Classical Rotating Strings and the Regge Slope

In conformal gauge the classical bosonic string is deceptively simple. The embedding fields obey a free two-dimensional wave equation,

$$(\partial_\tau^2-\partial_\sigma^2)X^\mu=0,$$

but they must also satisfy the Virasoro constraints

$$\dot X^2+X^{\prime 2}=0, \qquad \dot X\cdot X'=0.$$

For an open string with Neumann boundary conditions in all target-space directions, we also impose

$$X^{\prime\mu}(\tau,0)=X^{\prime\mu}(\tau,\pi)=0.$$

This page studies one of the most important classical solutions: a rigidly rotating open string. It is the cleanest classical derivation of the stringy Regge relation

$$J=\alpha' M^2.$$

This relation is the historical reason that strings were first taken seriously as models of hadrons, and it remains one of the best ways to remember what the parameter $\alpha'$ means physically.

Conventions

The open-string coordinate range is

$$0\leq \sigma\leq \pi.$$

The target-space metric is mostly plus,

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+),$$

and the string tension is

$$T=\frac{1}{2\pi\alpha'}.$$

We keep $\alpha'$ explicit throughout.

The rotating open-string ansatz

Consider a string rotating in the $X^1$-$X^2$ plane, with all other spatial coordinates set to zero. A convenient conformal-gauge parametrization is

$$X^0=A\tau,$$ $$X^1=A\cos\sigma\cos\tau, \qquad X^2=A\cos\sigma\sin\tau,$$

with

$$X^3=\cdots=X^{D-1}=0.$$

The constant $A$ has dimensions of length. It sets the size of the string. At a fixed target-space time

$$t=X^0=A\tau,$$

the spatial embedding is

$$\vec X(\tau,\sigma)=A\cos\sigma\, (\cos\tau,\sin\tau).$$

As $\sigma$ runs from $0$ to $\pi$, $\cos\sigma$ runs from $1$ to $-1$. Thus the string is a straight line segment of length $2A$ passing through the origin. As $\tau$ increases, this line segment rotates rigidly.

A classical open string rotating in the $X^1$-$X^2$ plane. The midpoint at $\sigma=\pi/2$ is at the origin, while the endpoints move at the speed of light.

The target-space angular velocity is not $1$; it is

$$\Omega=\frac{d\tau}{dX^0}=\frac{1}{A}.$$

The physical velocity of a point labelled by $\sigma$ is

$$\frac{d\vec X}{dX^0} =\frac{\partial_\tau\vec X}{\partial_\tau X^0} =\cos\sigma\,(-\sin\tau,\cos\tau),$$

so

$$\left|\frac{d\vec X}{dX^0}\right|=|\cos\sigma|.$$

The endpoints at $\sigma=0,\pi$ move at speed $1$, while the midpoint at $\sigma=\pi/2$ is at rest. This is a very stringy feature: the endpoints of the classical rotating open string are massless and move at the speed of light.

The speed is maximal at the two endpoints and vanishes at the midpoint.

Why the cosine?

For Neumann boundary conditions we need $X^{\prime\mu}=0$ at $\sigma=0,\pi$. The function $\cos\sigma$ has precisely this property:

$$\partial_\sigma\cos\sigma=-\sin\sigma, \qquad \sin 0=\sin\pi=0.$$

A sine profile would instead vanish at the endpoints and would be natural for Dirichlet boundary conditions.

Equations of motion and boundary conditions

The wave equation is immediate. For example,

$$\partial_\tau^2 X^1=-A\cos\sigma\cos\tau, \qquad \partial_\sigma^2 X^1=-A\cos\sigma\cos\tau,$$

so

$$(\partial_\tau^2-\partial_\sigma^2)X^1=0.$$

The same calculation applies to $X^2$, while $X^0=A\tau$ is trivially a solution.

