4
$\begingroup$

I'm a bit confused about the tensor density weight of Poisson brackets in general relativity and their covariance. It's perhaps related to being unclear as to what happens when I integrate a scalar density of some weight other than 1. Let's say I have the Poisson bracket of General Relativity in the 3+1 ADM formalism acting on some local scalar $f(x)$ on a space-time slice and some scalar quantity $G$. ($G$ could be the Hamiltonian, and $f(x)$ could be a scalar, but it could also be a scalar density, e.g .$\sqrt{g}$ which changes things but not the essence of what I'm asking). The Poisson bracket is given by \begin{align} \{f(x),G\}&=\int d^3y \Big[ \frac{\delta f(x)} {\delta g_{ab}(y)}\frac{\delta G}{\delta \pi^{ab}(y)} - \frac{\delta f(x)} {\delta \pi^{ab}(y)}\frac{\delta G}{\delta g_{ab}(y)}\Big] \\ &\stackrel{?}{=} \frac{\delta f(x)} {\delta g_{ab}(x)}\frac{\delta G}{\delta \pi^{ab}(x)} - \frac{\delta f(x)} {\delta \pi^{ab}(x)}\frac{\delta G}{\delta g_{ab}(x)} \end{align} with $g_{ab}$ the 3-metric and using the convention of taking its conjugate momenta $\pi^{ab}$ a tensor density of weight one (since we derive it from the Lagrangian density). 2 questions: The first is that the tensor weight of the first expression seems to be -2 (plus whatever comes with $f$, since I have the $d^3y$ on the top and the $\delta \pi^{ab}$ on the bottom. Since the left hand side is usually something like $\partial_tf(x)$, I would have expected it to have a tensor weight of 1. And this expression doesn't look like it will give diffeomorphism invariance, although I accept that it must (I guess one needs to consider how the 3-manifold sits in the 4-manifold for this).

There is some discussion of the invariance properties of the Poisson bracket here: Poisson brackets in curved spacetime, but I don't find it particularly enlightening. Anyone have a simple explanation?

$\endgroup$
2
$\begingroup$
  1. This is not limited to GR. More generally, given an $(r,s)$ tensor field $\phi(x)$, the conjugate momentum field $\pi(x)$ is an $(s,r)$ tensor density field. See also this related Phys.SE post$^1$.

  2. Given two scalar local functionals of the form $$ F =~ \int d^3x~\rho(x)~f(x)\qquad\text{and}\qquad G =~ \int d^3x~\rho(x)~g(x),\tag{A}$$ where $\rho(x)$ is a density field, and $f(x),g(x)$ are scalar fields, then the functional derivatives$^2$ $$\frac{\delta F}{\delta\phi(x)}~=~\frac{\partial [\rho(x) f(x)]}{\partial \phi(x)} -\frac{d}{dx^i} \frac{\partial [\rho(x) f(x)]}{\partial [\partial_i\phi(x)]}+\ldots\tag{B}$$ and $$ \frac{\delta G}{\delta\pi(x)}~=~\frac{\partial [\rho(x) g(x)]}{\partial \pi(x)} -\frac{d}{dx^i} \frac{\partial [\rho(x) g(x)]}{\partial [\partial_i\pi(x)]}+\ldots \tag{C}$$ are an $(s,r)$ tensor density field and an $(r,s)$ tensor field, respectively. Therefore the canonical Poisson bracket $$\{ F,G\}~=~ \int d^3x~\left(\frac{\delta F}{\delta\phi(x)}\frac{\delta G}{\delta\pi(x)}-\frac{\delta F}{\delta\pi(x)}\frac{\delta G}{\delta\phi(x)}\right)\tag{D}$$ is again a scalar local functional.

  3. In particular,$^3$ $$\begin{align} \{ f(x),G\}&=~ \int d^3y~\left(\frac{\delta f(x)}{\delta\phi(y)}\frac{\delta G}{\delta\pi(y)}-\frac{\delta f(x)}{\delta\pi(x)}\frac{\delta G}{\delta\phi(y)}\right)\cr &=~ \frac{\delta f(x)}{\delta\phi(x)}\frac{\delta G}{\delta\pi(x)}-\frac{\delta f(x)}{\delta\pi(x)}\frac{\delta G}{\delta\phi(x)}\end{align} \tag{E}$$ in terms of 'same-space' functional derivatives $$ \frac{\delta f(x)}{\delta\phi(x)}~:=~ \frac{\partial f(x)}{\partial \phi(x)} -\frac{d}{dx^i} \frac{\partial f(x)}{\partial [\partial_i\phi(x)]}+\ldots\tag{F} ,$$ cf. e.g. my Phys.SE answer here.

--

$^1$ If $\phi(x)$ is a tensor density field, then the conjugate momentum field $\pi(x)$ is a tensor field, i.e. then the roles are reversed.

$^2$ The ellipsis $\ldots$ denotes possible terms from higher-order space-derivatives.

$^3$ In this answer we use the convention that the Dirac delta distribution $\delta^3 (x,y)$ is density-valued $$\int d^3y~\delta^3 (x,y) f(y)~=~f(x). \tag{G}$$ Moreover, we use the convention that $$ \frac{\delta\phi(x)}{\delta\phi(y)}~=~\delta^3(x,y) \tag{H} .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.