# Is potential energy only acquired in a *Conservative Field*?

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1. Nov 25, 2015

### NikhilRGS

Is potential energy gained or lost only in a conservative field, or when work is done against or by conservative forces?

2. Nov 25, 2015

### .Scott

I would say yes. The potential energy is the energy that is gained when the object is brought to some sort of ground position or ground state. If that amount of energy is dependent on the path to the ground position, then the potential energy would be ill-defined.

3. Nov 25, 2015

### NikhilRGS

Thank you.
Is it totally safe and correct to say that there is no potential energy gained by a body when work is done on it against a non conservative force.

4. Nov 25, 2015

### Staff: Mentor

Yes, provided of course that there are no other forces acting on the body that are conservative.

If you push a block up an incline with friction, then some of your work goes into gravitational potential energy, and the rest is dissipated into thermal energy of the block, the incline and the surrounding air (if any).

5. Nov 25, 2015

### muscaria

In the case of force of friction this is fine. In the case of magnetic forces, no work is done! In general one must be careful how one defines the relationship between the potential energy function and the work function. In the "conservative" case (I don't find this to be an appropriate term in the common way one defines such forces) where the work function is a function only of the generalised coordinates $U (q_1,\cdots,q_n)$, you can identify the potential energy as $V(q_1,\cdots, q_n) = - U (q_1,\cdots,q_n)$. In the more general case where the work function depends also on the generalised velocities, i.e. $U (q_1,\cdots,q_n; \dot{q}_1, \cdots, \dot{q}_n)$ then the total energy may be identified as the sum of kinetic and potential energies provided you define $V$ as $$V=\sum_{i=1}^n\frac{\partial U}{\partial \dot{q}_i}\dot{q}_i - U (q_1,\cdots,q_n; \dot{q}_1, \cdots, \dot{q}_n).$$
In other words the potential energy is related to the work function via a Legendre transform with the generalised velocities taken as active variables of the transformation. Interestingly, the transformed variables are gauge field/vector potential components: $A_i = \frac{\partial U}{\partial \dot{q}_i}\dot{q}_i$.

6. Nov 25, 2015

### .Scott

No. To be safe, avoid the term "potential energy" when dealing with non conservative forces.

Your statement, "there is no potential energy gained by a body when work is done on it against a non conservative force" implies:
1) that there are two forces applied to the body, one conservative (involving potential energy) and one non conservative (stated explicitly);
2) the work applied to the body opposed the non conservative force; and
3) that same work was only against the non conservative force - not the conservative one.

So the assertion that there was no gain in potential energy would be correct - but only because the mere mention of potential energy suggest that there is a conservative force.

And since it is doubtful that that is what you meant to say - the statement is at best both unsafe and incorrect and at worse ambiguous.

7. Nov 25, 2015

### NikhilRGS

Can the work done by a non conservative force like friction 'in a round trip' ever be zero?

8. Nov 26, 2015

### muscaria

Given that the force of friction is always directed opposite to the displacement of an object, the work done $\textit{by}$ the object on the surroundings will always be positive (or equivalently, the work done $\textit{on}$ the object will always be negative). Thus as long as no extra external force is acting on the object, it will gradually slow down as energy is being transferred to the surroundings in the form of heat and the surroundings temperature is raised. So because the force of friction is always directed opposite to displacement, the work done in a round trip can never be zero. This holds for any closed curve which has at least one point with non-vanishing force of friction, i.e. the only way the work done can be zero is if the force of friction along the closed curve vanishes at every point along the curve.

9. Nov 26, 2015

### nasu

Actually the question is a little poorly formulated.
If there are no conservative forces, there is no potential energy to talk about. So the question about the variation (or non-) of a non-existing (or not defined) thing is not really meaningful.

If the system contains both conservative and non-conservative forces, whatever the PE does depends only on the work of the conservative forces. The work of the non-conservative ones has no effect on the PE.