Free Fall in Curved Spacetime: Does Minkowski Spacetime Exist?

[DanAil]

This topic has been discussed in the past on this forum, however there is one point that seems to be unclear.

One example of the setup is the following:
The universe is at a stage where all the matter is concentrated in a single black hole, except two spacecraft s A and B that orbit it far away in the same plane in a circular orbit but in opposite directions. As there are no stars to be seen, they can not determine if they are moving and consider themselves to be stationary. When their paths cross the first time they reset/sync their clocks. The question is what are the clocks going to show when they meet the second time?
The obvious conclusion is that clocks on A and B will display the same time as the orbits are symmetrical and any time dilation should be the same for both. Some consider this as a contradiction with the Special Relativity as the spacecraft A perceives itself as stationary and the moving clock on B should be running slower, while B expects exactly the opposite.

The answer is that the spacetime in the above example is not flat, so we cannot apply the Minkowski's metric in this case, which means that A and B should be not expecting their observations to be 'relative'. This answer is perfectly acceptable and ultimately the worldlines should be determining what the clocks will display.

The new question: In our universe there is no place with flat spacetime - even if we consider a very small local area. Any two objects A and B moving relative to each other without forces acting on them are actually in a free fall, influenced by some gravitational field. They may be 'falling' towards a planet or a star, orbiting a center of a galaxy or cluster of galaxies. Flat spacetime actually does not seem to exist.

Does this mean that the Minkowski spacetime should be never considered in the real world?

 

[Dale]

Does this mean that the Minkowski spacetime should be never considered in the real world?

Sure. It is an approximation and it can sometimes be a very good approximation. So it is a useful model even if it is never exact.

 

[Ibix]

Does this mean that the Minkowski spacetime should be never considered in the real world?

Not quite. It does mean that spacetime is always curved to some degree, yes. But there are plenty of situations where that curvature is negligible, and then Minkowski spacetime is a very useful approximation. Everything is simpler when you can work in Minkowski geometry. It's analogous to tiling your floor. The Earth is curved, but the error from neglecting that and pretending it's Euclidean is utterly negligible unless you're planning your country's road layout.

Your scenario with circular orbits does not fit into a Minkowski approximation because the curvature is important to the result: if you neglected curvature you would predict the spaceships would move in a straight line. But in scenarios like the cosmic ray muons neglecting gravity is a tiny correction, even though they are falling straight down.

 

[PeterDonis]

In our universe there is no place with flat spacetime - even if we consider a very small local area

Only if you assume our measurements are infinitely precise. But they're not. Over a small enough region of spacetime, our measurements cannot detect any curvature, so as a practical matter, we can consider such a region to be flat. It is pointless to argue that the region is not "really" flat because there is matter present somewhere. Yes, in theory that's true, but as the saying goes, in theory there is no difference between theory and practice, but in practice there is.

 

[jbriggs444]

Does this mean that the Minkowski spacetime should be never considered in the real world?

There is (I over-confidently state) no point on the surface of the Earth where the geoid is flat. Does this mean that we should discard all of our architectural blueprints plotted on flat paper with Euclidean assumptions because it is erroneous to build flat houses on curved ground?

 

[JandeWandelaar]

There is (I over-confidently state) no point on the surface of the Earth where the geoid is flat. Does this mean that we should discard all of our architectural blueprints plotted on flat paper with Euclidean assumptions because it is erroneous to build a flat houses on curved ground?

Hundertwasser did some non-Euclidean building:

 

[phinds]

Hundertwasser did some non-Euclidean building:

Yes, but that's not a great example because those curves do not follow space-time geodesics but rather are random.

 

[JandeWandelaar]

He used a passive, continuous diffeomorphism, judging by the looks.. The lawa of nature stay the same. A ball thrown in front of the houses will follow the same trajectory.

 

[DanAil]

Thank you for all the answers! As always, you provide great help clarifying the topic.

