0
$\begingroup$

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?

$\endgroup$
3
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Dec 28 '18 at 17:44
  • $\begingroup$ 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: $$\delta L = d/dt\sum_i(p_i \delta q_i) = 0$$ $\endgroup$ Dec 28 '18 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.