The Neumann boundary condition is also automatic:

$$X^{\prime 0}=0,$$

and

$$X^{\prime 1}=-A\sin\sigma\cos\tau, \qquad X^{\prime 2}=-A\sin\sigma\sin\tau.$$

Therefore

$$X^{\prime\mu}(\tau,0)=X^{\prime\mu}(\tau,\pi)=0.$$

Virasoro constraints

The nontrivial check is the constraint equation. We have

$$\dot X^0=A, \qquad \dot X^1=-A\cos\sigma\sin\tau, \qquad \dot X^2=A\cos\sigma\cos\tau,$$

so the spatial velocity squared is

$$|\dot{\vec X}|^2=A^2\cos^2\sigma.$$

The spatial tangent is

$$X^{\prime 1}=-A\sin\sigma\cos\tau, \qquad X^{\prime 2}=-A\sin\sigma\sin\tau,$$

so

$$|\vec X'|^2=A^2\sin^2\sigma.$$

With the mostly-plus metric,

$$\dot X^2=-A^2+A^2\cos^2\sigma=-A^2\sin^2\sigma,$$

and

$$X^{\prime 2}=A^2\sin^2\sigma.$$

Hence

$$\dot X^2+X^{\prime 2}=0.$$

Moreover,

$$\dot X\cdot X'=0,$$

because the velocity is perpendicular to the string tangent in the rotation plane. Thus the solution satisfies

$$T_{++}=T_{--}=0.$$

The Virasoro constraints hold pointwise on the worldsheet. The negative contribution from $X^0$ is essential.

This cancellation is the classical ancestor of a fact that will recur in quantization: the time coordinate is not an ordinary positive-norm oscillator. The constraints remove unphysical longitudinal and timelike degrees of freedom.

Spacetime Poincare currents

The conformal-gauge action for an open string is

$$S=\frac{T}{2}\int d\tau\int_0^\pi d\sigma\, \left(\dot X^2-X^{\prime 2}\right).$$

Flat target space has spacetime translation symmetry

$$\delta X^\mu=a^\mu,$$

and Lorentz symmetry

$$\delta X^\mu=\omega^\mu{}_{\nu}X^\nu, \qquad \omega_{\mu\nu}=-\omega_{\nu\mu}.$$

The corresponding translation current is

$$\mathcal P^{\tau\mu}=T\dot X^\mu, \qquad \mathcal P^{\sigma\mu}=-T X^{\prime\mu}.$$

The equation of motion is the conservation law

$$\partial_\tau\mathcal P^{\tau\mu} +\partial_\sigma\mathcal P^{\sigma\mu}=0.$$

The spacetime momentum is the charge on a constant-$\tau$ slice:

$$P^\mu=\int_0^\pi d\sigma\,\mathcal P^{\tau\mu} =T\int_0^\pi d\sigma\,\dot X^\mu.$$

For open strings with Neumann boundary conditions,

$$\mathcal P^{\sigma\mu}\big|_{\sigma=0,\pi}=-T X^{\prime\mu}\big|_{\sigma=0,\pi}=0,$$

so no spacetime momentum flows out of the endpoints. This is why the total momentum is conserved.

The Lorentz current is

$$\mathcal J^{\alpha\mu\nu} =X^\mu\mathcal P^{\alpha\nu} -X^\nu\mathcal P^{\alpha\mu}.$$

In particular,

$$\mathcal J^{\tau\mu\nu} =T\left(X^\mu\dot X^\nu-X^\nu\dot X^\mu\right),$$

and the spacetime angular momentum is

$$J^{\mu\nu} =\int_0^\pi d\sigma\,\mathcal J^{\tau\mu\nu} =T\int_0^\pi d\sigma\, \left(X^\mu\dot X^\nu-X^\nu\dot X^\mu\right).$$

Spacetime Poincare invariance gives worldsheet currents. Constant-$\tau$ integrals of their time components give the spacetime momentum $P^\mu$ and angular momentum $J^{\mu\nu}$.

Dirichlet directions

The discussion above assumes Neumann boundary conditions in the directions whose momentum is being conserved. Dirichlet boundary conditions fix the endpoint position and therefore break the corresponding target-space translations for a fixed D-brane background. Later, when D-branes become dynamical, the bookkeeping of conserved total momentum includes the brane as well.

Energy and angular momentum of the rotating string

For the rotating solution,

$$\dot X^0=A.$$

Therefore the energy is

$$E=P^0 =T\int_0^\pi d\sigma\,\dot X^0 =T\int_0^\pi d\sigma\,A =\pi T A.$$

The angular momentum in the rotation plane is

$$J=J^{12} =T\int_0^\pi d\sigma\, \left(X^1\dot X^2-X^2\dot X^1\right).$$

Now

$$X^1\dot X^2-X^2\dot X^1 =A^2\cos^2\sigma,$$

so

$$J=T A^2\int_0^\pi d\sigma\,\cos^2\sigma.$$

Using

$$\int_0^\pi d\sigma\,\cos^2\sigma=\frac{\pi}{2},$$

we get

$$J=\frac{\pi T A^2}{2}.$$

The energy density is constant in the coordinate $\sigma$, while the angular-momentum density is proportional to $\cos^2\sigma$.

