0
$\begingroup$

This is a step in Nakahara's Geometry, Topology and Physics, 2nd edition, 2003, on pages 7-8:

Given that $q_k ' = q_k +\epsilon f_k(q)$, we have that

$$\Lambda_{ij} = \frac{\partial q_i'}{\partial q_j} \simeq \delta_{ij} + \epsilon\frac{\partial f_i(q)}{\partial q_j}.$$

  1. First off, why is there an approximate equality? There are no terms left when we take the derivative of $q_k'$...?

  2. It is stated that the momentum $p_k'$ transforms as $$ p_i \rightarrow \sum_j p_j\Lambda_{ji}^{-1} \simeq p_i- \epsilon \sum_j p_j \frac{\partial f_j}{\partial q_i};$$ Is this obvious? Where can I derive this?

$\endgroup$
1
$\begingroup$
  1. The approximate equality is intended to remind you that Nakahara considers transformations with an infinitesimal parameter $\epsilon$ here, but you could as well take it as a full equality, it doesn't matter.

  2. The phase space is a cotangent bundle, where the coordinate $x$ are coordinates of the underlying manifold and the momenta $p$ lie in the cotangent space. By definition, elements of the cotangent space transform with the inverse Jacobian of a coordinate transformation.

$\endgroup$
  • $\begingroup$ so, for 2., it's just a definition? Is there no way I can make sense of it? $\endgroup$ – SuperCiocia Feb 1 '15 at 16:30
  • $\begingroup$ @SuperCiocia: Well, there is geometrical intuition behind the momenta being cotangents to the coordinates, and if you think of the cotangents as being spanned by $\mathrm{d}q^i$, it is immediately clear that they transform by the inverse Jacobian when changing to the basis $\mathrm{d}q'^i$. $\endgroup$ – ACuriousMind Feb 1 '15 at 16:32

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.