I cannot indicate an error in the following reasoning if it is done in the framework of classical physics.


Let's make the imaginary setup with two wedges and the ball, when the gravity in the left and right half-spaces is anti-parallel, namely:

$$g(x)=\begin{equation} \begin{cases} g, x \le 0\\ -g, x > 0 \end{cases}\,. \end{equation}$$

Let's put a ball on the lower wedge. Then, when released, the ball to the left starts to accelerate indefinetetly (dashed line), repeating the cycle.

I am aware that such a configuration of gravitational fields is impossible to create. From theoritical point of view I suspect there are some limitation by general relativity that the gravitational field should be smooth as well as such a system will start to emit strong gravitational waves losing energy and finally it will stabilize in some equilibrium.

But if we take classical physics what is it violated in such a configuration?

  • 2
    $\begingroup$ physics.stackexchange.com/q/584678 and en.wikipedia.org/wiki/Conservative_force $\endgroup$
    – BowlOfRed
    Commented Jul 10 at 16:32
  • 1
    $\begingroup$ Seems improbable to me. What are mass distributions which attracts the ball,- only this pair of wedges, or there's some more heavier bodies under/above wedges which attracts ball too ? In any case this setup feels highly unstable, wherever common COM is, it looks like ball will not move at all or will go to the lower gravitational potential (if any) and stay there. However, Gravity_train reasembles close idea. BUT, it's not perpetual engine, since you can't extract energy from it. (Try to extract kinetic energy from orbital satellite and see...) $\endgroup$ Commented Jul 10 at 18:17

2 Answers 2


If we let the $y$-axis point upwards then OP's 2D gravitational field is $$\vec{g}~=~ \begin{pmatrix} 0 \cr g~{\rm sgn}(x)\end{pmatrix}.$$ It has a non-zero curl $$(\vec{\nabla}\times \vec{g})_z~=~2g\delta(x),$$ and is hence not a conservative field.

  • $\begingroup$ Haha I remember these kind of problems in Griffiths, anyways short and sweet answer! $\endgroup$ Commented Jul 10 at 18:17
  • $\begingroup$ So basically is it prohibited to have non-conservative field in classical mechanics? Since if it is non-conservative it goes e.g. to thermodynamics (like friction), or relativaty (like magnetic)? $\endgroup$
    – Artem
    Commented Jul 11 at 15:03
  • $\begingroup$ A non-conservative field is not prohibited in CM per se; however any realization of OP's system (plus possible hidden sectors) is expected to obey total energy conservation. $\endgroup$
    – Qmechanic
    Commented Jul 11 at 18:38

The potential that yields your provided force field is discontinuous. One possible choice would be $$V(x,y) = \begin{cases} gy & x\leq 0 \\ -gy & x>0 \end{cases},$$ which has a dicontinuity at $x=0$ whenever $g>0$. Hence, the force in $x$-direction, i.e. $\partial_x V$ is not defined at $x=0$. You can solve Newton's equations only on one half of the anti-parallel gravitational system.

This is a purely mathematical argument. A physical argument would require a better (physically stable) model that yields a continuous potential. For instance: You could smoothen out $V(x,y)$ in $x$-direction introducing a large gradient around $x=0$. This will drastically slow down the particle at the transition point and ensure energy conservation.

  • $\begingroup$ So step-wise character of the acceleration prevents having a motion in both hemi-spaces? Since if I take a ball which is falling from the table, we also observe step-wise character of force change, but the ball is falling from the table - OK, nothing special happend? $\endgroup$
    – Artem
    Commented Jul 11 at 15:00
  • $\begingroup$ The discontinuity of the force is not (necessarily) the problem, but the one in the potential is. In your example, the potential is continuous, but not $C^1$, looks like one can solve the trajectory piecewise. $\endgroup$ Commented Jul 11 at 17:40

Your Answer

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

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