Let us consider two clocks,
clock A and clock B. Let us say that A is running correctly, whereas clock B
runs fast and slow. For one second clock B runs fast, that is it runs up two
seconds; then for the next two seconds clock B runs slow and runs only one
second. In a total of three seconds both the clocks run up three seconds.
A mapping of the times shown by the two clocks is given as below.
| Sl No. | Clock A (Seconds) | Clock B (Seconds) |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 0.5 | 1.0 |
| 3 | 1.0 | 2.0 |
| 4 | 2.0 | 2.5 |
| 5 | 3.0 | 3.0 |
| 6 | 3.5 | 4.0 |
| 7 | 4.0 | 5.0 |
| 8 | 5.0 | 5.5 |
| 9 | 6.0 | 6.0 |
With the above data, clock A is
convinced that clock B is running incorrectly.
Now, in the above table, the two clocks are synchronized at sl no 1, 5, 9. Let
us assume we wish to synchronize the clocks at sl no 3. In order to do this we
subtract 1 from all readings of clock A and 2 from all readings of clock B. The
resultant table is as below.
| Sl No. | Clock A (Seconds) | Clock B (Seconds) |
|---|---|---|
| 1 | -1 | -2 |
| 2 | -0.5 | -1.0 |
| 3 | 0.0 | 0.0 |
| 4 | 1.0 | 0.5 |
| 5 | 2.0 | 1.0 |
| 6 | 2.5 | 2.0 |
| 7 | 3.0 | 3.0 |
| 8 | 4.0 | 3.5 |
| 9 | 5.0 | 4.0 |
Now we interchange the columns "Clock A" and "Clock B"
| Sl No. |
New Sl No. |
Clock B (Seconds) | Clock A (Seconds) |
|---|---|---|---|
| 1 | -2 | -1 | |
| 2 | -1.0 | -0.5 | |
| 3 | 1 | 0.0 | 0.0 |
| 4 | 2 | 0.5 | 1.0 |
| 5 | 3 | 1.0 | 2.0 |
| 6 | 4 | 2.0 | 2.5 |
| 7 | 5 | 3.0 | 3.0 |
| 8 | 6 | 3.5 | 4.0 |
| 9 | 7 | 4.0 | 5.0 |
Now, we see that the first
mapping of clock A to clock B is identical to that of the second mapping seen
from clock B to clock A.
Can we say both these clocks are equivalent?
-Chandru - easyrelativity@yahoo.com
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