To give a context as to what I'm asking here ,I am talking about the energy of a two particle system (section 4.9 Taylor's Classical Mechanics) .

My question is what does $\nabla_1 U$=$\frac{\partial U}{\partial x_1}\hat{i}$+$\frac{\partial U}{\partial y_1}\hat{j}$+$\frac{\partial U}{\partial z_1}\hat{k}$ mean ?

What does the $x_1$,$y_1$ & $z_1$ mean here ?

If we are talking through the lens of basic principles ,what does it mean ?

I am not convinced that $gradient$ at $\vec{r_1}$ is $\nabla_1U$ ,I mean why can't we substitute thepoint (or) vector $r_1$ to the expression $\nabla U$?

If my mass is at $x_1$ then is $\partial{x_1}$ a small change in $x_1$ direction ? But isn't that same as a small change in x direction ?


1 Answer 1


First, excuse my English, I used google translator. Since the forces are conservative: $\nabla_1\times\vec{f_{12}}=0$ and $\nabla_2\times\vec{f_{21}}=0$. I can set the potentials $U_1$,$U_2$, such that:

$$\vec{\nabla_1}U_{1}(\vec{\mathfrak{r}})=\vec{f_{12}}~~~and~~~\vec{\nabla_2}U_{2}(\vec{\mathfrak{r}})=\vec{f_{21}}$$ $$\vec{\mathfrak{r}}=\vec{r_1}-\vec{r_2}$$

If you have trouble accepting this, you can use the Helmholtz equation to get an idea. Since vector fields in general satisfy the equation: $$\vec{f}=-\vec{\nabla}{U}+\vec{\nabla}\times\vec{H}$$ you can check that $\vec{\nabla}\times\vec{f}=0\leftrightarrow\vec{\nabla}\times\vec{H}=0$. Which makes it possible to define a potential function and get $\vec{f}$ as the negative of his gradient. The nabla indices are just because it occurs in the respective coordinates.

As for the meaning of $x_1,y_1,x_2,y_2$, they are the coordinates of each mass in relation to an origin. So that:


If you go back a little in the text of the book: $\vec{f_{12}}=-f~\mathfrak{\hat{r}}=-f~\mathfrak{\frac{\vec{r_1}-\vec{r_2}}{|\vec{r_1}-\vec{r_2}|}}$ and by the third law $\vec{f_{12}}=-\vec{f_{21}}$

To check $\vec{\nabla}_1U=\vec{\nabla}U$ just use the chain rule: $$-\vec{\nabla}_1U(\mathfrak{r})=-\bigg(\frac{\partial{U}}{\partial{x_1}}\hat{x}+\frac{\partial{U}}{\partial{y_1}}\hat{y}\bigg)=-\bigg(\frac{\partial{U}}{\partial{\mathfrak{r}_x}}\frac{\partial{\mathfrak{r}_x}}{\partial{x_1}}\hat{x}+\frac{\partial{U}}{\partial{\mathfrak{r}_y}}\frac{\partial{\mathfrak{r}_y}}{\partial{y_1}}\hat{y}\bigg)$$

replace $\mathfrak{r}_x$ and $\mathfrak{r}_y$ to get $$-\vec{\nabla}_1U(\mathfrak{r})=-\bigg(\frac{\partial{U(\mathfrak{r}_x,\mathfrak{r}_y)}}{\partial{\mathfrak{r}_x}}\hat{x}+\frac{\partial{U(\mathfrak{r}_x,\mathfrak{r}_y)}}{\partial{\mathfrak{r}_y}}\hat{y}\bigg)=-\vec{\nabla}U(\mathfrak{r})$$ So: $$\vec{\nabla}_1U=\vec{\nabla}U$$ if you do this for particle 2 you will see that a minus sign will appear confirming the third law. This is due to the difference in positions.

If you change the coordinates $x_1,y_1$ you are just changing the position $\vec{r_1}$ of the masses in relation to the origin.The interaction of particles depends on the separation vector $\vec{\mathfrak{r}}$ between them.

  • $\begingroup$ I do not think this answer is well-matched to the level of the question. For example the material on decomposition of an arbitrary field into solenoidal and conservative parts is not very relevant, and significantly more advanced. The answer to the specific question what does $\nabla_1$ mean is given but should be more prominent $\endgroup$
    – CWPP
    Aug 15, 2022 at 11:25

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