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I'm working on a classical mechanics problem in which the problem states that a particle of mass $m$ moves in a central field of attractive force of magnitude:

$$F(r, t) = \frac{k}{r^2}e^{-at}$$

  • $r$ is the distance from the force center,
  • $t$ is time,
  • and $k$, $a$ are constants.

I am concerned about the $t$ in the exponent. I did more than half of the question already (which involves finding lagrangian/hamiltonian equations) using a potential I'm not even sure is correct.

I calculated it by taking the integral of $F(r, t)$ with respect to $r$. It is just that there is a $t$ in the magnitude so I know I shouldn't be able to do that.

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Hi @Nosyt, and welcome to Physics Stack Exchange! I've updated you question and add some LaTeX. And if you have doubts in you calculations - add it. Without this it can't be checked on correcteness. – m0nhawk Oct 23 '12 at 5:29
A time dependent force is not necessarily conservative (usually it is not) so there might no potential exist. – Fabian Oct 23 '12 at 5:47
Nosyt, to clarify: were you given the potential $U(x,t)$, or were you given the force $F$? Or are you using $U$ for the force (which is quite confusing usage)? – David Z Oct 23 '12 at 5:49
This is a copy and paste of the question from a word file. My professor often makes mistakes :( 1. A particle of mass m moves in a central field of attractive force of magnitude (the one written above in latex) , where k and a are constants, t is the time , and r is the distance of m from the force center. (a) Find the Lagrangian and Hamiltonian functions, (b) Find Lagrange’s and Hamilton’s equations, ( c) Is H the total energy? Give reasons , and (d) Is H constant of Motion? Give reasons. – Nosyt Oct 23 '12 at 6:00
@Nosyt: please change it back into the more conventional $F$. – Fabian Oct 23 '12 at 9:02

There is no potential energy associated with your system as the force is time dependent. However, from your comment I understand that you want to know the

  • Lagrangian
  • Hamiltonian
  • Total energy

A particle which moves under the action of the time-dependent force follows Newton's equation of motion $$m \ddot r = F(r,t) + m r \omega^2$$ with $$F(r,t)=\frac{k}{r^2}e^{-at}$$ and $\omega$ the angular velocity.

This equation can be seen as the Euler-Lagrange equation of the Lagrangian $$L= \frac{m \dot r^2}{2} - \frac{l^2}{2 m r^2} - \frac{k}{r} e^{-a t}$$ with $l= m r^2 \omega$ the angular momentum (around the fixed axis of rotation) which is conserved.

The Hamiltonian is then given via a Legendre transform $$ H = p \dot r -L= \frac{p^2}{2m} + \frac{l^2}{2 m r^2} + \frac{k}{r} e^{-a t}$$ with $p= \partial L/\partial_{\dot r} = m \dot r .$

The Hamiltonian is also the total energy of the system (which is not conserved).

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Comment to the answer(v1): It should be emphasized that it's the angular momentum, not the angular velocity, that is a constant of motion. – Qmechanic Oct 23 '12 at 23:03
@Qmechanic: thank you for the comment. I edited the answer and made this fact explicit. – Fabian Oct 24 '12 at 7:50

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