There was a disagreement at our worksite at the end of the day. People my company hired to clean up some leaves were blowing onto a smallish vacant lot across the street, and the people next to the lot were highly annoyed.
The neighbors were dressed in worked clothes; perfectly normal people who were angry. However, they were black. Therefore, the white workers were apparently scared the whole time. One of them told me afterward that when the first neighbor left her house, "I thought she was going to stab me".
I'm sure the worker doesn't believe he is racist. He just thinks every angry black person is likely to stab him.
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Wednesday, December 21, 2016
Thursday, October 27, 2016
You can't tell a relativistic rest frame from measurements
So, after a year-and-a-half of no posts, and something like four years since my last post on relativity crankery, here we go again. Below the fold, I'm going to discuss whether seeing a smaller number on one clock versus another is an indication of which clock really moved in an LR interpretation. To do so, I'll use three scenarios, each with three clocks. The scenarios will be set up so that the clock which is "really at rest" changes between each scenario, but otherwise they will behave identically.
There are three clocks at rest on a planetoid, we will label them A1, B2, and C3. By whatever method you think applies, we'll say these three clocks ar all in the LR rest frame. They are close enough to synchronize without meaningful light-speed delay, and they do so. After synchronizing, they ignore each other, with A1 interacting with clocks B1 and C1, B2 interacting with clocks A2 and C2, and C3 interacting with clocks A3 and B3.
Scenario 1
From the viewpoint of A1, B1 passes by A1 (spacetime point S1) going at .5c; both set themselves to 0. C1 is traveling at .5c directly toward A1 along a path very close to B1 in the opposite direction. After traveling 10 ls (light-seconds) at this speed, B1 and C1 pass each other (spacetime point R1) and synchronize. B1 traveled for 20 seconds, and ticked off 17.32 seconds, between S1 and R1, so click C1, which has clicked through 17.32 seconds in this time, will also read 17.32 because of this synchronization. C1 ticks off 17.32 more seconds on the trip between R1 and the point where it compares itself with A1 (spacetime point Q1), while A1 ticks off another 20. So, C1 reads 34.64 seconds, A1 reads 40. Since B1 and C1 ticks at the same rate, B1 also reads 34.64 seconds, but since B1 is many light-seconds away from Q1, it can not be directly compared to A1 and C1.
From the viewpoint of B1, A1 passes by B1 (S1) going at .5c; both set themselves to 0. C1 is traveling at .8c directly toward B1 along a path very close to A1 in the same direction. A1 travels 8.66 ls in 17.32 seconds before C1 passes B1 (R1). Even though it has only clicked through 10.39 seconds, C1 synchronizes to the 17.32 seconds on B. In another 28.86 seconds (the solution to .8t = .5t + 8.66), C1 passes A1, and they compare times (Q1). In traveling 28.86 seconds at .8c, C1 has ticked off another 17.32 seconds, and reads 34.64. Meanwhile, A1 has traveled at .5c for 46.18 seconds (for a total distance of 23.09 ls), and reads 40. So, C1 reads 34.64 seconds (despite having clicked through only 27.71), A1 reads 40. B1 reads 46.18 seconds, but since B1 is many light-seconds away from Q1, it can not be directly compared to A1 and C1.
From the viewpoint of C1, B1 passes by A1 (S1) at a distance of 23.09ls away, B1 is coming toward C1 at a speed of .8c, and both A1 and B1 set themselves to 0 at S1. A1 is traveling at .5c directly toward C1 along a path very close to B1 in the same direction. B1 travels 23.09 ls in 28.86 seconds before it passes C1 (R1), and C1 then synchronizes to the B1's reading of 17.32 seconds. In another 17.32 seconds, A1 passes C1, and they compare times (Q1). In traveling 46.18 seconds at .5c, A1 has ticked off 40 seconds, and C1 reads 34.64 because of the synchronization at R1. Meanwhile, B1 has traveled at .8c for 46.18 seconds, and reads 27.71 seconds. So, C1 reads 34.64 seconds even though it ticked off 46.18, A1 reads 40. B1 reads 27.71 seconds, but since B1 is many light-seconds away from Q1, it can not be directly compared to A1 and C1.
