How to find the rank of the matrix $\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }$ for the Nambu-Goto string Lagrangian?

Question

In this case $$\mathcal{L}~=~-T\sqrt{-\dot{X^2}X'^2+(\dot{X}\cdot X')^2}.$$ I was reading some books and papers about the constraints in the Nambu-Goto action, and all of them say something like

the matrix has two zero eigenvalues corresponding to the eigenvectors $\dot{X^\mu}$ and $X'^\mu$,

and then is no more explanation, they simply write down the two primary constraints. So, if they were the only null-eigenvectors, the rank should be $R=D-2$ but if (for example) there is one more eigenvector the rank would be $D-3$.

In fact it is not hard to see that $\dot{X^\mu}$ and $X'^\mu$ are null-eigenvectors, this is the idea:

$$\begin{align}\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }\dot{X^\mu} =&\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }\dot{X^\mu} +\frac{\partial \mathcal{L}}{\partial \dot{X^\nu} } -\frac{\partial \mathcal{L}}{\partial \dot{X^\nu} }\cr =&\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }\dot{X^\mu} +\frac{\partial \mathcal{L}}{\partial \dot{X^\mu} }\delta_\nu^\mu -\frac{\partial \mathcal{L}}{\partial \dot{X^\nu} },\end{align}$$

$$\begin{align}\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }\dot{X^\mu} =&\frac{\partial }{\partial \dot{X^\nu} }\left( \frac{\partial \mathcal{L}}{\partial \dot{X^\mu} }\dot{X^\mu}\right)-\frac{\partial \mathcal{L}}{\partial \dot{X^\nu} }\cr =&\frac{\partial }{\partial \dot{X^\nu} }\left( \frac{\partial \mathcal{L}}{\partial \dot{X^\mu} }\dot{X^\mu}-\mathcal{L}\right)=0,\end{align}$$

and

$$\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }X'^\mu= \frac{\partial }{\partial \dot{X^\nu} }\left( \frac{\partial \mathcal{L}}{\partial \dot{X^\mu} }X'^\mu\right)=\frac{\partial }{\partial \dot{X^\nu} }\left( 0\right)=0.$$

That is all I have done, how can I be sure that there are no more eigenvectors?

By the way, in the case of the point-particle I got an explicit matrix $$\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }=-\frac{m}{(-\dot{X^2})^{3/2}}(g_{\mu\nu}\dot{X^2}-\dot{X}_\mu\dot{X}_\nu),$$ so you can be sure that the rank is in this case $D-1$. But in the case of the string I get an expression very large and I'm not sure if it is right.

How I can be sure about the rank of that matrix?

Also, it would be interesting to see the general case of

$$S=-T\int dt\, d^p\sigma \sqrt{-\det (g)}$$

with $X^\mu (t,\sigma ^1,\ldots ,\sigma ^p)$ where $g$ is the metric on $X^\mu$ induced from the ambient space-time metric (this is a $p$-brane ; $p=0$ particle; $p=1$ string) and try to find the rank. I suspect that the answer should be $R=D-(1+p)$. Perhaps you do not have to get a explicit formula for the matrix and there is a more clever way to find the Rank. What would be that way?

Answer 1

I) In this answer we will consider the standard Nambu-Goto (NG) string and show that the Hessian has co-rank 2. The target space (TS) metric $G_{\mu\nu}(X)$ has sign convention $(-,+,\ldots,+)$, and $c=1=\hbar$. The NG Lagrangian density is

$$\begin{align}{\cal L}_{NG} ~:=~&-T_0\sqrt{{\cal L}_{(1)}}, \cr {\cal L}_{(1)}~:=~&-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta}\cr ~=~&-\det\begin{pmatrix}a &c \cr c & d \end{pmatrix}\cr ~=~& c^2-a d\geq 0,\cr a~:=~&\dot{X}^2~\leq~0,\cr c~:=~&\dot{X}\cdot X^{\prime} ,\cr d~:=~&(X^{\prime})^2~>~0.\end{align}\tag{1}$$ The inequality ${\cal L}_{(1)}\geq 0$ is explained in e.g. Ref. 1. We will in the following only consider regular world-sheet points where $${\cal L}_{(1)}~>~0\tag{2}$$ is strictly positive.

