Einstein hates the merry-go-round. Any time he stands by, two little
horses, one after the other, maliciously look at the watch on his
wrist as if they expect to see something improper there. Einstein
knows what it is. If two Einsteins were standing by, any clock on the
merry-go-round would run slow by a factor of 1/gamma as it covered the
distance between the two Einsteins. And now the horses expect
Einstein's watch also to run slow as it covers the distance between
them. This expectation, based on the principle of relativity, is fatal
for Einstein. In order for his theory to survive, his watch must run
faster than the horses' clocks. Time contraction rather than time
dilation is necessary in this particular case.

Now the problem. Einstein's watch passes the first horse's clock and
both readings are set to zero. Then Einstein's watch passes the second
horse's clock and the readings are respectively t and t'. Given the
fact that the horses' clocks are not synchronized, could we still
consider a component of t', denoted by t1', such that t1'=gamma*t ?

Pentcho Valev

"Pentcho Valev" wrote in message
m...
Einstein hates the merry-go-round. Any time he stands by, two little
horses, one after the other, maliciously look at the watch on his
wrist as if they expect to see something improper there. Einstein
knows what it is. If two Einsteins were standing by, any clock on the
merry-go-round would run slow by a factor of 1/gamma as it covered the
distance between the two Einsteins.

Why two? Just one Einstein and one full round suffices.

And now the horses expect
Einstein's watch also to run slow as it covers the distance between
them.

The two horses are mistaken. But that's normal, horses are too stupid. :-)

This expectation, based on the principle of relativity, is fatal
for Einstein.

Please formulate what you think is the PoR. Anyway, it's fatal for the
horses' attempt to be confused with scientists.

In order for his theory to survive, his watch must run
faster than the horses' clocks. Time contraction rather than time
dilation is necessary in this particular case.

No, the Horse Theory didn't make it!

Now the problem. Einstein's watch passes the first horse's clock and
both readings are set to zero. Then Einstein's watch passes the second
horse's clock and the readings are respectively t and t'. Given the
fact that the horses' clocks are not synchronized, could we still
consider a component of t', denoted by t1', such that t1'=gamma*t ?

Assuming that the two merry-go-round clocks run at the same rate but
unsynchronized:
t'- C = t1' = t/gamma
with C the advance of the second clock on the first clock as determined in
the inertial frame.

Harald