Galilean's principle implies independence of time and dependence on relative distance

Question

Suppose a system of particles $q_1,\ldots,q_N$ of masses $m_1,\ldots,m_N$ that follow the equations of motion $$m_j\ddot{q}_j=f_j(q_k,\dot{q}_k)$$ in an inertial frame and satisfy the Galilean principle of relativity. Show the forces are of the form $f_j(q_k-q_l,\dot{q}_k-\dot{q}_l)$.

My idea: If we translate $q_j$ by $q_{l}, \tilde{q}_{j}=q_{j}-q_{l}$, we get $$\tilde{m_{j}}\left(\ddot{q}_{j}-\ddot{q_{l}}\right) = \tilde{f_{j}}\left(q_{k}-q_{l},\dot{q}_{k}-\dot{q}_{l}\right).$$ On the other side, $$\frac{m_{j}m_{l}}{m_{j}+m_{l}}\left(\ddot{q}_{j}-\ddot{q_{l}}\right) = \frac{m_{l}}{m_{j}+m_{l}}f_{j}(q_{k},\dot{q}_{k},t)+\frac{m_{j}}{m_{j}+m_{l}}f_{l}(q_{k},\dot{q}_{k},t) = f_{j}(q_{k},\dot{q}_{k},t)-\frac{m_{j}}{m_{l}}f_{l}(q_{k},\dot{q}_{k},t).$$ And $$f_{j}(q_{k},\dot{q}_{k},t)=\tilde{f_{j}}\left(q_{k}-q_{l},\dot{q}_{k}-\dot{q}_{l}\right)+\frac{m_{j}}{m_{l}}f_{l}(q_{k},\dot{q}_{k},t).$$ But from here I don't know how to proceed. I also taught about doing it by induction: If $n=2$ we translate the particles such that $q_1=-q_2$. Then $q_1-q_2=2q_2=-2q_1$ and $$m_1\ddot{q_1}=f_1(q_1,q_2,\dot{q}_1,\dot{q}_2,t)=\tilde{f_1}(q_1-q_2,\dot{q}_1-\dot{q}_2,t).$$ Suppose for $n>2$ it is satisfied. Then for $n+1$. We can write $q_{n}=-q_{n+1}$. And I believe we this we can lower the number of particles to $n$, but I don't know how to argument.

To prove that $f_j$ does not depend on $t$ I have no idea how to proceed.

Answer 1

It is given that the forces are of the form

$$m_j \ddot{q}_j(t) = f_j(\{q_k\},\{\dot{q}_k\})$$

and that the dynamics are invariant under Galilean transformations, meaning that if $\{q_k\}$ obey the equations of motion (EOM) then

$$q_k(t) \rightarrow q_k(t) + v_0 t + q_0.$$

Since such a transformation does not change the LHS of the EOM, then clearly this holds true only if the forces (the RHS) are likewise unchanged.

There exists a trivial one-to-one map from

$$(q_1,q_2,\ldots q_N)$$

to

$$ (q_2-q_1, q_3-q_2,\ldots q_N-q_{N-1},q_1).$$

There exists an equivalent and consistent map between the time derivatives. Thus, the original forces can be written with dependencies:

$$f_j(\{\Delta q_k\},\{\Delta \dot{q}_k\},q_1, \dot{q}_1).$$

(Note that these forces have different functional forms than the other forces with different arguments.)

Now, invoking Galilean invariance, choose $q_0 = -q_1, v_0 = -\dot{q}_1$. Then, the forces become: $$f_j(\{\Delta q_k\},\{\Delta \dot{q}_k\},0, 0).$$

Suppressing the functional dependence on the two constants, we see that it must be possible to write the forces as depending only on the relative positions and relative velocities of particles with indices differing by one. This is a special case of the desired result, and so we claim success.

One might object that in order to represent typical two-body interactions between identical particles this is a rather inconvenient way of doing things, since it makes it look like the interactions between $q_3,q_4$ are different from those between $q_3,q_5$. I'm not sure whether there's a good way to resolve this. As I mentioned in my not entirely correct comments, there's something funny about being able to write n-body interactions in terms of two-body differences like $q_k - q_l$.