Scenario 2
From the viewpoint of A2, B2 passes by A2 (spacetime point S2) going at .5c; both set themselves to 0. C2 is traveling at .5c directly toward A2 along a path very close to B2 in the opposite direction. After traveling 10 ls (light-seconds) at this speed, B2 and C2 pass each other (spacetime point R2) and synchronize. B2 traveled for 20 seconds, and ticked off 17.32 seconds, between S2 and R2, so click C2, which has clicked through 17.32 seconds in this time, will also read 17.32 because of this synchronization. C2 ticks off 17.32 more seconds on the trip between R2 and the point where it compares itself with A2 (spacetime point Q2), while A2 ticks off another 20. So, C2 reads 34.64 seconds, A2 reads 40. Since B2 and C2 ticks at the same rate, B2 also reads 34.64 seconds, but since B2 is many light-seconds away from Q2, it can not be directly compared to A2 and C2.
From the viewpoint of B2, A2 passes by B2 (S2) going at .5c; both set themselves to 0. C2 is traveling at .8c directly toward B2 along a path very close to A2 in the same direction. A2 travels 8.66 ls in 17.32 seconds before C2 passes B2 (R2). Even though it has only clicked through 10.39 seconds, C2 synchronizes to the 17.32 seconds on B. In another 28.86 seconds (the solution to .8t = .5t + 8.66), C2 passes A2, and they compare times (Q2). In traveling 28.86 seconds at .8c, C2 has ticked off another 17.32 seconds, and reads 34.64. Meanwhile, A2 has traveled at .5c for 46.18 seconds (for a total distance of 23.09 ls), and reads 40. So, C2 reads 34.64 seconds (despite having clicked through only 27.71), A2 reads 40. B2 reads 46.18 seconds, but since B2 is many light-seconds away from Q2, it can not be directly compared to A2 and C2.
From the viewpoint of C2, B2 passes by A2 (S2) at a distance of 23.09ls away, B2 is coming toward C2 at a speed of .8c, and both A2 and B2 set themselves to 0 at S2. A2 is traveling at .5c directly toward C2 along a path very close to B2 in the same direction. B2 travels 23.09 ls in 28.86 seconds before it passes C2 (R2), and C2 then synchronizes to the B2's reading of 17.32 seconds. In another 17.32 seconds, A2 passes C2, and they compare times (Q2). In traveling 46.18 seconds at .5c, A2 has ticked off 40 seconds, and C2 reads 34.64 because of the synchronization at R2. Meanwhile, B2 has traveled at .8c for 46.18 seconds, and reads 27.71 seconds. So, C2 reads 34.64 seconds even though it ticked off 46.18, A2 reads 40. B2 reads 27.71 seconds, but since B2 is many light-seconds away from Q2, it can not be directly compared to A2 and C2.
Scenario 3
From the viewpoint of A3, B3 passes by A3 (spacetime point S3) going at .5c; both set themselves to 0. C3 is traveling at .5c directly toward A3 along a path very close to B3 in the opposite direction. After traveling 10 ls (light-seconds) at this speed, B3 and C3 pass each other (spacetime point R3) and synchronize. B3 traveled for 20 seconds, and ticked off 17.32 seconds, between S3 and R3, so click C3, which has clicked through 17.32 seconds in this time, will also read 17.32 because of this synchronization. C3 ticks off 17.32 more seconds on the trip between R3 and the point where it compares itself with A3 (spacetime point Q3), while A3 ticks off another 20. So, C3 reads 34.64 seconds, A3 reads 40. Since B3 and C3 ticks at the same rate, B3 also reads 34.64 seconds, but since B3 is many light-seconds away from Q3, it can not be directly compared to A3 and C3.
