Peeter Joot [email protected]

Velocity volume element to momentum volume element 1.1

Motivation

One of the problems I attempted had integrals over velocity space with volume element d3 u. Initially I thought that I’d need a change of variables to momentum space, and calculated the corresponding momentum space volume element. Here’s that calculation. 1.2

Guts

We are working with a Hamiltonian e=

q

(pc)2 + e02 ,

(1.1)

where the rest energy is e0 = mc2 .

(1.2)

Hamilton’s equations give us uα = or pα = p

p α / c2 , e muα 1 − u2 / c2

(1.3) .

(1.4)

This is enough to calculate the Jacobian for our volume element change of variables ∂(u x , uy , uz ) dp x ∧ dpy ∧ dpz ∂(p x , py , pz ) 2 2 m c + p2y + p2z − p p − p p y x z x 1 2 2 2 2 dp x ∧ dpy ∧ dpz − p x py m c + p x + pz − pz py =  9/2 6 2 2 2 2 2 2 c m + (p/c) − p x pz − py pz m c + p x + py  −5/2 = m2 m2 + p2 / c2 dp x ∧ dpy ∧ dpz . (1.5)

du x ∧ duy ∧ duz =

1

That final simplification of the determinant was a little hairy, but yielded nicely to Mathematica. Our final result for the velocity volume element in momentum space, in terms of the particle energy is d3 u =

c6 e02 3 d p. e5

2

(1.6)

## Peeter Joot [email protected] Velocity ... - Peeter Joot's Blog

This is enough to calculate the Jacobian for our volume element change of variables dux < duy < duz = â(ux, uy, uz). â(px, py, pz) dpx < dpy < dpz. = 1 c6 (m2 + ...

#### Recommend Documents

Peeter Joot [email protected] Velocity ... - Peeter Joot's Blog
... momentum space, and calculated the corresponding momentum space volume element. Here's that calculation. 1.2 Guts. We are working with a Hamiltonian.

Peeter Joot [email protected] Change of variables in 2d phase ...
In [1] problem 2.2, it's suggested to try a spherical change of vars to verify explicitly that phase space volume is preserved, and to explore some related ideas. As a first step let's try a similar, but presumably easier change of variables, going f

Solutions to David Tong's mf1 Lagrangian problems. - Peeter Joot's Blog
dial decomposition. This is something that can be understood without the. Lagrangian formulation. To do so the missing factor is that before a conser-.

(INCOMPLETE) Geometry of Maxwell radiation ... - Peeter Joot's Blog
The first is the spatial pseudoscalar, which commutes with all vectors and bivectors. ... observe this is from the split of F into electric and magnetic field components. .... the differential operator to find eigenvectors for is the transverse gradi

Energy momentum tensor relation to Lorentz force. - Peeter Joot's Blog
In [Joot(b)] the energy momentum tensor was related to the Lorentz force in. STA form. .... So, to get the expected result the remaining two derivative terms must.