Vorde said:
The relativistic velocity formula comes very directly and very rationally from a postulate that violates common sense.
Vorde said:
The first postulate I'd completely agree with you, the second seems to me to be completely against common sense.
The velocity addition formula does not depend on the second postulate which is a requirement for Einstein's theory of Special Relativity. Even prior to him proposing that second postulate, the velocity addition formula can be derived just from the first postulate which you agreed makes common sense to you and the common belief at the time in an absolute ether rest frame, only in which light propagates at c.
We can illustrate how this works with an example involving three inertial observers who we show as colocated at the origin of a spacetime diagram depicting a single ether rest frame. We refer to the observers by their color. The black observer is stationary in the ether frame. The green observer is traveling to his right at 0.6c and the blue observer is traveling to his left at 0.6c.
Now we are going to see how each observer measures the speeds of each of the other observers using radar signals which will travel at c relative to the ether frame and not relative to each observer. However, because of the Principle of Relativity, we insist that when one observer measures the relative speed another observer, the other observer will get the same answer when measuring the relative speed of the first observer. And for that to happen we must come to the same conclusions that scientists came to prior to Einstein which is that clocks run slower when traveling through the ether. So on our spacetime diagram, we show dots for each observer that mark out one-year intervals of time.
Now a little review about radar signals. An observer aims a radar signal at another observer noting the time on his clock at which he sent it. Eventually that signal hits the other observer and reflects back to the first observer. When the reflected radar signal gets back to the first observer, he notes this second time. Now with these two time measurements, the observer does a little calculation. He adds the two times together and divides by two. This gives him the time at which he will say the "measurement" applies. He also subtracts the two times and divides that by two. This gives him a measurement of the distance (in compatible units where c=1). So he gets a distance applied at a particular time. To calculate the speed of the second observer, he uses that fact that they started out colocated so all he has to do is divide the distance by the applied time.
So let's see how this works first for the black observer measuring the speed of the green observer:
When the black observer's clock reads 1, he aims his radar gun at the green observer and he receives the echo back when his clock reads 4. The radar signals are shown in red and always propagate along 45-degree angles with respect to the ether frame. Black calculates (4+1)/2=2.5 as the time at which the measurement was made and he calculates (4-1)/2=1.5 as the distance that the green observer was away from him at his time of 2.5 and we can see that since he is at rest with respect to the ether, his measurement matches what the ether frame portrays as shown by the yellow line. He can also calculate the speed of the green observer with respect to him as 1.5/2.5=0.6c. Although I don't show it, he can make the same measurement of the blue observer and will get the same results since blue is symmetrically a mirror image of green.
Now let's see how green will measure the speed of black:
We see that the exact same explanation used by black to measure green applies for green measuring black. In fact, since we know that they must get the same answer, that is the reason why scientists concluded that green's clock must be slowed down while he is traveling through the ether. Note that the distance of 1.5 for the yellow line does not correspond to any such distance in the ether frame (but the green observer can't tell that).
Although I don't show it, we could also show that blue would get the same measurement of black's speed since green is symmetrically a mirror image of blue.
Now we'll take a look at how green would measure the speed of blue using exactly the same radar method as before:
Once again, green fires his radar signal at his clock time of 1 but this time, since he is waiting for the reflection from the blue observer, it takes until his clock time of 16 before he gets it. So his calculation of the time the measurement applies is (16+1)/2=8.5 and for the distance is (16-1)/2=7.5. So he concludes that at his time of 8.5, the distance to the blue observer was 7.5 and the speed of the blue observer is 7.5/8.5=
0.882c. Note that the length of the yellow line does not correspond to anything in the ether frame. And again because blue and green are mirror images, blue will get the same measurement of green's speed.
But now let's use the relativistic addition formula to see what speed we get when we add 0.6+0.6:
(0.6+0.6)/(1+0.6*0.6) = 1.2/(1+0.36) = 1.2/1.36 =
0.882c
This is the same answer we got using the ether frame and the radar measurement. So you see, it is not necessary to apply Einstein's second postulate to demonstrate the validity or the common sense of the relativistic velocity addition formula.