# What causes a force field to be “nonconservative?”

A conservative force field is one in which all that matters is that a particle goes from point A to point B. The time (or otherwise) path involved makes no difference.

Most force fields in physics are conservative (conservation laws of mass, energy, etc.). But in many other applications, the time paths DO matter, meaning that the force field is not "conservative."

What causes a force field to be "non-conservative?" Could you give some examples (probably outside of physics)?

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 I am not sure if this qualifies as a force field, but if there is friction in the system, the force is definitely not conservative – Thomas Rot Jul 9 '12 at 13:24 @ThomasRot: Good point. Almost all physics problems say, "neglect frictional forces." – Tom Au Jul 9 '12 at 13:41 Non-conservative only matters in macroscopic level, in atom level every type of force is conservative. – Shuhao Cao Jul 9 '12 at 14:16 @ShuhaoCaoL: Figured that might be the case. Thanks for the confirmation. – Tom Au Jul 9 '12 at 14:20

## migrated from math.stackexchange.comJul 9 '12 at 14:51

A force field $F_i(x)$ is conservative if for every curve $C$ from a point $y_1$ to a point $y_2$, we have $\int\limits_C F_i(x)\mathrm{d}x^i$, so that the energy difference between $y_1$ and $y_2$ is independent of the curve taken from one to the other. Equivalently, the integral around a closed curve must be zero, $\oint\limits_C F_i(x)\mathrm{d}x^i=0$ for every closed curve $C$. Alternatively, we require $\nabla\times F=0$, so that we can write $F=\nabla V$; that is, the curl of the force field is zero so that the force field can be expressed as a divergence. Generalizations of this elementary account to higher dimensions in terms of differential forms are possible.