I) First consider the point particle


If you choose the static gauge $$\tau=X^0$$ and replace it in the action you get


So now, you have an equivalent action with true degrees of freedom only. In fact, you can do the same with the light-cone gauge $$\tau=X^+$$ and obtain


II) Is it possible to do the same for the string? I have found no reference doing this. I have been trying replacing the conditions of the light-cone gauge. I think that the answer could be

$$\mathcal{L}\sim \dot{X}^I\dot{X}^I -{X^I}' {X^I}'.\tag{12.81} $$

This is eq.(12.81) in the 2nd edition of Zwiebach's book.

Additional Information: Basically you have to show that the gauge conditions $$n\cdot X=2\beta \alpha'n\cdot p \tau ,\tag{1}$$

$$n\cdot p=\frac{2\pi}{\beta}n\cdot \mathcal{P},\tag{2}$$

(where $\beta=2$ for open string and $\beta=1$ for closed string) imply $$\dot{X^2}= -X'^2\quad\text{and}\quad \dot{X}\cdot X'=0.$$

And with these relations you can easily reduce the action.

For the open string you can show that condition (2) implys $$\dot{X}\cdot X'=0.$$

This can be done using the boundary conditions of the open string.$^1$

But for the closed string I have found no way to show that. In Zwiebach page 180 there is an idea of this for open and closed strings.$^2$

To sum up, my question is if one can deduce the same for the closed string. What would be the procedure?

$^1$See Sundermayer, Constrained Dynamics page 218, or Hansen-Regge-Teitelboim, Constrained Hamiltonian Systems page 58.

$^2$ But in the case of closed string the full $\sigma=0$ line is constructed by requiring that at each point its tangent be orthogonal to $X'$. So it is like he is imposing $\dot{X}\cdot X'=0$, not deducing it.

  • 3
    $\begingroup$ It is possible. It's known as light cone quantization. $\endgroup$ – Prahar Mitra May 18 '15 at 22:30

Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective:

  1. The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian is of the form "Lagrange multipliers times constraints"$^1$ $$ H~:=~\int_0^{\ell}\! d\sigma~{\cal H}, \qquad {\cal H}~=~\lambda^{\alpha} \chi_{\alpha}, \qquad \alpha~\in~\{0,1\}.\tag{1} $$

  2. The two first class constraints read $$ \chi_0~:=~P\cdot X^{\prime}~\approx~0, \qquad \chi_1~:=~\frac{P^2}{2T_0}+\frac{T_0}{2}(X^{\prime})^2~\approx~0.\tag{2}$$ The two first class constraints $\chi_{\alpha}$ originate from (and generate) the world-sheet (WS) reparametrization invariance of the Nambu-Goto action.

  3. The corresponding Hamiltonian Lagrangian then reads$^1$ $$ L_H~:=~\int_0^{\ell}\! d\sigma~{\cal L}_H, \qquad {\cal L}_H ~:=~ P\cdot \dot{X}-{\cal H} . \tag{3} $$ The Euler-Lagrange eqs. for the Hamiltonian Lagrangian (3) are Hamilton's eqs. $$ \dot{X}^{\mu}~\approx~\lambda^0X^{\mu\prime}+ \frac{\lambda^1}{T_0}P^{\mu},\qquad \dot{P}^{\mu}~\approx~\left(\lambda^0 P^{\mu}+ T_0\lambda^1 X^{\mu\prime}\right)^{\prime},\tag{4} $$ together with the two constraints (2).

  4. We use light-cone (LC) coordinates $$ X^{\pm}~:=~\frac{X^0\pm X^1}{\sqrt{2}} ,\tag{5} $$ and Minkowski metric $$\eta_{\mu\nu} ~dX^{\mu}~ dX^{\nu} ~=~ -(dX^0)^2 + (dX^1)^2 + (dX^{\perp})^2$$ $$~=~ -2dX^+dX^- + (dX^{\perp})^2,\tag{6}$$ in the target space (TS). [The $\perp$ symbol denotes transversal coordinates.]