From the viewpoint of B3, A3 passes by B3 (S3) going at .5c; both set themselves to 0. C3 is traveling at .8c directly toward B3 along a path very close to A3 in the same direction. A3 travels 8.66 ls in 17.32 seconds before C3 passes B3 (R3). Even though it has only clicked through 10.39 seconds, C3 synchronizes to the 17.32 seconds on B. In another 28.86 seconds (the solution to .8t = .5t + 8.66), C3 passes A3, and they compare times (Q3). In traveling 28.86 seconds at .8c, C3 has ticked off another 17.32 seconds, and reads 34.64. Meanwhile, A3 has traveled at .5c for 46.18 seconds (for a total distance of 23.09 ls), and reads 40. So, C3 reads 34.64 seconds (despite having clicked through only 27.71), A3 reads 40. B3 reads 46.18 seconds, but since B3 is many light-seconds away from Q3, it can not be directly compared to A3 and C3.
From the viewpoint of C3, B3 passes by A3 (S3) at a distance of 23.09ls away, B3 is coming toward C3 at a speed of .8c, and both A3 and B3 set themselves to 0 at S3. A3 is traveling at .5c directly toward C3 along a path very close to B3 in the same direction. B3 travels 23.09 ls in 28.86 seconds before it passes C3 (R3), and C3 then synchronizes to the B3's reading of 17.32 seconds. In another 17.32 seconds, A3 passes C3, and they compare times (Q3). In traveling 46.18 seconds at .5c, A3 has ticked off 40 seconds, and C3 reads 34.64 because of the synchronization at R3. Meanwhile, B3 has traveled at .8c for 46.18 seconds, and reads 27.71 seconds. So, C3 reads 34.64 seconds even though it ticked off 46.18, A3 reads 40. B3 reads 27.71 seconds, but since B3 is many light-seconds away from Q3, it can not be directly compared to A3 and C3.
Conclusion
In scenario 1, where A1 was at rest, A1 read 40 seconds while C1 read 34.64 at Q1. In scenario 2, where B2 was at rest, A2 read 40 seconds while C2 read 34.64 at Q2. In scenario 3, where C3 was at rest, A3 read 40 seconds while C3 read 34.64 at Q3. Among the three clocks, it doesn't matter which clock you think is at rest. They always read the same. This is how SR (and LR for that matter)works. To do comparisons, you always have to pick a single rest frame for your calculations, but SR allows you to choose any single inertial frame, by any standard or by a purely arbitrary choice.
LR insists that one particular frame is the correct frame, and different versions of LR choose different frames. Despite choosing frames differently, every version of LR works, because every version of LR is actually SR (a working theory) with a predetermined method for choosing a frame.
Read more!
There are three clocks at rest on a planetoid, we will label them A1, B2, and C3. By whatever method you think applies, we'll say these three clocks ar all in the LR rest frame. They are close enough to synchronize without meaningful light-speed delay, and they do so. After synchronizing, they ignore each other, with A1 interacting with clocks B1 and C1, B2 interacting with clocks A2 and C2, and C3 interacting with clocks A3 and B3.
Scenario 1
From the viewpoint of A1, B1 passes by A1 (spacetime point S1) going at .5c; both set themselves to 0. C1 is traveling at .5c directly toward A1 along a path very close to B1 in the opposite direction. After traveling 10 ls (light-seconds) at this speed, B1 and C1 pass each other (spacetime point R1) and synchronize. B1 traveled for 20 seconds, and ticked off 17.32 seconds, between S1 and R1, so click C1, which has clicked through 17.32 seconds in this time, will also read 17.32 because of this synchronization. C1 ticks off 17.32 more seconds on the trip between R1 and the point where it compares itself with A1 (spacetime point Q1), while A1 ticks off another 20. So, C1 reads 34.64 seconds, A1 reads 40. Since B1 and C1 ticks at the same rate, B1 also reads 34.64 seconds, but since B1 is many light-seconds away from Q1, it can not be directly compared to A1 and C1.
