0

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?

Qmechanic
  • 201,751
Stepan
  • 209
  • This question sounds vaguely homeworky to me. Your answers lie here:

    https://en.wikipedia.org/wiki/Geodesic_deviation

     – Zo the Relativist Apr 29 '19 at 16:54
  • @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. https://youtu.be/7Zutjpu6Tv4?list=PLRlVmXqzHjURQIIebhT7UNTwGQHUEPlsb&t=290 – Stepan Apr 29 '19 at 17:07
  • 3
    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. – Zo the Relativist Apr 29 '19 at 17:16
  • @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. – Stepan Apr 29 '19 at 17:54
  • 1
    Free falling particles do indeed follow unique geodesics. And as long as the frame transformations are local and linear, SR is enough. – Cuspy Code Apr 29 '19 at 18:51
  • The problem though, is if I'm going to compare a frame that is in orbit to one that is not, then I am no longer local or linear. I have to pick a path that I will use to do the comparisons. In particular, since the question only gives us information about the paths, and not about the points used f or the comparisons, we don't even know if we're parallel translation along timelike,s pacelike, or null geodesics, even if we assume parallel transport along geodesics. – Zo the Relativist Apr 29 '19 at 22:41
  • Also, the whole point of curvature is that parallel transport in a circle along geodesics doesn't bring you back to your original vector. – Zo the Relativist Apr 29 '19 at 22:42

3 Answers3

6

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?

6

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.

spacetimeengineer
  • 776
  • 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. – Danu May 01 '19 at 09:53
3

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.

Deschele Schilder
  • 1