kinetic energy and conservative force field

The kinetic energy of a particle is a periodic function in time. Does it imply that the particle is in a conservative force field and there are no dissipative forces acting on it at any instance of time ?

EDIT (in view of comment by Willie)

Please consider the question as "Is the force acting on the particle conservative ?" Also consider the kinetic energy to be periodic with non zero period.

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Are you asking whether one can conclude that the particle is in a conservative force field, given that its KE is a periodic function in time? – luksen Aug 23 '11 at 9:56
@luksen : yes i mean whether it could be concluded. – Rajesh Dachiraju Aug 23 '11 at 10:17
Of course not. Let the force field be $\vec{F}(x,y,z) = y \hat{x}$. This force field is not conservative. The trajectory $(x,y,z)(t) = (0,0,t)$ is a solution to the Newtonian equation of motions. It has constant kinetic energy. In fact, given a particle trajectory you have no knowledge of the force field away from the trajectory, so how do you expect to conclude anything about the force field in the large? – Willie Wong Aug 23 '11 at 15:47
@Willie : thank you for the comment. I had a wrong choice of terminology. I'll edit the question. – Rajesh Dachiraju Aug 23 '11 at 16:48
...and what do you mean by a "conservative force" in this context now? – Willie Wong Aug 23 '11 at 17:05

No. I can take a ball and swing it back and forth periodically with my hand. The motion is periodic, but the situation is not conservative - my body generates a lot of heat.

A simple mathematical example is a forced, damped harmonic oscillator. It has a steady-state periodic solution that dissipates energy.

If you want to know whether a force field is conservative, take its curl. Time-independent force fields (force is a function of position but not time) are conservative iff their curl is zero.

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What has the body generating heat got to do with whether the situation is conservative or not? Simply holding your arms up in the air and doing no work still means your body generates excess heat within the muscles. – John McVirgo Aug 23 '11 at 19:22
@John if there's heat, energy is draining into heat, and not staying as kinetic or potential energy. I.E. the kinetic energy of the ball is not being stored in my arm, because my arm is not a perfect rubber band. – Mark Eichenlaub Aug 24 '11 at 1:33

There is quite a lot wrong with this question.

1. What is a conservative force field? As usually defined, a conservative force field is a field that is defined everywhere in space that is the gradient of some potential energy (which may be time dependent); equivalently we ask that the curl of the force field is zero. This requires knowing the force everywhere in space.
2. Suppose we know the trajectory $\vec{\gamma}(t)$ of a particle. Then by taking the second time derivative we can compute the acceleration, and using Newton's second law we can compute the force. Okay, so this gives us knowledge of the (possibly time-dependent) force field $\vec{F}$ and the space-time coordinates $(t,\vec{\gamma}(t))$: we have that $\vec{F}(t,\vec{\gamma}(t)) = m \ddot{\gamma}(t)$.
1. But what does this tell us? Absolutely nothing! At any given time, we are only given knowledge at one single point of the force. The force field could be anything at other points in space. Given the value of a function at a point, you can extend it arbitrarily! You can extend it so that the force, at that instant in time, is conservative; or you can extend it so that the force is not conservative. You can extend it anywhich way you want and still have a force compatible with the motion of your particle. (Another way to look at it using Mark Eichenlaub's answer: a particle executing simple harmonic motion could either be in a harmonic potential (so is in a conservative force field), or it could equally well be in the steady state of a driven, damped harmonic oscillator, in which case you have dissipation.
2. Okay, you say, let us suppose then the force field is time independent. This way by specifying $\gamma(t)$ you can specify what the force would be along every point visited by the particle. Now you run into two problems: (a) what if the trajectory visits the same point twice, but exhibits different acceleration at those two different times? Then it is impossible to find a well-defined time-independent force field that generate this motion. (b) what if the trajectory only visits a single curve of points? Then you have the problem as before: if a function is prescribed along a one-dimensional subset of three-dimensional space, you can essentially extend the function anyway you want. You can extend the force-field to be conservative, or you can extend it to be non-conservative.
3. And then there is the additional problem that just giving the kinetic energy is insufficient to specify the velocity of the trajectory: the kinetic energy can allow you to determine the modulus $|\dot{\gamma}|(t)$ of the particle, but it cannot allow you to determine the direction the particle is moving in. So associated with a kinetic energy there are infinitely many trajectories that intersects itself infinitely many times, and there are also infinitely many trajectories which trace out a smooth embedded curve in 3 dimensional space.
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