From the viewpoint of B1, A1 passes by B1 (S1) going at .5c; both set themselves to 0. C1 is traveling at .8c directly toward B1 along a path very close to A1 in the same direction. A1 travels 8.66 ls in 17.32 seconds before C1 passes B1 (R1). Even though it has only clicked through 10.39 seconds, C1 synchronizes to the 17.32 seconds on B. In another 28.86 seconds (the solution to .8t = .5t + 8.66), C1 passes A1, and they compare times (Q1). In traveling 28.86 seconds at .8c, C1 has ticked off another 17.32 seconds, and reads 34.64. Meanwhile, A1 has traveled at .5c for 46.18 seconds (for a total distance of 23.09 ls), and reads 40. So, C1 reads 34.64 seconds (despite having clicked through only 27.71), A1 reads 40. B1 reads 46.18 seconds, but since B1 is many light-seconds away from Q1, it can not be directly compared to A1 and C1.
From the viewpoint of C1, B1 passes by A1 (S1) at a distance of 23.09ls away, B1 is coming toward C1 at a speed of .8c, and both A1 and B1 set themselves to 0 at S1. A1 is traveling at .5c directly toward C1 along a path very close to B1 in the same direction. B1 travels 23.09 ls in 28.86 seconds before it passes C1 (R1), and C1 then synchronizes to the B1's reading of 17.32 seconds. In another 17.32 seconds, A1 passes C1, and they compare times (Q1). In traveling 46.18 seconds at .5c, A1 has ticked off 40 seconds, and C1 reads 34.64 because of the synchronization at R1. Meanwhile, B1 has traveled at .8c for 46.18 seconds, and reads 27.71 seconds. So, C1 reads 34.64 seconds even though it ticked off 46.18, A1 reads 40. B1 reads 27.71 seconds, but since B1 is many light-seconds away from Q1, it can not be directly compared to A1 and C1.
Scenario 2
From the viewpoint of A2, B2 passes by A2 (spacetime point S2) going at .5c; both set themselves to 0. C2 is traveling at .5c directly toward A2 along a path very close to B2 in the opposite direction. After traveling 10 ls (light-seconds) at this speed, B2 and C2 pass each other (spacetime point R2) and synchronize. B2 traveled for 20 seconds, and ticked off 17.32 seconds, between S2 and R2, so click C2, which has clicked through 17.32 seconds in this time, will also read 17.32 because of this synchronization. C2 ticks off 17.32 more seconds on the trip between R2 and the point where it compares itself with A2 (spacetime point Q2), while A2 ticks off another 20. So, C2 reads 34.64 seconds, A2 reads 40. Since B2 and C2 ticks at the same rate, B2 also reads 34.64 seconds, but since B2 is many light-seconds away from Q2, it can not be directly compared to A2 and C2.
From the viewpoint of B2, A2 passes by B2 (S2) going at .5c; both set themselves to 0. C2 is traveling at .8c directly toward B2 along a path very close to A2 in the same direction. A2 travels 8.66 ls in 17.32 seconds before C2 passes B2 (R2). Even though it has only clicked through 10.39 seconds, C2 synchronizes to the 17.32 seconds on B. In another 28.86 seconds (the solution to .8t = .5t + 8.66), C2 passes A2, and they compare times (Q2). In traveling 28.86 seconds at .8c, C2 has ticked off another 17.32 seconds, and reads 34.64. Meanwhile, A2 has traveled at .5c for 46.18 seconds (for a total distance of 23.09 ls), and reads 40. So, C2 reads 34.64 seconds (despite having clicked through only 27.71), A2 reads 40. B2 reads 46.18 seconds, but since B2 is many light-seconds away from Q2, it can not be directly compared to A2 and C2.
From the viewpoint of C2, B2 passes by A2 (S2) at a distance of 23.09ls away, B2 is coming toward C2 at a speed of .8c, and both A2 and B2 set themselves to 0 at S2. A2 is traveling at .5c directly toward C2 along a path very close to B2 in the same direction. B2 travels 23.09 ls in 28.86 seconds before it passes C2 (R2), and C2 then synchronizes to the B2's reading of 17.32 seconds. In another 17.32 seconds, A2 passes C2, and they compare times (Q2). In traveling 46.18 seconds at .5c, A2 has ticked off 40 seconds, and C2 reads 34.64 because of the synchronization at R2. Meanwhile, B2 has traveled at .8c for 46.18 seconds, and reads 27.71 seconds. So, C2 reads 34.64 seconds even though it ticked off 46.18, A2 reads 40. B2 reads 27.71 seconds, but since B2 is many light-seconds away from Q2, it can not be directly compared to A2 and C2.
