TransWikia.com

Why does a transformation on $p$ imply the inverse transformation on $q$ in phase space?

Physics Asked by Willem on July 20, 2021

I’m reading A conceptual introduction to Hamiltonian Monte Carlo. On pages 30-31 it is stated that applying a transformation to the momentum p in phase space, implies the opposite transformation to the parameters q:
$$p’= sqrt{M^{-1}}p$$
implies
$$q’= sqrt{M}q$$
for some Euclidean metric $M$, known as the mass matrix.

I don’t understand why this is. Suppose we are working in 1 dimension (so $M$ is just a number), and suppose $M = 4$. Then:
$$q’ = sqrt{M}q = 2q$$
The momentum would then be:
$$p = mv = m frac{dq}{dt} = frac{1}{2} m frac{dq’}{dt} = frac{1}{2}p’ iff p’ = 2p$$
To me it seems both p and q transform in the same way, getting multiplied by 2. So where did I go wrong?

One Answer

The author of this statistics paper aims for transformations $pmapsto p',,qmapsto q'$ that preserve Hamiltonian geometry. As far as I can tell from a quick look at Hamiltonian Monte Carlo this means he wants to preserve the energies expressed as $$ H=frac{1}{2}boldsymbol p^top M^{-1}boldsymbol p,,quad T=frac{1}{2}dot{boldsymbol q}^top Mdot{boldsymbol q},. $$ This looks quite familar to a physicist. Now take the square root of $M$ and $M^{-1},.$ Clearly, the invariance of $H$ and $T$ under the above transformations is compatible with $boldsymbol p'=sqrt{M^{-1}}boldsymbol p$ and $dot{boldsymbol q}'=sqrt{M}dot{boldsymbol q},$ and assuming that $M$ is constant we can undo the time derivative of the $dot{boldsymbol q}$ relation.

Answered by Kurt G. on July 20, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP