I am studying for my exam and wanted to clarify if the following I got from Taylor are true, because we have written something different in my lectures:

$$\nabla\times\vec F=0\ \ \Leftrightarrow\ \ \oint\limits_C\vec F\cdot d\vec r =0$$

If above holds for a force, then we are mathematically allowed to find a scalar potential field, such that: $$\vec F=-\nabla V$$

And if, in addition to above, the force is also non-dependent on velocity $\dot r$ and time $t$, then the mechanical energy of the system is conserved, right?

Is there any relationship between the force being dependent on $\dot r$ or $t$, and the force having a corresponding potential field? Can a force dependent on $\dot r$ also have a corresponding scalar potential field?


1 Answer 1


If the force is a function of $t$ or $v$, it is not possible to write: $$F = -\nabla V = - \left (\frac{\partial V}{\partial x} , \frac{\partial V}{\partial y} , \frac{\partial V}{\partial z} \right )$$ because for the same point $x,y,z$ it would be possible different values for $F$. The above expression states that $F = F(x,y,z)$


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