ticks 'n bricks

symmetry of clocks in special relativity

the "paradox"

In special relativity, clocks slow down by a factor of 1/sqrt(1-v^2), sometimes called the "time dilation factor" or "gamma". But according to the principle of relativity as used in special relativity, any uniformly moving observer will see the same effects if they consider themselves to be at rest. Therefore, if two observers are moving, each relative to the other, then each will "see" the other observer's clock "running slow".

But isn't this a paradox? After all, how can it be that both A's clock is ticking slower than B's clock and B's is ticking slower than A's?

the analogy

No, it's not a paradox, and the resolution is because of the fact that the relationship of one space-time coordinate system to another in special relativity is mathematically a kind of rotation. It is the same effect one would get if one had two x-y (instead of space-time) coordinate systems, and there was an angle between the two x axes.

Consider this diagram, of two x-y coordinate systems, one with coordinate axes rendered in blue, the other with coordinate axes rendered in red.

Now, let's build a brick wall along each of these two x axes. The analogy here is, each brick in the wall is analogous to a clock tick along the timeline of a moving observer in special relativity. Consider a person using the red coordinate system to compare x-length of red and blue bricks.

We can see that somebody using the red coordinate system will think the x-extent of the blue bricks is less than the x-extent of the red bricks, and at the very same time, looking at the very same bricks, somebody using the blue coordinate system will think the x-extent of the red bricks is less than the x-extent of the blue bricks.

the application to relativity

The coordinate axes that are laid out by the rules for two people in relative motion have an analogous effect, although the "rotation" of the coordinate axes have a slightly different mathematical formula, which causes the axes to "scissor" together.

To illustrate this, consider this pair of diagrams, both showing the exact same situation with a pair of observers in motion, one drawn with the coordinates of the red observer perpendicular, and the other with the coordinates of the blue observer perpendicular. The thicker, more horizontal, arrowed lines are the time axes for the corresponding colorcoded observer, and the thinner, more vertical lines are the space axes.

The marks on each line show that the relative speed of the two is enough to give them a time dilation factor of 2. We know this from the diagram, because the first tick out on any time axis lines up with a half-tick on the other time axis, when you go parallel to the matching color space axis. This is true no matter what axis you start on, and no matter which diagram you use. The situation is perfectly symmetrical. Now, to get a "brick dilation factor" of 2, you'd need to rotate by 60 degrees. For motion in special relativity, a "time dilation factor" of 2 means the relative speed is about 0.8660 of lightspeed.

Note that if you start with the left diagram above, and imagine "prying apart" the blue axes to make them be at right angles, you see that this implies that you "squish together" the red axes to keep the relationships the same. And indeed, that's just what is seen on the diagram to the right.

What this means is that the blue time direction is "at an angle" to the red time direction, and likewise the blue space direction is at an angle to the red space direction. Thus, just like the bricks, the two ticks in the red timewise direction land on a line that's (stretched from this perspective, but) only 1 blue tick long. And taking that back along the blue space direction, we see that that 1 blue second (according to the blue observer) corresponds to half-a-tick on the red clock.

Note that the counterintuitive parts (that each thinks the other's clock is ticking slower) are explained in just exactly the same way that each bricklayer thinking the other's bricks are shorter; namely because the coordinate axes are at an angle to one another.

The key point is, the meaning of "same time as" differs depending on which space axis (red or blue) you follow, just as the meaning of "same x coordinate as" differs, depending on which y-axis you followed in the brick example, the red one or the blue one.

the twin paradox

In the twin paradox, one of two twins makes a trip out at some constant high speed, and then back. Using simple time dilation, each twin thinks the other "must" end up younger, because each twin measures the other's clock as ticking slower. How does this notion of rotation of coordinate frames resolve this disagreement? For one take on that issue, there is a discussion of the unaccelerated twin paradox, which has some more diagrams showing how multiple frames of reference act together to make one twin age more than the other, so that both will get consistent rather than contradictory results.

