A particle moves under a conservative force with potential energy \(V(\boldsymbol{r})\). The point \(\boldsymbol{r}=\mathbf{0}\) is a position of equilibrium, and the axes are so chosen that \(x, y, z\) are normal co- ordinates. Show that, if \(V\) satisfies Laplace's equation, \(\boldsymbol{\nabla}^{2} V=0\) (see \(\left.\S 6.7\right)\), then the equilibrium is necessarily unstable, and hence that stable equilibrium under purely gravitational and electrostatic forces is impossible. (Of course, dynamic equilibrium stable periodic motion - can occur. Note also that the two- dimensional stable equilibrium of Problem 11 does not contradict this result because there is another force imposed, confining the charge to the horizontal plane.)

Step by step solution

01

Calculate the force from the potential energy

In order to calculate the force F acting on the particle, we need to use the gradient of the potential energy V(𝚿) with respect to its position. This can be represented as a vector: \[\boldsymbol{F} = -\boldsymbol{\nabla}V.\]

02

Calculate Laplace's equation

We are given that the potential energy V satisfies Laplace's equation, which can also be represented as a scalar: \[\boldsymbol{\nabla}^{2}V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} = 0.\]

03

Determine the stability of the equilibrium

To determine the stability of the equilibrium, we need to find the nature of the equilibrium by analyzing the second derivatives of potential energy in x, y, and z directions. If all the second derivatives are positive, the equilibrium is stable; if at least one of them is negative, the equilibrium is unstable. According to the Laplace's equation, the sum of these second derivatives is zero. That implies, at least one of them must be negative, as all three cannot be positive (which would result in a non-zero sum): \[\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} = 0,\] Therefore, the equilibrium is necessarily unstable.

04

Conclusion

We have shown that if the potential energy V satisfies Laplace's equation, then the equilibrium is necessarily unstable, and hence stable equilibrium under purely gravitational and electrostatic forces is impossible.

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