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Classical mechanics thought experiment:

  • We throw a ball upwards with a velocity of $10~m/s$ we know that the ball will fall on ground in 2 seconds.
  • We throw another ball now with a velocity of $~10\sqrt{2}~ m/s$ at an angle of 45 degrees, we know that this ball will too fall on the ground in 2 seconds.
  • We know that this happens because the vertical component of the velocity controls for how long will the ball remain airborne and both the balls have the same vertical velocity.

Can this be applied in the following scenario too?

Simple relativity thought experiment:

  • An observer (observer A) is traveling a spaceship at half the speed of light, while another observer (observer B) is stationary on earth.
  • Both of you have photon clocks with photons bouncing between the sides of a cylinder, where each oscillation accounts for one second in its own frame of reference.
  • Now as observer A travels past earth, due to relative motion, we say that the distance traveled by the photon in clock is more, hence the time dilates to allow the speed of light to be constant.

My question:

  • In the thought experiment we say that the photon in the spaceship traveled a larger distance due to the inherent velocity of the spaceship it was in, relative to earth.
  • But shouldn't the duration the photon takes to travel between the ends of the clock (flat mirrors) must be affected only by the velocity it has in THAT direction.
  • If we break the velocity of the photon in space (relative to earth) into velocities mutually perpendicular components of the net velocity vector, we would observe that the time the photon takes to move between the two ends of the clock are affected only by the component in that direction.
  • Since the photon in both the clocks have the same velocity and travel the same distance (vertically between) how do they differ?

I am familiar with Einstein-velocity addition for one dimension: $u = \frac{v + u^{\prime}}{1 + \frac{vu^{\prime}}{c^2}}$ but couldn't find anything for two dimensions. For context I am studying classical mechanicn in school and have no knowledge about relativity except what I learnt from the internet.

Thanks!!

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But shouldn't the duration the photon takes to travel between the ends of the clock (flat mirrors) must be affected only by the velocity it has in THAT direction.

Yes correct. This photon is a clock hand that moves inside a clock. The clock hand moves in "that direction", which we call the y-direction, at speed:

$$ v_y = \sqrt {c^2-v_x^2 } $$

Or, if the photon is shot at angle $ \alpha $ then

$$v_y = c * sin \alpha$$ $$v_x = c * cos \alpha$$

,where y-direction is "that direction" and x-direction is the other direction.

Shoot a photon straight up. Now you have a clock that moves at speed zero, and a clock hand that moves inside the clock at speed c.

Shoot a photon at 45 degree angle. Now you have a clock that moves at speed 0.70 c, and a clock hand that moves at speed 0.70 c inside the clock.

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  • $\begingroup$ oh right, yeah i think i got it. thanks!! $\endgroup$ Feb 6, 2021 at 0:35

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