Two Synchronised Clocks Problem

Question

Two synchronised clocks are both moved from point $A$ to point $B$ at the same time through two different paths. One clock is moved at the speed of light through a one-lightsecond-long path. The other one is moved at the speed of $1\ m/s$ through a 1-m long path. Both clocks arrive at the point $B$ at the same time.

Question :

Are clocks synchronised after the procedure? If not, which one is ahead of another and is it because of different travelled distances, different speeds or both?

I was reading a book about spacetime and suddenly had this question in my mind but the book didn't have an answer, so I will be happy if someone will answer it!

Edit :

Considering the fact that time stops functioning properly when something moves at the speed of light, take the speed of the first clock as a speed close to the speed of light or $1/2$ of it and its distance as a distance that will be travelled by it in one second.

Answer 1

Due to the way you have set up the problem this turns out to be very easy to answer. If a clock travels at a constant speed and covers a distance $\ell$ in a time $t$ the elapsed time on the clock is calculated using an equation called the metric. Specifically in special relativity the equation is the Minkowksi metric, and the elapsed time, $\tau$, is given by:

$$ c^2\tau^2 = c^2t^2 - \ell^2 \tag{1}$$

If the (constant) speed of the clock is $v$ then the distance it travels in the time $t$ is just $\ell = vt$, and we can use this to substitute for $\ell$ in equation (1) to get:

$$ c^2\tau^2 = c^2t^2 - v^2t^2 $$

and dividing through by $c^2$ and taking the square root we get:

$$ \tau = t\sqrt{1 - v^2/c^2} \tag{2} $$

We generally use the symbol $\gamma$ (known as the Lorentz factor) to mean:

$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

and using this equation (2) becomes:

$$ \tau = \frac{t}{\gamma} \tag{3} $$

which some of you will recognise as the equation for time dilation in special relativity, and indeed this is one way of deriving that equation.

In your problem you have set things up so all the clocks take the same time of one second to get from A to B, so we get the elapsed time by setting $t = 1$ in equation 2 to get the elapsed time:

$$ \tau = \sqrt{1 - v^2/c^2} \tag{4} $$

And there is your answer. The elapsed time depends on $v$, and decreases as $v$ increases. Indeed as $v \to c$ the elapsed time goes to zero. So the clocks will not be synchronised because the clock that travelled faster will have a smaller elapsed time.

This result applies as long as the clocks travel at a constant speed. Obviously a clock moving at $c/2$ will have to move in some sort of squiggly path to take a second to get the one metre from A to B, but the shape of the path doesn't matter - all that matters is that the speed is constant.

It is possible to extend the calculation to clocks that do not move at a constant speed. Suppose the speed of the clock is $v(t)$ where $v$ varies with $t$. In that case we consider the small time $dt$ during which the clock moved a distance $\ell = v(t)dt$. Then equation (1) gives us:

$$ d\tau = dt \sqrt{1 - v(t)^2/c^2} $$

And the proper time is given by integrating this:

$$ \tau = \int_0^t dt \sqrt{1 - v(t)^2/c^2} $$

Taking $t$ as a constant then gives us equation (2).

Answer 2

The one light second you are talking about is the time according to us the light will take to cover a certain path in one second. The clock moving at the speed of light itself will not even measure time. This is what we call time dilation. A simple analogy to it is the time required for light to reach earth from sun. It takes around 490 seconds for sunlight to reach us but the sunlight itself will not experience any time.Here 490 seconds are seconds according to earth. So or a clock moving at speed of light no time must have passed neither the clock will read 1 second. I think the clocks will not be synchronised. Passage of time is affected by speed and gravity (curvature in time creates gravity).

Answer 3

If two clocks start synchronised at a point A in spacetime and travel via different asymmetrical paths to point B in spacetime, then they will be out of synch when they meet.

In the scenario you imagine, the faster clock will show the less time- it is an example of the twin paradox. Strictly speaking the result is not due to the speed of the clocks (which is entirely relative), nor to the distance traveled (ditto) but to the fact that the faster clock must experience a greater turnaround acceleration if it is to return to meet the other clock. It is the acceleration (or more generally a change of inertial reference frame) which causes the twin paradox effect (which is explained in lots of other answers on this site).

Answer 4

Building on @JohnRennie 's answer, here the visualization
(a "position-vs-time graph" or "spacetime diagram") of the calculation
for three clocks traveling at constant speed (constant-velocity for half the trip, then the oppositely-directed constant-velocity for the remaining half of the trip):
$|\vec v_{Alice}|=0c$, $\qquad$ $|\vec v_{Bob}|=(3/5)c$, $\qquad$ and $\qquad$ $|\vec v_{Carol}|=(4/5)c$,
with corresponding
$\gamma_{Alice}=1$, $\qquad$ $\gamma_{Bob}=(5/4)$, $\qquad$ and $\qquad$ $\gamma_{Carol}=(5/3)$.

These speeds are chosen for mathematical convenience (since their Doppler factors are rational [here: 1,2, and 3], we end up with Minkowski-triangles associated with Pythagorean Triples, leading to calculations with fractions).

So, the clocks that were synchronized at event O arrive unsynchronized at event Z.

By counting the "light-clock diamonds" (equal-area parallelograms, formed by the light-signals in their standard-issue light-clocks), we see that Alice's clock elapses 10 ticks, Bob's 8 ticks, and Carol's 6 ticks.

These various elapsed times ("the wristwatch 'proper times' along a worldline) are due to the fact that these are along different worldlines from O to Z, akin to @WillO 's comment about the odometers.

Here is a slightly more-general situation involving piecewise-inertial worldlines ( the diagram from What is preventing this statement to be true: the twin brother who accelerated away is really decelerating and Twins Paradox: Why is one frame considered to be the accelerating frame )

While velocities and accelerations (done for various periods) are needed to execute the various worldlines, the "effect" is ultimately about the worldlines themselves.