  5. We fix the gauge symmetry by imposing two corresponding gauge-fixing conditions $$ X^+(\tau,\sigma)~=~f(p^+(\tau))\tau\qquad\text{and}\qquad P^+(\tau,\sigma)~=~p^+(\tau), \tag{7}$$ i.e. the LC gauge, where $f>0$ is some positive function, usually taken to be a linear or constant function. We have dropped a conventional multiplicative constant in the condition (7b) for simplicity.

  6. In the LC gauge (7), the Hamiltonian Lagrangian density (3) becomes $$ {\cal L}_H ~=~P\cdot \dot{X} -\lambda^0\left\{-X^{-\prime} p^+ +\chi^{\perp}_0 \right\} -\lambda^1\chi_1 $$ $$~\sim~P\cdot \dot{X} - \lambda^{0 \prime} X^- p^+ -\lambda^0\chi^{\perp}_0 -\lambda^1 \left\{-\frac{p^+ P^-}{T_0} + \chi^{\perp}_1\right\}. \tag{8}$$ The $\sim$ symbol means here equal modulo total derivative terms. Also we have introduced the hopefully natural short-hand notation $$ \chi^{\perp}_0~:=~P^{\perp}\cdot X^{\perp\prime}, \qquad \chi^{\perp}_1~:=~\frac{(P^{\perp})^2}{2T_0}+\frac{T_0}{2}(X^{\perp\prime})^2.\tag{9}$$

  7. Next decompose the two coordinates $X^-$ and $P^-$ in mean-value coordinates $$ x^-(\tau)~:=~ \frac{1}{\ell}\int_0^{\ell}\! d\sigma~X^-(\tau,\sigma),\qquad p^-(\tau)~:=~ \frac{1}{\ell}\int_0^{\ell}\! d\sigma~P^-(\tau,\sigma),\tag{10}$$ and coordinates $$Y^-~:=~X^- - x^-,\qquad R^-~:=~P^- - p^-,\tag{11}$$ with zero means.

  8. The kinetic term in the Hamiltonian Lagrangian (8) simplifies to $$ \int_0^{\ell}\! d\sigma~P\cdot \dot{X} ~=~\int_0^{\ell}\!d\sigma\left\{\left(-f(p^+)-\tau\frac{df(p^+)}{d\tau}\right) P^- -p^+\dot{X}^- + P^{\perp}\cdot \dot{X}^{\perp}\right\}$$ $$~=~- \ell f(p^+)p^- - \ell p^+\dot{x}^- +\int_0^{\ell}\!d\sigma~ P^{\perp}\cdot \dot{X}^{\perp}.\tag{12} $$ In the last equality of (12) we have dropped the term $\frac{df(p^+)}{d\tau}$, because $p^+$ is a constant of motion, since $x^-$ is a cyclic coordinate (in the Hamiltonian sense), cf. eq. (16) below.

  9. Note that in the Hamiltonian Lagrangian (8) the variables $Y^-$ and $R^-$ only appear in one place each inside $X^-$ and $P^-$, respectively. Integration over $Y^-$ and $R^-$ implies that $$\lambda^{0 \prime\prime}~\approx~ 0 \quad\text{and}\quad \lambda^{1 \prime}~\approx~ 0,\tag{13}$$ respectively. To understand the extra derivatives in eq. (13), see e.g. this Phys.SE post for the analogous argument for the Polyakov string action. Pertinent boundary/periodicity conditions then implies that $$\lambda^{0 \prime}~\approx~ 0.\tag{14} $$ So both the two Lagrange multipliers $\lambda^0$ and $\lambda^1$ are global constants.

  10. Integration over the constant $\lambda^1$-mode imposes the LC energy condition (2b) $$ p^- ~\approx~ \frac{T_0}{p^+\ell}\int_0^{\ell}\! d\sigma ~\chi^{\perp}_1,\qquad \chi^{\perp}_1~:=~\frac{(P^{\perp})^2}{2T_0}+\frac{T_0}{2}(X^{\perp\prime})^2.\tag{15}$$ Then $p^-$ is no longer an independent variable. The LC Hamiltonian reads $$ H^{LC}~:=~f(p^+)\ell p^-+\lambda^0 \int_0^{\ell}\! d\sigma~\chi^{\perp}_0 ~=~\int_0^{\ell}\! d\sigma~{\cal H}^{LC}, \tag{16}$$ $$ {\cal H}^{LC}~:=~\lambda^0\chi^{\perp}_0 +T_0\frac{f(p^+)}{p^+} \chi^{\perp}_1.\tag{17}$$ It is tempting to introduced the (possibly confusing) short-hand notation $$ \lambda^1~:=~T_0\frac{f(p^+)}{p^+}~>~0 \tag{18}$$ in eq. (17), so that the LC Hamiltonian (17) superficially has the same form as the original Hamiltonian (1). The $\lambda^1$ in eq. (18) should not be confused with the Lagrange multiplier zero-mode $\lambda^1$, which at this stage has been integrated out.

