Register to reply

Poisson Bracket for 1 space dimension field

by Bobdemaths
Tags: bracket, dimension, field, poisson, space
Share this thread:
Bobdemaths
#1
Jan23-13, 03:42 AM
P: 3
Hi,

Suppose you have a collection of fields [itex]\phi^i (t,x)[/itex] depending on time and on 1 space variable, for [itex]i=1,...,N[/itex]. Its dynamics is defined by the Lagrangian

[itex]L=\frac{1}{2} g_{ij}(\phi) (\dot{\phi}^i \dot{\phi}^j - \phi ' ^i \phi ' ^j ) + b_{ij}(\phi) \dot{\phi}^i \phi ' ^j [/itex]

where [itex]\dot{\phi}^i [/itex] denotes the time derivative of the field [itex]{\phi}^i [/itex] and [itex]\phi ' ^i [/itex] denotes its space derivative, and where [itex]g_{ij}(\phi) [/itex] is a symmetric tensor, and [itex]b_{ij}(\phi) [/itex] an antisymmetric tensor.

One easily computes that the momenta conjugate to the fields [itex]\phi^i (t,x)[/itex] are [itex]\pi_i = A_i + b_{ij} \phi ' ^j[/itex], where [itex]A_i = g_{ij} \dot{\phi}^j [/itex].

Now I would like to show that the (equal time) Poisson Bracket [itex]\{A_i,A_j\}[/itex] is
[itex]\{A_i(t,x),A_j(t,y)\}=(\partial_i b_{jk} + \partial_j b_{ki} + \partial_k b_{ij} ) \phi ' ^k \delta(x-y)[/itex]
using the canonical relation [itex]\{\phi ^i(t,x) , \pi_j (t,y)\}=\delta_j^i \delta(x-y)[/itex].

I tried to write [itex]A_i = \pi_i - b_{ij} \phi ' ^j[/itex], and then use [itex]\{\phi ' ^i(t,x) , \pi_j (t,y)\}=\delta_j^i \delta ' (x-y)[/itex]. But then I can't get rid of the [itex]\delta ' [/itex], and I don't get the [itex]\partial_k b_{ij} [/itex] term.

Am I mistaken somewhere ? Thank you in advance !
Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
Bobdemaths
#2
Jan23-13, 12:36 PM
P: 3
OK I found the answer. To get rid of the derivatives of Dirac deltas, I used the identity [itex]f(x) \delta ' (x-y) = f(y) \delta ' (x-y) - f'(x) \delta (x-y)[/itex] which follows from the convolution between delta and [itex](f \times \varphi)'[/itex] ([itex]\varphi[/itex] is a test function).
This also produces the remaining wanted term.


Register to reply

Related Discussions
Poisson bracket Advanced Physics Homework 0
Poisson bracket Quantum Physics 14
Poisson bracket Classical Physics 0
Poisson bracket and Lax pairs General Math 0
Poisson Bracket Introductory Physics Homework 1