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I am trying to read of synchronization of two clocks in same inertial frame in special relativity. Suppose we have two synchronized clocks in an inertial frame placed at positions $x_1$ and $x_2$ in that frame. Suppose two observers at $x_1$ and and $x_2$ try to measure speed of some object moving in between $x_1$ and $x_2$(not at the midpoint or any close) with a constant speed. Now both observers, watch the object travel the distance $ \Delta x$. Let the object be closer to $x_1$. The observer at $x_1$ records times at which the object enters the $\Delta x$ region and another when it leaves the region. Lets call them $t_1$ and $t'_1$ and similarly the observer at $x_2$ records $t_2$ and $t'_2$. enter image description here

Now the time difference between $t_1$ and $t'_1$ is should not be same as the difference between $t2$ and $t'2$ since the light travels at a finite speed. So the velocities as measure by $x_1$ and $x_2$ observers should be different because the time intervals measured by them are different. This would imply both the observers measure different speeds. But I am wondering how is this possible? How can an object have two different speeds in same inertial frames? Am I missing something?

(PS: sorry for bad figure)

beginner
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    Special Relativity does not describe what an observer would see with his own eyes (due to the finite value of the speed of light), but what actually happens in spacetime. The visible picture is altered not only by Lorentz transformations, but also because light speed is finite and we can only see light traveling from objects. Have a look at a great computer game by MIT developers called "Slower speed of Light". When a player moves through the virtual world with near-light speed, he sees distant objects become even more distant (despite the Lorentz contraction) because light has finite speed. – Prof. Legolasov Sep 02 '15 at 19:06
  • Sorry I am not getting you. Can you elaborate your comment and relate it to question? As far as I know or have read, special relativity is all about switching between different frames and analyzing what happens then. – beginner Sep 02 '15 at 19:11
  • Imagine that you are doing Newtonian physics and instead of light you have sound. Following your logic, one would still end up with different velocities, which is impossible. So the problem is not with special relativity: there is a mistake in your calculations - you forgot to take into account that light/sound propagates with finite speed. – Prof. Legolasov Sep 02 '15 at 19:16
  • So how can one proceed to do the right calculation here? – beginner Sep 02 '15 at 19:20
  • Related: http://physics.stackexchange.com/questions/111078/time-dilation-all-messed-up/111089#111089 – Ellie Sep 02 '15 at 22:31

4 Answers4

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Suppose as the object enters $\Delta$x, it hits an antenna which sends a signal to $x_1$. $x_1$ proceeds to record the time. It also does the same for $x_2$, who also records the time. A similar signal is sent when it leaves $\Delta$x.

Yes the light signal will take a longer time to reach $x_2$, and thus the signals will reach $x_2$ at different times, but for someone in the same frame as $x_1$ and $x_2$ the time difference $x_1$ and $x_2$ measure will be the same.

The spec.relativity bit comes in when you look at the reference frame of the object. The events will no longer be simultaneous.

Shaurya Bhave
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There is a fundamental flaw in your setup. While the actual times the entrance signals may reach clocks 1 and 2 could be different, $t_1'-t_1$ will be the same as $t_2'-t_2$, because the clocks are at rest with respect to each other.

In fact, consider there are two events: entrance and exit. SR says that $$c^2\Delta t^2 - \Delta x ^2 = c^2\Delta t'^2-\Delta x'^2$$ for all inertial reference frames observing the two events. You agree that your observers in the same reference frame as each other measure the same $\Delta x$ as each other. Then they must measure the same $\Delta t$ or the equality wouldn't hold.

Bill N
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In order to measure the velocity in your setup, you have to measure the actual time difference for the object entering and exiting the region $\Delta x$, not the time difference seen by a distant observer receiving signals of the entrance and exit. There are (at least) two ways to do this.

For example, station toll-takers at the locations where the object will enter and exit the region $\Delta x$, call them $x_i$ and $x_f$. These are in the same reference frame as $x_1$ and $x_2$. Give them clocks and get all four clocks synchronized. Then $x_i$ will record the time $t_i$ when the object passes through his toll gate and enters $\Delta x$, and $x_f$ will record the time $t_f$ when the object passes through her toll gate and exits $\Delta x$. Afterwards everyone shares the two times, and since we know the distance $x_f - x_i = \Delta x$ we can calculate velocity. This is the velocity in the reference frame of $x_1$ and $x_2$.

To do the calculation the way you want to do it, the observers $x_1$ and $x_2$ need to know how far they are from the points I called $x_i$ and $x_f$, and they need to use that information to figure out the delay between when the object passed $x_i$ and when they actually observed it at their distant locations or received the signal that the event had occurred. Although $t_1 \ne t_2$, if the distances and the speed of the signal propagation are known, both observers at $x_1$ and $x_2$ will compute the same time $t_i$ and similarly the same time $t_f$. So while it is true that $t^\prime_1 - t_1 \ne t^\prime_2 - t_2$ except in a few special geometries, both observers can and must compute the same $t_f - t_i$ before they can compute the correct velocity.

Anonymous Penguin
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There are 6 times in this inertial frame. Observers won't use t1/t1' or t2/t2' to calculate the true speed, rather, they need use them to calculate the orginal $t_{enter}$ and $t_{exit}$ , which will be the same for both observers.

user2165
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