Poisson Bracket for 1 space dimension field


by Bobdemaths
Tags: bracket, dimension, field, poisson, space
Bobdemaths
Bobdemaths is offline
#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 !
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Bobdemaths
Bobdemaths is offline
#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.


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