II) Momenta are $$\begin{align} P_{(1)\mu} ~:=~&\frac{\partial {\cal L}_{(1)}}{\partial \dot{X}^{\mu}}\cr ~=~& 2c X^{\prime}_{\mu} -2 d \dot{X}_{\mu} , \cr P_{(1)}\cdot \dot{X}~=~&2{\cal L}_{(1)},\end{align}\tag{3} $$ $$ P_{\mu} ~:=~\frac{\partial {\cal L}_{NG}}{\partial \dot{X}^{\mu}} ~=~ -\frac{T_0}{2{\cal L}_{(1)}^{\frac{1}{2}}} P_{(1)\mu}. \tag{4}$$

III) The original Hamiltonian density vanishes identically

$$ {\cal H}_0~:=~ P\cdot \dot{X} - {\cal L}_{NG}~=~0 , \tag{5}$$

as we would expect for a reparametrization-invariant theory. This means that whatever primary constraints there are, there will not be any secondary constraints. We find two primary constraints

$$ P_{(1)}\cdot X^{\prime} ~=~0 \quad\Rightarrow \quad \chi_0~:=~P\cdot X^{\prime} ~=~0,\tag{6}$$

$$ P_{(1)}^2~=~ -4d {\cal L}_{(1)}\quad\Rightarrow \quad \chi_1~:=~\frac{P^2}{2T_0}+\frac{T_0}{2}(X^{\prime})^2~=~0.\tag{7} $$

The two primary constraints (6) & (7) then form a first class Poisson algebra

$$\begin{align}\{\chi_0(\sigma),\chi_0(\sigma^{\prime})\}_{PB} ~=~&\left[ \chi_0(\sigma)+\chi_0(\sigma^{\prime})\right] \delta^{\prime}(\sigma-\sigma^{\prime})\cr ~=~&\{\chi_1(\sigma),\chi_1(\sigma^{\prime})\}_{PB},\cr \{\chi_0(\sigma),\chi_1(\sigma^{\prime})\}_{PB} ~=~&\left[ \chi_1(\sigma)+\chi_1(\sigma^{\prime})\right] \delta^{\prime}(\sigma-\sigma^{\prime}). \end{align}\tag{8} $$

Equivalently, if we define

$$ \chi_{\pm} ~:=~ \frac{\chi_1\pm\chi_0}{2} ~=~ T_0Y_{\pm}^2 ,\tag{9}$$

with

$$\begin{align} Y^{\mu}_{\pm} ~:=~&\frac{1}{2T_0}P^{\mu}\pm\frac{1}{2}X^{\mu\prime},\cr Y_{\pm,\mu}~:=~&G_{\mu\lambda}Y^{\lambda}_{\pm}\cr ~=~&\frac{1}{2T_0}P_{\mu}\pm\frac{1}{2}G_{\mu\lambda}X^{\lambda\prime}, \end{align}\tag{10}$$

which satisfies

$$\begin{align} \{Y_{\pm,\mu}& (\sigma), Y_{\pm^{\prime},\nu}(\sigma^{\prime})\}_{PB}\cr ~=~& \frac{1}{4T_0}\left[ \pm G_{\mu\nu}(\sigma)\pm^{\prime} G_{\mu\nu}(\sigma^{\prime})\right]\delta^{\prime}(\sigma-\sigma^{\prime}) \cr &+\frac{1}{4T_0}\left[\pm \partial_{\nu}G_{\mu\lambda}\mp^{\prime}\partial_{\mu}G_{\nu\lambda}\right]X^{\lambda\prime} \delta(\sigma-\sigma^{\prime}) \cr ~=~& \frac{\pm 1 \pm^{\prime} 1}{4T_0}\left[ G_{\mu\nu}(\sigma) + G_{\mu\nu}(\sigma^{\prime})\right]\delta^{\prime}(\sigma-\sigma^{\prime})\cr & - \frac{\pm 1 \pm^{\prime} 1}{8T_0}\Gamma_{[\mu,\nu]\lambda}X^{\lambda\prime} \delta(\sigma-\sigma^{\prime})\cr & + \frac{\pm 1 \mp^{\prime} 1}{4T_0}X^{\lambda\prime} \Gamma_{\lambda,\mu\nu} \delta(\sigma-\sigma^{\prime}), \end{align}\tag{11}$$

then we classically get a direct sum of two copies of the Witt algebra

$$\begin{align}\{\chi_{\pm}(\sigma),&\chi_{\pm^{\prime}}(\sigma^{\prime})\}_{PB}\cr~=~& \frac{\pm 1 \pm^{\prime} 1}{2} \left[ \chi_{\pm}(\sigma)+\chi_{\pm^{\prime}}(\sigma^{\prime})\right] \delta^{\prime}(\sigma-\sigma^{\prime}).\end{align} \tag{12}$$

