# Light Clock Problem

1. Jan 10, 2015

### Quarlep

Lets suppose theres two obsrver O observer is in rest and A is moving with a constant velocity. A observer has a light clock which every click it makes sound.So A observer sees everything normal. Lets suppose A observer heard 10 tics but O sees him moving so O observer will heard 5 tics.But both of them cant be true.
Which one is true ?

2. Jan 10, 2015

### ShayanJ

This is wrong. Both of them hear the same number of ticks. Just the delay between the ticks is longer for O.

3. Jan 10, 2015

### Staff: Mentor

Why not?

If $N_A$ is the number of ticks A hears and $N_O$ is the number of ticks that O hears then those are different quantities and there is no logical reason that they must be equal. You can certainly set up scenarios where they are, but you can also set up scenarios where they are not.

Last edited: Jan 10, 2015
4. Jan 10, 2015

### stevendaryl

Staff Emeritus
What you need to realize is that if neither observer accelerates, then you can't separate time dilation from the problem of clock synchronization.

Let's set observer $O$ up with a partner, $O'$ who is at rest relative to $O$, and a distance of $100$ light-seconds (a light second is the distance light travels in one second). Observer $A$ travels from $O$ to $O'$. If $A$ is traveling at 86.6% of the speed of light, then it will take her 115 seconds to get to $O'$. For simplicity of calculations, let's assume that the light clocks "tick" once per second. Here's the sequence of events from the point of view of both $O$ and $A$

From the point of view of $O$
1. Event $e_1$: $O$ starts counting "ticks" of his light clock at the moment $A$ passes by. $A$ starts counting "ticks" of her light clock.
2. Event $e_2$: $O'$ starts counting "ticks" of his light clock. In $O$'s frame, $e_1$ and $e_2$ are simultaneous (to arrange this requires synchronizing clocks, and knowing when $A$ will pass $O$, but let's assume that has happened)
3. Event $e_3$: $A$ passes $O'$. According to $O$, this event happens $115$ seconds after $e_2$. By this time, $A$ has counted 57.5 seconds worth of "ticks" of her light clock, and $O'$ has counted 115 seconds worth of "ticks". So $O$ and $O'$ conclude that $A$'s light clock is "ticking" at 1/2 the rate of their light clocks.
So does that objectively mean that $A$ is the one whose clock is running slower? No, because let's look at things from the point of view of $A$ now. From $A$'s point of view, $O$ and $O'$ have not correctly synchronized their clocks. From her point of view, $O'$ starts counting 172.5 seconds BEFORE $O$ does. So she sees the following sequence of events:

1. Event $e_2$: $O'$ starts counting ticks of his light clock.
2. Event $e_1$: $O$ and $A$ pass each other, and both start counting ticks of their respective clocks. From A's point of view, this event happens 172.5 seconds after event $e_2$. So at this time, $O'$ has already counted to $86.25$ seconds (172.5 seconds have passed, but $O'$ is counting at half the rate.
3. Event $e_3$: $A$ and $O'$ reach each other. According to $A$, this event happens 57.5 seconds after event $e_1$ and a full 230 seconds after event $e_2$. So $A$ has counted up to $57.5$, while $O'$ has counted up to $115$ (half of the 230 seconds since he began counting).