Over a small enough region of spacetime, our measurements cannot detect any curvature, so as a practical matter, we can consider such a region to be flat.

As of June 2022 the atomic clocks can 'detect' the curvature near the surface of the planet Earth (measuring gravitational time dilation) in about 1 millimeter difference of elevation - it is already starting to impact our daily lives :).

Your scenario with circular orbits does not fit into a Minkowski approximation because the curvature is important to the result

The spacecraft s could be orbiting with a velocity close to the speed of light a billion lightyears away from the center of the black hole, and thus not being able to detect any curvature with their technology. From their point of view they have a definitive prove that they are in a flat spacetime.

The challenge does not seem to be with the approximation, but in the fact that there are two independent methods that seem to contradict:

1. When we use Minkowski spacetime the reciprocity of the time dilation in the Special Relativity is obvious.

2. When we consider the curved space time (as the example in the initial post), it seems that we 'loose' the reciprocity and the time the clocks show is calculated using the worldlines, providing an definitive result that is not relative.

If in reality there is no flat spacetime, then should we be using the first method?

 

[jbriggs444]

1. When we use Minkowski spacetime the reciprocity of the time dilation in the Special Relativity is obvious.

2. When we consider the curved space time (as the example in the initial post), it seems that we 'loose' the reciprocity and the time the clocks show is calculated using the worldlines, providing an definitive result that is not relative.

Apples and oranges.

In 1, you are talking about time dilation. That is a coordinate effect. A matter of convention rather than of physical reality. Two trajectories that only intersect once. With the requirement to use a conventional synchronization standard in order to compare the elapsed coordinate time of the one against the elapsed coordinate time of the other.

In 2 you are talking about differential aging. That is a physical effect. A real physical measurement. You have two trajectories that intersect twice. One can measure the elapsed proper time along one path and the elapsed proper time on the other path and compare.

Nothing at all to do with curvature or the lack thereof.

 

[PeterDonis]

The spacecraft s could be orbiting with a velocity close to the speed of light a billion lightyears away from the center of the black hole

Please show your work. I do not think any such orbits as you describe exist in the scenario you posed. You can't just wave your hands and conjure up a circular orbit of any desired speed at any desired radius.

and thus not being able to detect any curvature with their technology

If they can detect that the black hole is present and that its presence is affecting their trajectory, then they can detect curvature.

If they can't, then why is the black hole even there in your scenario? What's the point?

 

[PeterDonis]

If in reality there is no flat spacetime, then should we be using the first method?

Both of your methods can be applied whether spacetime is flat or curved. So there is no contradiction that needs to be resolved.

 

[Dale]

If in reality there is no flat spacetime, then should we be using the first method?

You can use the first method whenever the assumptions on which it is based are good approximations.

You are speaking as though an approximation is true or false and that it not the way to look at approximations. You look at them as good or bad, an approximation may still be good even though it is false.

 

[JandeWandelaar]

The spacecraft s could be orbiting with a velocity close to the speed of light a billion lightyears away from the center of the black hole

Could they? Won't they just pass the hole?

 

[Ibix]

From their point of view they have a definitive prove that they are in a flat spacetime.

No they don't, because they meet each other again. That's definitive proof that they are not in flat spacetime.

At the time they pass each other (and for some time before and after) they can use SR as an excellent approximation, and will struggle to detect any deviation. But if they watch each other through telescopes they will eventually notice the other showing changing lateral motion. Eventually they may see gravitationally lensed duplicate images. Then they will see the ship return. All of this is due to the curvature of spacetime, and shows that while spacetime is locally very close to flat, globally it is not flat.

The spacecraft s could be orbiting with a velocity close to the speed of light a billion lightyears away from the center of the black hole, and thus not being able to detect any curvature with their technology.

This isn't right. Orbital velocity is proportional to $1/\sqrt r$, so if you want flat spacetime you need very low orbital velocity.

