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I was trying to go through a simple exercise and was getting tripped up on the mathematical intuition of my elementary physics. I thought I should be able to get any old central force from the central potential by noting the curl of the force is zero thus there is some potential field which satisfies:

$F_C = -\nabla U_C$.

Where the C is a subscript for central. Now let's take the basic central potential and account for the central object being displaced from the origin by a vector $\vec{x}$. That is to say that

$U_C(\vec{r}) = \frac{k}{|\vec{r}-\vec{x}|}$.

Where k is just some factor. Since this is only a radial potential then I only need concern myself with the $\hat r$ component of the gradient. However going through the exercise as described you arrive at

$F_C = \frac{|\vec{r}|-|\vec{x}|cos(\theta)}{|\vec{r}-\vec{x}|^3} \hat r$

I believe I stumbled upon the subtly of the problem. The $\nabla$ is not the $\nabla$ of my coordinate system, but the $\nabla$ of the central bodies coordinate system, which I will now denote as $\nabla'$, thus it is only correct to say

$F_C = -\nabla' U_C$.

Could anyone comment on the issues at hand. The potential should not change from being conservative to no longer being conservative by simply shifting frames, so why do I have to use a special frame's $\nabla$?

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  • $\begingroup$ if in doubt (and unexperienced) work using Cartesian coordinates. That being said: You simply miscalculate the gradient. F_c is not directed in the direction $\hat{r}$ but $\hat{r-x}$. If you want to compute the Gradient in spherical coordinates you have to expand the absolute value in the potential which includes a term $\cos{theta}$ and thus an additional contribution. A simpler way would be to work in appropriate curvilinear coordinates, but you dont have to do that. Given that you do the calculations right (for reference: en.wikipedia.org/wiki/Curvilinear_coordinates ) $\endgroup$ – Bort Jul 30 '15 at 10:46
  • $\begingroup$ Thanks, I completely forgot after expanding that I had a $\theta$, which then required me to consider the $\theta$ component of the gradient. Even though a $\theta$ is starring at me in my incomplete force... I should have noticed that after only getting things to work when things where collinear and the $\theta$ term kills off the $\hat \theta$ term. $\endgroup$ – Novice C Jul 30 '15 at 10:55
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( I turned my comment into an answer to signal that this is question is answered)

if in doubt (and unexperienced) work using Cartesian coordinates. That being said: You simply miscalculate the gradient. $F_c$ is not directed in the direction $r$ but $r−x$. If you want to compute the Gradient in spherical coordinates you have to expand the absolute value in the potential which includes a term $\cos\theta$ and thus an additional contribution. A simpler way would be to work in appropriate curvilinear coordinates, but you dont have to do that. Given that you do the calculations right (for reference: en.wikipedia.org/wiki/Curvilinear_coordinates )

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