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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)?

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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.

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    $\begingroup$ 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. $\endgroup$ Feb 23, 2020 at 19:53
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    $\begingroup$ Exactily, $x(t)$ it's just a parametrization of the path $\endgroup$ Feb 23, 2020 at 19:54
  • $\begingroup$ 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 $\int_{\mathbf{x}_1}^{\mathbf{x}_2} 3 dV(\mathbf{x}(t))$? $\endgroup$
    – MattHusz
    Oct 19, 2021 at 14:46
  • $\begingroup$ @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. $\endgroup$ Oct 19, 2021 at 17:02
  • $\begingroup$ @DavideMorgante yep you're right my mistake. Thanks for explaining! $\endgroup$
    – MattHusz
    Oct 19, 2021 at 20:37

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