# Light clock question

liometopum
I am trying to understand the calculations for a relativisitic light clock and am doing something wrong.

I am the stationary observer.
A light clock goes by at c/2, or 93000 miles per second.
The length of the light clock 'rod', which is perpendicular to the line of motion is 186,000 miles so it takes one second in the other (.5c) frame of reference.
Photon fires off and travels along the rod and hits the receiver.

From my point of view:
The base of the triangle is 93,000 miles, which is c/2.
The light clock rod is 186,000 miles
The hypotenuse is 207954 miles.

The light should travel the 207,954 miles (hypotenuse) distance.

The ratio of the hypotenuse to the vertical is 207954/186000= 1.118. So it takes 1.118 seconds for the light to travel that distance.

Is this right so far?

Next, the light should hit the receiver at the same moment whether the light travels straight up from the other frame of reference or from my frame of reference.

I should be viewing the time to be slower for the moving frame, by 1/1.118 or 89.445 percent of my clock.

However, using the standard formula for SR at .5c, the clockrate should be about 86.6 percent of my clockrate.

Something is amiss. What?

## Answers and Replies

I am trying to understand the calculations for a relativisitic light clock and am doing something wrong.

I am the stationary observer.
A light clock goes by at c/2, or 93000 miles per second.
The length of the light clock 'rod', which is perpendicular to the line of motion is 186,000 miles so it takes one second in the other (.5c) frame of reference.
Photon fires off and travels along the rod and hits the receiver.

From my point of view:
The base of the triangle is 93,000 miles, which is c/2.
The light clock rod is 186,000 miles
The hypotenuse is 207954 miles.
If you want to figure out the length of the base and hypotenuse, start from the assumption that the light takes a time of T to get from bottom to top, that implies the light clock moves sideways a distance of v*T = 0.5c*T in that time so that's the length of the base. And the light must move at c, so the distance traveled by the light in time T must be c*T, so that's the length of the hypotenuse. With the height as 1 light-second (easier to use units of light-seconds and seconds, so c=1, rather than miles and seconds), we have:

(0.5c*T)^2 + (1 light-second)^2 = (cT)^2
0.25c^2 * T^2 + (1 light-second)^2 = c^2 * T^2

0.75c^2 * T^2 = (1 light-second)^2
T^2 = (1/0.75) second^2
T = 1/sqrt(0.75) seconds

So, this implies the length of the hypotenuse is 1/sqrt(0.75) light-seconds = 215,100 miles, and the length of the base is 0.5/sqrt(0.75) light-seconds = 107,550 miles.

The problem with your numbers is that it's impossible to find a time T such that c*T = 207954 miles (required by the fact that the light must travel along the hypotenuse at c) and that 0.5c*T = 93,000 miles (required by the fact that the light clock is moving at 0.5c)

liometopum
Thanks very much! I see that I need to use the time of the other frame, not my time. So instead of multiplying .5 by 1 second, I multiply by the relativistic time, in this case 1/.866.