1
$\begingroup$

Consider the following figure

enter image description here

where $R=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=|\mathbf{r}-\mathbf{r}'|$ is the module of the $\mathbf{R}$ vector depends not only on the location of the $P$ point but also on the location $P'$ where the $dV'$ volume is located (fixed once located in the volume $\mathcal{V}$). Obviously if you change $P'=(x',y',z')$ it will also change $\mathbf{R}$. Since the potential \begin{equation} \psi(\mathbf{r})=-G\iiint_{\mathcal{V}} \frac{\rho(x',y',z')dx'dy'dz'}{|\mathbf{r}-\mathbf{r}'|} \end{equation}

we calculate the gradient of the quantity

\begin{equation} \boldsymbol{\nabla}_{(\mathbf r)}\frac{1}{|\mathbf r-\mathbf r'|} \end{equation}

Calculating, respectively, the partial derivatives $\partial_x=\partial/\partial x$, $\partial_y=\partial/\partial y$ and $\partial_z=\partial/\partial z$ compared to the function $1/|\mathbf{r}-\mathbf{r}'|$, we will have

\begin{align*} \frac{\partial}{\partial x}\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}} & = \frac{\partial}{\partial x}\left((x-x')^2+(y-y')^2+(z-z')^2\right)^{-\frac 12}= && \\ &=\left(-\frac 12\right)\Bigl[\ldots\ldots\Bigr]^{-\frac32}\cdot 2\cdot (x-x')= && \\ &=-\frac{x-x'}{R^3} && \\ \end{align*}

Similarly $$\frac{\partial}{\partial y}\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}=-\frac{y-y'}{R^3}$$ $$\frac{\partial}{\partial z}\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}=-\frac{z-z'}{R^3}$$

Hence

$$ \boldsymbol{\nabla}_{(\mathbf r)}\frac{1}{|\mathbf r-\mathbf r'|}=-\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}. $$ from which

\begin{align} \boldsymbol{\nabla}_{(\mathbf{r})}\psi(\mathbf{r}) & = {\mathbf \nabla}_{(\mathbf{r})}\left(-G\iiint_{\mathcal{V}} \frac{\rho(x',y',z')\,dx'dy'dz'}{|\mathbf{r}-\mathbf{r}'|}\right) = && \tag{*}\\ &= -G\iiint_{\mathcal{V}}\left( {\mathbf \nabla}_{(\mathbf{r})}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)\rho(x',y',z')\,dx'dy'dz'= && \\ &= -G\iiint_{\mathcal{V}} \left(-\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\right)\rho(x',y',z')\,dx'dy'dz' = && \nonumber\\ &= G\iiint_{\mathcal{V}} \frac{\rho(x',y',z') \, dx'dy'dz'}{R^2}\,\mathbf{\widehat R}=\mathbf{g}(\mathbf{r}) && \nonumber\\ \nonumber \end{align}

I ask with such kindness, considering that I have to prove that $\mathbf{g}(\mathbf{r})=-\boldsymbol{\nabla}\psi(\mathbf r)$ I was not able to find the error of a less missing sign in the various steps of the $(^*)$.

I hope you will appreciate my effort and my question is clear.

$\endgroup$
  • $\begingroup$ You should be careful about the derivative of $1/R$, and the explicit definition of $\mathbf{R}$. Is it $\mathbf{r}-\mathbf{r}'$ or $\mathbf{r}'-\mathbf{r}$? $\endgroup$ – Oktay Doğangün Jul 18 '18 at 20:26
  • $\begingroup$ @OktayDoğangün $\mathbf R=\mathbf r-\mathbf r'$ certainly. If it had been the other way round, the minus sign would have been found. Thank you very much for your comment. $\endgroup$ – Sebastiano Jul 18 '18 at 20:29
  • $\begingroup$ Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. Additionally, it is unclear what you are actually doing because except for the tag "newtonian-gravity" there is nothing to go on what some of your symbols - like the potential - actually mean. $\endgroup$ – ACuriousMind Jul 19 '18 at 19:40
  • $\begingroup$ @ACuriousMind For me it was important to understand better also the question of the user who gave me an unclear answer. Evidently I realized that the spirit of this site is different from how I think. Thank you so much. $\endgroup$ – Sebastiano Jul 19 '18 at 19:44
2
$\begingroup$

The expression for the gravitational field induced by the collection of masses $P'$ on the point P reads

$\mathbf{g}(\mathbf{r})=G\iiint_{\mathcal{V}} \left(-\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\right)\rho(x',y',z')\,dx'dy'dz'$.

This solves the sign problem as you can easily see from line 3 of (*). Had the minus sign in the expression not been there, the gravitational field would be pointing along the direction of $\mathbf{R}$ which would make the force repulsive, instead of attractive, as gravity is to our knowledge so far.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.