0
$\begingroup$

The following was stated in a general relativity lecture series:
"Full space is needed in GR. Minkowski space is sufficient for SR. However, Minkowski space is sufficient to compute properties of free falling bodies. SR is sufficient to do frame transformations between

(a) an object falling on a black hole,

(b) orbiting a black hole and

(c) travelling by inertia in intergalactic space.

Even more so, a gyroscope cannot help one to distinguish orbits (a), (b) and (c)"

Is this statement correct?

$\endgroup$
  • $\begingroup$ This question sounds vaguely homeworky to me. Your answers lie here: en.wikipedia.org/wiki/Geodesic_deviation $\endgroup$ – Jerry Schirmer Apr 29 at 16:54
  • $\begingroup$ @JerrySchirmer I've got my PhD in chemistry and don't take physics classes. As a layman watching youtube courses I want to know if I can trust the speaker on this statement. youtu.be/… $\endgroup$ – Stepan Apr 29 at 17:07
  • 3
    $\begingroup$ The statement strikes me as pretty wrong from a GR perspective. In particular, reference frames in general relativity are attached to a spacetime point, and if you are going to compare frames at two different points, you're going to have to specify a parallel translation scheme, and that choice is non-unique. And thanks to geodesic deviation, even nearby points are going to have arbitrarily distant frames after some finite time. $\endgroup$ – Jerry Schirmer Apr 29 at 17:16
  • $\begingroup$ @JerrySchirmer "that choice is non-unique" I understand the problem of instability. E.i. two arbitrary close points can divert to infinite distance due to a gravitational capture of one of them. But isn't a free fall unique for a given point of a spacetime? We don't bring QM uncertainty considerations in the picture. $\endgroup$ – Stepan Apr 29 at 17:54
  • 1
    $\begingroup$ Free falling particles do indeed follow unique geodesics. And as long as the frame transformations are local and linear, SR is enough. $\endgroup$ – Cuspy Code Apr 29 at 18:51
6
$\begingroup$

This basically sounds wrong to me. If you want accurate information, you'd be better off looking at books rather than videos on youtube. It's a little difficult to tell for sure whether this person is confused or is just not expressing themselves with sufficient precision. Again, that's one of the advantages of written communication: precision.

Full space is needed in GR.

This is already pretty weird. Are these the speaker's actual words, or is this a paraphrase or translation from another language? I don't think any relativist would refer to "full space" in contradistinction to Minkowski space. Even with the context, it isn't clear what this person means here.

However, Minkowski space is sufficient to compute properties of free falling bodies.

This doesn't really make much sense. Unless there's some missing context, they haven't described what they mean by "properties." A single pointlike, neutral, free-falling particle doesn't really have any interesting properties. It has some mass, and it's at rest in its own inertial (free-falling) frame. Check back later -- it still has that mass, and it's still at rest relative to itself.

To make this a nontrivial statement, you would have to have, say, two such particles. And then this statement becomes clearly false. For example, two such particles can follow geodesics that coincide at event A and then coincide again later at event B. That would be impossible in Minkowski space.

SR is sufficient to do frame transformations between [...]

This sounds like the lecturer doesn't understand that frames in GR are local, so, e.g., we can't have a frame that encompasses both a and b from their list. Or they may mean something different, but again it's impossible to tell because they aren't speaking precisely.

Related: How do frames of reference work in general relativity, and are they described by coordinate systems?

$\endgroup$
5
$\begingroup$

Not exactly. I wouldn't say that General Relativity reduces to Special Relativity for free-falling frames because it implies that General Relativity loses meaning in these inertial situations. It's true however that all spacetime manifolds are locally Minkowskian but all this means is that any small piece of a spacetime manifold will approach a Minkowskian (flat) spacetime as you zoom into it further and further. This concept deeply relates to the equivalence principle.

$\endgroup$
  • $\begingroup$ I think one has to be careful when saying that "spacetime manifolds are locally Minkowskian". After all, curvature is a local quantity which cannot be dispensed with, even in a small patch. I get what you are trying to say, but I'm not sure what the correct mathematical way of expressing that would be. $\endgroup$ – Danu May 1 at 9:53
3
$\begingroup$

As stated in the answer by Ben Crowell, the spacetime manifolds that are described in the citation are only locally Minkowskian. This means that if we consider the global situation, frame dragging effects come into existence (which are proved to exist around our Earth in freely falling satellites, using gyroscopes; see here). So maybe locally the spacetime can be described by the Minkowski metric, globally it certainly can't. Otherwise, the free-falling satellite wouldn't have been able to detect the aberrations due to the frame-dragging effect.

The same holds for the situations a), b), and c) you describe.

$\endgroup$
  • $\begingroup$ Can you change "the previous answer" to a link, or name the author of the answer you mean? The order of the other answers shifts as the vote totals change. $\endgroup$ – rob Apr 30 at 0:24
  • $\begingroup$ Done! I didn't realize the order of the answers shifts. $\endgroup$ – descheleschilder Apr 30 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.