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  Deriving the Poisson bracket relation of the Ashtekar variables

+ 2 like - 0 dislike

I'm trying to figure out how to calculate the orthogonality of Ashtekar variables with respect to the ADM hypersurface metric and conjugate momentum.

$$\{{A_a}^i(x), {E^b}_j(y)\} = 8 \pi \beta \delta^i_j \delta^b_a \delta(x,y)$$

as is given in Kiefer's book on p.127 eqn 4.120.

I've been assuming the configuration variables of these Poisson brackets are the ADM hypersurface metric $\gamma_{ab}$ and conjugate momentum $\pi^{ab}$.

My rationality for this assumption:

Immediately after the equation I'm trying to solve, the book states that ${A_a}^i$ and ${E^b}_j$ will be the new configuration variable and canonical momentum. This implies that different configuration variables were previously being used.

The Poisson bracket $\{f,g\}$ applied to the ADM formalism is defined in terms of the spatial metric $\gamma_{ab}$ and conjugate momentum $\pi^{ab}$ as follows: $\{f,g\} = {{\partial f}\over{\partial \gamma_{ab}}} {{\partial g}\over{\partial \pi^{ab}}} - {{\partial f}\over{\partial \pi^{ab}}} {{\partial g}\over{\partial \gamma_{ab}}}$. This definition is found in Kiefer p.112 eqn.4.64, Romano page 14 eqn 2.33, and Alcubierre p.81 eqn.2.7.14.

Romano has a similar statement, footnote 11 on page 26, but with ${A_a}^i$ and ${E^b}_j$ already chosen to be the configuration variables: $\{{A_a}^I(x), {E^b}_j(y)\} = \delta^b_a \delta^i_j \delta(x,y)$. Giulini also has a similar statement at p.26 eqn.5.23.

What I've gathered so far:

Substituting $\{{A_a}^i(x), {E^b}_j(y)\}$ into the Poisson bracket definition gives us:

$${{\partial}\over{\partial \gamma_{ab}}} {A_a}^i(x) {{\partial}\over{\partial \pi^{ab}}} {E^b}_j(y) - {{\partial}\over{\partial \pi^{ab}}} {A_a}^i(x) {{\partial}\over{\partial \gamma_{ab}}} {E^b}_j(y)$$

Now ${A_a}^i = {A_a}^{i\hat{t}}$ is the timelike portion of the self-dual connection, and $${A_\alpha}^{IJ} = {^+}{\omega_\alpha}^{IJ} = {1\over2}({\omega_\alpha}^{IJ} + {i\over2} {\epsilon^{IJ}}_{KL} {\omega_\alpha}^{KL})$$ is the self-dual of the spin connection.

And the spin connection ${\omega_\alpha}^{IJ}$ is defined as the Minkowski coordinate connection to cancel $\nabla_\alpha {e_\mu}^I$, as $${{\omega_\alpha}^I}_J = {\Gamma^\mu}_{\nu\alpha} {e_\mu}^I {e^\nu}_J - {e^\mu}_J \partial_\alpha {e_\mu}^I$$

For ${\Gamma^\mu}_{\nu\alpha}$ the affine connection of the spacetime metric.

With all this said, I would think to chain rule ${{\partial {A_a}^i}\over{\partial \gamma_{cd}}} = {{\partial {A_a}^i}\over{\partial {e_f}^J}} {{\partial {e_f}^J}\over{\gamma_{cd}}}$ and then substitute the first term for the derivatives of the definitions of ${A_\alpha}^{IJ}$ and ${\omega_\alpha}^{IJ}$ above.

The second term stumps me though. For $\gamma_{cd} = {e_c}^j {e_d}^k \delta_{jk}$ how would you calculate ${{\partial {e_f}^i}\over{\partial \gamma_{cd}}}$?

My next thought is to simplify the inverse of the derivative: \begin{align} {{\partial \gamma_{cd}}\over{\partial {e_f}^i}} &= {\partial\over{\partial {e_f}^i}} ({e_c}^j {e_d}^k \delta_{jk})\\ &= ({\partial\over{\partial {e_f}^i}} {e_c}^j) {e_d}^k \delta_{jk} + {e_c}^j ({\partial\over{\partial {e_f}^i}} {e_d}^k) \delta_{jk}\\ &= \delta^j_i \delta^f_c {e_d}^k \delta_{jk} + {e_c}^j \delta^k_i \delta^f_d \delta_{jk}\\ &= \delta^f_c e_{di} + \delta^f_d e_{ci} \end{align} ...but how do you solve the inverse of a rank-4 tensor?

Is there a better approach?



Kiefer, Claus. "Quantum Gravity."

Romano. "Geometrodynamics vs Connection Dynamics."

Giulini, "Ashtekar Variables in Classical General Relativity.

Alcubierre, Miguel. "Introduction to 3+1 Numerical Relativity."

This post imported from StackExchange Physics at 2015-10-20 22:02 (UTC), posted by SE-user thenumbernine
asked Oct 19, 2015 in Theoretical Physics by thenumbernine (10 points) [ no revision ]
retagged Oct 20, 2015

1 Answer

+ 0 like - 0 dislike

Why not use $\gamma_{cd} = {e_c}^j {e_d}^k \delta_{jk}$, differentiate this with respect to $\gamma_{ab}$ and by the chain rule deduce the result? That is, we know $$ \frac{\partial\gamma_{cd}}{\partial\gamma_{ab}}=\delta^{ab}_{cd} $$ but we also know $$ \frac{\partial\gamma_{cd}}{\partial\gamma_{ab}}=\frac{\partial({e_c}^j {e_d}^k \delta_{jk})}{\partial\gamma_{ab}}\sim 2e\delta\frac{\partial e}{\partial\gamma} $$ Set these two expressions equal, and you've got your result, right?

This post imported from StackExchange Physics at 2015-10-20 22:02 (UTC), posted by SE-user Alex Nelson
answered Oct 20, 2015 by Alex Nelson (100 points) [ no revision ]
Thanks for the tip! That looks like the approach I've got listed. The $\approx$ is bugging me -- I'm looking for something with a $=$.

This post imported from StackExchange Physics at 2015-10-20 22:02 (UTC), posted by SE-user thenumbernine
@thenumbernine It's just the product rule: $\delta_{jk}({e_{c}}^{j} (\partial {e_{d}}^{k}/\partial_{ab}) + {e_{d}}^{k} (\partial {e_{c}}^{j}/\partial_{ab}))$, then set it equal to $\delta^{(a}_{c}\delta^{b)}_{d}$, and you're done.

This post imported from StackExchange Physics at 2015-10-20 22:02 (UTC), posted by SE-user Alex Nelson

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