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The isolated system of particles is being observed. In the coursebook, $\vec F_\mu$ is by definition the vector sum of forces of all other particles acting on $\mu$-th particle. Usually, potential energy is defined to be a gradient of some scalar function $U$ in respect to chosen coordinate system, i.e. $$\vec F=-\bigg(\frac{\partial U}{\partial x} \vec e_x+ \frac{\partial U}{\partial y}\vec e_y+\frac{\partial U}{\partial z}\vec e_z \bigg).$$

But in this case, the definition is following:

$$\vec F_\mu=-\bigg(\frac{\partial U}{\partial x_\mu} \vec e_x+ \frac{\partial U}{\partial y_\mu}\vec e_y+\frac{\partial U}{\partial z_\mu}\vec e_z \bigg),$$

where $U=U(\vec r_1, \vec r_2 ... \vec r_n$). I have been trying to find similar definitions in other textbooks and I have stumbled upon this definition in Chaichan, Merches and Tureanu (2012), also. It is written in both books as $$\vec F_\mu=-grad_\mu U.$$

Can somebody help me with physical intuition? I would understand it if it was $$\vec F_\mu=-\bigg(\frac{\partial U}{\partial x} \vec e_x+ \frac{\partial U}{\partial y}\vec e_y+\frac{\partial U}{\partial z}\vec e_z \bigg)\Bigg|_\mu$$ but this is actually taking partial derivatives in respect to coordinates of an observed particle?

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It's not taking partial derivatives with respect to an observed particle's position, but rather the space of all possible positions of that particle. Think of the potential energy as being defined prior to the particle having an actual path.

Really, at heart, these things are defined on a phase space not on ordinary physical space.

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  • $\begingroup$ So if there are no constraints, it would be same as the whole space? $\endgroup$
    – Caneholder123
    Commented Jun 26, 2015 at 15:34
  • $\begingroup$ I'm familiar with the notion of phase space, but that's as far as it goes. If the phase space in classical mechanics is consisted of positions and momenta of all particles, would this derivative than be in respect to three-dimensional subspace $(x_\mu, y_\mu, z_\mu)$ of this phase space? $\endgroup$
    – Caneholder123
    Commented Jun 26, 2015 at 15:54
  • $\begingroup$ @Caneholder123: yes. You're taking the derivative with respect to one configuration variable,w tih the other five held constant. $\endgroup$
    – Zo the Relativist
    Commented Jun 26, 2015 at 16:47
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In your last equation $U$ is a function or $n$ variables. Which of these does your $x$, $y$, and $z$ in that equation represent?

To find the contribution to the potential energy due to the action of forces on a particular particle, one has to take the partial of the potential with respect to the variables representing the position of that particle. I honestly don't know what your $x$, $y$, and $z$ are supposed to stand for.

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  • $\begingroup$ It represents the $U$ that is a sum of contributions of all particles. $x$, $y$ and $z$ represent the coordinates of some chosen coordinate system, and $\mu$ is index denoting position of particle $\mu$. But in that case, it would be external field $U=U(t)$ and not $U=U(\vec r_1, \vec r_2 ... \vec r_n)$ and the system would be just this particle $\mu$? $\endgroup$
    – Caneholder123
    Commented Jun 26, 2015 at 15:32
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    $\begingroup$ @Caneholder123: aaaaaaah, I think I understand what you're misunderstanding. $\mu$ is a placeholder for $x$, $y$, or $z$ in that second equation. It is not a particle index. $F_{\mu} = - \nabla_{\mu}U$ is shorthand for $ F_{x} = \frac{\partial U}{\partial x}$, etc. $\endgroup$
    – Zo the Relativist
    Commented Jun 26, 2015 at 16:49

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