Life is shorter than you might think
Remember when you were in elementary school, and a year seemed like an eternity. A decade later, in college, the years just zoomed by. This lead to the following proposition:
The rate at which time appears to pass by is proportional to how much time you’ve experienced.
So when you are 40 time feels as if it’s ticking away twice as quickly as when you are 20.
This proposal follows a theme about the way we sense a lot of things. We measure sounds in decibels, a logarithmic scale, for this reason.1 Somebody paid $30k per year is elated to get a christmas bonus of $5k, while a hot-shot programmer at Amazon with a take home pay of $300k would be insulted. When a family in the suburbs remodels they will not be satisfied with adding 100 square feet to their “modest” 2000 square foot home. Conversely, a crampt city dweller could only dream of having such a cavernous extra room. It’s as if we feel relative changes of circumstance, not absolute changes.
Apparently this is how we sense sound, space, and wealth. Why not time?
Math!
Let \(t\) denote actual time, and let \(\tau(t)\) denote the time experienced by an individual, let’s call it “subjective time”. The hypothesis is that a year that goes by at age 40 feels half as long as when you are 20. This last sentence can be mathematized as
\[\tau(40) - \tau(39) \approx \frac{1}{2} (\tau(20) - \tau(19))\]More generallly, there is no reason to only consider increments of a year, and factors of two. We can consider arbitrarily small increments of time so that the differences turn into derivatives. In general, the hypothesis of this post is that
\[\tau'(\lambda t) = \lambda^{-1} \tau'(t)\]for all \(\lambda > 0\). This restricts \(\tau'(t)\) to a function of the form
\[\tau'(t) = \frac{A}{t}\]for some coefficient \(A>0\).
proof
Consider the case where $$t=1$$. Then the equation reads $$ \tau'(\lambda) = \frac{t'(1)}{\lambda} $$ for all $$\lambda > 0$$. This defines $$\tau'$$ for all postive values in the form $$\tau'(t) = A / t$$ for some unknown $$A>0$$.Integration of \(\tau'\) then give us
\[\tau(t) = A \log(t) + B\]for another unknown \(B\).
There is no clear way to choose \(A\) and \(B\). However, the model is not completely useless.
How much longer do I have?
If you are old enough to read this sentence then you can ask this of the model.
Assuming I live 80 years, how long will my remaining days feel?
If you are \(t\) years old then, naively, the answer to this question should be \(\tau(80) - \tau(t)\). Unfortunately, this answer depends on the choice for \(A\) and \(B\). It’s also singular as \(t \to 0\).
By design \(\tau\) is subjective, and that’s reflected by free parameters \(A\) and \(B\). To get an objective answer, one must ask the right question. The following would be better
Assuming I live 80 years, how long will my remaining days feel in proportion to how my last year felt?
If you are \(t\) years old, the model will spit out an objective answer:
\[D(t) = \frac{\tau(80) - \tau(t)}{\tau(t) - \tau(t-1)}.\]The numerator is how much subjective time you have remaining and the denominator is how much subjective time elapsed during the last year. The constuction of the above function is objective because it is independent of the unknowns \(A\) and \(B\). In particular,
\[D(t) \equiv \frac{ \log(80) - \log(t) }{ \log(t) - \log(t-1)}\]For me this means I have 28 more “subjective years”, where by “subjective years” I mean the feeling of a year at my current age, \(t=37\). Depressingly less than my actual time left on this earth, which is would be 43 real years.
Final thoughts
It’s weird to have a math blog talk about subjective experience like this. However subjective experience is what you feel, and so it matters. My guess is that when I’m 80 the 28 “subjective years” will feel like a better measure of time than the 48 “real” years.
He said “It’s all in your head”
And I said “So’s everything” but he didn’t get it
- Fiona Apple, Paper Bag