Notes on string theory: Energy-momentum tensor

In general in Quantum Field Theory (QFT), provided we are working in flat space, we have the following formula:

    \[ T^{ab} = \frac{2}{\sqrt{- \gamma}} \frac{\delta S}{\delta \gamma_{ab}} \]

In String Theory (ST), we vary S_{P} slightly with respect to the background metric and we arrive at the following,

    \[ T^{ab} = \frac{-4 \pi}{\sqrt{- \gamma}} \frac{\delta S_P}{\delta \gamma_{a b}} \]

In sum, when it comes to the constraints on T^{ab} for physical fluctuations, I think two important comments are necessary:

1. Diff invariance on the worldsheet implies \nabla_{a} T^{ab} = 0.

2. Weyl invariance implies T_{a}^{b} = \gamma^{ab}T_{ab} = 0.

One will find that the tracelessness of the EM tensor is a direct consequence of the symmetries imposed, such that: T_{a}^{a} = 0.

Due to diff invariance, it should also be highlighted that the EM tensor is covariantly conserved,

    \[ \nabla_{a} T^{ab} \rvert_{x^\mu \ satisfies \ EoM} = 0 \]

Although, in a past entry, we have already found the equations of motion by varying S_P with respect to our dynamical fields \gamma and X^{\mu}, I think it is worthwhile, for pedagogical purposes, to pursue the same steps but this time focusing explicitly on computing the EM tensor with slightly different notation. In fact, I think what follows is quite a useful exercise and the end result is also nice to look at.

Invoking the Polyakov action we have, written in a slightly different way than in previous discussions,

    \[ S_{P} = - \frac{1}{4 \pi \alpha^{\prime}} \int d\tau d\sigma \sqrt{- \gamma} \gamma^{ab} \partial_{a} X^{\mu} \partial_{b} X_{\mu} \]

Where, in what follows, we will set \gamma^{ab} \partial_{a} X^{\mu} \partial_{b} X_{\mu} = \partial X^{2}.

Notice, also, that what we’re actually working with is a 1+1 QFT. We also have our metric, \gamma, so you can think of this action describing a (1+1)-dim theory of scalar fields coupled to gravity.

We now compute the EM tensor in this theory, recalling \delta \gamma^{ab} = - \delta \gamma_{cd} \gamma^{ca} \gamma^{db} and \delta \gamma = \gamma \cdot \gamma^{ab} \delta \gamma_{ab}.

    \[ T^{ab} = \frac{4 \pi}{\sqrt{- \gamma}} \frac{1}{-4 \pi \alpha^{\prime}} \frac{\delta}{\delta \gamma_{ab}} \int [\frac{1}{2} (-\gamma)^{\frac{1}{2}} -\partial \gamma^{ab} \delta \gamma_{ab} (\partial X^{2}) - \sqrt{\gamma} \partial^{a} X^{\mu} \partial^{b} X_{\mu} \delta \gamma_{ab}] \]

    \[ \implies -\frac{1}{\alpha^{\prime}}[\partial^{a}X^{\mu}\partial^{b}X_{\mu} - \frac{1}{2} \gamma^{ab}\partial_{c}X^{\mu}\partial^{c}X_{\mu}] \]

Which gives the EM tensor. From this configuration, the tracelessness can easily be found:

    \[ T_{a}^{a} = -\frac{1}{\alpha^{\prime}} [ (\partial X)^{2} - \frac{1}{2} \gamma^{ab} \gamma_{ab} (\partial X)^2] = 0 \]

This reflects the fact that when we do a Weyl transformation, it does not effect the coordinates and matter fields of our theory.

What is nice about varying our action in this theory in this way is that we discover the explicit result,

    \[ 0 = \frac{\delta S}{\delta \gamma_{ab}} = T^{ab}= 0 \]

What is this saying? We see, and can communicate directly, that the EoM is, or represents, the vanishing of the EM tensor.

Additionally, I want to emphasis one more result. Again, using slightly different notation, let’s go back and think about varying S_P with respect to X^{\mu}.

    \[ \delta S_P = - \frac{2}{4 \pi \alpha^{\prime}} \int d\tau d\sigma \sqrt{- \gamma} \gamma^{ab} \nabla_{a} X^{\mu} \cdot \nabla_{b} \delta X_{\mu} \]

Now, we integrate by parts such that the covariant derivative \nabla_{b}\delta X_{\mu} acts on the rest of the action,

    \[ = \frac{1}{2 \pi \alpha^{\prime}} \int d\tau d\sigma \nabla_{b}[\sqrt{-\gamma}\gamma^{ab} \nabla_{a} X^{\mu}] \delta X_{\mu} + boundary \ terms \]

    \[ \implies \nabla_{b}[\sqrt{-\gamma}\gamma^{ab}\nabla_{a}X^{\mu}] = 0 \]

Notice, approached and written this way, we see an explicit expression for the d’Alembertian,

    \[ \sqrt{-\gamma} \Box X^{\mu} = 0 \]

Which makes sense, as our fields satisfy the massless Klein-Gordon equation in 2-dimensions on a curved background. Perhaps this is obvious, and perhaps the result foreseeable, but I think it is still nice to look at.