  11. The gauge-fixed LC Hamiltonian Lagrangian then becomes $$ L^{LC}_H~=~ - \ell p^+\dot{x}^- +\int_0^{\ell}\!d\sigma~ P^{\perp}\cdot \dot{X}^{\perp} - H^{LC},\tag{19} $$ where the LC Hamiltonian $H^{LC}$ is given in eq. (16). The remaining dynamical variables are the transversal variables $X^{\perp}$ and $P^{\perp}$ and the zero-modes $x^-$ and $p^+$. The non-zero fundamental Poisson brackets read $$ \{x^-, p^+\}_{PB}~=~-\frac{1}{\ell}, \qquad \{X^{\mu\perp}(\sigma), P^{\perp}_{\nu}(\sigma^{\prime})\}_{PB}~=~\delta^{\mu}_{\nu}\delta(\sigma-\sigma^{\prime}).\tag{20} $$ This answers OP's question about the true degrees of freedom. On top of that there is a zero-mode Lagrange multiplier $\lambda^0$, to be discussed below.

  12. Now let's discuss the role of the zero-mode Lagrange multiplier $\lambda^0$. For the open string with free ends, we should impose Neumann boundary conditions $$\frac{\partial{\cal L}_H}{\partial X^{\prime}_{\mu}}~=~0\quad\text{for}\quad \sigma~\in~\{0,\ell\}, \tag{21} $$ which for the coordinate $\mu=+$ imply that the constant mode $\lambda^0=0$ must vanish (if $p^+\neq 0$).

  13. For the remainder of this answer, we will discuss the closed string. Integration over the constant mode Lagrange multiplier $\lambda^0$ imposes the so-called level-matching constraint/condition (LMC) $$\int_0^{\ell}\!d\sigma~\chi^{\perp}_0~\approx~0, \qquad \chi^{\perp}_0~:=~P^{\perp}\cdot X^{\perp\prime}. \tag{22}$$ Conversely, an integration over a particular transversal string mode would assign a (quantum) average value to $\lambda^0$. However, to keep a nice clean classical picture, cf. eoms (23) below, we prefer to postpone the integrations over a particular transversal string mode and the zero-mode $\lambda^0$ for later.

  14. The LC Hamilton's eqs. read $$ \dot{X}^{\perp}~\approx~\lambda^0X^{\perp\prime}+ \frac{\lambda^1}{T_0}P^{\perp},\qquad \dot{P}^{\perp}~\approx~\lambda^0P^{\perp\prime}+ T_0\lambda^1X^{\perp\prime\prime},\tag{23} $$ where $\lambda^1$ is given by eq. (18). Eliminating the transversal momenta $P^{\perp}$ yields $$ \ddot{X}^{\perp}-2\lambda^0 \dot{X}^{\perp\prime} + (\lambda^0+\lambda^1)(\lambda^0-\lambda^1) X^{\perp\prime\prime}~\approx~0. \tag{24} $$

  15. Let's introduce new WS coordinates $$ \sigma^{\pm}~:=~ \sigma \pm \lambda^{\pm}\tau ~\equiv~(\sigma + \lambda^0\tau) \pm \lambda^1\tau, \qquad \lambda^{\pm}~:=~\lambda^1\pm \lambda^0, \tag{25}$$ along the characteristics of the PDE (24). The LC Hamilton's eqs. (23) become $$ P^{\perp}~\approx~T_0(\partial_+ - \partial_-)X^{\perp},\qquad (\partial_+ - \partial_-)P^{\perp}~\approx~T_0(\partial_+ + \partial_-)^2X^{\perp}.\tag{26} $$ The eom (24) factorizes $$ \partial_+\partial_-X^{\perp}~\approx~0, \tag{27} $$ with full solution being a sum of a left- and a right-mover $$ X^{\perp}~\approx~ X^{\perp}_L(\sigma^+)+X^{\perp}_R(\sigma^-). \tag{28} $$ Periodicity conditions impose further conditions on the left- and right-movers, cf. Refs. 1-5.

