1
$\begingroup$

It's a well known fact that classical mechanics isn't a deterministic theory if you only include the positions and masses of various particles as part of the initial conditions. You also need to include the velocities/momenta of the various particles, which are formulated in terms of time-derivatives. This is somewhat surprising because it is natural to think that velocities/momenta aren't really intrinsic to a time, they are just time-derivatives of position.

This trend is also present in quantum physics, where (for example) the Hamiltonian of a physical system needs to be specified for the laws of physics to be deterministic. The Hamiltonian is equivalent to the total energy, where the total energy depends on other "time-derivative properties" like momentum.

Another strange feature of these "extra" quantities that are needed for determinism to work is that they are frame-relative (unlike mass/charge, for example). The velocity (and therefore momentum and kinetic energy) are all frame-relative. Not all time-derivative quantities are frame relative. For example, acceleration is frame-independent in classical physics. But, acceleration doesn’t help in delivering determinism. Surprisingly, the time-derivative quantities that give us determinism are frame-relative ones.

Question: Is it generally true that these kind of "time-derivative" properties that need to be included for determinism to work are always sensitive to an inertial frame? Or, can determinism be formulated in terms of time-derivative properties that not frame-relative? I am interested in this question both in the case of classical mechanics and quantum mechanics.

$\endgroup$
4
  • 5
    $\begingroup$ Why are you concerned that the velocity is frame-relative when the position is as well? $\endgroup$
    – G. Smith
    Jun 13, 2020 at 21:36
  • $\begingroup$ Good point. I was thinking that in spacetime theories like GR (or “Galilean space-time” for Newtonian mechanics) there are objective frame independent facts about the spacetime manifold and the fundamental fields that live on that manifold. So perhaps we can ultimately understand “positions” in a frame independent way. I just find it surprising that some time-derivative facts are frame independent (such as acceleration in Galilean space-time) and some aren’t, but the ones that are needed for determinism are frame relative. $\endgroup$
    – Taro
    Jun 13, 2020 at 21:45
  • $\begingroup$ Acceleration is frame-independent under a Galilean boost, but it isn’t under a Lorentz boost. And it isn’t frame-independent under rotations. $\endgroup$
    – G. Smith
    Jun 13, 2020 at 21:51
  • $\begingroup$ That's right, so I guess this puzzle is better motivated in non-relativistic physics where Galilean transformations are what matters? In any case, if these temporal derivatives are inevitably frame-relative, I would take that to be an answer to my question. The answer would be "no", the time-derivative-quantities that are necessary for determinism are inevitably frame-relative. $\endgroup$
    – Taro
    Jun 13, 2020 at 21:54

2 Answers 2

1
$\begingroup$

There are three points I'd like to clarify.

  • You can formulate the deterministic time-evolution laws of classical mechanics as first-order partial differential equations and thus, you wouldn't need to specify any time derivatives of quantities that specify the state of the system at a given instant of time. This is exactly what Hamiltonian formalism does. A state of the system at a given instant of time is given by $(q,p)$ where $q$ are canonical coordinates and $p$ are canonical momenta. Both $\dot{p}$ and $\dot{q}$ are then determined by the equations of motion which are given by the Hamilton's equations.

  • The time evolution of a system in quantum mechanics is governed by first order partial differential equations. So, you don't need to specify the initial values of the first order time derivatives of the state (in the Schrödinger formulation) or of the operators (in the Heisenberg formulation) in order to evaluate the deterministic time evolution of the system. The momenta that show up in the Hamiltonian of a quantum mechanical system are the canonical momenta (operators) of the same character described above. So, they should not be treated as time derivatives of the canonical positions (operators). In fact, you'd run into troubles if you try to do that (even more so than you'd in classical mechanics due to the non-commutative nature of momenta and positions).

  • As pointed out in the comments, the frame-dependence of our descriptions doesn't really rely on the usage of time derivatives in the specification of the initial state. Thus, even in the canonical formulation of mechanics (classical or quantum), our description of the system does depend on the frame. In classical mechanics, the value of the canonical positions and the canonical momenta are frame dependent and so is the quantum state of a system. However, this doesn't mean that the laws of mechanics (either classical or quantum) are frame-dependent, and that's what matters. You can formulate the laws of mechanics in a formally frame-independent way, however, the truth is that you'd always need to choose a frame of reference to actually do any experiment. And thus, you'd be able to assign concrete values to the observables only in a frame dependent manner (when you compare notes, it might turn out that some of them take frame-dependent values, for example, the value of mass. But you'd always actually measure them in a particular frame even if it might turn out that their value turns out to be the same in all frames.) The frame-independence of the laws of physics lies in the fact that the time evolution of these frame-dependent values of observables would follow the same law in each of the (inertial) frames which you can independently verify in each of the frames without appealing to some other frame.

$\endgroup$
1
$\begingroup$

If we have the 4-vectors for position and momentum, the results are not frame-dependent in classical (relativistic) mechanics.

In relativistic QM, if a spinor is the solution of an Hamiltonian, it is also not frame-dependent.

$\endgroup$

Your Answer

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

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