The equivalent problems are also found in Marion problem 7-22, and other formal classical mechanics textbook. Here what i want to know why instructor solution and some websites gives this kinds of approach.

A particle of mass m moves in one dimension under the influence of a force \begin{align} F(x,t) = -k x^{-2} exp(-t/τ) \end{align} where k and τ are positive constants.
what i want to do is compute its lagrangian.

\begin{align} & F(x,t) = - \frac{\partial U(x,t)}{\partial x}, \end{align}

First i know that this kinds of approach (The $F=-\nabla U$ is only holds for conservative forces, $i.e$, path-independent, position dependence potential), is vaild indeed

Many similar textbook, and their solution used above relation to rederive $U$, and compute $L$ and $H$.

I think their purpose is to show that even if $H$ is written as $H=T+U$, this $H$ is different from $E$. $i.e$, even though $H$ is written as $H=T+U$, since the force is depend on time explicitly, $H$ is no longer $E$.

I understand the procedure but not accept the first part. In general (?) i know that time dependent force is non-conservative force. Am i wrong? Does $F=-\frac{\partial U}{\partial x}$ kinds of thing always holds?

cf, from wiki

Mathematical description[edit] A force field F, defined everywhere in space (or within a simply-connected volume of space), is called a conservative force or conservative vector field if it meets any of these three equivalent conditions:

  1. The curl of F is the zero vector: \begin{align} \nabla \times \vec{F} = \vec{0}. \, \end{align}

  2. There is zero net work (W) done by the force when moving a particle through a trajectory that starts and ends in the same place:
    \begin{align} W \equiv \oint_C \vec{F} \cdot \mathrm{d}\vec r = 0.\, \end{align}

  3. The force can be written as the negative gradient of a potential, \begin{align} \Phi: \vec{F} = -\nabla \Phi. \, \end{align}


1 Answer 1


If you are wondering if $F=\frac{-\partial U}{\partial x}$ holds, you may check the equivalent condition that you posted, $\nabla\times\vec{F}=0$. In this case the curl will come out to be zero, so you may then proceed to write the force as the negative gradient of a potential. This appears to work because the del operator is not taking time into account, only the path. That is where the hamiltonian then comes in.

In this case, they are not saying that $H\neq E$ because of the time dependence. Since $H=U+T$ and $E=U+T$, here $H=E$ (though this is not always the case). Since H contains time explicitly, H is not a conserved quantity. Since $H=E$, energy is therefore not conserved even though the force appeared to be conservative. That is what they were demonstrating. If $H\neq E$, then H not being a conserved quantity would be irrelevant to whether energy was conserved or not.

  • $\begingroup$ I find it difficult to find this force conservative, what about the net work when the particle goes back and forth to the starting place? the net work certainly will not be zero (the force changes with time along the path, it is a function of t so it changes when you are coming back) $\endgroup$
    – user83548
    Commented May 17, 2016 at 16:47
  • $\begingroup$ physics.stackexchange.com/questions/27896/… $\endgroup$
    – user83548
    Commented May 17, 2016 at 16:56
  • $\begingroup$ The curl is only defined for 3D vector fields. I don't think it makes sense to use the curl condition you mention for this 1D force. $\endgroup$
    – Paul T.
    Commented May 17, 2016 at 17:16
  • 1
    $\begingroup$ Bruce, you are correct, this force is not conservative. This is being shown here when they say H = E and H is not conserved because it contains time. As for the curl only being defined only in three dimensions, you may take the curl in one dimension with two of the components being zero. While I could be incorrect, I think that going to one dimension this way and showing the curl to be zero still implies a conservative force (though in this case the time dependence makes it not conservative). $\endgroup$ Commented May 17, 2016 at 17:38
  • 1
    $\begingroup$ the actual definition is (2). (1) and (3) are equivalent to (2) only when the field is a function of position only. $\endgroup$
    – user83548
    Commented May 18, 2016 at 15:34

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