2

Given light-cone coordinates, and some worldline parameterised by some arbitrary parameter $\tau$.

$$x^{+} = \frac{1}{\sqrt{2}}(x^0+x^1),$$ $$x^{-} = \frac{1}{\sqrt{2}}(x^0-x^1).\tag{2.50}$$

And the lagrangian:

$$L=-m\sqrt{-\dot{x}^2},\tag{11.3}$$

$$p_\mu=\frac{\partial L}{\partial\dot{x}^\mu}= \frac{m\dot{x}_\mu}{\sqrt{-\dot{x}^2}}.\tag{11.4}$$

In the light-cone gauge; $$x^+ = \frac{1}{m^2}p^+\tau.\tag{11.7+29}$$

Which implies; $$\frac{\partial}{\partial \tau} = \frac{p^+}{m^2}\frac{\partial}{\partial x^+}\rightarrow \frac{p^+p^-}{m^2}.\tag{11.33}$$

My textbook then hypothesised the hamiltonian of the system to be:

$$H= \frac{p^+p^-}{m^2}=\frac{p^Ip^I+m^2}{2m^2},\tag{11.34}$$

where the index $I$ denotes transverse coordinates.

I am wondering where eq. (11.34) comes from as I've been trying to show this from the definition of the hamiltonian.

$$H= \dot{x}^\mu p_\mu - L,$$ $$H= \frac{\sqrt{-\dot{x}^2}}{m}p^\mu p_\mu + m\sqrt{-\dot{x}^2}.$$

Plugging in from the system that $p^\mu p_\mu = -m^2$ we obtain zero which is different to what my text says.

(Using the differential relation ship I think its possible to show only the $p^+$ contribution to the momentum is non-zero and we can reduce the hamiltonian down to a similar form they have, however since $p^-$ is zero in this argument it is still zero.)

The textbook is: A First course in string theory, Barton Zwiebach, second edition, page 222. The lagrangian formalism is on page 216-217.

Qmechanic
  • 220,844
jensen paull
  • 7,016

1 Answers1

1

  1. Perhaps the cleanest$^1$ derivation is to consider the Hamiltonian Lagrangian$^2$ $$L_H~:=~ p_{\mu} \dot{x}^{\mu} - \underbrace{\frac{e}{2}(p^2+m^2)}_{\text{Hamiltonian}},\tag{1}$$ where $e>0$ is an einbein field, where dot denotes differentiation wrt. the world-line (WL) parameter $\tau$, and where $$ p^2~:=~ \eta^{\mu\nu} p_{\mu} p_{\nu} ~=~-2p^+p^- + {\bf p}_{\perp}^2, \tag{2}$$ cf. e.g. my Phys.SE answer here. The einbein $e$ is a Lagrange multiplier that imposes the mass-shell constraint $$ p^2+m^2~\approx~0. \tag{11.6}$$ The Hamiltonian (1) vanishes on-shell due to the WL reparametrization invariance, cf. e.g. this & this Phys.SE posts.

  2. Next go to the light-cone gauge $$x^+~=~\frac{p^+}{m^2}\tau. \tag{11.7+29}$$ The gauge fixing (11.7+29) breaks WL reparametrization invariance, so that the emerging notion of Hamiltonian/energy is no longer necessarily zero on-shell. This fact is the main answer to OP's underlying question.

    In light-cone quantization it is usually assumed that $p^+>0$ is strictly positive. We also assume that the point particle is massive $m^2>0$.

  3. If we integrate out $p^−$ and $e$, we get $$\begin{align} \left. L_H\right|_{x^+=\frac{p^+}{m^2}\tau}\quad=\quad & -p^+\cdot \dot{x}^- +{\bf p}_{\perp}\cdot \dot{\bf x}_{\perp}\cr & +p^+p^-(e-\frac{1}{m^2})\cr & -\frac{e}{2}({\bf p}_{\perp}^2+m^2)\cr \quad\stackrel{p^-,~e}{\longrightarrow}\quad & -p^+\cdot \dot{x}^- +{\bf p}_{\perp}\cdot \dot{\bf x}_{\perp}\cr & - \underbrace{\frac{{\bf p}_{\perp}^2+m^2}{2m^2}}_{\text{Hamiltonian}} .\end{align}\tag{3}$$ The Hamiltonian (3) is the right-hand expression of eq. (11.34), cf. OP's question.

    NB: Note that the middle expression of eq. (11.34) only holds on-shell.

References:

  1. B. Zwiebach, A first course in String Theory, 2nd edition, 2009; sections 11.1 + 11.3.

$^1$ It is a clean derivation in the sense that eqs. (1) and (3) are fully self-contained & consistent Hamiltonian off-shell formulations of $2D+1$ and $2D-2$ variables, respectively. In contrast, the derivation in Ref. 1 is less systematic, relies on EOMs, and so on.

$^2$ In general, the Hamiltonian Lagrangian is a Lagrangian defined in the phase space of a Hamiltonian theory, such that its Euler-Lagrange (EL) eqs. are Hamilton's eqs.

In contrast, the ordinary Lagrangian is defined in the configuration space.

In the presence of constraints, the above definitions have to be appropriately modified.

We assume a $D$-dimensional Minkowski spacetime with metric $\eta_{\mu\nu}$ of sign convention $(−,+,\ldots,+)$. Also we put $c=1$ for simplicity.

Qmechanic
  • 220,844