0

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
  • 29,127
Nick Maslov
  • 33

1 Answers1

0

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
  • 3,636