3

I met the word smearing function (or test function) when I was learning ADM formalism in GR books. What makes me scratch my head is the reason of introducing such a smearing function when we calculate the Poisson bracket of constraints.

For example, when we are calculating time evolution of a primary constraint $\phi^a=0$, we should in principle have $$\int s_a\partial_t\phi^a d^3x=\int s_a\left[\phi^a,\mathcal{H}\right] d^3x\simeq0$$ where $\mathcal{H}$ is the Hamiltonian density, "$\simeq$" simbol means that this equality meets on shell, and $s_a$ is the corresponding smearing function which goes to $0$ very fast on the boundary and is $C^\infty$ (infinitely differentiable)

My question is then,

  1. why we shall introduce such a smearing function rather than do the Poisson bracket directly?
  2. suppose two another smearing functions $m^a$, $n^b$, and an existing vector field depending on space and time $\psi^a(\textbf{x},t)$, which is also $C^\infty$ and vanishes on the boundary. We have $$m^an^b=\psi^ap^b$$ $$m^a\psi^b=n^aq^b$$ Here, may I ask whether $p^b$ and $q^b$ are still smearing functions?

------------------------------------------------second edition-----------------------------------------------

Let me give an example to the second question. Suppose we have a primary constraint $T_{ab}\psi^a=0$. Here $T_{ab}$ is a tensor and $\psi=\psi^a(\textbf{x},t)$ is the vector field mentioned above. Now, when we are calculating the secondary constraints, since calculating Poisson bracket makes us have two smearing function, we find, let's say, such a term $$\int m^an^bT_{ab}d^3x$$ could we now use $m^an^b=\psi^ap^b$ so that this term becomes $$\int m^an^bT_{ab}d^3x=\int\psi^ap^bT_{ab}d^3x$$ and thus vanishes because of the primary constraint?
Thanks for the attention

Chunhui
  • 67
  • https://physicstravelguide.com/advanced_notions/quantum_field_theory/smearing_functions – MatterGauge Jun 02 '22 at 13:10
  • @Felicia Thanks for sharing. Actually I have seen this page. It helps me about the physics reason about it. But I'm not sure what does the "singular" in the sentence "These are in general rather singular if we calculate them naively" refers to. It still really helps to understand rather briefly – Chunhui Jun 02 '22 at 13:57

1 Answers1

1

In this answer we address OP's first question:

  1. In Hamiltonian field theory, in order to avoid multiplying two distributions together, the Poisson bracket is defined for local functional (rather than point-wise), cf. e.g. my Phys.SE answer here. We therefore turn the constraint $\phi^a(x)$ into a local functional $$\Phi^a[s]~=~\int_Mdx~ s(x)\phi^a(x),$$ where $s(x)$ is a test function. So instead of saying that we have a constraint for each point $x\in M$, we have a constraint for each test function $s$, which is a more well-defined/smooth way of implementing the constraint.
Qmechanic
  • 201,751
  • Thanks for the answer and link! 1. May I ask, why two distributions cannot be miltiplied together, but two functionals can? I can understand the physics reason but I'm not clear about it in mathematical aspect:( 2. I was thinking about my second question. So here "p" and "q" seems to meet the inquirements of "smearing function", but the point is that they have to depend on time. It is still weird though, let me update my question to make it clearer! – Chunhui Jun 02 '22 at 13:42
  • In the theory of distributions, it is not always possible to define a product (although check out the theory of Colombeau).
    • – Qmechanic Jun 02 '22 at 15:48