Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Deriving the Poisson bracket relation of the Ashtekar variables

+ 2 like - 0 dislike
905 views

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?

Thanks.

Sources:

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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...