Note in particular that the $+$ and the $-$ sector in eq. (12) Poisson mutually commute! The total Hamiltonian density becomes of the form "Lagrange multipliers times constraints"

$$ {\cal H}~=~\lambda^{\alpha} \chi_{\alpha}, \qquad \alpha~\in~\{0,1\} .\tag{13}$$

IV) The Hessian reads

$$H_{(1)\mu\nu}~:=~\frac{\partial^2 {\cal L}_{(1)}}{\partial \dot{X}^{\mu}\partial \dot{X}^{\nu}} ~=~2 X^{\prime}_{\mu}X^{\prime}_{\nu} -2d G_{\mu\nu}\tag{14} ,$$

$$H_{(1)\mu\nu}X^{\prime\nu}~=~0, \qquad H_{(1)\mu\nu}\dot{X}^{\nu}~=~P_{(1)\mu},\tag{15} $$

$$\begin{align} H_{\mu\nu}~:=~& \frac{\partial^2 {\cal L}_{NG}}{\partial \dot{X}^{\mu}\partial \dot{X}^{\nu}}\cr ~=~&-\frac{T_0}{2{\cal L}_{(1)}^{\frac{1}{2}}}H_{(1)\mu\nu}+\frac{T_0}{4{\cal L}_{(1)}^{\frac{3}{2}}}P_{(1)\mu}P_{(1)\nu}\end{align}\tag{16} ,$$ $$\begin{align} -\frac{{\cal L}_{(1)}^{\frac{3}{2}}}{T_0}H_{\mu\nu} ~=~&\frac{1}{2}{\cal L}_{(1)}H_{(1)\mu\nu}+\frac{1}{4}P_{(1)\mu}P_{(1)\nu}\cr ~=~& (c^2-ad)(X^{\prime}_{\mu}X^{\prime}_{\nu} -d G_{\mu\nu})\cr &- (c X^{\prime}_{\mu} -d \dot{X}_{\mu})(c X^{\prime}_{\nu} -d \dot{X}_{\nu}).\end{align}\tag{17} $$

V) It is easy to check that $\dot{X}$ and $X^{\prime}$ are two zero-modes for the Hessian $H_{\mu\nu}$.

Now consider an arbitrary zero-mode

$$Z~\notin~ {\rm span}_{\mathbb{R}}(\dot{X},X^{\prime}).\tag{18}$$

We would like to find two real numbers $\alpha,\beta\in\mathbb{R}$ such that the vector $$V ~:=~ Z - \alpha \dot{X} - \beta X^{\prime} \tag{19}$$ is orthogonal to $\dot{X}$ and $X^{\prime}$, i.e. $$ V\cdot \dot{X}~=~0\qquad\text{and}\qquad V\cdot X^{\prime}~=~0. \tag{20}$$ It is easy to see that this is possible if ${\cal L}_{(1)}\neq 0$, which is satisfied in regular worldsheet points, cf. ineq. (2). It then follows from eq. (18) that $V\neq 0 $. And since $\dot{X}$ is non-spacelike, then eq. (20) implies that $V$ is space-like.

VI) Finally the quadratic form reads

$$ 0~=~Z^{\mu}H_{\mu\nu}Z^{\nu}~\stackrel{(14)}{=}~\frac{T_0 d V^2}{{\cal L}_{(1)}^{\frac{1}{2}}}~>~0.\tag{21} $$ Contradiction.

Hence $\dot{X}$ and $X^{\prime}$ are the only two zero-modes. They go hand in hand with the two first class constraints (6) and (7).