But if after reading that, you still don't understand why the twin paradox is not really a paradox, then here's the same paradox, in bricks instead of ticks.

the bricklaying twin paradox

In this analogy to the twin paradox, two twin bricklayers (a red twin and a blue twin) each make a brick wall starting from the same place and heading off in a straight line. Each wall starts out at a 60-degree angle to the other wall. That means that each thinks the other's bricks are short, and they use half as many bricks as their twin for each bit of wall. This is just like in the twin paradox, each astronaut twin thinks the other's clock ticks half as many times as their own.

bricks (start)

The blue and red bricklayers each start building a wall, with a 60 degree angle between the two walls.

ticks (start)

The blue and red astronauts each get in a spacecraft, and start out with the two spacecraft moving apart at 0.8660 lightspeed.

bricks (turnaround)

The blue bricklayer, after building a kilobrick-long wall, starts building the wall angling back toward the red bricklayer's wall, with an interior angle of again 60 degrees, so the brick dilation factor is still 2. The red bricklayer keeps extending a straight wall. So eventually the walls meet again.

ticks (turnaround)

The blue astronaut, after heading out for a kilosecond, turns around and heads back towards the red astronaut at the same speed, so the time dilation factor is still 2. The red astronaut doesn't change course. So eventually the spacecraft meet again.

bricks (red counts blue)

At the meeting, the red bricklayer (who didn't change wall angle) figures that each brick on the red wall equals two bricks on the blue wall, so the expectation is that the red bricklayer will have used half as many bricks as the blue bricklayer between the points where the red and blue walls meet. And ineed, that's what the red bricklayer finds.

ticks (red counts blue)

At the meeting, the red astronaut (who didn't change course) figures that each tick on the red clock equals a half-tick on the blue clock, so the expectation is that the red astronaut's clock will read twice the elapsed time. And indeed that's what the red astronaut finds.

bricks (blue counts red)

And the blue bricklayer (who did change wall direction) figures that each brick on the blue wall equals two bricks on the red wall. But the calculation gives 2 kilobricks times the brick dilation factor of 2, or 4 red kilobricks. But there is only 1 red kilobrick in the red wall.

The diagram below shows why just multiplying by the brick dilation factor counts some of the red bricklayer's bricks twice. Specifically, we see that the blue bricklayer counted 3 "extra" or "phantom" kilobricks. When you subtract out the 3 kilobricks of red wall between the perpendicular from the outgoing blue wall to the perpendicular to the returning blue wall, you get 2 blue kilobricks of wall, times the brick dilation factor of 2 gives 4 kilobricks, minus the 3 kilobricks counted twice, gives a 1-kilobrick red wall, which is just the right answer.

ticks (blue counts red)

Now the blue astronaut (who did change course) also knows that each tick on the blue clock equals only one-half tick on the red clock. But the calculation gives 2 kiloseconds blue time spent, divided by the time dilation factor of 2, gives only 1 kilosecond elapsed red time. But there are actually 4 kiloseconds of elapsed red time.

And the diagram below shows why just dividing by the dime dilation factor skips some of the red twin's clock ticks. The skipped ticks needed to be accounted for also. The red ticks between the (perpendicular) space axis before the turnaround and the (perpendicular) space axis after the turnaround never get counted. The formula for this is given in Feynman's Lectures on Physics, volume I, page 15-6. It basically works out to be

(x1-x2)u/sqrt(1-u^2)
((.8660)*0.9897)/sqrt(1-0.9897^2)  ==  6

The reason this time is skipped instead of counted twice is because the axes "scissor" together instead of going in the same direction, and this in turn is due to the so-called "metric" of spacetime in special relativity.

So anyway, that's 1 kilosecond out plus 1 kilosecond back plus 6 kiloseconds skipped, divided by the time dilation factor of 2, gives 4 kiloseconds, which is just the right answer.

the moral

Bricks or ticks, if you are the one that changes course, you are the one that has to take account of the extra bricks or missing ticks.

Last edited Thu Nov 30 15:18:20 1995 - Last generated Thu Nov 30 15:18:21 1995