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The action for a string in this background $$G_{IJ}\tag{1}$$ can be written as the Nambu-Goto action

$$S_{NG}=\int d\sigma^1d\sigma^2\sqrt{g}\quad\quad\Rightarrow\quad\mathcal{L}=\sqrt{g}\tag{2}$$

where the induced two-dimensional metric is

$$g_{ab}=G_{IJ}\partial_aX^I\partial_bX^J.\tag{3}$$

This action represents the worldsheet area of the string, a two dimensional Riemannian manifold. This area is minimal if the Euler-Lagrange equations are satisfied but also if an equivalent equation is satisfied: the Hamilton-Jacobi equation, which have this form (see footnote at page 13 in Drukker)

$$G^{IJ}\left(\frac{\delta S}{\delta X^I}\right)\left(\frac{\delta S}{\delta X^J}\right)=G_{MN}\partial_1X^{M}\partial_1X^{N}\tag{4}$$

(in this form), where $$\partial_a=\frac{\partial}{\partial\sigma^a}\,,\quad\sigma=1,2.\tag{5}$$

I know that the Hamilton-Jacobi equation is

$$\frac{\partial S}{\partial t}+H\left(\frac{\partial S}{\partial x},x\right)=0.\tag{6}$$

How this expressions translates into the previous one?

EDIT:

Let me show you what I have. From (2) and the expression of the determinant

$$g=\frac{1}{2}\varepsilon^{ab}\varepsilon^{cd}g_{ac}g_{bd}\tag{7}$$

$$ P_I^a=\frac{\partial\mathcal{L}}{\partial\partial_aX^I}=\frac{1}{\sqrt{g}}\varepsilon^{ab}\varepsilon^{cd}\partial_cX^JG_{IJ}g_{bd}\tag{8}$$

right?

Then

$$\mathcal{H}=P_I^a\partial_aX^I-\mathcal{L}\tag{9}=\sqrt{g}.$$

why is not zero?

EDIT 2

Let us start with an equivalent action, Polyakov

$$S_P=\frac{1}{2}\int d^2\sigma\sqrt{-h}h^{ab}\partial_aX^I\partial_bX^JG_{IJ}.\tag{10}$$

The momentum is

$$P_I^a=\frac{\partial\mathcal{L}_P}{\partial\partial_aX^I}=\sqrt{-h}h^{ab}\partial_bX^JG_{IJ}.\tag{11}$$

Let us choose

$$h_{ab}=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\,\quad\quad\Rightarrow\sqrt{-h}=1.\tag{12}$$

The Hamiltonian is then,

$$\mathcal{H}_P=\frac{1}{2}\int d^2\sigma h_{ab}P^a_IP^b_JG^{IJ}.\tag{13}$$

Due to reparametrization invariance

$$h_{ab}P^a_IP^b_JG^{IJ}=0,\tag{14}$$

or

$$G^{IJ}P_I^\sigma P_J^\sigma=\partial_\tau X^I\partial_\tau X^JG_{IJ}.\tag{15}$$

Is this correct?

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1 Answer 1

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  1. Since we assume that the target-space (TS) metric $G_{IJ}$ does not depend explicitly on the world-sheet (WS) coordinates $(\tau,\sigma)$, the relevant Hamilton-Jacobi (HJ) equation is the time-independent formulation $$H(x, \frac{\partial W}{\partial x})~=~E\tag{A}$$ in terms of Hamilton's characteristic function $W$ rather than Hamilton's principal function $S$. Because of WS reparametrization invariance, the rhs. $E=0$ of the HJ equation (A) vanishes, cf. e.g. this Phys.SE post.

  2. In fact due to WS reparametrization invariance, the Legendre transformation of the Nambu-Goto (NG) action is singular. We encounter 2 primary constraints $$ \frac{1}{2T_0}P^2\mp \frac{T_0}{2}(X^{\prime})^2~=~0 \qquad\text{and}\qquad P\cdot X^{\prime} ~=~0, \tag{B}$$ cf. e.g. this Phys.SE post. [Here the $\mp$ sign corresponds to Euclidean (Minkowskian) TS signature, respectively. Note that the TS metric induces a WS metric of the same$^1$ signature. The constraints (B) can alternatively be deduced from the equivalent Polyakov action, cf. my Phys.SE answer here and links therein.] The HJ theory is usually not developed systematically for constrained systems, but we can view $$ \frac{1}{2T_0}\left(\frac{\delta W}{\delta X}\right)^2\mp\frac{T_0}{2}(X^{\prime})^2~=~0 \qquad\text{and}\qquad \frac{\delta W}{\delta X}\cdot X^{\prime} ~=~0 \tag{C}$$ as the appropriate analog of the HJ equation/eikonal equation. The first equality in eq. (C) corresponds to OP's eq. (4). In Ref. 1 the boundary of the WS is a Wilson-loop parametrized by $\tau$.

  3. Concerning OP's Hamiltonian density (9): Note that OP's $a$-index should by definition only be a temporal WS index, not a spatial WS index. Then the Hamiltonian density (9) indeed vanishes. In particular one does not sum over the $a$-index here. In eq. (11) OP is introducing polymomenta a la De Donder & Weyl. There is a similar issue with OP's eq. (13).

References:

  1. N. Drukker, D.J. Gross & H. Ooguri, Wilson Loops and Minimal Surfaces, arXiv:hep-th/9904191.

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$^1$ Eq. (12) is inconsistent with OP's Riemannian TS signature.

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  • $\begingroup$ Thanks. But I think the result I was trying to get is more general than the expression you get in (C). I added a reference 'cause it seems thiis result is actually well-known for Riemannian metrics. I was trying to derive it from $P_I^a\partial_aX^I=\mathcal{L}$ which is the same as $\mathcal{H}=0$. $\endgroup$ Commented Jun 10, 2019 at 14:28
  • $\begingroup$ thanks again. Why do the index "a" should be temporal? couldnt it be spatial? I mean, if the idea is to vanish the Hamiltonian, I could choose also the other, right? What does mean choosing only one of them? Actually, in that paper, authors choose the spatial WS coodinate. Thus $$\mathcal{H}=P^\sigma\cdot{Y}'-\mathcal{L}$$. $\endgroup$ Commented Jun 13, 2019 at 18:07
  • $\begingroup$ Also, due to reparametrization invariance, in general, it means that eq.(4) in my question is actually the vanishing Hamiltonian, right? $\endgroup$ Commented Jun 13, 2019 at 18:11
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jun 14, 2019 at 7:33

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