# Conservation of momentum

I've got following problem.

There are $N$ particles in an isolated system. The equation of motion for a particle $i$ is $$m_i\ddot{\vec{x}_i}(t)=\vec{F}_i(\vec{x}_1(t),...,\vec{x}_n(t),\dot{\vec{x}_1}(t),...,\dot{\vec{x}_n}(t),t),$$

and the force has a potential

$$\vec F_i=-\nabla_{\vec{x}_{i}}V(\vec{x}_1(t),...,\vec{x}_n(t)).$$

The potential is translation-invariant.

Now, the total force is $\vec F=\sum_i \vec F_i=0$. As a proof it is stated that for an arbitrary unit vector $\vec e$ holds:

$$\vec e\cdot\vec F=-\frac{d}{d\lambda}V(\vec{x}_1(t)+\lambda\vec e,...,\vec{x}_n(t)+\lambda\vec e)|_{\lambda=0}=0.$$

This last equation I do not understand. Could anyone possibly give me a clue?

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Let $\vec y_i = \vec x_i + \lambda \vec e$

You have : $\frac{dV(\vec y_1, \vec y_2, ....\vec y_n)}{d\lambda} = \sum_i \vec \nabla_iV(\vec y_1, \vec y_2, ....\vec y_n) . \frac{d \vec y_i}{d \lambda}$

( chain rule for partial derivatives)

Here $\vec \nabla_iV$ is the vector which has components $(\vec \nabla_iV)_j = \frac{\partial V}{\partial (y_i)_j}$

But : $\nabla_i V(\vec y_1, \vec y_2, ....\vec y_n)|_{\lambda=0} = \nabla_i V(\vec x_1, \vec x_2, ....\vec x_n) =- \vec F_i$

Moreover, $\frac{d \vec y_i}{d \lambda} = \vec e$, and this equality is true also for $\lambda = 0$

Finally, we have :

$$\frac{dV(\vec y_1, \vec y_2, ....\vec y_n)}{d\lambda}|_{\lambda=0} = \sum_i -\vec F_i. \vec e = - \vec F. \vec e$$

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Thank you very much! I get it now. – fallingEntropy Jul 21 '13 at 10:17

The term $\lambda\vec e$ added to each argument in the potential is just a translation under which the potential is known to be invariant. So the derivative must be zero.

But if you use the chain rule for partial derivatives you get the result $\vec e\cdot\vec F$ and this can only be zero for all $\vec e$ if $\vec F$ is zero

This gives conservation of momentum, not conservation of force.

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Thank you! Yes, I actually meant conservation of momentum, you are right. – fallingEntropy Jul 21 '13 at 10:17