# Where do I go from here to show that linear momentum is conserved under all instances of translation symmetry?

I've worked through a simple derivation of symmetries implying conservation laws from an invariant Lagrangian.

Namely a quantity $$Q$$ is conserved in the equation below, where $$i$$ is a degree of freedom, $$p$$ is the generalised momentum and $$f(q)$$ is a function determining the coordinate shift, such that $$\delta q_i=f_i(q)\delta$$ (each coordinate shifts by an amount proportional to $$\delta$$ and $$f(q)$$, a function of position, is the proportionality factor. $$Q=\sum_ip_if_i(q)$$

But where do I go from here to show that linear momentum is conserved under all instances of translation symmetry? I can write a Lagrangian for a given instance and show it is the case, but how do I generalise?

• Dec 28 '18 at 16:24
• Hint: To show that the $j$'th momentum $Q=p_j$ is conserved, pick the symmetry generator $f^i(q)=\delta^i_j$. Dec 28 '18 at 17:12
• @Qmechanic, do you mean $f^i(q_j) = \delta^i_j$? Dec 28 '18 at 18:33

Question: Consider the following quantity $$Q = \sum_ip_if_i(q)$$ where $$f_i(q)$$ is defined as $$\delta q_i = f_i(q)\delta$$. It is given that for arbitrary translations in space(given by the arbitrary choices of $$f_i(q)$$), $$Q$$ is conserved. How do we show that the linear momentum is then conserved?
Solution: Just take the time derivative of $$Q$$. We get, $$\frac{dQ}{dt} = \sum_i\frac{d}{dt}(p_if_i(q)) = \sum_i \frac{dp_i}{dt}f_i(q)$$ Here, we have assumed that the $$f_i(q)$$ are constant factors, dependent only on the initial positions of the particles, before the translation. Since $$\frac{dQ}{dt}=0$$, the last equation implies, $$\frac{dp_i}{dt}=0$$ as the $$f_i(q)$$ may be arbitrary as per the assumption in the question.
• Thanks for your help! But if $f_i(q)$ is something like $2q$, wouldn't that mean that the conserved quantity Q is $\sum_i2p_iq_i$, rather than just $\sum_ip_i$? That would conserve something other than just linear momentum, despite the Lagrangian being invariant under a translation, or am I misinterpreting everything? Dec 28 '18 at 17:44
• Also, here is the working thus far if it's useful: Calculating how much the Lagrangian changes when $q_i$ and $\dot q_i$ shift in a translation (by $\delta q(i) = f_i(q) \delta$ and $\delta \dot q_i = d/dt(\delta q(i)$ respectively): $$\delta L = \sum_i((\partial L/\partial \dot q_i)\delta \dot q_i+(\partial L/\partial q_i)\delta q_i) = 0$$ Using Euler-Lagrange equation: $$\delta L = \sum_i(p_i\delta \dot q_i + \dot p_i\delta q_i) = 0$$ Using product rule: $$\delta L = d/dt\sum_i(p_i \delta q_i) = 0$$ Dec 28 '18 at 17:58