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As has been discussed, the Nambu-Goto action can be interpreted as providing the area of the worldsheet mapped by the string in spacetime. For what it is, I think this action is quite nice. There is a lot we can do with it, and there are a lot of ways in which one can play around with it and produce some fairly neat results. But there is also a problem. Similar to the action for the point particle, with the NG action we ended up with a square root unfriendly to quantisation. In other words, it is generally not adequate for the study of the quantum physics of strings (though it is possible to quantise the NG action). It may also be added that, for the NG action, it takes quite a bit of work to arrive at the wave equations, momenta identities, and other useful things. So we want an action that does not contain a square root – that is polynomial – and that is generally easier to work with.

This means that we need to come up with another action – the Polyakov action.

Currently, I do not know of a forward derivation that works directly from the Nambu-Goto action to the Polyakov action. It would seem a highly non-trivial feat. But there are a number of things one can do. As is common in majority of textbooks and leading literature, the standard procedure is to leverage the point particle analogue as a guiding principle, such that one may think of whether the NG action can be modified with some sort of auxiliary field. The answer is that it can indeed be modified in such a way. The result is the Polyakov action. Most texts describe this and show the result – it is taken more or less as a given in Polchinski, for example. Emphasis is then placed on showing how one can work backward from the Polyakov action to the NG action, showing they are classically equivalent with the former still providing the correct quantum theory.

In the next post, we will take this generally textbook approach to the Polyakov action. We will also begin to study the equations of motion and other interesting things.

But since I don’t find that approach entirely satisfying, in this entry I want to introduce the Polyakov action slightly differently. I want to take a more extended approach through General Relativity (this is one of a number of ways that one can arrive at the Polyakov action).

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We start by invoking GR in m-dimensions, with the inclusion of scalar fields and the cosmological constant. As this is just a simple scalar field theory the action is given as follows,

[ S[g, phi] = int d^{4}x sqrt{-g} (mathcal{R} – frac{1}{2} g^{mn} partial_{m} phi partial_{n} phi – Lambda) ]

Recall that this action is diffeomorphism invariant. It also has global symmetry, such that $phi rightarrow phi + a$.

What we want to do is replace $partial_{m} phi partial_{n} phi$ with many scalar fields (a linear combination with the products of many scalar fields). So we’re going to take the conventional route of employing $i$ and $j$ indices, with the inclusion of $M_{ij}$ in the action. Here, $M_{ij}$ is not a function of $phi$ – it is just a matrix of numbers. In other words, $M_{ij} neq f(phi)$. This is what we get,

[ S[g, phi] = int d^{4}x sqrt{-g} (mathcal{R} – frac{1}{2} g^{mn} partial_{m} phi^{i} partial_{n} phi^{j} M_{ij} – Lambda) ]

Several comments: first, notice the $i$ and $j$ indices – $phi^i$ and $phi^j$ – represent the internal space of our theory where $phi$ can take values. Secondly, the global symmetry that we had before remains: $phi^i rightarrow phi^i + a^i$. But there is also another global symmetry that we can write, because if we transform the scalars by any matrix $phi^i rightarrow Lambda_{j}^{i} phi^i$ such that $Lambda^{T}MLambda = M$, we have another global symmetry for the action.

In addition to the above, if we choose a particular $M$ such that it equals identity, $M = mathbb{I}$, where $M$ is a $D times D$ matrix and where $i = 1, … , D$, the global symmetry group can be written: $M =

mathbb{I} rightarrow SO(D)$.

Another matrix one might consider is where,

[ M = begin{pmatrix}

-1 & … & 0 \

0 & … & mathbb{I}_{(D-1)+(D-1)} end{pmatrix} ]

Now the global symmetry group can be written as $SO(1, D-1)$. (In QFT, this choice for $M$ is often avoided due to the desire for positive energy. If this symmetry group is used, one of the fields will have the incorrect sign for the kinetic term). What is nice about this, I think, is that if we combine $SO(1, D-1)$ with the previous expression for the global symmetry $phi^i rightarrow Lambda^i + a^i$, it looks like a translation. So the result is translations plus the Lorentz group together to form the Poincaré group.

But even more interestingly, if we maintain this choice of $M$ and we choose $phi^i rightarrow phi^i + a^i$, what we’re looking at for the internal space is something that particularly resembles the Minkowski space.

From this, we can return to the action written above. When we vary the metric, the equations of motion can be found as:

[ frac{delta S}{delta g_{mn}} = 0 rightarrow mathcal{R}_{m} – frac{1}{2} g_{mn} mathcal{R} + Lambda g_{mn} = T_{mn} ]

Where, $mathcal{R}_{m} – frac{1}{2} g_{mn} mathcal{R}$ is the Einstein tensor. On the right-hand side, $T_{mn}$ is the energy-momentum tensor of the matter theory. The energy-momentum tensor can be written as,

[ T_{mn} = M_{ij} (partial_{m} phi^{i} partial_{n} phi^{j} – frac{1}{2} g_{mn} g^{rs} partial_{r} phi^{i} partial_{s} phi^{j}) ]

**M = 1**

One of the interesting and fun things to think about with respect to the present theory is for different cases of $M$. In the case for $M=1$ – if there is only 1-dimension, as illustrated in a past post on the point particle, the space only has time. So, $X^0 rightarrow tau$. We can also establish that the metric $M_{ij} rightarrow eta_{mu nu}$, as we’ll be working in a sort of Minkowski configuration. So we will also replace the previous notion with notation more appropriate for the context, as our scalar fields no longer carry $i$ indices: $phi_(tau)^{i} rightarrow X_(tau)^{mu}$.

