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"Chart" is the technical term from Riemannian geometry. It is all spelled out in chapter 2 of the reference I provided ( https://www.physicsforums.com/threads/acceleration-and-the-twin-paradox.779110/page-7 ). Although the notes are from a course on general relativity, that specific chapter is only about Riemannian geometry, which is applicable to any branch of physics, including SR, QM, and even classical mechanics. If you want to learn about non-inertial frames in SR then you have to study the basics of charts and Riemannian manifolds.CKH said:When we talk about an inertial frame, we can assign that a 4-D coordinate system (why is this now called a "chart", perhaps more general somehow?).
Yes.CKH said:That coordinate system covers all of space and time (in SR). So an inertial frame is not something that changes with time, as you say.
All of this is correct, but ...CKH said:The traveler experiences change over time as measured by the clock he carries with him. When he is moving inertially, he is at rest in some inertial frame and his clocks ticks at the same rate as clocks in that frame. When he moves with acceleration, at each instant of his time, he is at rest in a different inertial frame. Hyperplanes of simultaneity in each of these different frames intersect the worldline of the home clock at a different point.
This doesn't necessarily follow. You can define a chart this way, but any other smooth one-to-one mapping which assigns a constant coordinate to the traveler is equally valid. Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.CKH said:So, for the traveler, the simultaneous reading on the home clock changes because his motion becomes coincident with different inertial frames as his motion changes.
Any diffeomorphism from open sets in the manifold to open sets in R4 is a valid chart. See the reference above. The only requirement beyond being a valid chart is that in X's rest frame the spatial coordinates for X are constant. This leaves an immense amount of freedom for defining a non-inertial object's rest frame, and no standard convention.CKH said:This seems to be the central issue here. Can you explain what you mean in more detail, in SR where spacetime is flat? ...
I need understand what we mean by the "non-inertial rest frame", if we consider it as a 4-D coordinate system. We are trying to describe 4-D rest frame for the traveler (so over the whole trip). I can understand how a traveler in uniform motion has a 4-D rest frame with coordinate axes. When his motion is not inertial, how do you define his 4-D frame? Can you create a coordinate system for this frame over all spacetime?
Here is an example of an alternative method for constructing a non-inertial rest frame which is also commonly used: http://arxiv.org/abs/gr-qc/0104077
There is a MCIF. There is not an instantaneous 3D chart since that would not be a mapping from an open set of the manifold to an open set in R4.CKH said:However, to me it does makes sense to say there is an instantaneous chart at each point along on the traveler's world line.
More than that, you have to make sure that the resulting mapping is a diffeomorphism.CKH said:Yes, the frames have to be connected using the traveler's clock, position and orientation.
The chapter I posted is a gentle introduction to Riemannian geometry, which is used for non-inertial frames in SR also. GR does not "own" Riemannian geometry. I agree that we should stick with only SR and not consider gravity. All of my above comments have been restricted to SR (and in fact, I have only been considering the instantaneous turn around scenario except when explicitly responding to a comment from someone else about the gradual turn around). So, there is no need to go on about GR vs SR, I am not using GR.CKH said:Your reference looks like a nice introduction to general relativity which I want to study, but that is not an issue here since we have no gravitational fields in SR.
Yes, that is all correct.CKH said:There is a diagram in Minkowski spacetime in a earlier post that assumes the simplified case with instant acceleration of the traveler. It shows the world line of the home clock and that of the traveler in the rest frame of the home clock (the base frame). The world line of the home clock is a straight vertical line segment (between these events). The world line of the traveler is a dogleg ">" segment.
The wordline for an object is the path of the object in the Minkowski spacetime. Along a world line, a tangent ray (pointing upward toward positive time) is the positive time axis of the object in its rest frame at that point. In Minkowski space, clock ticks have equal length regardless of the direction of time. So the integrated length of a world line is the elapsed time along that world line (for a clock on that world line).
Actually, a straight line is the longest timelike interval. Any timelike dogleg path is shorter. The "triangle inequality" is reversed for timelike intervals.CKH said:The shortest path from the event where the traveler leaves home and the event where he returns home is a straight line so such a worldline takes the least time. The dogleg path is longer in time on a clock following that path because the dogleg is longer.
You may want to read chapter 1 of the reference I posted also: http://preposterousuniverse.com/grnotes/grnotes-one.pdf It gives a good introduction to Minkowski space in a way that prepares you for more in the future.CKH said:OK, now here's where I'm losing it, the equation for proper time, i.e. the time on a clock for a world line.
I don't get why squared values of spatial distance are subtracted from squared values of time to get proper time, if what I said above is true in Minkowski space about path lengths. So I do not have a firm grip on Minkowski space.
That is fine, but then you should drop the topic of non-inertial frames altogether and concentrate on understanding Minkowski spacetime and four-vectors.CKH said:Hill first, mountain later.
Not necessarily. Doing so can lead to mappings that are not diffeomorphisms, even in flat spacetime.CKH said:Because spacetime is flat here, you can also use MCIFs along the traveler's worldline and determine the simultaneous readings on the home clock for the traveler.
Please read the material provided. Until then, you are just not prepared. You may also simply want to stick to inertial frames until you have a firm grasp on four-vectors, the spacetime interval, and so forth.CKH said:A 4-D inertial frame doesn't do anything (it's fixed). As the traveler moves along his world line, he is momentarily at rest in different inertial frames. Thus over a short period of time, the traveler moves from one inertial rest frame to a different one. The traveler's hyperplane of simultaneity changes, so in the travel's simultaneous reading on the home clock changes over that period of time. So as you say, obviously his notion of simultaneity changes and thus the simultaneous reading on the home clock.
In the case where the traveler instantly reverses his velocity, the traveler "jumps" into a new inertial frame, causing the simultaneous reading on the home clock to "jump".
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