Simple question about inertial frames of reference

[LBrandt]

You are correct, I misread your statement. Sorry. But in that case, I'm a little confused. I was under the impression that a stationary observer would be considered to be in an INERTIAL frame of reference. I realize that the Earth is moving, but at it is moving at constant velocity. My original understanding was that an inertial frame of reference would be one in which there is either no movement or movement at constant velocity, whereas a non-inertial frame of reference would be one in which there is either acceleration, deceleration or change of direction.

Louis

 

[JesseM]

You are correct, I misread your statement. Sorry. But in that case, I'm a little confused. I was under the impression that a stationary observer would be considered to be in an INERTIAL frame of reference. I realize that the Earth is moving, but at it is moving at constant velocity. My original understanding was that an inertial frame of reference would be one in which there is either no movement or movement at constant velocity, whereas a non-inertial frame of reference would be one in which there is either acceleration, deceleration or change of direction.

The problem is that statements like "no movement or movement at constant velocity" are meaningless unless you specify a coordinate system in which to define "movement" in terms of change in position coordinate / change in time coordinate...no matter what object you are considering, you can always define some coordinate system where the object's coordinate position as a function of coordinate time is either unchanging or changing at a constant rate. An inertial observer is only one who is at rest or moving at constant speed relative to inertial coordinate systems, so that definition is of little use to you unless you've already come up with a definition of what it means for a coordinate system to qualify as "inertial". In Newtonian physics gravity is treated as a force so it is possible for someone to feel G-forces and still be at rest in an inertial frame, but in general relativity things work differently. In GR gravity is instead understood in terms of curved spacetime, and no coordinate system covering a large region of curved spacetime (say, a coordinate system covering an entire orbit of the Earth) is considered "inertial", but according to the equivalence principle there is the concept of a "local inertial frame" in a region of spacetime small enough that the effects of spacetime curvature (tidal forces) can be considered negligible and the laws of physics look the same as they do in flat SR spacetime, and such a local inertial frame is always the rest frame of an observer in free-fall who feels no G-forces.

 

[LBrandt]

Thanks again for the quick reply. I certainly learn more from this forum than I contribute, and that's because I'm still basically a novice when it comes to special and general relativity. However, I thought that one of the principles that is usually mentioned when discussing the twin paradox is the notion that the stay-at-home twin represents the inertial frame member of the paradox and the traveling twin (the accelerating one), the non-inertial member.
Again, I'm just learning.

Louis

 

[JesseM]

However, I thought that one of the principles that is usually mentioned when discussing the twin paradox is the notion that the stay-at-home twin represents the inertial frame member of the paradox and the traveling twin (the accelerating one), the non-inertial member.

The twin paradox is usually understood to be a problem occurring in the flat spacetime of special relativity, with no gravity. You could have a variant of the twin paradox in curved spacetime, but then no coordinate system covering the entire path of each twin from beginning to end would be an inertial one. In general relativity, even if one twin is moving inertially in a local sense (i.e. at each point on his worldline, he is at rest in a locally inertial frame in the neighborhood of that point, and thus is experiencing no G-forces) that doesn't necessarily mean he'll have aged more than a different twin who moves non-inertially between their two meetings.

 

[LBrandt]

I'll accept that. Thanks.

 

[yuiop]

In GR gravity is instead understood in terms of curved spacetime, and no coordinate system covering a large region of curved spacetime (say, a coordinate system covering an entire orbit of the Earth) is considered "inertial", but according to the equivalence principle there is the concept of a "local inertial frame" in a region of spacetime small enough that the effects of spacetime curvature (tidal forces) can be considered negligible and the laws of physics look the same as they do in flat SR spacetime, and such a local inertial frame is always the rest frame of an observer in free-fall who feels no G-forces.

Louis, please avert your eyes, as what I am going to say next might be confusing at this stage!

I just thought I would clarify that even an accelerating observer experiencing proper acceleration can consider themselves to be in flat SR spacetime for a sufficiently local region and the time dilation of nearby passing clocks traveling at relative velocities that are near the speed of light will be just the regular SR time dilation factor. If you are still reading Louis (and I know you are ) the important thing is that measurements by an observer in a gravitational field of distant events require more than just SR to analyse. In the twins example, we tend to ignore the fact that the stay at home twin on the Earth is not really inertial and treat the effect of the Earth's gravitational field on time dilation as negligible relative to the time dilation due to the high speeds involved.

 

[JesseM]

I just thought I would clarify that even an accelerating observer experiencing proper acceleration can consider themselves to be in flat SR spacetime for a sufficiently local region and the time dilation of nearby passing clocks traveling at relative velocities that are near the speed of light will be just the regular SR time dilation factor.

Wouldn't the accelerating observer be locally equivalent to an accelerating observer in flat SR spacetime, not an inertial one, though? Are you saying that in some sense even an accelerating observer in SR can make local measurements of passing clocks and find that they are running slow by the normal SR time dilation factor seen in inertial frames? If so, it seems to me this would depend on exactly how you define the local measurements of an accelerating observer (what kind of ruler they use to measure distances, how they synchronize clocks, etc.)

 

[yuiop]

Are you saying that in some sense even an accelerating observer in SR can make local measurements of passing clocks and find that they are running slow by the normal SR time dilation factor seen in inertial frames?

Yes.

If so, it seems to me this would depend on exactly how you define the local measurements of an accelerating observer (what kind of ruler they use to measure distances, how they synchronize clocks, etc.)

There are a number of ways to do this. One is to assume that a sufficiently nearby clock is running at approximately the same rate and ignore the slight error. Another is to speed up or slow down a nearby clock so that it remains synchronised and allow for the adjustment after carrying out the experiment. Finally this last method should give fairly accurate results.

Measurement method in an accelerating frame using a single clock:

Place one mirror a short distance above the clock at and measure its radar distance using the single clock. (The clock and mirror are both co accelerating and at rest relative to each other). Let's say the the radar distance is one pico-light second. Place another (co-accelerating) mirror below the clock and adjust its distance so that the radar distance is also one pico-light second. The distance to the two mirrors can be calibrated precisely using an interferometer. Arrange things so that when a test particle passes the upper mirror, a light signal is sent to the central clock to start it and when the particle passes the lower mirror a second light signal is sent to stop the central clock. The elapsed time on the central clock is a good representation of the time it took the particle to transfer from the first mirror to the second mirror because the stop/start signal delays of one pico second cancel each other out. You now have a method to measure the velocity of passing particles in an accelerating reference frame. If the velocity is measured to be v, then the time dilation factor should just be sqrt(1-v^2) to a reasonable degree of accuracy.