Poisson Bracket for 1 space dimension field

[Bobdemaths]

Hi,

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

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

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

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

Now I would like to show that the (equal time) Poisson Bracket \{A_i,A_j\} is
\{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)
using the canonical relation \{\phi ^i(t,x) , \pi_j (t,y)\}=\delta_j^i \delta(x-y).

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

Am I mistaken somewhere ? Thank you in advance !

 

[Bobdemaths]

OK I found the answer. To get rid of the derivatives of Dirac deltas, I used the identity f(x) \delta ' (x-y) = f(y) \delta ' (x-y) - f'(x) \delta (x-y) which follows from the convolution between delta and (f \times \varphi)' (\varphi is a test function).
This also produces the remaining wanted term.