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This question already has an answer here:

The book "The Dancing Wu Li Masters" page 189 talking about General Relativity says "A geodesic is not always a straight line".

Is that true? What is a definition of "straight line" that makes sense in this context?

The naive definition of "straight" is a three dimensional Euclidean sense. However, that is problematic. You'd need to project our four dimensional universe onto a "flat" three dimensional space. Presumable this would be done by choosing a pretty big inertial frame of reference and setting the Time axis to a constant. But choose a different frame, or different constant, and suddenly "straight" means something different.

Perhaps it makes more sense to define "straight" by projecting our four dimensional universe onto a four dimensional Euclidean space. But I think this will not work. Consider how difficult it is to project a curved 2D space onto a flat 2D space, such as making a map of the Earth. Except in special cases, the "straight lines" on the map do not correspond to any useful concept of "straight" on the globe. So I think in 4D the result will be not any better.

I am starting to think that the statement "A geodesic is not always a straight line" is simply nonsense, because there is no concept of "straight" except for the geodesic itself.

Perhaps five dimensions? Is there a projection of our four dimensional universe onto a five dimensional Euclidean space that would give us a sensible definition of "straight"?

Sorry if this is off-topic or too "philosophical".

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marked as duplicate by Danu, user36790, John Rennie, Community Oct 12 '15 at 3:51

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    $\begingroup$ you can measure curvature without going in the upper dimension: on Earth you could draw large triangles and mesure de sum of their angles. Or large disks and compare perimeter to surface. And same in volume. $\endgroup$ – Fabrice NEYRET Oct 9 '15 at 19:39
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    $\begingroup$ I think this old answer of mine is quite relevant. $\endgroup$ – Danu Oct 9 '15 at 19:48
  • $\begingroup$ No, straight lines make little sense if any in a generic curved space. The statement probably just reflects the possibility of curved spaces. $\endgroup$ – Solenodon Paradoxus Oct 9 '15 at 19:50
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    $\begingroup$ It's worth noting that the book is intended for a lay audience who are assumed be using a very simple notion of straightness on order of placing a ruler on a map of the region of space under discussing and seeing if they can line it up with the whole path of a light beam. $\endgroup$ – dmckee Oct 9 '15 at 20:11
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    $\begingroup$ I would also point out that the "Dancing Wu Li Masters" is decidedly not a good reference for learning about quantum physics. The book is basically "quantum mysticism" and so is not considered mainstream physics. That said, the question itself is good, so no problem. $\endgroup$ – march Oct 9 '15 at 22:47
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I note that the Wikipedia article on The Dancing Wu Li Masters starts with:

The Dancing Wu Li Masters is a 1979 book by Gary Zukav, a popular new age work about mysticist interpretations of quantum physics.

I don't think anything more needs saying. Insofar as the term straight line has any meaning in curved spacetime it means a geodesic, so the two terms mean the same. Whether the phrase A geodesic is not always a straight line has some new age mysticist meaning I don't know, but it has no meaning in physics.

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I think the comment by Danu and his linked answer is far better than anything I can write here, so I will keep this short. Why not just accept that a straight line has relevence only in 3 D space and is replaced by the generalised concept of a geodesic in higher dimensional spaces?

I would normally quote Wikipedia, but Danu does a better job of extending the idea to curved space, in my opinion.

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Imagine the following scenario: you have a space telescope trained on a distant star, stabilized by gyroscopes so that it points in a constant direction. It orbits far enough away from its host body that tidal and relativistic effects on precession are negligible. Also, the space between the telescope and the star is empty.

Since the star is far away and its relative speed is small, the line between the telescope and the star is a good benchmark for a straight line: there are no material or gravitational influences on particles traveling through the intervening space, which is flat.

The telescope has its "cross hairs" centered on the distant star, when a massive object passes between the two. For simplicity, imagine its a black hole so we don't have to worry about observation of this interloper. The black hole lenses the light from the distant star, making it appear to move: the point of light will appear first to move one direction, possibly splitting into multiple images, one of which will then move back into the original location of the image. Of course, the distant star hasn't itself moved, and the telescope hasn't rotated; they both still measure the same straight line that they did before.

However, due to the gravitational influence of the passing black hole - that is, the curvature of space around the black hole - the geodesics which the light followed from the star to the telescope did change and stray off of the "straight line" path. As measured compared to the Euclidean line between the star and telescope, the geodesic was bent.

Of course I'd be putting words in the author's mouth to say that this is exactly what he meant, but I do believe this is a likely (and fairly correct, in the linguistic sense) explanation of the phenomena he references.

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  • $\begingroup$ I think you are making the same mistake as the book author. Saying the light strayed off the "straight line" path implies that in some sense the "straight line" is still "there", but the light decided not to go that way. But actually the light always takes the shortest path between two points. No shorter path exists. You may argue "sure it doesn't exist in physical spacetime, but it still exists in some theoretical superset of spacetime." But that is a problem. Correct me if I am wrong, but I do not think such a superset can be even defined in any sensible or useful way. $\endgroup$ – John Henckel Oct 12 '15 at 17:02

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