Let us consider a system whose evolution is given by dependency of $N$-dimensional real-values vector on time: $\vec r = \vec r (t)$ (or in terms of components $x_i = x_i (t)$).

Let us also assume that second order derivative of the vector depends on $\vec r$:

$$ \ddot {\vec r} = \vec f(\vec r), $$

or in terms of the components:

$$ \ddot x_i = f_i(x_1, x_2, \dots, x_N). $$

Can we show that in this case, there is some function of $\vec r$ and $ \dot {\vec r}$ that will be constant in time no meter what initial conditions we started from?

$$ E(\vec r, \dot {\vec r}) = \text{const} $$

I am obviously trying to make an analogy with the energy conservation law of classical mechanics. So, in the end we probably need to show that

$$ E(\vec r, \dot {\vec r}) = T(\dot {\vec r}) + V(\vec r). $$

  • 1
    $\begingroup$ Homogenity In time leads to conservation of energy. And the proof can be found in any analytical classical mechanics book like goldstein. The condition is that lagrangian should not have explicit dime dependence $\endgroup$ – crabNebula Oct 2 '20 at 14:52
  • $\begingroup$ physics.stackexchange.com/q/72852 $\endgroup$ – crabNebula Oct 2 '20 at 14:53
  • $\begingroup$ You must have a symplectic structure and "time homogeneity", otherwise you are just looking for a generic integral of motion of a generic system of differential equations (which, in general, may not have any physical meaning). $\endgroup$ – Quillo Oct 2 '20 at 15:08

I hope my answer it's not too basic, I also hope I am not overlooking something..

You have a generic system (which may not even have any obvious physical interpretation) of $N$ second-order ODEs, $$ \ddot{x}_i = f_i(x_1...x_N) \, . $$ This can be written as $2N$ equations, $$ \dot{x}_i = p_i \\ \dot{p}_i = f_i(x_1...x_N) $$ Now, for the system to be Hamiltonian, you have to be able to write it as $$ \dot{x}_i = \partial_{p_i} H \\ \dot{p}_i = -\partial_{x_i} H $$ This is possible only if $f$ is such that $$ \partial_{x_i} H = -f_i \, . $$ In order to find $H$, notice that it should also be true that $$ \partial_{p_i} H = p_i \, , $$ so that, if you assume that $H$ is separable (I am not sure if this is necessary, but it's the easy way) as $H= T(p) + V(x)$, you have $$ H = \sum_i (p_i^2)/2 + V(x_1...x_N) $$ It seems that only condition you have to ask is $f_i = -\partial_{x_i} V$, i.e. the force field is that gradient of something. If you are able to find the potential $V$ that generates $f_i$, then you can construct $H$ as $H=T+V$. Once you find $H$ you are guaranteed that the value $E$ of $H$ (defined by the initial conditions) is conserved (because you have no explicit time into $f_i$).

  • 1
    $\begingroup$ thanks for the answer. Do I get it right, to be Hamiltonian is a sufficient but probably not necessary condition for the conservation law? In other words, if a system is Hamiltonian than there is a function of coordinates and momenta (Hamiltonian) that is constant but it might be the case that for non Hamiltonian systems there is also another function of positions and momenta that is constant? $\endgroup$ – Roman Oct 5 '20 at 13:27
  • $\begingroup$ Yes, it that case you just have a generic integral of motion that is not the "usual" Hamiltonian. However, if you have such an integral of motion (i.e. the conserved quantity) and your $f$ does not allow for the construction of the "usual" H, then I don't know how to interpret that integral of motion... For sure there are non-Hamiltonian systems that allow for one (or more) conserved quantities. Finding these quantities in the general case it's difficult and there no general techniques (or, at least, I am not aware of any of them). $\endgroup$ – Quillo Oct 5 '20 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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