# Conservative Forces & Conservation of Energy?

I'm trying to relate them, I'm trying to find the key relation that would show how the conservative forces serve conservation of energy. How would they relate?

Also, how are non-conservative forces related to conservation of energy?

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Note that with both conservative and non-conservative forces, energy is conserved. However, the potential energy of a non-conservative field is not possible to express as a function simply of co-ordinates. – Meow May 20 '14 at 15:42
What do it mean to say that energy is conserved based on the forces? How would we know that energy is conserved due to the forces regardless being conservative for not?\ – XCIX May 21 '14 at 2:23
Even without the notion of forces, you can have conservation of energy: physics.stackexchange.com/questions/68591/… – jinawee Jun 13 '14 at 8:21

Let $T=\frac12mv^2$. Then $$\mathrm dT= mv\dot v\,\mathrm dt=m\dot vv\,\mathrm dt= F\,\mathrm dx$$ In case of conservative forces, $$F\,\mathrm dx=-\mathrm dV$$ and thus $$\mathrm d(T+V)= 0$$ This misses some physically interesting cases (in particular the Lorentz force) so let's see if we can do better: \begin{align*} \mathrm d(T+V) &= F\,\mathrm dx + \frac{\partial V}{\partial x}\mathrm dx + \frac{\partial V}{\partial v}\mathrm dv \\&= F\,\mathrm dx + \frac{\partial V}{\partial x}\mathrm dx + \mathrm d\left(\frac{\partial V}{\partial v} v\right) - \frac{\mathrm d}{\mathrm dt}\frac{\partial V}{\partial v} v\,\mathrm dt \end{align*} which implies $$\mathrm d\left(T + V - \frac{\partial V}{\partial v}v\right) = \left( F + \frac{\partial V}{\partial x} - \frac{\mathrm d}{\mathrm dt}\frac{\partial V}{\partial v} \right)\,\mathrm dx$$ You obviously get a conserved quantity if $$F = -\frac{\partial V}{\partial x} + \frac{\mathrm d}{\mathrm dt}\frac{\partial V}{\partial v}$$ This is the non-standard notion of velocity-dependent conservative forces.
The more conventional approach would be the switch to the Lagrangian description, which essentially goes from $$\frac{\mathrm d}{\mathrm dt}\frac{\mathrm dT}{\mathrm dv}=\dot p_\text{kinetic}= F_\text{Newton}= -\frac{\partial V}{\partial x} + \frac{\mathrm d}{\mathrm dt}\frac{\partial V}{\partial v}$$ to $$\frac{\mathrm d}{\mathrm dt}\frac{\partial(T-V)}{\partial v}=\dot p_\text{canonical}= F_\text{Lagrange}= \frac{\partial(T-V)}{\partial x}$$ which agree as $\partial T/\partial x=0$.