Lightcone Coordinates and the Stress Energy Tensor

String Theory without the Baggage, Episode 24.

No snow on the valley floor yet, but at least that means sunshine!

Last time we studied the equations of motion - and the associated boundary conditions - for 𝑋^𝜇. With boundary conditions in mind, let's revisit the equations of motion for 𝛾_𝑎𝑏.

Hopefully you remember that 𝛾_𝑎𝑏 has no kinetic terms in the Polyakov action. Hence its variation,

\(\frac{\delta S}{\delta \gamma_{mn}} = \frac{\delta }{\delta \gamma_{mn}}\frac{\kappa}{2} \int d^{2}\sigma \, \sqrt{-\gamma}\gamma^{ab}\partial_{a}X^{\mu}\partial_{b}X_{\mu},\)

amounts to a constraint equation

\(\frac{\delta S}{\delta \gamma_{mn}} =\frac{\kappa}{2} \int d^{2}\sigma \, \sqrt{-\gamma} \;T^{mn}= 0.\)

Here 𝑇^a𝑏 represents the stress-energy tensor on the worldsheet.

You might also remember that a variation in the worldsheet coordinates 𝛿𝜎 induces a variation in 𝛾_𝑎𝑏:

\(\delta\gamma_{ab} = \nabla_{a}\delta\sigma_{b} + \nabla_{b}\delta\sigma_{a}.\)

Let us revisit the variation of action by considering a coordinate variation specifically.

\(\delta S =\frac{\kappa}{2} \int d^{2}\sigma \, \sqrt{-\gamma} \;T^{ab}\left( \nabla_{a}\delta\sigma_{b} + \nabla_{b}\delta\sigma_{a} \right).\)

With the result you proved, the variation becomes

\(\delta S =\frac{\kappa}{2} \int d^{2}\sigma \, T^{ab}\left( \partial_{a}(\sqrt{-\gamma}\;\delta\sigma_{b}) + \partial_{b}(\sqrt{-\gamma}\;\delta\sigma_{a}) \right).\)

This is a nontrivial observation that you should prove as an exercise. We've already seen in it done implicitly a prior episode:

Gory Details: a Review of Coordinate Isometries

We can then integrate by parts and simplify,

\(\delta S = \kappa \int d^{2}\sigma \, \left( \partial_{a}T^{ab} \;\sqrt{-\gamma}\;\delta\sigma_{b}\right) = 0.\)

The surface term of course vanishes by the constraint equation, so we are left with an additional fact about the stress-energy tensor, namely

\( \partial_{a}T^{ab} = 0.\)

Let us see where this leads.

Using Lightcone Coordinates

Last time we solved the wave equation using lightcone coordinates:

\(z = \frac{1}{2}(\tau + \sigma),\quad \bar{z} = \frac{1}{2}(\tau - \sigma).\)

Let's see what the metric is under this change of coordinates.

Inverting the coordinate transformation, we find

\(\tau = z + \bar{z},\quad \sigma = z - \bar{z}.\)

Therefore,

\(d\tau = dz + d\bar{z},\quad d\sigma = dz - d\bar{z}.\)

Now, the line element on the worldsheet is:

\(ds^{2} = -d\tau^{2} + d\sigma^{2}.\)

Computing,

\(ds^{2} = -( dz + d\bar{z})^{2} + ( dz - d\bar{z})^{2} = -4 dzd\bar{z}.\)

Hence we have¹

\(\gamma_{z\bar{z}} = \gamma_{\bar{z}z} = -2,\)

and

\(\gamma_{zz} = \gamma_{\bar{z}\bar{z}} = 0.\)

Once can check by explicit computation that the inverse metric is

\(\gamma^{-1}\rightsquigarrow -\frac{1}{2}\left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right).\)

The stress-energy tensor reads

\(T_{ab} \propto \partial_{a}X^{\mu}\partial_{b}X_{\mu} - \frac{1}{2}\gamma_{ab}\gamma^{mn}\partial_{m}X^{\mu}\partial_{n}X_{\mu}.\)

By definition of a tensor it is covariant under coordinate transformations. Let's look at it in lightcone coordinates.

\(T_{zz} = \partial X^{\mu}\partial X_{\mu},\quad T_{\bar{z}\bar{z}} = \bar{\partial} X^{\mu}\bar{\partial} X_{\mu}.\)

The diagonal term is interesting. Carefully minding all the minus signs and factors of two we find

\(T_{z\bar{z}} = \partial X^{\mu}\bar{\partial} X_{\mu} - \frac{1}{2}(-2)(-\frac{2}{2})\partial X^{\mu}\bar{\partial} X_{\mu} = 0.\)

Let's now check the conservation condition for T,

\(\partial^{a}T_{ab} = \gamma^{ac}\partial_{c}T_{ab} = \gamma^{az}\partial T_{ab} + \gamma^{a\bar{z}}\bar{\partial} T_{ab} = 0.\)

Given that the metric is a constant here we see that

\(\partial^{a}T_{ab} = \gamma^{ac}\partial_{c}T_{ab} = \gamma^{\bar{z}z}\partial T_{\bar{z}b} + \gamma^{z\bar{z}}\bar{\partial} T_{zb}.\)

This gives us two equations. Let's first check the one for b = z:

\(\partial^{a}T_{az} = \gamma^{ac}\partial_{c}T_{ab} = \gamma^{\bar{z}z}\partial T_{\bar{z}z} + \gamma^{z\bar{z}}\bar{\partial} T_{zz}.\)

As we've already seen, the off-diagonal vanishes, hence the first term is zero. So we find

\(\bar{\partial} T_{zz} = 0.\)

Similarly,

\(\partial T_{\bar{z}\bar{z}} = 0.\)

A quick look at the Dirichlet variables from last time for 𝑋^𝜇, together with how T_ab is constructed shows that

\(T_{zz} = T_{zz}(z),\quad T_{\bar{z}\bar{z}} = T_{\bar{z}\bar{z}}(\bar{z}).\)

That is, T_zz depends only on the holomorphic or left-moving coordinate z. The antiholomorphic case follows in parallel. As with our derivatives, we will adopt the simplified notation,

\(T_{zz} = T(z),\quad T_{\bar{z}\bar{z}} = \bar{T}(\bar{z}).\)

In summary, the three constraint equations we had on 𝑋^𝜇 in the 𝜎 and 𝜏 variables get repackaged into two constraint equations the separately act on X_L and X_R.

This fact will have profound implications for the study of physics on the worldsheet, as we’ll see next time.

Before we close let us reframe the constraint equation T^ab = 0 in terms of our new left and right moving variables.

Given that

\(X^{\mu}(z,\bar{z}) = X_{L}^{\mu}(z) + X_{R}^{\mu}(\bar{z}),\)

we see that

\(T_{zz} = \partial X^{\mu}\partial X_{L\,\mu} = \partial X_{L}^{\mu}\partial X_{L\,\mu} = 0,\quad T_{\bar{z}\bar{z}} = \bar{\partial} X^{\mu} \bar{\partial}X_{R\,\mu} = \bar{\partial}X_{R}^{\mu} \bar{\partial}X_{R\,\mu} = 0.\)

or, in short, the constraint equations are just

\(|\dot{X}_{L}|^{2} = 0,\quad |\dot{X}_{R}|^{2} = 0.\)

1. Remember that there are two terms here.