  16. On-shell, the LC Hamiltonian (16) becomes $$ H^{LC}~\approx~T_0\int_0^{\ell}\! d\sigma\left[\lambda^+(\partial_+X_L^{\perp})^2 + \lambda^-(\partial_-X_R^{\perp})^2 \right]$$ $$~=~T_0\int_0^{\ell}\! d\sigma\left[ \lambda^0 \left\{(\partial_+X_L^{\perp})^2-(\partial_-X_R^{\perp})^2\right\} + \lambda^1 \left\{(\partial_+X_L^{\perp}\}^2+(\partial_-X_R^{\perp})^2\right\}\right],\tag{29} $$ where $\lambda^1$ is given by eq. (18). Notice that the implicit dependence of $\lambda^0$ in eq. (29) always appears in the combination $\sigma + \lambda^0\tau$ [due to the new WS coordinates $\sigma^{\pm}$, cf. eq. (25)]. Since the string is $\ell$-periodic, we can shift the $\sigma$-integration in LC Hamiltonian (29) to get rid of the implicit $\lambda^0$-dependence. Thus the $\lambda^0$ in front of the LMC (22) is the only actual $\lambda^0$-dependence, as it should be. Integration over $\lambda^0$ enforces the LMC (22).

  17. Finally, let's return to OP's question. Classically, the orthogonally condition $$ \dot{X}\cdot X^{\prime}~\approx~0,\tag{30}$$ that OP asks about, is equivalent to picking the zero-mode $\lambda^0=0$ to be zero, cf. eqs. (2a) and (4a). This is what happens in the open string. In the closed string, we are supposed to integrate over $\lambda^0$. However, we can get away with working in a $\lambda^0=0$ "gauge" if we additionally impose the level-matching constraint (22) by hand. This latter approach is often taken in string theory textbooks.


  1. B. Zwiebach, A first course in String Theory, 2nd edition, 2009.

  2. J. Polchinski, String Theory, Vol. 1, 1998.

  3. R. Blumenhagen, D. Lust and S. Theisen, Basic Concepts of String Theory, 2012.

  4. K. Sundermeyer, Constrained Dynamics, Lecture Notes in Physics 169, 1982.

  5. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.


$^1$ Here is a proof that our starting point in this answer, the Hamiltonian Lagrangian (3), describes the NG string, at least classically. If we integrate out the $P^{\mu}$ momenta in the Hamiltonian Lagrangian (3), we get the Lagrangian density$^2$

$$ {\cal L}~=~T_0\frac{\left(\dot{X}-\lambda^0 X^{\prime}\right)^2}{2\lambda^1} -\frac{T_0\lambda^1}{2} (X^{\prime})^2.\tag{i}$$

Integrating out next the auxiliary variables $\lambda^0$ leads to $$ \left. {\cal L}\right|_{\lambda_0} ~=~-\frac{T_0{\cal L}_{(1)}}{2(X^{\prime})^2\lambda^1} -\frac{T_0\lambda^1}{2}(X^{\prime})^2,\tag{ii}$$ where $$ {\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta} ~=~(\dot{X}\cdot X^{\prime})^2-\dot{X}^2(X^{\prime})^2~\geq~ 0.\tag{iii} $$

