0
$\begingroup$

Consider a system of 4 particles in an external conservative field. So force acting on each particle is derived from potential energy $U(x,y,z)$of the particle+field system:

Total (external) force on the system of four particles is then: $$ \overrightarrow {F_{tot}} = \overrightarrow {F_1}^{ext} +\overrightarrow {F_2}^{ext} + \overrightarrow {F_3}^{ext} + \overrightarrow{F_4}^{ext} = -\overrightarrow{\nabla_1}(U_1) - \overrightarrow{\nabla_2}(U_2) - \overrightarrow{\nabla_3}(U_3) - \overrightarrow{\nabla_4}(U_4) $$

Why do we need different subscripts, namely 1, 2, 3 and 4, for $\nabla$ operators ?

It means, for example, $$ \overrightarrow{\nabla_1} = {\partial \over\partial x_1} \hat i + {\partial \over\partial y_1} \hat j + {\partial \over\partial z_1} \hat k$$

But each of $x_1,y_1,z_1$ can have all values from $-\infty$ to $+\infty$; same for $x_2,y_2,z_2,x_3,y_3,...z_4$

All the particles are moving in the SAME 3-dimensional space.

$\endgroup$
9
  • $\begingroup$ The field has a potential (energy per unit charge), say $V(x,y,z)$, a function of the space coordinates $x,y,z$ and in general of time $t$. A particle $P_{k}$ located at $\mathbf{r}_{k}=(x_{k},y_{k},z_{k})$ has potential energy $U_{k}(x_{k},y_{k},z_{k})=q_{k}\cdot V(x_{k},y_{k},z_{k})$ where $q$ the charge of the particle (for example electric in electromagnetic field or mass in gravitational field). $\endgroup$
    – Voulkos
    Commented Apr 27, 2016 at 11:12
  • $\begingroup$ The field force on this particle is $\mathbf{F}_{k}=-q_{k}\cdot\left[\boldsymbol{\nabla}V(x,y,z)\right]_{\left(x=x_{k},y=y_{k},z=z_{k}\right)}$. So, you must find by partial differentiation the gradient of the potential $\boldsymbol{\nabla}V(x,y,z)=\left(\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z}\right)$ as function of the variables $x,y,z$ and then replace these variables by their values $x_{k},y_{k},z_{k}$ respectively. $\endgroup$
    – Voulkos
    Commented Apr 27, 2016 at 11:13
  • $\begingroup$ @Frobenius: Thanks for comments. I got your point. Can we say like this also ? : Say particle no. 2 has occupied some $(x_2,y_2,z_2)$ at some fixed time $t_0$. So remaining 3 particles can not occupy location $(x_2,y_2,z_2)$ at the same time $t_0$. i.e. particle no. 2 can not occupy $(x_k,y_k,z_k)$ where $k = 1, 3, and 4$. And to distinguish the range of these space co-ordinates, we need to put subscripts on del operators. $\endgroup$
    – atom
    Commented Apr 28, 2016 at 3:57
  • $\begingroup$ I think that we must not use such a picture. It's already wrong to say that every particle $k$ has its own potential $U_{k}$ : it's like to say that there exist 4 fields, each field acting on one particle but not on the other 3. By the way my favorite Physicist is P.A.M. Dirac whose photo as child you use in your profile. $\endgroup$
    – Voulkos
    Commented Apr 28, 2016 at 5:03
  • $\begingroup$ "The Strangest Man-The Hidden Life of Paul Dirac, Mystic of the Atom [Farmelo]" i.imgur.com/g3yJiWm.jpg $\endgroup$
    – Voulkos
    Commented Apr 28, 2016 at 5:19

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.