If in reality there is no flat spacetime, then should we be using the first method?

My analogy to tiling a floor was not an idle one. You can do it using Euclidean geometry because the curvature of the Earth is undetectably small on those scales. Absolutely you should use Euclidean geometry because using spherical geometry is much harder. The difference in correct results will be on the scale of micrometers, but the probability of error and the time taken will be much higher.

So you may treat the walls of your kitchen as parallel, even though you know that they lie along great circles and eventually would meet if you extended them far enough. If you are laying out a city, though, you would need to keep in the back of your mind that if your city is large enough Euclidean rules will no longer apply. Similarly, your spaceships may use SR when they are relatively near each other, but would be wrong to attempt to describe a full orbit with it.

 

[Ibix]

This isn't right. Orbital velocity is proportional to $1/\sqrt r$, so if you want flat spacetime you need very low orbital velocity.

Actually, I'll modify this. With a sufficiently large black hole, one with a Schwarzschild radius of billions of light years, you could be a billion light years away and still be just above the photon sphere, so orbiting near the speed of light. But you'd be able to see yourself in your rear view mirror, and you'd meet the other rocket at arbitrarily short intervals, all of which would be a hint that you weren't in flat spacetime.

 

[DanAil]

In 1, you are talking about time dilation. That is a coordinate effect. A matter of convention rather than of physical reality. Two trajectories that only intersect once.

In 2 you are talking about differential aging. That is a physical effect. A real physical measurement. You have two trajectories that intersect twice.

Good point. Noted. Thank you!

If they can detect that the black hole is present and that its presence is affecting their trajectory, then they can detect curvature. If they can't, then why is the black hole even there in your scenario? What's the point?

The point was to have a scenario where the trajectories intersect again, although the spacecraft s consider themselves to be stationary in flat spacetime - using the black hole is one example that could cause this.

In reality, the probability of any two objects in free fall to meet again is not zero. They might be not able to directly detect curvature but their trajectories could bend ever so slightly every time when passing near mass, and intersect again.

You are speaking as though an approximation is true or false and that it not the way to look at approximations. You look at them as good or bad, an approximation may still be good even though it is false.

Thank you @Dale ! You are right about the approximation.

 

[PeterDonis]

The point was to have a scenario where the trajectories intersect again, although the spacecraft s consider themselves to be stationary in flat spacetime

This is impossible. In flat spacetime two inertial (free-fall) trajectories cannot intersect more than once. So if the spacecraft are both in free fall and they meet twice, from that fact alone they know they are not in flat spacetime.

 

[PeroK]

The point was to have a scenario where the trajectories intersect again, although the spacecraft s consider themselves to be stationary in flat spacetime - using the black hole is one example that could cause this.

The point is that if spacetime is curved, the people in the spacecraft can detect this and will never measure spacetime to be flat.

If they measure the spacetime curvature within the spacecraft over a short period, they may find it small enough to be neglected for practical purposes.

But, if they place small objects at rest at different locations within the spacecraft , then over time those objects will drift apart. Which demonstrates that the spacetime is curved.

 

[JandeWandelaar]

But what if the gravity field is homogeneous? Like that of an infinite massive plane? The objects in the ship will stay where they are. An infinite massive plane is quite unrealistic, but an observer in a spaceship can think they are in free space or falling freely in such a homogenous field.

 

[Nugatory]

But what if the gravity field is homogeneous? Like that of an infinite massive plane? The objects in the ship will stay where they are. An infinite massive plane is quite unrealistic, but an observer in a spaceship can think they are in free space or falling freely in such a homogenous field.

The infinite plane is unphysical (and off-topic in this thread) but can be discussed in another thread: https://www.physicsforums.com/threa...al-field-possible-in-gr.1016569/#post-6648179

 

[Dale]

But what if the gravity field is homogeneous? Like that of an infinite massive plane?

Then the two trajectories can never meet twice. That only happens if spacetime is curved.