References:

1. B. Zwiebach, A first course in String Theory, 2nd edition, 2009; p. 109-110.

2. E. Kiritsis, String Theory in a Nutshell, 2007; p.15.

Answer 2

I) In this alternative answer we resolve the singular Hessian $H_{\mu\nu}$ of the Nambu-Goto string action by introducing two auxiliary variables from the onset, thereby indirectly showing that the Hessian $H_{\mu\nu}$ must have co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. Consider the extended Nambu-Goto Lagrangian density$^1$

$$\begin{align} {\cal L}(X,e,b) ~:=~&-\frac{{\cal L}_{(1)}(X,b)}{2e}-\frac{eT_0^2}{2}, \qquad e~>~0,\cr {\cal L}_{(1)}(X,b)~:=~&-b^2+2bc-a d ,\cr a~:=~&\dot{X}^2~\leq~0,\cr c~:=~&\dot{X}\cdot X^{\prime} ,\cr d~:=~&(X^{\prime})^2~>~0. \end{align}\tag{1}$$

II) The eoms for $e$ and $b$ are

$$ (eT_0)^2~\approx~{\cal L}_{(1)}(X,b) \qquad\text{and}\qquad b~\approx~c, \tag{2} $$ respectively. [The equation of motion (eom) means the EL equation. The $\approx$ sign means here equal modulo the eoms.] If we integrate out the auxiliary variables $e$ and $b$, we get the usual Nambu-Goto Lagrangian density

$$\begin{align}{\cal L}_{(1)}(X,b\!=\!c) ~=~&-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta} \cr ~=~&-\det\begin{pmatrix}a &c \cr c & d \end{pmatrix}\cr ~=~& c^2-a d~\geq~ 0, \cr {\cal L}_{NG}(X)~:=~&{\cal L}\left(X,e\!=\!\frac{\sqrt{{\cal L}_{(1)}(X,b\!=\!c)}}{T_0} ,b\!=\!c\right)\cr ~=~& -T_0\sqrt{{\cal L}_{(1)}(X,b\!=\!c)}.\end{align}\tag{3} $$

III) Momenta are

$$ P_{(1)\mu} ~:=~\frac{\partial {\cal L}_{(1)}}{\partial \dot{X}^{\mu}} ~=~ 2b X^{\prime}_{\mu} -2 d \dot{X}_{\mu} ,\tag{4} $$ $$ P_{\mu}~:=~\frac{\partial {\cal L}}{\partial \dot{X}^{\mu}} ~=~ -\frac{1}{2e} P_{(1)\mu}. \tag{5}$$

It is possible to solve for the velocities

$$ \begin{align} \dot{X}_{\mu}~=~&\frac{b}{d}X^{\prime}_{\mu}-\frac{1}{2d}P_{(1)\mu}\cr ~=~&\frac{b}{d}X^{\prime}_{\mu}+\frac{e}{d}P_{\mu}.\end{align}\tag{6}$$

This indirectly shows that the original Nambu-Goto Hessian $H_{\mu\nu}$ (in the $X$-sector only) must have co-rank (less or equal to) 2. The introduction of auxiliary variables $e$ and $b$ has made the correspondence $\dot{X} \leftrightarrow P$ bijective. [The full Legendre transformation is singular in the auxiliary sector, since $\dot{e}$ and $\dot{b}$ do not appear in ${\cal L}$.]

IV) Removing the velocity dependence (6), the Lagrangian density becomes

$$ {\cal L}_{(1)}~=~-\frac{e^2P^2}{d}.\tag{7} $$

The Hamiltonian density becomes of the form "Lagrange multipliers times constraints"

$$\begin{align} {\cal H}~:=~& P\cdot \dot{X} - {\cal L}\cr ~=~&\frac{b}{d}\chi_1 +\frac{e}{2d}\chi_2, \cr H :=& \int \! d\sigma~{\cal H}. \end{align}\tag{8}$$

Note that the auxiliary variables $e$ and $b$ play the role of Lagrange multipliers in the Hamiltonian formulation. The two first-class constraints are

$$ \begin{align} \chi_1~:=~&P\cdot X^{\prime} ~\approx~0, \cr \chi_2~:=~&P^2+T_0^2(X^{\prime})^2~\approx~0.\end{align}\tag{9} $$

The first half of Hamilton's eqs. reproduces eq. (6). The second half of Hamilton's eqs. yields the Nambu-Goto eom.

References:

1. B. Zwiebach, A first course in String Theory, 2nd edition, 2009; p. 109-110.

2. E. Kiritsis, String Theory in a Nutshell, 2007; p.15.

--

$^1$ The Gaussian integration over the auxiliary variable $b\equiv b_M$ looks naively unstable in Minkowski signature. One should Wick rotate $\tau_E=i\tau_M$ to Euclidean signature to get a Lagrangian density $-{\cal L}_M={\cal L}_E>0$ bounded from below with $-ib_M=b_E\in\mathbb{R}$.

