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?


1 Answer 1

  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.

  • $\begingroup$ so, for 2., it's just a definition? Is there no way I can make sense of it? $\endgroup$ Commented Feb 1, 2015 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
    Commented Feb 1, 2015 at 16:32

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