# Work of a conservative force

Given a conservative force $$\mathbf{F}=-\nabla V$$, how do I show that $$W=V(\mathbf{x}_2)-V(\mathbf{x}_1)\quad ?$$ I can start as follows $$W=\int_{\mathbf{x}_1}^{\mathbf{x}_2} \mathbf{F}\cdot d\mathbf{x} = \sum_{i=1}^3 \int_{\mathbf{x}_{1,i}}^{\mathbf{x}_{2,i}} F_i dx_i = -\sum_{i=1}^3 \int_{\mathbf{x}_{1,i}}^{\mathbf{x}_{2,i}} \partial_i V dx_i=???$$ The integration is on any path connecting $$\mathbf{x}_1$$ and $$\mathbf{x}_2$$.

Is there a way to compute it with elementary calculus (integration by substitution, FTC, and so on)?

One simple way is as follows: given $$\mathbf{x}\equiv \mathbf{x}(t)$$ we can write

$$\int_{\mathbf{x}_1}^{\mathbf{x}_2}\mathbf{F}(\mathbf{x}(t))\cdot d\mathbf{x} = \int_{\mathbf{x}_1}^{\mathbf{x}_2} \left(F_1(\mathbf{x}(t))dx + F_2(\mathbf{x}(t))dy+F_3(\mathbf{x}(t))dz\right) = \\ = \int_{\mathbf{x}_1}^{\mathbf{x}_2} (F_1(\mathbf{x}(t))\dot{x}+F_2(\mathbf{x}(t))\dot{y}+F_3(\mathbf{x}(t))\dot{z})dt$$

since

$$dx = \frac{dx}{dt}dt = \dot{x}\,dt$$

Given that the force is conservative $$F_i = dV/dx_i$$ and so the integral becomes

$$\int_{\mathbf{x}_1}^{\mathbf{x}_2} \left(\frac{\partial V}{\partial x}(\mathbf{x}(t))\dot{x}+\frac{\partial V}{\partial y}(\mathbf{x}(t))\dot{y}+\frac{\partial V}{\partial z}(\mathbf{x}(t))\dot{z}\right)dt = \int_{\mathbf{x}_1}^{\mathbf{x}_2}\frac{d}{dt}V((\mathbf{x}(t))dt = V(\mathbf{x}(t_2))-V(\mathbf{x}(t_1))$$

since $$\mathbf{x}_1 = \mathbf{x}(t_1)$$ and so on. The only mathematical tool we used is the chain rule.

• Thank you. This is exactly what I was looking for. I consider $t$ just as a parameter of the path, since the curve $\mathbf{x}(t)$ is generic, not necessarily a trajectory. Feb 23, 2020 at 19:53
• Exactily, $x(t)$ it's just a parametrization of the path Feb 23, 2020 at 19:54
• I'm having a bit of trouble understanding that last step. $(\partial V/\partial x)\dot{x}=\partial V/\partial t$. But, we get one of these for each coordinate, so why don't we end up with $\int_{\mathbf{x}_1}^{\mathbf{x}_2} 3 dV(\mathbf{x}(t))$? Oct 19, 2021 at 14:46
• @MattHusz I don't understand your reasoning. By definition $dV(x_i(t))/dt$ total derivative gives $\partial_i V(x_i(t)) \dot{x}^i$ just by virtue of the chain rule. Oct 19, 2021 at 17:02
• @DavideMorgante yep you're right my mistake. Thanks for explaining! Oct 19, 2021 at 20:37