Does Galilean symmetry imply that all systems are monogenic?

[hgandh]

The equations of motions for a closed system consisting of $N$ particles are:
$$m_i \vec x_i'' = \sum_{j \neq i}^N \vec F(\vec x_i, \vec x_i', \vec x_j, \vec x_j')$$
$$ i = 1,..., N$$
Now if we impose the requirement that this closed system be symmetric under Galilean transformations, do we get the requirement that
$$\vec F_i = - \frac {\partial V} {\partial x_i} + \frac {d} {dt} (\frac {\partial V} {\partial x_i'})$$
?

 

[mitochan]

I do not think so. Even if force does not come from potential V, e.g. friction, Galilean transformation stands.

 

[hgandh]

I do not think so. Even if force does not come from potential V, e.g. friction, Galilean transformation stands.

The underlying interactions at the microscopic level that give rise to friction are derivable from potentials.

 

[anuttarasammyak]

Let me see. In your last equation how do $x_i$ and $x'_i$ relate ?

 

[hgandh]

Let me see. In your last equation how do $x_i$ and $x'_i$ relate ?

prime denotes derivative with respect to time

 

[mitochan]

I see. Then how do $\mathbf{F}$ in your first equation and $\mathbf{F_i}$ in your last equation relate? What are the variables of $V$ ?

 

[mitochan]

I assume
$$F_i :=\sum_{j=1}^{N} F(x_i, \dot{x}_i,x_j, \dot{x}_j)$$
where Newton's third law says
$$F(x_i, \dot{x}_i,x_j, \dot{x}_j)=-F(x_j, \dot{x}_j,x_i, \dot{x}_i)$$
so $$F(x_i, \dot{x}_i,x_i, \dot{x}_i)=0$$

Symmetry of translation and Galelian transformation does not change acceleration in LHS so in RHS
$$F(x_i, \dot{x}_i,x_j, \dot{x}_j)= F(x_i+Vt+X, \dot{x}_i+V,x_j+Vt+X, \dot{x}_j+V)$$
for any V and X. In appropriate choices of V and X, it becomes
$$F(0,0,x_j-x_i,\dot{x}_j-\dot{x}_i)=\mathbf{F}(x_j-x_i,\dot{x}_j-\dot{x}_i)$$
Thus defined RHS function $\mathbf{F}$ has two parameters, relative coordinate and relative speed.
So as a result
$$F_i =\sum_{j=1}^{N} \mathbf{F}(x_j-x_i,\dot{x}_j-\dot{x}_i)$$

$$\mathbf{F}(x_j-x_i,\dot{x}_j-\dot{x}_i)=-\mathbf{F}(x_i-x_j,\dot{x}_i-\dot{x}_j)$$
$$\mathbf{F}(0,0)=0$$

I have no idea how to relate this to potential form. EM, whose Lorentz force depends on relative velocity, holds vector and scalar potentials and follow not Galilean but Lorentz transformation.

In the case
$$\mathbf{F}=\mathbf{F}(x_j-x_i)=\mathbf{F}(r)$$
$$\phi(R):=-\int_0^R F(r)dr$$
would give potential if integration could not change by any integration path.