Finally integrating out

$$\lambda^1~>~0 \tag{iv}$$

then yields the standard NG Lagrangian density

$${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}}. \tag{v}$$

[We have assumed that the auxiliary variable $\lambda^1>0$ is positive (iv) to get rid of a negative square root branch. It is imperative that the negative square root branch is not present. After Wick rotation, it would lead to an unstable exponentially growing (rather than an exponentially suppressed) Boltzmann factor. ] Conversely, if we Legendre transform the Lagrangian density (i), we get back the Hamiltonian density (1), cf. my Phys.SE here. Moreover, we should mention the well-known fact that if we integrate out the full WS metric $h_{\alpha\beta}$ in the Polyakov Lagrangian density

$$ {\cal L}_P~=~-\frac{T_0}{2} \sqrt{-h} h^{\alpha\beta} \partial_{\alpha}X \cdot\partial_{\beta}X ~=~\frac{T_0}{2} \left\{\frac{\left(h_{\sigma\sigma}\dot{X}- h_{\tau\sigma}X^{\prime}\right)^2}{\sqrt{-h}h_{\sigma\sigma}} - \frac{ \sqrt{-h}}{h_{\sigma\sigma}}(X^{\prime})^2 \right\}, \tag{vi} $$

then we get the standard NG Lagrangian density (v), cf. this Phys.SE post. [We choose the branch for the square root $\sqrt{-h}$ that has the same sign as $h_{\sigma\sigma}$ in order to make the kinetic term $\dot{X}^2$ positive definite.] Conversely, the Polyakov Lagrangian density (vi) can be derived from the Polyakov (P) De Donder-Weyl (DDW) Lagrangian density

$$ \begin{align}{\cal L}_{P,DDW}~=~&P^{\alpha} \cdot \partial_{\alpha}X +\frac{h_{\alpha\beta}P^{\alpha}\cdot P^{\beta}}{2T_0\sqrt{-h}} \cr ~=~& P^{\tau}\cdot \dot{X} +P^{\sigma}\cdot X^{\prime}+ \frac{(P^{\sigma} + \lambda^0 P^{\tau})^2}{2T_0 \lambda^1} - \frac{\lambda^1}{2T_0} (P^{\tau})^2\end{align} \tag{vii} $$

by integrating out the polymomenta $P^{\alpha}=(P^{\tau};P^{\sigma})$. See also e.g. my Phys.SE answer here. In the second equality of eq. (vii), we have identified

$$ \lambda^0~=~\frac{h_{\tau\sigma}}{h_{\sigma\sigma}} ~=~-\frac{h^{\tau\sigma}}{h^{\tau\tau}}, $$ $$\lambda^1~=~\frac{\sqrt{-h}}{h_{\sigma\sigma}} ~=~ \frac{-1}{\sqrt{-h}h^{\tau\tau}}~\geq~0 \quad\Leftrightarrow\quad h~:=~\det\left(h_{\alpha\beta}\right)_{\alpha\beta}~=~-\left(\lambda^1 h_{\sigma\sigma}\right)^2~\leq~0 . \tag{viii}$$

Similarly, in the Lagrangian picture, the Polyakov Lagrangian density (vi) is equal to the Lagrangian density (i) under the identification (viii). The point is that only 2 out of the 3 degrees of freedom in the WS metric $h_{\alpha\beta}$ enter the Polyakov action due to Weyl symmetry at the classical level. Therefore the WS metric $h_{\alpha\beta}$ can be replaced with only 2 variables $\lambda^0$ and $\lambda^1$. The correspondence (vi) $\leftrightarrow$ (i) establishes a more refined equivalence between the Polyakov and the Nambu-Goto Lagrangian formulations than just integrating out the full WS metric $h_{\alpha\beta}$ by brute force.

Finally, if we only integrate out

$$P^{\sigma}~\approx~-T_0\lambda^1X^{\prime}-\lambda^0P^{\tau}\tag{ix} $$

in the Polyakov De Donder-Weyl Lagrangian density (vii) but keep the $P^{\tau}\equiv P$ variable, then the Polyakov De Donder-Weyl Lagrangian density (vii) becomes the Hamiltonian Lagrangian density (3), i.e. our starting point in this answer. This shows that the Nambu-Goto and the Polyakov Hamiltonian formulations are equivalent, cf. this Phys.SE post.

$^2$ The Gaussian integration over the auxiliary variable $\lambda^0\equiv \lambda^0_M$ 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 $-i\lambda^0_M=\lambda^0_E\in\mathbb{R}$. In other words, the product $\lambda^0_M\tau_M=\lambda^0_E\tau_E$ should remain invariant under Wick rotation.


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