Answer 3

A general proof applying to reparametrization-invariant actions $$S = \int d^2 u \mathcal{L}(x^{\mu},x_{,i}^{\mu}), \ \ \ \ i = 0,1, \ \ \ \mu = 0,1,\dots, D- 1,$$ is to note that part of the derivation of Noether's theorem (the derivation where you assume the volume element varies also), schematically derived as follows (see the reference below for details) for an infinitesimal variation of the parameters $\tilde{u}^i(u) = u^i + \varepsilon^i(u)$ by noting \begin{align} \delta S &= \delta \int d^2 u \mathcal{L}(x^{\mu},x_{,i}^{\mu}) \\ &= \int d^2 u [\dfrac{ \partial \mathcal{L}}{\partial x^{\mu}} \delta x^{\mu} + \dfrac{ \partial \mathcal{L}}{\partial x^{\mu}_{,i}} \delta x^{\mu}_{,i} + \mathcal{L} \partial_i \varepsilon^i ] \\ &= \int d^2 u \{ \partial_i [ (\mathcal{L} \delta^i_j - \dfrac{ \partial \mathcal{L}}{\partial x^{\mu}_{,i}} x^{\mu}_{j})\varepsilon^j ] - ( \dfrac{\partial \mathcal{L}}{\partial x^{\mu}} - \dfrac{\partial }{\partial u^i} \dfrac{\partial \mathcal{L}}{\partial x_{,i}^{\mu}}) x^{\mu}_{,i} \} \\ &= 0, \end{align} implies, for variations of the $u^i$ which vanish at the boundaries, that the operators \begin{align} L_{\mu}(\partial x, \partial^2 x) = \dfrac{\partial \mathcal{L}}{\partial x^{\mu}} - \dfrac{\partial }{\partial u^i} \dfrac{\partial \mathcal{L}}{\partial x_{,i}^{\mu}} \end{align} satisfy the identities \begin{align} L_{\mu}(\partial x, \partial^2 x) x^{\mu}_{,i} = 0 , \ \ \ i = 0,1 \end{align} automatically, which means we only have $D - 2$ equations of motion for $D$ coordinates, so that the solution depends on two arbitrary functions of the parameters $u^0,u^1$.

For actions like the Nambu-Goto action not depending on the $x^{\mu}$ the operators $L_{\mu}$ take the form $$L_{\mu}(\partial x, \partial^2 x) = \dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}}{\partial x'^{\mu}}) $$ and satisfy the identities \begin{align} L_{\mu}(\partial x, \partial^2 x) \dot{x}^{\mu} &= 0 \\ L_{\mu}(\partial x, \partial^2 x)x'^{\mu} &= 0 \end{align} which when expanded become, say, for $\dot{x}^{\mu}$: \begin{align} 0 &= L_{\mu}(\partial x, \partial^2 x) \dot{x}^{\mu} \\ &= [\dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}}{\partial x'^{\mu}})]\dot{x}^{\mu} \\ &= [\dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}(\dot{x},x')}{\partial x'^{\mu}})]\dot{x}^{\mu} \\ &= [ \dfrac{\partial^2 \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\nu} \partial \dot{x}^{\mu}} \ddot{x}^{\nu} + \dots + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}(\dot{x},x')}{\partial x'^{\mu}})]\dot{x}^{\mu} \\ &= [\dfrac{\partial^2 \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\nu} \partial \dot{x}^{\mu}} \dot{x}^{\mu} ] \ddot{x}^{\nu} + \dots \end{align} and since this must hold regardless of the choice of $x^{\mu}$, the coefficients of each derivative of $x^{\nu}$ must be zero automatically, implying the Hessian satisfies $$[\dfrac{\partial^2 \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\nu} \partial \dot{x}^{\mu}} \dot{x}^{\mu} ] = 0$$ and similarly starting from $x'^{\mu}$, giving the rank of the Hessian to be $D- 2$. Furthermore analyzing the components of $$t_{ij} = \mathcal{L} \delta^i_j - \dfrac{ \partial \mathcal{L}}{\partial x^{\mu}_{,i}} x^{\mu}_{j} = 0$$ you get both primary constraints for the Nambu-Goto action. This is all done in:

Reference:

1. Barbashov, B. M., & Nesterenko, V. V. (1990). Introduction to the relativistic string theory; Sec. 3, 7, Appendix B.