# Time Dilation - Light clock experiment [duplicate]

In the light clock experiment of the time dilation theory, why does the light travel in triangles for the light clock in motion when the outside observer is viewing it. I'm not able to understand why does the light travel a longer distance for the light clock in motion as compared to the stationery light clock. The distance between the mirrors in both the light clock is the same. The only difference is that one is in motion and the other is not. If the distance between the mirrors in both the light clocks is the same, then why does light have to travel in triangles for the light clock in motion when the outside observer is viewing it. Why can't it travel straight as it does in the stationery light clock. Please explain. I'm unable to understand the concept of time dilation.

• If you're looking at your own light clock, then the light beam just goes up and down, but if you're looking at the light clock of someone whizzing past you it appears that the light beam of their light clock is traveling in "triangles". So to you it appears that the light has to travel a greater distance to bounce from one mirror to the other than it does with your own light clock. On the other hand, the owner of the other light clock sees the light beam of his own clock just going up and down, and sees the light beam of your light clock traveling in "triangles". – user93237 Jul 21 '17 at 23:46
• If you're in a frame where the clock is moving rightward and the light beam is moving vertically, how will the light from the bottom ever hit the mirror at the top? – WillO Jul 22 '17 at 2:54
• A note : The light clock for the interpretation of time dilation is a "a posteriori" construction since it uses a result of Special Relativity : that the dimension of the clock normal to the velocity ( i.e. its height $\,h\,$ ) is invariant between the two frames (that of the rest frame of the clock and the frame of the moving observer). – Frobenius Jul 22 '17 at 8:24

Let's move away from special relativity and work with a nonrelativistic example. Imagine Alice is on a train moving past Bob. Alice throws a ball straight up in the air and catches it. She sees the ball follow a straight line path straight up then straight down. The displacement of the ball in her reference frame is $(x,y)=(0,v_0t-\frac{1}{2}gt^2)$. Bob sees the train moving with speed $v_t$ and everything else moving with the train. So he observers the ball follow a parabolic path $(x,y)=(v_tt,v_0t-\frac{1}{2}gt^2)$. Bob doesn't see Alice catch the ball at $x=0$, he sees her catch it at $x=\frac{2v_0v_t}{g}$.
Now, going back to the path of light bouncing between mirrors. Just like with the ball, the light is moving straight up and down in the frame of the mirrors, but the mirrors themselves are moving, adding their sideways motion to the light. This imparts a sideways velocity to the light. In Newtonian physics, this would combine with the vertical velocity, $c$ of the light to create a total speed $\sqrt{c^2+v^2}$. Instead, in special relativity, the speed of light is fixed at $c$, making the vertical component of the velocity $\sqrt{c^2-v^2}$. It is because of this decreased vertical component of the speed of light that time dilation occurs.
• @adosar no citation necessary. The horizontal component of the light has to be the same as the horizontal velocity of the mirrors, otherwise the light won't hit the mirror again. Velocity is just $v^2=v_x^2+v_y^2$. According to special relativity, the speed of light in every direction to every inertial observer is $c$, so just solve $c^2=v^2+v_y^2$ for $v_y$. – Johnathan Gross Feb 4 '20 at 20:28