If a clock is moved along a crooked path half as fast, and thus for twice as long, is the total time lost by the clock due to time dilation halved?

Question

There is a type of relativity thought experiment where, in order to eliminate the problems caused by signal delay due to the finite speed of light, a lattice of clocks and rods (either automated or manned by technicians) is used to collect data on a strictly local basis about what happens where and when. It is essential that the clocks in the lattice are synchronized at the outset and a simple way to do this would be to travel around the lattice visiting each clock in turn and synchronizing it with a master clock that you carry with you.

The problem with this method is that the motion of the master clock will affect how fast it runs (there are velocities and accelerations) according to relativity, presumably slowing it down rather than speeding it up since moving clocks run slow, for sure, and I think accelerated clocks also run slow. Thus the clock that is synchronized with the master clock last will not be synchronized with the clock that was synchronized with the master clock first.

If the lattice is small, the effects will be tiny, and not even detectable, let alone problematic. In any case, if in the thought experiment a spaceship, at rest, is 300,000 kilometers long (one light second), but when going full speed is half that distance in length due to the Lorentz contraction, small differences between the times indicated by the lattice clocks would be negligible. The effects are not small or subtle.

Having said that, it would be convenient if there was a way to synchronize the clocks perfectly (or near as darn it), regardless of the size of the lattice and I'm wondering whether it would suffice to move the master clock very slowly over a very long period of time, to reduce the accelerations and speeds as much as one needs to, in order to get the total of the effects on the running speed of the master clock down to as low as one desires, effectively to zero, even with a big lattice, by which I mean a lattice with the same number of clocks, but with longer rods joining them: I don't want to complicate the question by varying the number of clocks, nor the number of ranks, files, and columns (if that's the right way to put it). I guess here I'm imagining taking an infinite amount amount of time to move the master clock all around the (finite) lattice.

Clearly the percentage slowing of the master clock will be less with smaller velocities and therefore smaller accelerations, but the amount of time that those effects operate will equally clearly be greater.

Not being very good a the math of relativity, I can't even figure out the effect of reducing the velocity on the total amount the master clock is set back. And the math of acceleration in relativity is a closed book to me.

The equation for gamma, according to https://en.wikipedia.org/wiki/Time_dilation#Time_dilation_caused_by_gravity_or_acceleration, which is what you must multiply the aging rate at rest by to get the aging rate at velocity v is 1/(1-v^2/c^2)^0.5 but I haven't figured out how to use it to find out whether the increase in journey time is the greater effect than the decrease in aging rate. Math is not my strong point. The article doesn't give an equation for acceleration.

So my question is, does the total amount of time that the master clock gets set back by go down rapidly as the velocities and accelerations go down, even though the journey time goes up correspondingly? Dare one hope that it is as simple as halve the accelerations and velocities and the result is half the time deficit?

Answer 1

it would be convenient if there was a way to synchronize the clocks perfectly (or near as darn it)

The easier way is to equip the lattice clocks with radio transmitters and receivers and synchronize using light pulses.

However, slow clock transport is interesting in its own right.

does the total amount that the master clock get set back go down rapidly as the velocities and accelerations go down, even though the journey time goes up correspondingly?

To travel a distance $L$ at a constant speed $v$ requires a time $t=L/v$. The accelerations to turn do not affect time dilation provided the speed is constant. During time $t$ the clock loses a time $$\Delta t=(\gamma-1)t$$ where $\gamma$ is the time dilation factor as defined in the question.

Now, if we are interested in the behavior at very small $v$ then we can do a first-order Taylor series expansion about $v=0$. Recall the first order expansion about $v=0$ is $$ f(v)= f(0) + f'(0)\ v + O(v^2)$$ where $O(v^2)$ is all of the terms of order $v^2$ or higher. Those terms become negligible as $v$ becomes small. Using Mathematica to evaluate the Taylor series gives $$\Delta t=\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)\frac{L}{v}=\frac{Lv}{2c^2}+O\left(v^2\right) \approx \frac{Lv}{2c^2}$$ so for small $v$ the time error is proportional to $L$ and $v$. For a fixed $L$ it can therefore be made arbitrarily small by using a small $v$.

This could be reasonable for a small lattice, but if the lattice is a 3D lattice then $L$ would grow as the cube of the lattice size, $s$ (assuming lattice spacing remains constant). Since $L\propto s^3$, to keep a fixed $\Delta t$ would require $v \approx 2c^2 \ \Delta t/L \propto s^{-3}$ to decrease by the inverse cube of the lattice size. So $t=L/v\propto s^3/s^{-3}=s^6$ would grow as the 6th power of the lattice size! That would rapidly become infeasible even for modest size lattices. The radio approach would become rapidly more appealing.

Answer 2

If a clock is moved along a crooked path half as fast, and thus for twice as long, is the total time lost by the clock due to time dilation halved?

Yes, that is a correct approximation for slowly moving clocks.

For clocks moving at any speed, this is exactly true: halving the kinetic energy of the clock = halving the time dilation of the clock.

(here time-dilation means number of ticks lost per hour, where hour means hour according to a static clock)

Obviously the halving of the kinetic energy must be done by slowing down the motion of the clock, not by sawing off some parts of the clock.

Accelerations should be infinite, because otherwise we need calculus. Acceleration of clock has no effect on clock. Oh, the clock travels at constant speed but changes direction, never mind then, acceleration can be anything.

(But why not calibrate the master clock so that it runs correctly at speed 0.9 c, and then make the clock travel at speed 0.9 c)

Answer 3

Let's say we have a stack of two 1 kg spinning disks. And two grams of matter-anti-matter-gun-powder loaded between said disks. The gunpowder is at the center part of the disks, so the gunpowder does not have angular momentum.

Now when the gun-powder explodes two grams of gun-powder 'disappears' and each disk gains one gram of mass, and some linear velocity.

Conservation of angular momentum tells us that the stack spins now 1-1/1000 times slower. This is the so called time dilation effect.

Now we halve the amount of gunpowder. This time the stack spins 1-1/2000 times slower. The slowing down effect was halved .

But the linear speed of the disks is slowed down by $1/\sqrt{2}$ , as we have learned in high-school, velocity increases as the square root of the energy.

Time dilation decreased more than speed decreased, which we wanted to prove.

We kept the amount of gunpowder small, so that we can use the non-relativistic velocity-kinetic-energy formula.