Scenario 3
From the viewpoint of A3, B3 passes by A3 (spacetime point S3) going at .5c; both set themselves to 0. C3 is traveling at .5c directly toward A3 along a path very close to B3 in the opposite direction. After traveling 10 ls (light-seconds) at this speed, B3 and C3 pass each other (spacetime point R3) and synchronize. B3 traveled for 20 seconds, and ticked off 17.32 seconds, between S3 and R3, so click C3, which has clicked through 17.32 seconds in this time, will also read 17.32 because of this synchronization. C3 ticks off 17.32 more seconds on the trip between R3 and the point where it compares itself with A3 (spacetime point Q3), while A3 ticks off another 20. So, C3 reads 34.64 seconds, A3 reads 40. Since B3 and C3 ticks at the same rate, B3 also reads 34.64 seconds, but since B3 is many light-seconds away from Q3, it can not be directly compared to A3 and C3.
From the viewpoint of B3, A3 passes by B3 (S3) going at .5c; both set themselves to 0. C3 is traveling at .8c directly toward B3 along a path very close to A3 in the same direction. A3 travels 8.66 ls in 17.32 seconds before C3 passes B3 (R3). Even though it has only clicked through 10.39 seconds, C3 synchronizes to the 17.32 seconds on B. In another 28.86 seconds (the solution to .8t = .5t + 8.66), C3 passes A3, and they compare times (Q3). In traveling 28.86 seconds at .8c, C3 has ticked off another 17.32 seconds, and reads 34.64. Meanwhile, A3 has traveled at .5c for 46.18 seconds (for a total distance of 23.09 ls), and reads 40. So, C3 reads 34.64 seconds (despite having clicked through only 27.71), A3 reads 40. B3 reads 46.18 seconds, but since B3 is many light-seconds away from Q3, it can not be directly compared to A3 and C3.
From the viewpoint of C3, B3 passes by A3 (S3) at a distance of 23.09ls away, B3 is coming toward C3 at a speed of .8c, and both A3 and B3 set themselves to 0 at S3. A3 is traveling at .5c directly toward C3 along a path very close to B3 in the same direction. B3 travels 23.09 ls in 28.86 seconds before it passes C3 (R3), and C3 then synchronizes to the B3's reading of 17.32 seconds. In another 17.32 seconds, A3 passes C3, and they compare times (Q3). In traveling 46.18 seconds at .5c, A3 has ticked off 40 seconds, and C3 reads 34.64 because of the synchronization at R3. Meanwhile, B3 has traveled at .8c for 46.18 seconds, and reads 27.71 seconds. So, C3 reads 34.64 seconds even though it ticked off 46.18, A3 reads 40. B3 reads 27.71 seconds, but since B3 is many light-seconds away from Q3, it can not be directly compared to A3 and C3.
Conclusion
In scenario 1, where A1 was at rest, A1 read 40 seconds while C1 read 34.64 at Q1. In scenario 2, where B2 was at rest, A2 read 40 seconds while C2 read 34.64 at Q2. In scenario 3, where C3 was at rest, A3 read 40 seconds while C3 read 34.64 at Q3. Among the three clocks, it doesn't matter which clock you think is at rest. They always read the same. This is how SR (and LR for that matter)works. To do comparisons, you always have to pick a single rest frame for your calculations, but SR allows you to choose any single inertial frame, by any standard or by a purely arbitrary choice.
LR insists that one particular frame is the correct frame, and different versions of LR choose different frames. Despite choosing frames differently, every version of LR works, because every version of LR is actually SR (a working theory) with a predetermined method for choosing a frame.
Read more!
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