# Leading and lagging clock times in Lorentz Transforms

• I
• exmarine
In summary, the author argues that "leading clocks lag" when compared to trailing clocks, but this conclusion is incorrect because the relative velocity between the frames must be positive.f

#### exmarine

Someone posted this link to a paper I really appreciated.

http://www.hindawi.com/journals/physri/2015/895134/

But doesn’t the author have the wrong sign on the relative velocity in his Lorentz Transform associated with his figure 2b? And if so, doesn’t that reverse his conclusion that “leading clocks lag”? His leading clock is #3 in the figure, and its time would then be LARGER than that of the trailing clock #1? So wouldn’t the correct rule be “trailing clocks lag”?

He never really shows the time phase LT calculation, but he does show a negative relative velocity in the first column on page 3. But the particle is moving to the right along the observer’s positive x’-axis. So it seems to me that the relative velocity should be positive in the LT from the particle’s x-axis frame to the observer’s x’-axis. The LT must approach the Galilean Transform for very small velocities, and that would be (x’=x + vt). This produces a later time in clock #3 than in clock #1, not the earlier time indicated in his figure 2b.

Thanks.

I think it's correct. There are two frames: $F$ (as shown in figure 2b), in which the particle is moving at velocity $\vec{v}$ (to the right, in the picture), and $F'$ (as shown in figure 2a), in which the particle is at rest. The clocks in both figures are (I assume) clocks that are synchronized to show $t'$, the time as measured in frame $F'$.

From the point of view of frame $F'$, the clocks are synchronized; they all show the same time. From the point of view of frame $F$, the clocks are offset, according to the Lorentz transformation:

$t' = \gamma (t - \frac{vx}{c^2})$

So clocks with a larger value of $x$ show a smaller value for $t'$.

Thank you very much for looking at this. But you think for very large time (1≪t), the observer’s coordinate (x’) for the location of the particle at (x=0) will be very large NEGATIVE? Don’t think so. It is moving to the right as indicated in his figure 2b, so it will be very large positive (1≪x’). The relative velocity must be positive in the LT, and that switches the conclusion about which clock is lagging.

I think I see my error. I was looking at the wrong clock. Sorry...