Eliminate $A$ between

$$E=\pi T A$$

and

$$J=\frac{\pi T A^2}{2}.$$

From $A=E/(\pi T)$,

$$J=\frac{\pi T}{2}\frac{E^2}{\pi^2T^2} =\frac{E^2}{2\pi T}.$$

Since

$$T=\frac{1}{2\pi\alpha'},$$

we find

$$\frac{1}{2\pi T}=\alpha'.$$

In the rest frame of the string,

$$M^2=E^2,$$

and therefore

$$J=\alpha'M^2.$$

This is the classical open-string Regge trajectory.

The rotating open string gives a linear relation between spin and mass squared. Quantum effects shift the intercept but not the leading large-$M^2$ slope.

Interpretation of the Regge slope

The relation

$$J=\alpha'M^2$$

says that $\alpha'$ is the slope of the leading open-string Regge trajectory. Equivalently,

$$\alpha'=\frac{1}{2\pi T}$$

is the inverse string tension up to the conventional factor $2\pi$.

This is physically intuitive. If the string tension is large, it costs a lot of energy to make a long string, so the slope in the $J$-$M^2$ plane is small. If the tension is small, long strings are cheap, and large angular momentum can be obtained at smaller mass.

Classically $A$ is continuous, so the curve is continuous. Quantum mechanically the angular momentum is quantized, and the leading trajectory becomes a sequence of states. The intercept is also shifted by the quantum zero-point energy. Schematically one expects

$$J=\alpha'M^2+a,$$

where $a$ is a quantum intercept. In the bosonic open string, this same intercept is responsible for the tachyon in the spectrum. We will derive this from oscillator quantization rather than assume it.

Open versus closed strings

The result above is for an open string. Closed strings have left- and right-moving oscillators and, on the leading trajectory, a different effective slope. In the quantum theory the closed-string leading Regge trajectory has slope $\alpha'/2$ rather than $\alpha'$. This distinction will become clear once we derive the open- and closed-string mass formulas.

Summary

The rigidly rotating open string

$$X^0=A\tau, \qquad X^1=A\cos\sigma\cos\tau, \qquad X^2=A\cos\sigma\sin\tau$$

satisfies the conformal-gauge equations of motion, the Neumann boundary condition, and the Virasoro constraints. Its conserved charges are

$$E=\pi T A, \qquad J=\frac{\pi T A^2}{2}.$$

Eliminating $A$ gives

$$J=\frac{E^2}{2\pi T}=\alpha'M^2.$$

Thus $\alpha'$ is the Regge slope of the classical open string. The next lecture note begins the quantization of the closed string, where the same conformal-gauge equations become oscillator mode expansions and Virasoro constraints on the string Hilbert space.

Exercises

Exercise 1. Check the rotating solution

Verify explicitly that

$$X^0=A\tau, \qquad X^1=A\cos\sigma\cos\tau, \qquad X^2=A\cos\sigma\sin\tau$$

satisfies

$$(\partial_\tau^2-\partial_\sigma^2)X^\mu=0$$

and the Neumann boundary condition at $\sigma=0,\pi$.

Solution

For $X^0=A\tau$,

$$\partial_\tau^2X^0=0, \qquad \partial_\sigma^2X^0=0.$$

For $X^1$,

$$\partial_\tau^2X^1=-A\cos\sigma\cos\tau, \qquad \partial_\sigma^2X^1=-A\cos\sigma\cos\tau.$$

Thus

$$(\partial_\tau^2-\partial_\sigma^2)X^1=0.$$

The same calculation gives

$$(\partial_\tau^2-\partial_\sigma^2)X^2=0.$$

The boundary derivatives are

$$X^{\prime 0}=0,$$ $$X^{\prime 1}=-A\sin\sigma\cos\tau, \qquad X^{\prime 2}=-A\sin\sigma\sin\tau.$$

At $\sigma=0$ and $\sigma=\pi$, $\sin\sigma=0$, so

$$X^{\prime\mu}=0.$$

Exercise 2. Endpoint speed

Show that the physical velocity of a point on the string is

$$\left|\frac{d\vec X}{dX^0}\right|=|\cos\sigma|.$$

What are the speeds of the endpoints and of the midpoint?

