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So we can't use more than one clock to measure light speed, but what about using just one? A beam is split in two opposite directions, one toward the sensor, and the other toward a mirror which reflects it back toward the sensor.

             Time
|        <3y-a-  ^  -x-b> |
|        <3y-b-  ^  <y-b- |
           Distance
|        3X      ^  X     |

The distance between the mirror and source is X. The distance between the source and sensor is 3X. So the distance between the mirror and sensor is 4X. Beam A travels the 3x distance in 3y time, Beam B travels the X distance in x time, then travels the X distance back past the source in y time, then travels to the sensor in 3y time.

  • $x$ represents travel time away from the sensor
  • $y$ represents travel time toward the sensor
  • $x + y$ is the round trip between event and mirror
  • $3y$ is the trip from event to sensor
  • $a = 3y $ (Total travel time of beam A is 3y)
  • $b = 4y + x $ (Total travel time of beam B is 4y + x)
  • $c_y = \frac {a}{3} $
  • $c_x = b - \frac {4}{3} a$
  • Multiply time A by 4/3 and subtract it from time B. You now have the speed of light in each direction.

In theory, any difference should prove that you are moving relative to whatever determines the absolute speed of light.

Qmechanic
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Arlen Beiler
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2 Answers2

5

Close but no cigar.
You forget that the clock doesn't know when the beam is sent out. It only knows that the second beam arrives $\Delta t$ time after the first beam, and travelled a distance of $x+y$ longer forwards and back.
Of course trying to sync the clock and the beam involves sending information back or forth which itself has to travel at the speed of light, and thus spoils the experiment.

Somnambulist
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0

If you can find a way to measure the one-way speed of light without comparing the times of two events happening in two different locations, you will have solved it. Considering the meaning of "one-way" that seems unlikely to say the least. But any such arrangement will require an assumption about the relationship between the times on the two clocks, which is equivalent to simply postulating that the two-way speed equals the one-way speed.

RC_23
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