From conservation of energy, show that for a conservative force field, $\ \vec F=-\vec \nabla \phi $ and $\ \vec \nabla X \vec F $

Attempt ::

If the force is conservative, i.e. total energy is conserved, then the work done is equal to minus the change in potential energy,

$\ d\phi = -dW = -F.dr = -F_idx_i$

$\ d\phi = \frac{\partial \phi}{\partial x_i} dx_i=(\vec \nabla \phi)_idx_i$

Thus, $\ \vec F=- \vec \nabla \phi$

Is this correct?

I couldn't start with the next part of question. Any hints?

  • 2
    $\begingroup$ It seems like you're assuming what you're actually trying to prove, that is, $\text d W = -\text d \phi$. Hint for $F=\nabla \phi$: $\intop ^{\mathbf r} _{\mathbf r _0} \mathbf F \cdot \text d \mathbf r$ is path independent, so it defines a function. $\endgroup$
    – pppqqq
    Dec 5, 2016 at 16:05
  • $\begingroup$ There is a proof on wikipedia... $\endgroup$ Dec 5, 2016 at 16:36

2 Answers 2


Do you mean "show that $\nabla \times \vec{F}=0$" ? If so, it's just a matter of taking the curl of $\nabla \phi$ and noting that $\frac{\partial^2}{\partial_{x_i}\partial_{x_j}}=\frac{\partial^2}{\partial_{x_j}\partial_{x_i}}$.


$T_2 - T_1 = \Delta K = \int \mathbf F \cdot \mathbf{dr}$ states that the net work done on an object is equal to its change in kinetic energy. The right hand side encodes that work done by both conservative and non conservative forces. The former type can always be written as the gradient of some scalar function, the potential energy. Viz. $$\int \mathbf F \cdot \mathbf{dr} = \int (\mathbf F_{\text{cons}} + \mathbf F_{\text{non-cons}}) \cdot \mathbf{dr} = \int \mathbf F_{\text{cons}} \cdot \mathbf{dr} + \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr}.$$ Now, $\mathbf F_{\text{cons}} = - \nabla \phi$, where $\phi$ is the scalar function alluded to previously. Then $$\Delta K = -\int \nabla \phi \cdot \mathbf{dr} + \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr} = - \Delta \phi + \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr} $$ Rearranging gives $$\Delta K + \Delta \phi = \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr} $$ If there are no non-conservative forces at play (as you consider here) then the right hand side is identically zero and you have conservation of mechanical energy. If there are, it is not zero and corresponds to a dissipative term.

The latter statement is that given $\mathbf F = - \nabla \phi$, $\nabla \times \mathbf F = 0$ follows. To see this, we do an explicit computation: $$(\nabla \times \nabla \phi)_i = \epsilon_{ijk} \partial_j \partial_k \phi = 0 \,\,\,\text{for all}\,\,\, i$$ as the contraction is between an antisymmetric levi civita symbol and a symmetric $\partial_j \partial_k$ pair so thereby vanishes.


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