1
$\begingroup$

Yes, I know the title seems stupid since the most important property of potential is that it's actually path independent. But I have a point.

I just want to know is it possible to define a function like regular potentials with the difference that they carry the properties of both conservative and non-conservative fields and then derive the well-known ordinary "Potentials" as a special case of the first function which is just path-independent?

The reason I pursue such a thing is there are situations which there exists a non-conservative field in the system and we are studying the system by using Lagrangian or Hamiltonian formalisms. Then of course there wouldn't be such terms as $V$(Potential energy) inside the Lagrangian or Hamiltonian functions while the non-conservative field must somehow show its effects. So there should be a function similar to $V$ which shows the effects of the non-conservative field. Besides I always wondered how can we study non-conservative fields in quantum mechanics since our only chance to show the effect of fields in the "Schrödinger equation" is just manipulating the $V$ term which can only be associated to conservative fields(I know in quantum mechanical levels most of the fields are conservative but I expect quantum mechanics as a universal theory be able to describe every field whether being conservative or non-conservative).

Now the question is:

Is it possible to assign a (possibly path-dependent) function to every field and then derive the regular "Potential" as a special case for conservative fields(of course in order to study non-conservative fields in Hamiltonian and Lagrangian formalisms of classical mechanics or quantum mechanics easier)?

$\endgroup$
4
  • 2
    $\begingroup$ (I seem to be posting this a lot lately) Might be useful: Helmholtz decomposition $\endgroup$ – AccidentalFourierTransform Nov 28 '16 at 18:05
  • 1
    $\begingroup$ I went to talk recently enough by Michael Berry, which was about "curl forces", forces with non-zero curl, and which consequently aren't derivable from a potential. He pointed out that lots of things change, like the dynamics won't be Lagrangian or Hamiltonian, there can be a break between symmetry and conserved quantities, Symmetry without a conserved quantity, or a conserved quantity but no symmetry. That and there can be no obvious way to quantise the theory, without a hamiltonian or Lagrangian that is. $\endgroup$ – snulty Nov 28 '16 at 18:55
  • 1
    $\begingroup$ As a quote from the end of a Berry paper (2012) "Classical dynamics with curl forces, and motion driven by time-dependent flux" - "Finally, there is the question of whether quantum mechanics can be defined for curl forces. We do not know the answer in general, but remark that in the case of localized time-dependent flux the motion could perhaps be quantized on the multiply-connected space of the plane minus the origin, via the Hamiltonian (5.5)." $\endgroup$ – snulty Nov 28 '16 at 18:58
  • 1
    $\begingroup$ I don't know enough about this to be able to comment much more on it, only that I was at one of his talks on this stuff, and that's parts of what I recall. Apologies if I may have misremembered parts! $\endgroup$ – snulty Nov 28 '16 at 19:00
1
$\begingroup$

Yes. Let there be given a covector field $F$. Consider an arcwise connected component $M$ of the manifold with a fiducial point ${\bf r}_0\in M$. For each point ${\bf r}_1\in M$ and oriented curve $\gamma$ from ${\bf r}_0$ to ${\bf r}_1$, define the potential-like functional $$U[\gamma]~:=~-\int_\gamma\! \langle \mathrm{d}{\bf r}, F\rangle~=~-\int_\gamma \! \mathrm{d}\lambda ~\langle \dot{\bf r}, F\rangle, \qquad \dot{\bf r}~:=~\frac{\mathrm{d}{\bf r}}{\mathrm{d}\lambda}.\tag{1}$$ This definition (1) reduces to a potential function $U({\bf r}_1)$ for conservative fields.

$\endgroup$
4
  • $\begingroup$ Thank you. That was really helpful. However since I'm not much familiar with the language of topology and differential geometry, can you also add an "Analysis"-friendly definition of the concept? By the way would you also write the generalized "Schrödinger equation" or the generalized classical "Newton's second law($-\frac{\partial V}{\partial x}=m\frac{\mathrm{d}^2 x}{\mathrm{d}t^2}$)" for both conservative and non-conservative fields in general using the quantity you defined? $\endgroup$ – Hamed Begloo Nov 29 '16 at 10:25
  • $\begingroup$ Thank you again. Just one more thing: Does it mean that I can replace the function defined in your post with the $V$ term in the "Schrödinger equation" or "Newton's second law" to obtain the equation for any arbitrary field whether being conservative or non-conservative? $\endgroup$ – Hamed Begloo Nov 29 '16 at 11:50
  • $\begingroup$ Newton's second law, which deals with forces, is not affected by the presence (or not) of potentials. $\endgroup$ – Qmechanic Nov 30 '16 at 10:20
  • $\begingroup$ I don't talk about that version of "Newton's second law" which contains the forces.In the formalisms of analytical mechanics we can reach the equations governing "Time evolution" of the system/particle. In quantum mechanics it is $-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{{\mathrm{d} x^2}}\psi(x) + V(x)\psi(x) = E\psi(x)$ and in classical mechanics it is $-\frac{\partial V}{\partial x}=m\frac{\mathrm{d}^2 x}{\mathrm{d}t^2}$.Now my question is: Can I replace the $V$ term in these questions with the function you defined to derive the general equation which holds for non-conservative fields as well. $\endgroup$ – Hamed Begloo Dec 1 '16 at 17:18

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.