0
$\begingroup$

In, Nonlinear Dynamics And Chaos, 2nd edition page 160, by Steven H. Strogatz, he writes

Let’s be a bit more general and precise. Given a system $$\dot x =f(x),$$ a conserved quantity is a real-valued continuous function $E(x)$ that is constant on trajectories, i.e., $$dE/dt =0.$$ To avoid trivial examples, we also require that $E(x)$ be non-constant on every open set. Otherwise a constant function like $E(x) ≡ 0$ would qualify as a conserved quantity for every system, and so every system would be conservative! Our caveat rules out this silliness.

What does this condition translate to physically?

I get that he is saying that, for example, if a satellite orbiting a star has its distance increased from $r_1$ to $r_2$, then $E(r_2)-E(r_1)$ must be non-zero. And for this specific case, where we can calculate the energies explicitly, it is correct that $E(r_2)-E(r_1)$ must be non-zero.

But how can we say this in general, before calculating energy function, for all conservative systems?

$\endgroup$
1
  • $\begingroup$ I think it is just a classification issue? A conserved quantity should be a meaningful notion, otherwise why would we have it? For instance, the number of fairy unicorns flying alongside any elementary particle is a conserved quantity (it is always zero), but we don't want to therefore conclude that this is a quantity we should track or care about. $\endgroup$ Commented Jul 15 at 14:21

1 Answer 1

2
$\begingroup$

If $E$ does not depend on $x(t)$, then $\frac{dE}{dt}=0$ doesn't give you any information about $x(t)$. Since the entire goal of classical mechanics is to solve for $x(t)$, this means such an $E$ would be useless. As an example, the value of $1$ doesn't change with time, but that does not help you understand the motion of a pendulum.

On the other hand, a standard energy formula like $E=\frac{1}{2}m \dot{x}^2 + V(x)$ manifestly does depend on $x$, and so $\frac{dE}{dt}=0$ does let you infer something about $x$. For example, when $V(x)=0$, conservation of energy tells you that the velocity must be constant (Newton's first law).

$\endgroup$
1
  • $\begingroup$ Strogatz initially derived the equation you provided and then said the above statement, which got me confused as to why make an extra clarification. But now I realize that in fact he was going backward, that is, by showing that his statement is much more general and allows for many more types of energies and not just the one he derived and you mentioned $\endgroup$ Commented Jul 16 at 4:38

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.