Answer 2

Take the Galilean's transformation $(q_{j},t)\mapsto(q_{j},t+t_{0})$ then, by the principle of relativity $$f_{j}(q_{k},\dot{q}_{k},t+t_{0})=f_{j}(q_{k},\dot{q}_{k},t).$$ Since $t_{0}\in\mathbb{R}$ is arbitrary then $$f_{j}(q_{k},\dot{q}_{k},t+t_{0})=f_{j}(q_{k},\dot{q}_{k},t),\quad\forall t_{0}\in\mathbb{R}.$$ Take $t$ fixed and $t_{0}=-t$ then $$f_{j}(q_{k},\dot{q}_{k},t) =f_{j}(q_{k},\dot{q}_{k},t+t_{0}) =f_{j}(q_{k},\dot{q}_{k},t-t) =f_{j}(q_{k},\dot{q}_{k},0) =\tilde{f}_{j}(q_{k},\dot{q}_{k}).$$ Therefore, the equations of motion are independent of time. Now, I will show that the forces depend on their relative positions.Apply the Galilean's transformation $q_{k}(t)\mapsto q_{k}(t)+a=\tilde{q}_{k}.$ Notice that $$\dot{\tilde{q}}_{k}=\dot{q}_{k}+a$$ and $$\ddot{\tilde{q}}_{k}=\ddot{q}_{k}.$$ Therefore, by the principle of relativity, we have the force $$f_{j}\left(q_{k},\dot{q}_{k}\right)=m_{j}\ddot{q}_{j}(t)=m_{j}\ddot{\tilde{q}}_{j}(t)=f_{j}\left(\tilde{q}_{k},\dot{\tilde{q}}_{k}\right)=f_{j}(q_{k}+a,\dot{q}_{k}).$$ However, since $a\in\mathbb{R}^{3}$ is arbitrary then $$f_{j}\left(q_{k},\dot{q}_{k}\right)=\tilde{f}_{j}\left(q_{k},\dot{q}_{k}+a\right),\quad\forall b\in\mathbb{R}^{3}.$$ Take $q_{k}$ fixed and $b=q_{k}-q_{n}$ then $$f_{j}\left(q_{k},\dot{q}_{k}\right) =f_{j}\left(q_{k}-q_{n},\dot{q}_{k}\right) =f_{j}(q_{1}-q_{n},\ldots,q_{n}-q_{n},\dot{q}_{k}) =f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},0,\dot{q}_{k}) =\tilde{f}_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{k});$$ this is, $$f_{j}\left(q_{k},\dot{q}_{k}\right)=\tilde{f}_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{k}).$$ Now, we apply the Galilean's transformation $$q_{k}\mapsto q_{k}+bt=\tilde{q}_{k}$$ then $$\dot{\tilde{q}}_{k}=\dot{q}_{k}+b, \ddot{\tilde{q}}_{k}=\ddot{q}_{k}$$ and, by the principle of relativity, $$f_{j}\left(q_{k}+bt,\dot{q}_{k}+b\right) = f_{j}(\tilde{q}_{k},\dot{\tilde{q}}_{k}) = m_{j}\ddot{\tilde{q}}_{k} = m_{j}\ddot{q}_{k} = f_{j}(q_{k},\dot{q}_{k}).$$ From the previous case $$f_{j}\left(q_{k}+bt,\dot{q}_{k}+b\right) = f_{j}\left(q_{1}+bt-q_{n}-bt,\ldots,q_{n-1}+bt-q_{n}-bt,\dot{q}_{k}+b\right) = f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{k}+b).$$ Hence $$f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{k}+b)=f_{j}(q_{k},\dot{q}_{k}).$$ Since $b\in\mathbb{R}^{3}$ is arbitrary then $$f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{k}+b)=f_{j}(q_{k},\dot{q}_{k}),\quad\forall b\in\mathbb{R}^{3}.$$ Fix $q_{k}$ and take $b=-q_{n}$ then $$f_{j}(q_{k},\dot{q}_{k}) = f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{k}+b) = f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{1}-\dot{q}_{n},\ldots,\dot{q}_{n}-\dot{q}_{n}) = f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{1}-\dot{q}_{n},\ldots,,\dot{q}_{n-1}-\dot{q}_{n});$$ this is $$f_{j}(q_{k},\dot{q}_{k}) =f_{j}(q_{1}-q_{n},\ldots,q_{n-1}-q_{n},\dot{q}_{1}-\dot{q}_{n},\ldots,,\dot{q}_{n-1}-\dot{q}_{n}) =f_{j}(q_{k}-q_{n},\dot{q}_{k}-\dot{q}_{n}).$$ Since there was nothing special about taking $n$, take any $l\in\left\{ 1,\ldots,n\right\}$ and similarly $$f_{j}(q_{k},\dot{q}_{k})=f_{j}(q_{k}-q_{l},\dot{q}_{k}-\dot{q}_{l}).$$ Finally, by considering the Galilean's transformation $$q_{j}(t)\mapsto Aq_{j}(t)$$ where $A\in O_{3}(\mathbb{R})$, invoking the principle of relativity and considering that $$\frac{d}{dt}A\left(q_{k}-q_{l}\right)=A\left(\dot{q}_{k}-\dot{q}_{l}\right)$$ and taking l fixed (take $l=n$ for example, as in the previous proof) then $$\tilde{f}_{j}\left(A\left(q_{k}-q_{l}\right),A\left(\dot{q}_{k}-\dot{q}_{l}\right)\right) =m_{j}A\left(\ddot{q}_{j}-\ddot{q}_{l}\right) =Am_{j}\left(\ddot{q}_{j}-\ddot{q}_{l}\right) =Af_{j}(q_{k}-q_{l},\dot{q}_{k}-\dot{q}_{l}).$$