I wanted to check my understanding of special relativity and set up one experiment that could lead to deriving the Lorentz factor. Usually, the deriviation for the Lorentz factor is explained using a light clock forming a 90° angle with the direction in which the observer is moving.


Here, we see that the direction in which the light ray is moving, is perpendicular to the moving direction. And if you calculate the time it takes for the light in both of the light clocks(in each reference frame), the relationship between the two turns out to be exactly $1/\sqrt{(1-v^2/c^2)}$, which is the right Lorentz factor.

However, you could follow a similar logic and consider the case in which the light clock is leaning on its side and shooting light rays in the direction of motion.


Let's suppose that a moving observer is moving from left to right horizontally. He or she holds a light clock that lay horizontally with the direction of motion. The distance of the light clock is a and the observer is moving at the speed $v$ relative to the stationary observer. The stationary observer can measure his or her time with the identical light clock. Time on each reference frame is defined as the time it takes for the light to bounce back to the observer, so we would be able to say that we can find the relationship of the time of the stationary observer and the time of the moving observer which is the Lorentz factor. I wonder why my result $c/(c^2-v^2)$ does not match the real Lorentz factor. Were my assumptions wrong? Why does it work when the light clock is perpendicular to the moving direction, but doesn't when it is facing the direction of motion?


1 Answer 1


Your calculations assume that the observers agree on the length of Alice's clock.

For the vertical clock, that's the right assumption. Here is why: Alice can put little dabs of paint on the top and bottom of her light clock, which brushes against Bob's identical light clock as it passes, leaving marks. If according to Bob, her light clock is shorter or longer than his, he'll be able to verify that by the locations of the marks --- and she'll have to agree because the marks are there for all to see. Then they'll agree on whose clock is shorter, and can use this to reach agreement on who's moving, contrary to the first principle of relativity: There's no objective way to determine which inertial observer is in motion.

No analogous experiment forces Alice and Bob to agree on the length of a horizontal light clock, so they don't have to agree. And in fact your calculations shows that they can't agree on the length of the horizontal clock, because if they did, they'd caclulate the wrong value for time dilation.

  • $\begingroup$ Thank you, but I think I am a bit confused. Following your logic, the lengths of clocks must be the same when the clocks are placed vertically. But length is determined by time and velocity. We can derive t' from the relationship between the length(the length of the clocks that they agree on) and the speed of light. However, because t' is now different from t, the length that they agreed on before will now be different(the length of the clocks). So at first they were able to agree on the length of the clocks, but now they can't. $\endgroup$ Commented May 17 at 13:48
  • $\begingroup$ Only the length in direction of motion ist changed, not the length perpendicular to the motion, Willo explained it with the paint dots quiet well $\endgroup$
    – trula
    Commented May 17 at 14:03
  • $\begingroup$ Yes, that is true, however, according to the logic that I have suggested in the comment section, it seems as if the length of the clocks cannot remain the same. But I think there is something wrong with my argument, but I am not sure what it is. $\endgroup$ Commented May 17 at 14:13
  • $\begingroup$ @IamEinstein : I don't completely understand your comment because you haven't told me what t and t' are. But I suspect your mistake is to think that the time it takes for light to travel from the bottom to the top of a moving clock is a measure of the length of that clock. Not so, because the light is not moving along the clock. $\endgroup$
    – WillO
    Commented May 17 at 14:17
  • $\begingroup$ is it? sorry but can you give me a more detailed explanation? (It represents the time of the stationary observer while t' represents the time of the moving observer) $\endgroup$ Commented May 17 at 14:18

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