- Blackout
- Faster Than Light
- Hex Board
- Invariants
- Listening To OEIS
- Logic Gates
- Penrose Maze
- Syntactic Sugar
- Terminal Colors

- Traffic Engineering with Portals
- Algebra and Data Types
- What's a Confidence Interval?
- Uncalibrated quantum experiments act clasically
- Pixel to Hex
- Linear vs Binary Search
- There and Back Again
- Tree Editor Survey
- Rust Quick Reference
- The Prisoners' Lightbulb
- Notes on Concurrency
- It's a blog now!

To travel back in time is to exceed the speed of light, and to exceed the speed of light is to travel back in time.

I’ve heard on occassion that you cannot go faster than the speed of light without also going back in time, so I worked out why this is. Follow along with me, if you will. We’ll only need one thing: Lorentz transformations, which tell you how to translate from one observer’s space-time coordinate system to another, in the framework of special relativity.

Ok, so we’re *starting* with the assumption that someone (say Amy) is
going faster than the speed of light. Except… fast according to
whom? One of the central tenants of special relativity is that every
(non-accelerating) observer is entitled to use their own space-time
coordinate system, according to which they are stationary. For
example, if you’re by the side of the highway and I’m in a car driving
past you, you can say that you’re standing still and I’m going 60mph
to the East, and I can say that *I’m* sitting still and *you’re* going
60mph to the West, and neither of us is wrong.

(Ok, so actually we’re both in Earth’s gravitational field, which bring general relativity instead of just special relativity into the picture and complicates matters. But make it a space highway and now this is 100% accurate with no complications.)

So back to the question: if Amy is going faster than the speed of
light, that need to be faster than the speed of light *according to
someone* (or something). So, let’s be more specific:

Bob is on Earth. From Bob’s frame of reference (i.e., from his space-time coordinate system), Amy travels from the Earth to the Moon in 90ms, putting her speed at about 1.5c (where c is the speed of light).

Call the distance that Amy traveled `d`

(around 40,000km), and the
time it took her `t`

(90ms). Then we have that:

ct < d

Now here’s the trick. We’re going to suppose that a *third* person is
travelling in the opposite direction—going from the Moon to the
Earth—at a particular speed, and then prove that from this third
person’s frame of reference (i.e., according to their space-time
coordinate system) Amy is travelling backwards in time.

Call this person Candy. To pick out her speed relative to Bob, we’re
going to start with the fact that `ct < d`

. Multiply both sides by `c`

and divide by `d`

to get that `c`

. Thus
we can pick a speed ^{2}t/d < c`v`

for Candy such that:

c^{2}t/d < v < c

Notice that this speed is by construction less than `c`

: Candy is a
perfectly normal traveller going sub-luminal speeds.

Now we should ask what Amy’s speed is according to Candy. This is exactly what Lorentz transformations are for: they say how to convert from one frame of reference (Bob’s) to another (Candy’s). A Lorentz transformation tells you how to convert both distance and time, but we’ll just look at the time conversion. The conversion is that:

t' = γ(t - vd/c^{2})

where `t'`

is Amy’s travel time according to Candy, and
`γ = 1/sqrt(1-v`

is called
the Lorentz factor. Here’s the shocking reveal: ^{2}/c^{2})`t'`

is negative, so
according to Candy, Amy went back in time!

How can we prove that it’s negative? We just need to do some algebra.
Here are the steps; the last step uses the way that we chose Candy’s
speed `v`

:

t' = γ(t - vd/c^{2}) = γt - γvd/c^{2}= γd/c^{2}(c^{2}t/d - v) < 0

Thus if you go faster than the speed of light in *some* frame of
reference, then there is *another* frame of reference in which you
have gone backwards in time.