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alan123hk
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I believe this does not belong to the homework category. I hope I won't be mistaken.
I am reading a book to self-study special relativity, the following is an example mentioned in the book.
When clock C' and clock C1 meet at times t'=t1=0, both clocks read zero. The Observer in reference frame S determines that the readings of the C1 clock and the C2 clock are the same because they are precisely aligned with each other. However, for the observer in the reference frame S', the two clocks C1 and C2 are not synchronized, when the reading of C1 is 0, the reading of C2 is not zero. Assume that the moving speed of the reference frame S' in the positive x direction relative to the reference frame S is v. The distance between C1 and C2 measured by the observer in the reference frame S is L. So when C' and C1 meet at time t'= t1=0, how does the observer in the reference frame S' think of the reading (Δ) of the clock C2 ?
The answer mentioned in the book is to use the Lorentz transformation formula, as shown below.
$$ t' = \gamma \left( t-\frac {v} {c^2} x \right) ~~~~~\Rightarrow~~~~~t'=0= \gamma \left( \Delta-\frac {v} {c^2} L \right)~~~~~\Rightarrow~~~~~\Delta=\frac {v} {c^2}L $$
But unfortunately I don't understand the above reasoning logic. Is the above answer correct ?
I may be wrong, but I feel that the following way is more reasonable.
$$ t' = \gamma \left( t-\frac {v} {c^2} x \right) ~~~~~\Rightarrow~~~~~t'=\Delta= \gamma \left(0-\frac {v} {c^2} L \right)~~~~~\Rightarrow~~~~~\Delta=-\gamma\frac {v} {c^2}L $$
This seems to be really wrong, why dose the answer have a minus sign.
What is wrong with my derivation process above ?
Thanks for helping.
I am reading a book to self-study special relativity, the following is an example mentioned in the book.
When clock C' and clock C1 meet at times t'=t1=0, both clocks read zero. The Observer in reference frame S determines that the readings of the C1 clock and the C2 clock are the same because they are precisely aligned with each other. However, for the observer in the reference frame S', the two clocks C1 and C2 are not synchronized, when the reading of C1 is 0, the reading of C2 is not zero. Assume that the moving speed of the reference frame S' in the positive x direction relative to the reference frame S is v. The distance between C1 and C2 measured by the observer in the reference frame S is L. So when C' and C1 meet at time t'= t1=0, how does the observer in the reference frame S' think of the reading (Δ) of the clock C2 ?
$$ t' = \gamma \left( t-\frac {v} {c^2} x \right) ~~~~~\Rightarrow~~~~~t'=0= \gamma \left( \Delta-\frac {v} {c^2} L \right)~~~~~\Rightarrow~~~~~\Delta=\frac {v} {c^2}L $$
But unfortunately I don't understand the above reasoning logic. Is the above answer correct ?
I may be wrong, but I feel that the following way is more reasonable.
$$ t' = \gamma \left( t-\frac {v} {c^2} x \right) ~~~~~\Rightarrow~~~~~t'=\Delta= \gamma \left(0-\frac {v} {c^2} L \right)~~~~~\Rightarrow~~~~~\Delta=-\gamma\frac {v} {c^2}L $$
This seems to be really wrong, why dose the answer have a minus sign.
What is wrong with my derivation process above ?
Thanks for helping.
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