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I'm trying to compute a result shown in my classical mechanics lecture on my own. Namely, consider that a system composed of $n$ particles follows a law of force

$m_k\ddot{\vec{x_k}} = \vec{F_k}(\vec{x_i};\dot{\vec{x_i}};t)$, (1)

where we write $i$, but we are actually considering all positions and velocities of the system as arguments of $F_k$. Now, if the force can be written as a time-independent potential

$F_k = \frac{\partial}{\partial \vec{x_k}}V(\vec{x_1},\dots, \vec{x_n})$, (2)

we can write the Galilei invariance characterised by a rotation $R$ and a translation $\vec{b}$ as follows:

$R\frac{\partial}{\partial \vec{x_k}}V(\vec{x_1},\dots, \vec{x_n}) = \frac{\partial}{\partial \vec{x_k'}}V(\vec{x_1'},\dots, \vec{x_n'})$, (3)

where $\vec{x_i'} = R\vec{x_i} + \vec{b}$ is the Galilei-Transformation of $\vec{x_i}$. In my lecture, it was stated that because of this we can write:

$V(\vec{x_1},\dots, \vec{x_n}) = V(\vec{x_1'},\dots, \vec{x_n'})$. (4)

Now, I'm struggling to prove why (4) follows from (3). I tried using a naive chain rule on the inverse Galilei-transformation $\vec{x_k'} \mapsto \vec{x_k}$ but I always end up with:

$R\frac{\partial}{\partial \vec{x_k}}V(\vec{x_1},\dots, \vec{x_n}) = R^T\frac{\partial}{\partial \vec{x_k}}V(\vec{x_1'},\dots, \vec{x_n'})$. (5)

And I haven't yet found a way to imply (3) using (5). Now, I'm not sure if my calculations are right, or if I have missed something to prove that (3) follows from (5). I'd rather say that I've been misusing the chain rule.

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1 Answer 1

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The potential $V$ is a scalar quantity, so it is invariant under coordinate transformations. As for why 3 implies 4, I think you have to be more careful about how you're defining your gradient.

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