Special relativity holds a special (*ahem*) place in most ~~physicist’s~~ physicists’ hearts. I myself fondly remember learning special relativity from the first edition of Taylor and Wheeler’s Spacetime Physics obtained from my local county library (this edition seemed a lot less annoying than the later edition I used at Caltech.) One of the fun things I remember calculating when I learned this stuff was what “right in front of your nose” meant in different frames of reference.

Suppose you’ve got your own handy dandy inertial frame all set up. You’ve got your rules and ~~you’re~~ your clocks and are all ready to do some measurements. So you perform a measurement and find your nose (yes, your nose) at the origin when your clock strikes midnight on January 7, 2009. At the same time, you also measure that a bug, is just as centimeter and does a backflip. The bug is right in front of your nose, at midnight on January 7, 2009, and it does a flip. Really. In the picture above, your nose at that time is O and the bug doing the backflip is at A.

But now consider the view of you from a rocket passing by (rockets are essential tools for relativity, you know.) Of course this rocket put the origin of its spacetime digram so that the space origin is your nose and its time origin is midnight on January 7, 2009. Cool enough. But because this is special relativity the rocket will see something else. In fact the rock records that the bug doing the backflip has occurred at an earlier time, and at a father location away. This is point B on the diagram above.

What I find cool about this is as follows. On the above plot of drawn the curve x^{2}-t^{2}=1 (time and space can both be measured using centimeters: to convert time to seconds divide by the speed of light.) This represents the set of points that the bug event A could have been transformed into via a Lorentz transform, i.e. if you chose a point on this line, then there is some rocket moving at some relative velocity which will record the bug backflip at this coordinate in its inertial frame. Great. But this also means the opposite. There is a reference frame in which a point in your current inertial frame has a very large value of position and a very large (say negative) value of time.

In other words, there is a reference frame in which what is “right under your nose” is far far away, and just seconds after the big bang (let’s ignore cosmology for now.) There is an event which is a centimeter away from you, which someone who is traveling very very fast will see as ocuring at the same time in that person’s reference frame.

What is this speed (again ignore cosmology) such that, the age of the universe is just under your nose (which we take to be a centimeter away)? Well a simple calculation tells you that this speed is R/sqrt(1+R^{2})c where R is the time back you want to examine times the speed of light divided by the distance away of under your nose (1 centimeter.) For an time of 13.7 billion years (which is about 430 times 10^{17} seconds), this gives an R of about 1.3 times 10^{28} (R is dimensionless.) Now try to punch in this value of R into the above formula. Whoops, not so good for your calculator. But we can re-express the fraction of the speed of light that you need to travel by rewriting the speed as c/sqrt(1/R^{2}+1), which can be Taylor expanded for small values of 1/R as c (1-1/(2R^{2}). In other words, you need to move just under the speed of light, where by just, I mean, about 10^{-56} fraction slower. In other words 99.99…99 percent of the speed of light with a total of about 56 9s!

So, right under your nose, someone could see an event that happened the age of the universe ago. Where, by someone, we mean someone much much much faster than even Usain Bolt.