial coordinates. Only those coordinate systems that can be reached from an inertial coordinate system by the action of the Poincaré group are inertial coordinates (it should be noted that the Lorentz transformations are a special case of the Poincaré transformations that preserve the origin).

If you want to work in polar coordinates you should be careful as objects that are Lorentz covariant are not necessarily diffeomorphism covariant which is the appropriate notion of covariance in general relativity. That being said, there is no problem working in polar coordinates in Minkowski space, you must just be careful since you are not in an inertial coordiante system and so some assumptions may not hold.

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