Solution

The physical velocity is the derivative with respect to target-space time $X^0$, not worldsheet time $\tau$:

$$\frac{d\vec X}{dX^0} =\frac{\partial_\tau\vec X}{\partial_\tau X^0}.$$

Since $\partial_\tau X^0=A$ and

$$\partial_\tau\vec X =A\cos\sigma(-\sin\tau,\cos\tau),$$

we get

$$\frac{d\vec X}{dX^0} =\cos\sigma(-\sin\tau,\cos\tau).$$

Therefore

$$\left|\frac{d\vec X}{dX^0}\right|=|\cos\sigma|.$$

At $\sigma=0$ and $\sigma=\pi$, the speed is $1$. At $\sigma=\pi/2$, the speed is $0$.

Exercise 3. The Virasoro constraints

For the rotating solution, prove that

$$\dot X^2+X^{\prime 2}=0, \qquad \dot X\cdot X'=0.$$

Solution

We have

$$\dot X^0=A, \qquad |\dot{\vec X}|^2=A^2\cos^2\sigma,$$

so

$$\dot X^2=-A^2+A^2\cos^2\sigma=-A^2\sin^2\sigma.$$

Also

$$X^{\prime 0}=0, \qquad |\vec X'|^2=A^2\sin^2\sigma,$$

so

$$X^{\prime 2}=A^2\sin^2\sigma.$$

Thus

$$\dot X^2+X^{\prime 2}=0.$$

Finally,

$$\dot{\vec X}=A\cos\sigma(-\sin\tau,\cos\tau),$$

and

$$\vec X'=-A\sin\sigma(\cos\tau,\sin\tau).$$

Their Euclidean dot product is zero, and $X^{\prime 0}=0$, so

$$\dot X\cdot X'=0.$$

Exercise 4. Derive the momentum current

Starting from

$$S=\frac{T}{2}\int d\tau d\sigma\, (\dot X^2-X^{\prime 2}),$$

show that the spacetime translation current can be written as

$$\mathcal P^{\tau\mu}=T\dot X^\mu, \qquad \mathcal P^{\sigma\mu}=-T X^{\prime\mu}.$$

Then show that current conservation is the string equation of motion.

Solution

The Lagrangian density is

$$\mathcal L=\frac{T}{2}(\dot X^2-X^{\prime 2}).$$

The canonical momentum density conjugate to $X_\mu$ is

$$\frac{\partial\mathcal L}{\partial\dot X_\mu}=T\dot X^\mu.$$

The spatial flux is

$$\frac{\partial\mathcal L}{\partial X'_\mu}=-T X^{\prime\mu}.$$

Therefore the translation current is

$$\mathcal P^{\tau\mu}=T\dot X^\mu, \qquad \mathcal P^{\sigma\mu}=-T X^{\prime\mu}.$$

Current conservation gives

$$0=\partial_\tau\mathcal P^{\tau\mu} +\partial_\sigma\mathcal P^{\sigma\mu} =T\partial_\tau^2X^\mu-T\partial_\sigma^2X^\mu.$$

Thus

$$(\partial_\tau^2-\partial_\sigma^2)X^\mu=0.$$

Exercise 5. Classical Regge relation

For the rotating open string, compute

$$E=T\int_0^\pi d\sigma\,\dot X^0$$

and

$$J^{12}=T\int_0^\pi d\sigma\, (X^1\dot X^2-X^2\dot X^1).$$

Eliminate $A$ and show that $J=\alpha'M^2$.

Solution

Since $\dot X^0=A$,

$$E=T\int_0^\pi d\sigma\,A=\pi T A.$$

Next,

$$X^1=A\cos\sigma\cos\tau, \qquad X^2=A\cos\sigma\sin\tau,$$

and

$$\dot X^1=-A\cos\sigma\sin\tau, \qquad \dot X^2=A\cos\sigma\cos\tau.$$

Therefore

$$X^1\dot X^2-X^2\dot X^1 =A^2\cos^2\sigma(\cos^2\tau+\sin^2\tau) =A^2\cos^2\sigma.$$

Thus

$$J=T A^2\int_0^\pi d\sigma\,\cos^2\sigma =\frac{\pi T A^2}{2}.$$

Using $A=E/(\pi T)$,

$$J=\frac{E^2}{2\pi T}.$$

Since $T=1/(2\pi\alpha')$,

$$J=\alpha' E^2.$$

In the rest frame $M^2=E^2$, so

$$J=\alpha'M^2.$$