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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?

Kyle Kanos
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Nick Maslov
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1 Answers1

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The question appears a bit vague to me, but I shall assume that the problem you want to address is the following:

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.

P.S: If I have misinterpreted the question, on made any blunders, feel free to point it out.

Lelouch
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  • 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? – Nick Maslov 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: $$\delta L = d/dt\sum_i(p_i \delta q_i) = 0$$

     – Nick Maslov Dec 28 '18 at 17:58