I did not say SR cannot be tested. In post 103, said: “Your setup only allows for the tick rates wrt to C's clocks to be measured. It is not possible for A to measure B's tick rate wrt to A's clocks.” That does not “falsify SR,” just your set up is inadequate for such a measurement; however, I now extended your set up to show that the “MacM version of SR” contradicts itself, so it is false. (I apologize in advance for being careful, using defined symbolic terms, etc. to specify in detail everything, but you should read anyway and try to find a SPECIFIC point where I have not followed the procedures that your agreed to in post 93.)
To show that your POV is false, I need two clocks, A & B and two inertial frames C & E, which have respective motion at speed Vce along a common “X-axis” line. The origin of E, designated as E(0,0), is already far to the right of the origin of C and the separation between these origins is increasing. All motions I will use are along this same X-axis line. Before proceeding to my proof, either read blue text or skip it and refer back as needed.
]I expended some effort to clearly understand your version of SR and then summarize various cases. One was case (2b), which you quoted in your post 93 as follows:
“When there are two clocks, A & B, with constant relative velocity, both of which were once at rest in a frame, and they have no acceleration now wrt to that frame, then the Time Dilation, TD, of each is calculated ignoring the other as in (2a). This results in their respective “physical time dilations” PTDa & PTDb. If either were to have its time dilation correctly measured by observers, which are at rest wrt to the other, they would also measure the same PTDs. For example, if observers stationary wrt to clock A measure correctly the TDb they find DTb = PTDb.” {This is same as your first post quoted above, but more procisely stated in your terms and my math symbols.}
And you replied in post 93 with the single word: “correct.”
Just to remind others and you, if needed, case (2a) was a single clock now steadily moving wrt a frame that it was once at rest in earlier. We all agreed that such a steadily moving clock is ticking more slowing than other clocks that are stationary in that frame where the moving clock was once at rest and that conventional SR calculates the reduced tick rate, TD, correctly using its speed wrt that frame.)
Here is the current (also called “first”) situation:
Clock A is at rest at frame C’s Cartesian coordinate system origin, but B has steady speed along C’s negative X-axis equal to VBc . I.e. clock B is steadily crossing C’s “Y-axis” grid lines and clock A is stationary at C(0,0). Thus we all agree that it would seem correct to calculate the Time Dilations of clock B, TDb, compared to clock A’s tick rate via the standard SR equation using the speed VBc.
Here is what had happened a little earlier:
Both A & B were stationary at the origin of frame E, I.e. were at E(0,0), so were co-moving away from C(0,0) at speed Vce. (If you forgot what that is, see 2nd paragraph again.) Then both accelerated (in separate rockets, a & b) in the negative X direction (towards C’s origin) exactly the same until their C coordinates was very large and positive, for example the point C(100000, 0). At this point they are co-moving towards every more negative X locations with speeds wrt to C(0,0) of VAc = VBc or since identical, just Vc, which happens to be VBc as given just above in the “current situation,” I.e. Vc = VBc at this point. And of course, their common speed wrt to E(0,0) or Ve = Vc + Vce = VBc + Vce. At this point clock B stops accelerating and just forever coasts, but clock C begins to de-accelerate so that its speed is zero just as it reaches C(0,0), where it remains “forever.” I.e. this is how the “first situation” came to exist.
Now since both A & B started from rest together in frame E, we realize that the procedure (2b) (See blue text) must be used to compute their time dilations separately and then take the difference to find TDb, the Time Dilation of B wrt to the tick rate of clock A which is now stationary in frame C at C(0,0). I.e. we use the standard SR equation to calculate B’s actual “physical tick rate” with respect to the place where it was stationary with clock A, namely E(0,0) or their “common rest frame.”
To do this calculation for clock B, we use B’s speed wrt to that common rest frame. I.e. we use speed of VBe = Ve = VBc + Vce, from bold of prior paragraph. I will designate the results of this calculation, with VBe = Vce + Vce as PTDbe, the Physical Time Dilation of clock B wrt any of frame E’s stationary and synchronized clocks where clock B & A were once mutually at rest at E(0,0).
Likewise I use the std SR equation to calculate PTDae with the speed of clock A wrt to E(0,0) which is simply Vce as C is at C(0,0) forever now. Now in accordance with MacM’s agreed procedure, the Time Dilation of clock B wrt to the tick rate of Clock A, which is stationary at C(0,0) or “TDb” is just the difference in their Physical Time Dilations evaluated by std SR equations when their speeds wrt their “common rest frame” or in this case wrt E(0,0) is the speeds used in the calculations. I.e. TDb = PTDbe – PTDae. (This is a positive time dilation as PTDbe > PTDae since in either frame and certainly in frame E clock B is moving faster than clock A so has greater dilation.)
SUMMARY: Now two “MacM approved” but different methods have been used to calculate TDb. To distinguish them I will rename the one of the “first situation” one, which used speed VBc, TDb1 and call the other TDb2. TDb2 used two different speeds wrt to E(0,0), THEIR COMMON REST FRAME, to separately calculate the Physics Time Dilations of clocks, A & B and then subtracted these two PTDs to get TDb2.
The standard SR equations for PTDs are non-liner so it is highly unlikely that TDb1 = TDb2, but I suspect that for some particular choice of Vce, this is possible. I am too lazy to do all this numerically. (Perhaps Pete will – he sometimes, at least years ago, did enormous amounts of SR calculations.) I will just note that if by rare chance and the first Vce I selected did result in TDb1 =TDB2, then I would just select a different value for Vce. I.e. replace frame E in my story with frame F.
Also to further drive the stake thru MacM’s version of SR, I note that I have not yet told the entire history of clocks A & B. They were both once at mutual rest in frame F, side-by-side at F(0,0), before accelerating and then mutually coming to rest in frame E at E(0,0). Thus frame E is not the only “common rest frame.” It just happens to be their last one. As clocks A & B were NEVER at mutual rest in frame C, it is not a “common rest frame,” yet the calculation of TDb1 was done in accord with MacM’s agreed procedure and by far the simplest to do. (Also in accord with JamesR’s & Pete’s too.) In fact as MacM likes to point out, TDb1 is the one case for which there is “empirical validation” of the procedure in Earth based experiments . Also, there could have been a “common mutual rest” frame G before F, etc. for an infinite set of “MacM approved” and mutually conflicting TDb which use their speed wrt to a “common rest frame.”.