Sunday, November 8, 2015

My time is different than yours

joão pestana

I will assume that you know the pythagorean theorem. You know that sing-along in which the square of the hypotenuse is equal to the sum of the squares of the other two sides of a right triangle. That literally means that if you draw 3 squares, each having a side equal to each of the 3 different sides of your triangle, and you sum the areas of the smaller 2 then you get the exact area of the bigger one!

Figure 1 — A right triangle with each side labeled a, b and c.

If you want to be mathematical — and I know you do — we simply state the previous as the lovable equation `a^2=b^2+c^2` and you can check the figure 1 to identify each side (`a` is the hypotenuse). But we don't discuss mathematics today, even if it's geometry. I want to talk about the theory of relativity — yes, Albert Einstein's relativity!

Let's begin with a story and a challenge. There is a master clockmaker who is such an amazing craftsman that he has the godlike power to build identical clocks who always mark precisely the same time. His ability is such that he offers with each clock a lifetime warranty. You can get your money back if one of his clocks differs 1 second from all the rest. It's impossible to meddle with the mechanics. Can you think of any way of taking advantage of his generous lifetime warranty and get your money back?

It might surprise you, but the answer is yes. All you need to do is take the clock with you on a quick 31 thousand years ride at the speed of sound. Of course, if you can't wait that long, you can speed things up by speeding up! You only need 32 years orbiting the Earth at the International Space Station which has an average velocity 23 times higher than the speed of sound. I believe this last scenario is the most favorable one in order to deceive the master clockmaker.

For a long time, the Russian cosmonaut Sergei Avdeyev held the record for time dilation experienced by a human being. Due to his 747 cumulative days of space flight, he aged roughly 0.02 seconds less than a person who never left the Earth. According to Einstein's special relativity, the clock of a person with greater velocity ticks slower, but if you're thinking that's just a theory... Without taking time dilation effects into consideration, GPS and other systems alike would be useless. Satellites' clocks do run slower than earthbound ones and we need to account for the discrepancy in time measurement to avoid triangulation errors.

If I'm running and you're sitting or if I hike the mountains and you go swimming on the sea, our clocks run differently. The first effect is due to the different velocities and the second one is a consequence of gravity — or spacetime distortion. The bottom line is that my time is different than yours and I can show you why using only Pythagoras' theorem.
Figure 2 — Representation of a moving car at different positions and periods.

Suppose you're driving your car and you throw a ball into the air. If you can't imagine, just look at figure 2. As you move along, because your car is traveling, the ball goes up at different positions for each period. From your point of view, the ball only goes up, because both you and the ball are moving along with the car at the same velocity. For someone else who is just watching you pass by, the ball travels not a straight line up, but rather it describes a curve.

Long before Einstein, Galileo Galilei already described this type of event in his Dialogo sopra i due massimi sistemi del mondo (Dialogue Concerning the Two Chief World Systems) published in 1632 using ships as an example. The principle of relativity is far from a novelty. What Einstein added was the idea that the speed of light is always measured to be the same by everyone. It's as if light was the ball that you're throwing and it would always describe the same arch, whether you're standing still or inside a moving car.

Figure 3 — Stationary train carriage with a light beam emitting at A and reflecting at B.

How can we then demonstrate the effect of time dilation using what we've discussed so far? It's quite simple and we need only one more example. Imagine that you're inside a train carriage as in figure 3 and that there is an emissor of light at the floor on point A such as a lightbulb and a reflector at point B such as a mirror. How long does the beam take to make the trip from A to B and back to A? It travels at the speed of light `c` and all we need to do is divide the total distance `bar(AB)+bar(BA)` by the velocity. Let's say that the height of the train carriage is `h` and thus `bar(AB)+bar(BA)=2h`.

`Delta t_(i n s i d e)=(2h)/c`

Now suppose that the carriage is moving at a constant velocity `v` and that a friend of yours is watching it pass by on the train platform. Because you're inside the train, nothing has changed. You still see the light beam travel from A to B and back to A in a vertical line and the time in which that happens is still `Delta t_(i n s i d e)`. Remember the example with the ball and the car? Well, it wasn't in vain. On the platform, as your friend watches the light beam go up and down, it does not describe a vertical line. It actually resembles a triangle as in figure 4.

Figure 4 — Moving train carriage at a constant velocity.

Yes, this is the reason we can apply Pythagoras' theorem. So let's do it! The distance that the light beam describes from A to B is not just the height of the train carriage, but the hypotenuse of a right triangle in which the height is just one of its sides. The other side is the distance that the train traveled since the light was emitted at A up until the point it was received at B and that is simply `Delta x`. Let's call this distance `Z^2=(Delta x)^2+h^2` and it's larger than `h`.

`Delta x=v (Delta t_(o u t s i d e))/2`

If `Delta t_(o u t s i d e)` is the time measured by your friend on the platform that the light beam takes to complete the trip from A to B and back to A, then half of that is just the time to reach B starting at A. We want to relate your friend's time with your time `Delta t_(i n s i d e)`. Let's review what we know of the problem so far.

`{(Delta t_(i n s i d e)=(2h)/c), (Delta x=v (Delta t_(o u t s i d e))/2),(Z^2=(Delta x)^2+h^2):}`

If we write `h=(c Delta t_(i n s i d e))/2` and `Z=(c Delta t_(o u t s i d e))/2`, we can write the Pythagoras' expression as:

`((c Delta t_(o u t s i d e))/2)^2=(v (Delta t_(o u t s i d e))/2)^2+((c Delta t_(i n s i d e))/2)^2`

And solving that equation for `Delta t_(o u t s i d e)` yields the expression:

`Delta t_(o u t s i d e)=(Delta t_(i n s i d e))/sqrt(1-(v/c)^2)`

Which I've used to determine the values I expressed in the fourth paragraph of this article. As you can see, you can relate the time experienced by a stationary observer — your friend on the platform — to that of a moving observer — you inside the train — by the relation between the velocity of the moving observer and the speed of light. You may notice that if the train stops `Delta t_(o u t s i d e)=Delta t_(i n s i d e)`, but as its velocity increases `Delta t_(o u t s i d e) > Delta t_(i n s i d e)`.

We can indeed get our money back from the master clockmaker by taking the clock on a fast trip, but we should check the warranty clause to see if it's not void for relativistic time dilations. I would guess that a master clockmaker would have this problem anticipated.