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Consider a uniform, infinitely thin disc of surface mass density $\sigma$ and radius $R$ placed in the $xy$-plane with its center as the origin. The gravitational potential at a point on the axis of the disc ($z$-axis) at a distance $z$ from the center of the disc is given by: $$\phi(z)= -2\pi G \sigma \left(\sqrt{R^2 +z^2}-z\right). $$

The $z$-component of the field at this point is given by $$f_z(z)=-2\pi G \sigma\left(\frac{z}{\sqrt{R^2+z^2}}-1\right).$$

Given the symmetry of the problem, shouldn't $\phi$ be symmetric functions of z? Here, we clearly have $\phi (-z) \neq \phi(z)$.
Also, the field should be antisymmetric, i.e $f_z(-z) = -f_z(z)$, which is not satisfied.

Qmechanic
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bhoutik
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1 Answers1

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Given the usual derivation of this potential and force, the correct interpretation of $z$ is the distance from the center of the disc, not the $z$-coordinate. There's the same issue with the gravitational potential of a point mass: $\phi(r) = GM/r.$ If you try to interpret $r$ as a coordinate that can have negative values, then you'll get the wrong answer.

To interpret $z$ as a coordinate, you can use absolute value of $z$: $$\phi(z) = -2G\sigma\left(\sqrt{R^2 + z^2} -|z|\right).$$

Mark H
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  • I do not think the same problem exists for point mass as the range of 'r' in spherical polar coordinates is 0 to $\infty$. r co-ordinate never takes a negative value. – bhoutik Jan 05 '23 at 04:46
  • @bhoutik I was using that as an example of how misinterpreting a variable can lead to wrong answers. You are right that $r$ is never negative. My argument is that $z$ is the same in this instance: a measure of distance, not of position. – Mark H Jan 16 '23 at 21:37