1
$\begingroup$

I've edited my question to clear any possible confusing parts:

This an exercise from the book "The Theoretical minimum", I'm paraphrasing.

In classical mechanics, dynamical laws must be reversible and deterministic.

Consider the function N(n) where:

N is a point on a infinite line, and n symbolizes time.

Here are some possible dynamical laws of this system:

  1. N(n+1) = N(n) + 1

  2. N(n+1) = N(n) - 1

  3. N(n+1) = N(n) + 2

  4. N(n+1) = N(n)²

Determine which equations are allowable. (deterministic and reversible)

--

The solutions are on:

http://www.madscitech.org/tm/slns/l1e3.pdf

What I don't understand is why equation 4, N(n+1) = N(n)² is not allowable.

If N is 2 we get 4 and then 16. If we "reverse the arrows" we get 16, 4 and 2. If N is 1, it will never change.

How is this not deterministic and reversible?

Thanks in advance!

$\endgroup$
2
  • $\begingroup$ perhaps you can rephrase what your question exactly is? Are you saying $N$ is a function of $n$, which itself is time (usually $t$)? What exactly does 'allowable' mean in this case? As it stands now it's a bit hard to understand what you're asking... $\endgroup$
    – JMJ
    Commented Nov 28, 2013 at 21:36
  • $\begingroup$ Hey sorry I thought I might be a bit confusing. I'm very new to physics. Yes N is a function of n, which is time. It isn't t because t is for continuous evolution of time, whereas here time evolves in a stroboscopic manner. (I'm paraphrasing the book). Allowable means that the system is both deterministic and reversible. My question is, how is the equation N(n+1) = N(n)² not allowable. Hope this helps. $\endgroup$
    – user34914
    Commented Nov 28, 2013 at 21:41

2 Answers 2

1
$\begingroup$

Is it because there are two arrows into each state (which happen to not be shown in the solution)? Both $N(n)$ and $-N(n)$ go to $N(n+1)=N^2(n)$ under time advancement. Thus, under time reversal, you're not sure one to go back to.

$\endgroup$
0
$\begingroup$

Actually, if you know that your current state is $N(n)=16$, there's no way for you to discern under that law whether $N(n-1)=4$ or $N(n-1)=-4$. There isn't a constraint stating that $N(k)\ge0$ in the problem statement.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.