Another Twins Paradox question

[Al68]

Jesse,

I was only referring to the fact that the ship's observer could look at his clock when he arrived at the star system, it would read 5.77 years. He could then use radar to measure his distance to earth, it would be 10 light years. He could calculate how far apart the Earth and star system were in his frame before he decelerated as 0.866c * 5.77 years = 5 light years. This would be approximate, neglecting the distance he would have needed to decelerate.

Please don't read more into this than I intended. I'm not under the impression that the ship could observe Earth's position in real time. And as far as calculating the magnitude of the ship's decceleration relative to Earth's apparent position, and Earth's apparent position relative to the ship as a funtion of ship's time, as observed by the ship, I don't think I'll try it right now. Especially since this was my point. These are the kinds of questions I have. I'm only asking them because they are not covered in most textbooks. And textbooks are not interactive (neither are most professors for that matter).

And I think it is normally considered reality that, theoretically, a ship's crew could end up 10 light years from earth, at rest with earth, after an elapsed proper time of 5.77 years by the ship's clock. Is this not considered objective reality by physicists?

Thanks,
Alan

 

[JesseM]

Jesse,

I was only referring to the fact that the ship's observer could look at his clock when he arrived at the star system, it would read 5.77 years. He could then use radar to measure his distance to earth, it would be 10 light years. He could calculate how far apart the Earth and star system were in his frame before he decelerated as 0.866c * 5.77 years = 5 light years. This would be approximate, neglecting the distance he would have needed to decelerate.

OK, but now you're comparing distances in two different inertial reference frames. So he did not "observe" that the Earth moved from 5 light years to 10 light years, he just figured out that the distance to the Earth is different in two different inertial frames. The fact that each frame happened to be his instantaneous rest frame at that moment is irrelevant, it doesn't imply something like "the Earth moved from 5 light years to 10 light years from his point of view" unless you define his "point of view" as a non-inertial coordinate system which defines distances at each moment to be identical to the distance in his instantaneous inertial rest frame at that moment.

And I think it is normally considered reality that, theoretically, a ship's crew could end up 10 light years from earth, at rest with earth, after an elapsed proper time of 5.77 years by the ship's clock. Is this not considered objective reality by physicists?

My impression is that the only thing a physicist would call a truly "objective reality" is something that is coordinate-independent, like the proper time along a worldline between two events on that worldline, or the answer to whether one event lies in the past light cone of another, or what the readings on two clocks will be at the moment they meet at a single point in spacetime. Any statement that's true in one coordinate system but not true in another can't be considered a simple physical truth, such statements can only be true relative to a particular choice of coordinate system (which means you have to specify which coordinate system you're talking about when making the statement).

 

[Al68]

OK, how about from the (non inertial) frame of the ship. Again, I know the ship cannot observe Earth in real time. When the ship accelerates away from earth, then cuts its engines, the ship's crew and Earth will disagree about what time the engines were cut. And they will disagree about the magnitude of the ship's acceleration. If the ship's crew calculates that their velocity relative to Earth is 0.866c after acceleration, will Earth also agree that the ship's velocity is 0.866c relative to earth? Is velocity invariant between inertial reference frames?

Thanks,
Alan

 

[pervect]

OK, how about from the (non inertial) frame of the ship. Again, I know the ship cannot observe Earth in real time. When the ship accelerates away from earth, then cuts its engines, the ship's crew and Earth will disagree about what time the engines were cut. And they will disagree about the magnitude of the ship's acceleration. If the ship's crew calculates that their velocity relative to Earth is 0.866c after acceleration, will Earth also agree that the ship's velocity is 0.866c relative to earth? Is velocity invariant between inertial reference frames?

Thanks,
Alan

The short answer is yes, everyone in an inertial coordinate system will agree that the velocity is .866 c after the acceleration stops, as long as inertial coordinates are used.

The procedure I would recommend to get a meaningful velocity from the coordinate system of an accelerting spaceship is rather involved, but it will give the same answer as the velocity measured by inertial observers, which is why I recommend it.

Let us say that we have an object, O, and that we want to measure it's velocity relative to the accelerating spaceship, using the coordinate system (viewpoint) of the spaceship.

Basically, one constructs a second space-ship, that maintains a "constant distance" away from the first.

This second space-ship is placed at the location of the object O.

Spaceship #2, at the location of object O, then measures the velocity of object O, using its own local set of clocks and rulers. (These will be different from the clocks and rulers in spaceship #1).

It can be shown that object O will measure the same velocity for spaceship #2 as spaceship #2 measures for object O, as long as measurements are made with standard clocks and rulers.

Technically one constructs a tetrad of "orthonormal basis vectors" at the second space-ship (even more exactly, an orthonormal tetrad of one-forms).

To make this work, one has to have a meaningful concept of what it means for spaceship #2 to be "stationary" (maintaining a constant distance) with respect to spaceship #1. If space-time is flat, or even if it's non-flat and static, this can be defined by the fact that light signals exchanged between spaceship #2 and spaceship #1 will always have the same travel time.

If the space-time is not static, there isn't any way that I'm aware of to define a meaningful concept of a "stationary" distant spaceship, and this procedure for defining relative velocity breaks down.

 

[JesseM]

OK, how about from the (non inertial) frame of the ship.

There is no single standard way to construct a coordinate system for a non-inertial object. If you specify how the coordinate system is constructed, then you can figure out how things will behave in that coordinate system.

Again, I know the ship cannot observe Earth in real time. When the ship accelerates away from earth, then cuts its engines, the ship's crew and Earth will disagree about what time the engines were cut. And they will disagree about the magnitude of the ship's acceleration.

It depends whether you're defining "acceleration" in terms of change in coordinate velocity, or in terms of the acceleration felt by the ship itself (which means acceleration at a given moment is the rate of change of velocity as seen in the ship's instantaneous co-moving frame at that moment). Often in relativity "acceleration" is used to refer to the latter, so "constant acceleration" would mean a constant G-force experienced by passengers, not a constant rate of change of velocity as seen in a single inertial frame. But I assume you were talking about the former type of acceleration, in which case different inertial frames will indeed disagree about the ship's acceleration.

If the ship's crew calculates that their velocity relative to Earth is 0.866c after acceleration, will Earth also agree that the ship's velocity is 0.866c relative to earth? Is velocity invariant between inertial reference frames?

Between the two rest frames of the objects who are in relative motion, yes. Other frames besides the rest frame of the ship and the rest frame of the Earth won't see the distance between the ship and the Earth increasing at 0.866 light-years per year, though.