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According to Wikipedia "The gravitational potential $V$ at a distance $x$ from a point mass of mass $M$ can be defined as the work $W$ that needs to be done by an external agent to bring a unit mass in from infinity to that point: $$V(\vec{x}) = \frac{1}{m}\int^x _{\infty} \vec{F}\cdot d\vec{x} = \frac{1}{m}\int^x _{\infty}G\frac{Mm}{x^2}dx=-\frac{GM}{x}$$

where $G$ is the gravitational constant, and $\vec{F}$ is the gravitational force."


When multiple point masses situated at $\vec{x}_1, \ldots ,\vec{x}_n$ with masses $M_1, \ldots ,M_n$ respectively, $V(\vec{x})$ is defined as $$\sum _{i=1}^n -G\frac{M_i}{|\vec{x}-\vec{x}_i|},$$ that is, the sum over the gravitational potential regarding each individual point mass.

In this case, can $V$ still be considered the work necessary to bring a unit mass from infinity to $\vec{x}$? If so, why? If not, then why is $V$ defined as such and what interpretation can be given to the more general formula?

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  • $\begingroup$ why would it not still be the work required to bring a fiducial mass from infinity to $\vec x$? $\endgroup$
    – JEB
    Aug 17, 2021 at 18:13
  • $\begingroup$ @JEB when dealing with a single point mass situated at $x_0$ the path $C$ in the line integral $\int_C \vec{F} \cdot d\vec{x}$ is a stright line from $\vec{X}$ to $\infty$ in the direction of the vector $\vec{x}-\vec{x}_0$ (or, at least, it has as an endpoint a point in the direction of the vector $\vec{x}-\vec{x}_0$.) When deadling with multiple point masses it isn't even clear which path $C$ to take to integrate. $\endgroup$
    – Leo
    Aug 17, 2021 at 18:24
  • $\begingroup$ @JEB perhaps the idea is that, when dealing with a single point mass situated at $x_0$, the path $C$ chosen does not matter as long as its endpoint "is at infinity". Then I would immediately understand why the formula for $V$ when multiple point masses are involved is still the work needed to bring a unit mass from infinity. Yet I do not know if the first assertion is correct. $\endgroup$
    – Leo
    Aug 17, 2021 at 18:27

2 Answers 2

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We're worried about $V(\vec x)$, which is found by integrating on a path from infinity to $\vec x$:

$$ V(\vec x) \equiv \int_{\infty}^{\vec x}\vec F(\vec x')\cdot d\vec x'$$

but $\vec F=-\nabla V$, so:

$$ V(\vec x) =-\int_{\infty}^{\vec x}\nabla V \cdot d\vec x'=-(V(\infty)-V(\vec x))=-(0-V(\vec x))=V(\vec x)$$

Or: because the force is the gradient of $V$, the integral is independent of path. You can deform any curved path to a straight path when integrating over $\vec F$, and correct that integral by adding the enclosed area integral over $\nabla \times \vec F$, but that is always zero because:

$$ \nabla \times \nabla V = 0$$

everywhere.

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  • $\begingroup$ Perhaps I've done a poor job at explaining myself. It's not that choosing a different path between two points in $\mathbb{R}^n$ concerns me (this does not affect the integral, since the force is conservative). My issue is that $\infty$ is not a point, and the "way" in which the path goes to infinity can vary greatly, and I wonder if such variations matter. I explained my concern more thoroughly in here. If your answer above addresses this, please let me know. $\endgroup$
    – Leo
    Aug 17, 2021 at 22:42
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Yes, the Newtonian gravitational potential for multiple point masses is simply the sum of the gravitational potentials for each mass.

This works because Newtonian gravitation is linear. This means that the total force is equal to the sum of the forces from each mass. The same approach does not work in General Relativity because the Einstein field equations are non linear.

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  • $\begingroup$ Even understanding that the total force is equal to the sum of the forces from each mass, it is still not clear to me that $V$ is the work necessary to bring a unit mass from infinity to $\vec{x}$. Is the idea that $\int \sum \vec{F_i} d\vec{x}=\sum \int \vec{F}_id\vec{x}=\sum V_i$? If so, it's not clear to me which paths the integrals would be integrating along. $\endgroup$
    – Leo
    Aug 17, 2021 at 18:37
  • $\begingroup$ @Leo exactly, that is what linear means. Regarding the path, that doesn’t matter. For a conservative force that integral only depends on the starting and ending point, not the path. $\endgroup$
    – Dale
    Aug 17, 2021 at 19:20

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