Is the Light Clock Calibration the Key to Understanding Relativity?

In summary: Hi PeterDonis,Thanks for your reply. We are agreed on the light clock. The fact that it takes light longer to travel a longer path is not surprising. Likewise if you calibrate a clock to that path's length its not surprising it takes longer to tick over. But that proves nothing regarding time dilation.If instead you have a clock that ticks over each time light travels one metre, and that is the same in both frames, then the clock ticks at the same rate in both frames. Note that in the stationary frame the clock already ticks over each time light travels one metre, assuming the mirrors are one metre
  • #36
DaleSpam said:
Calculate how long it takes for a horizontal clock to tick given its length L and speed v.

DaleSpam said:
Calculate how long it takes for a horizontal clock to tick given its length L and speed v.

Time is distance over c. And distance in this case is the length of the clock plus the distance it has moved. Or for the return leg, less the distance moved.
So if the distance between mirrors is, say, one metre, and the train/clock advances 0.1 metres, the distance the light travels is 1.1 metres, divided by c gives the clock tick over time. Or 0.9metres divided by c for the return leg. I don't see what this proves other than the there and back leg is 2 metres, which is the same as the stationary frame
 
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  • #37
DAC said:
Time is distance over c. And distance in this case is the length of the clock plus the distance it has moved. Or for the return leg, less the distance moved.
Write it out algebraically, in terms of ##L## and ##v##. What does ##L## have to be in order for the horizontal clock to tick in sync with a vertical clock of length ##L_0## ?
DAC said:
So if the distance between mirrors is, say, one metre, and the train/clock advances 0.1 metres, the distance the light travels is 1.1 metres, divided by c gives the clock tick over time. Or 0.9metres divided by c for the return leg. I don't see what this proves other than the there and back leg is 2 metres, which is the same as the stationary frame
Is that so? Are you sure about it? How did you determine that? What speed would lead to a 1.1 m trip out and a .9 m trip back?
 
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  • #38
DaleSpam said:
Write it out algebraically, in terms of ##L## and ##v##. What does ##L## have to be in order for the horizontal clock to tick in sync with a vertical clock of length ##L_0## ?Is that so? Are you sure about it? How did you determine that? What speed would lead to a 1.1 m trip out and a .9 m trip back?
DaleSpam said:
Write it out algebraically, in terms of ##L## and ##v##. What does ##L## have to be in order for the horizontal clock to tick in sync with a vertical clock of length ##L_0## ?Is that so? Are you sure about it? How did you determine that? What speed would lead to a 1.1 m trip out and a .9 m trip back?
OK you win, anyway don't you guys provide answers rather than questions. Keep up the good work.
 
  • #39
And on that note I think we can close the thread
 

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