yogi said:
Jesse - Hurkyl Let me address part of what you have both said - it has to do with the propriety of R as a reference frame - here i will again direct you to Einstein's statement (1905 Part 4) - that it doesn't make any difference what path is followed - the equation for the time difference only concerns the relative velocity v between the clock that remains at rest after sync and the clock that is put in motion - I interpret this to mean that the moving clock (in our case R) could zig zag all over the place as long as its heading velocity remains v - if we believe Einstein's statement then should we not be able to disregard any radial acceleration that R incurs during the round trip flight
Are you referring to this part of section 4 of the
1905 paper?
It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.
If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be 1/2*tv^2/c^2 second slow. Thence we conclude that a balance-clock7 at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.
If you're saying that this justifies treating the circular-moving clock as having its own rest frame, you misunderstand what he was saying there. Einstein was saying that time dilation
as seen in an inertial reference frame is dependent only on velocity in that frame. So if something moves on a curved path at constant speed as seen in an inertial frame, you can calculate the time elapsed just by multiplying the coordinate time in that frame by \sqrt{1 - v^2/c^2}. But Einstein is not claiming there that the object moving on the curved path itself has its own rest frame where the usual rules of relativity apply; these rules only apply in frames "in uniform translatory motion", ie no acceleration (and changing direction always involves acceleration).
yogi said:
Now if we follow hurkyl's logic that we will see R running at half the assigned C rate if we continue to use timing experiments between R and C at all points of the path, and vice versa, can we not conclude that the spaceship is a good inertial frame - it shouldn't really make a difference in the outcome if the path were comprised of many straight segments that approximate the circle and the readings were always taken at the midpoint of the seqments
But making the path into a series of straight line segments wouldn't help either, because R would not have a
single inertial rest frame throughout the journey, it would have a series of different inertial rest frames. This is just like the twin paradox, where for convenience you usually assume that the traveling twin's path is just made up of two straight line segments.
yogi said:
Now I do not see that this experiment leads to the notion of a universal time - but if both R and C run at the same speed at all times as a necessary consequence that they are not being acted upon by any forces or factors that would change the rate at which they run - then we do have a condition where they run at different rates as judged by each other - but unlike apparent time dilation - R judges C as fast and C judges R as slow
Usually in the context of relativity, when people say things like "A judges B to be running slow" they just mean that in the inertial frame where A is at rest, B is running slow. But again, R does not have a single inertial rest frame throughout the journey. It is certainly true that in any frame where R is
momentarily at rest, then in that frame C will be running slower than R
at that moment, but then that frame will see R's velocity change throughout the journey and at some point it will have to see R running slower than C. And again, there is no physical reason to prefer one inertial frame's description of the whole situation to another's, so you cannot say that R was running slower than C at every moment, you can only say that R's
average rate of ticking over an entire orbit was slower than C's (this will be true in every inertial frame).
yogi said:
Again are we getting the right numerical answer for the wrong reason.
WHY do you think it is wrong? You keep objecting to the idea that every frame is equally valid but you never spell out any reasons, and you seem to have agreed in your last post that as long as all the laws of physics have equations with the mathematical property of lorentz-invariance, this insures that every inertial frame will see the same laws of physics and will make the same predictions about all physical questions like what two clocks read at the moment they meet. If you agree with this, it seems that you must agree that no inertial frame's analysis is
physically preferred over any other's, so all you're left with is some kind of aesthetic preference for one frame's description over all others, no different than someone who prefers cartesian coordinates to polar coordinates.