As for the metric $g_{mn}$, in the case of $M=1$ it only has one degree of freedom, and the convention I’ve employed (used for convenience) is simply to define $g = – e^{2}$. As there is only a time component, the idea then is to define $e$ as positive with a minus sign in front. This $e$ term is of course also a field – in fact, one can think of it as an auxiliary field – and it should be clear that $X$ is also a field. This is quite explicit in the new action,

[ S[e, X] = int dtau e(-frac{1}{2} (frac{1}{-e^2} dot{X}^{mu} dot{X}^{nu} eta_{mu nu} – Lambda) ]

Where $partial_{tau} X^{mu} equiv dot{X}^{mu}$.

We can tidy things up a bit. The action begins to look a bit more familiar,

[ S[e, X] = int frac{1}{2} int dtau (frac{1}{-e^2}

dot{X}^{mu} dot{X}^{nu} eta_{mu nu} – 2eLambda) ]

Now, the goal here in the $M=1$ case is to compute the equations of motion for the field $e$, with the hope that something interesting might ‘pop out’. We could, indeed, use the energy-momentum tensor instead; however, the above action is quite simple. So, we can compute the EoM directly:

[ frac{delta S}{delta e} = 0 rightarrow (-e^{-2} dot{X}^{mu} dot{X}^{nu} eta_{mu nu} – 2Lambda) = 0 ]

The key realisation is that $e$ can be integrated out since it does not have a kinetic term. This also means we can try to find the EoM, solving for $e$, and then substitute this value back into the action. Let’s do that.

Setting $dot{X}^{mu} dot{X}^{nu} eta_{mu nu}$ to be $dot{X}^2$ for simplification purposes,

[ -e^{-2} = frac{2 Lambda}{dot{X}^2 } ]

[ e^{2} = frac{- dot{X}^2}{2 Lambda} ]

[ implies e = frac{sqrt{- dot{X}^2}}{sqrt{2 Lambda}} ]

Thus, the effective action with just $X$ remaining (as $e$ has been integrated out):

[ S_{eff}[X] = frac{1}{2} int dtau -sqrt{2 Lambda} sqrt{- dot{X}^2} – sqrt{2 Lambda} sqrt{- dot{X}^2} ]

[ = sqrt{2 lambda} int dtau sqrt{dot{X}^2} ]

This should begin to look quite familiar. Preserving this result and putting it to one side for the moment, if we consider the non-relativistic limit, we can arrive at an expression for $sqrt{2 Lambda}$ that simplifies the above a bit more. Recall, as a first step, that the metric $eta_{mu nu}$ is mostly plus. So, we can write,

[ dot{X}^2 = -(frac{dtau}{dtau})^2 + (frac{dvec{x}}{d tau})^2 ]

We can now write the action,

[ S = – sqrt{2 Lambda} int dtau sqrt{1- (frac{dvec{x}}{d tau})^2} ]

Approximating the term under the square root, we get: $~ 1 – frac{1}{2}

(frac{dvec{x}}{d tau})^2$. So,

[ S = int dtau (-sqrt{2 Lambda} + frac{sqrt{2 Lambda}}{2}

(frac{dvec{x}}{d tau})^2 + …) ]

Ignoring the linear term, the more interesting question concerns the term $

frac{sqrt{2 Lambda}}{2} (frac{dvec{x}}{d tau})^2$. What is this? What does this term resemble? The answer is that it resembles the action of a particle – that is, the kinetic term.

If, in other words, we interpret $sqrt{2 Lambda} = m$ and plug this back into the action, look what happens:

[ S_{eff}[x] = -m int dtau sqrt{dot{X}^2} ]

This is the action of a particle propagating in flat spacetime.

Replacing the metric $eta_{mu nu}$ with a general metric $G_{mu nu}(X)$ that depends on the field $X$,

[ S[X] = -m int dtau sqrt{- dot{X}^{mu} dot{X}^{nu} G_{mu nu}} ]

This, as we know from a previous post, is the action for a particle moving in a curved spacetime. It is a nice result, given it connects with past entries and considerations, allowing us also to take a slightly different approach. There is also more we can do with this, before we completely exhaust the $M=1$ case. But we want to derive the Polyakov action for a string! All is not lost, with all of this hard work being important to complete just such a task.

**M=2**

The next idea is to attempt to translate the general field theory described above to the case of $M=2$. Consider, also, that we now have the coordinates $(X^0, X^1) rightarrow (tau, sigma)$. This should already be familiar, given past discussions. Additionally, the scalar fields are still $phi^i rightarrow X^{mu}(tau, sigma)$. Likewise, we should note $M_{ij} rightarrow eta_{mu nu}$.

Now, even though we can more or less translate everything we’ve already accomplished to the $M=2$ case, we do have one problem. We need to update our equations of motion, and this also entails that we study what the Riemann tensor is in the updated context. This requires that we recall our knowledge about the Riemann tensor in m-dimensions, including its symmetry relations and components, where the number of degrees of freedom is given by:

[ R_{mu rightarrow sigma}^{lambda} = frac{1}{12} M^2 (M^2 – 1) ]

So, when $M=2 rightarrow frac{1}{12} (2)^2 (2^2 – 1) = 1$. This means that we only have one parameter or degree of freedom.

Since we’re working in the case of $M=2$, we can write out what is below, where we have included the correct symmetry properties for the metric such that $g_{mn} rightarrow h_{mn}$:

[ R_{mnpq} = alpha (g_{mp}g{qn} – g_{mq}g_{pn}) ]

Where $alpha$ is the degree of freedom or, one might say, the constant of proportionality. We need to find this constant of proportionality, if we’re to move forward. The key is this: if $alpha$ is proportional to the Ricci scalar – the only other scalar in our theory – we can try to produce the Ricci scalar on the left-hand side and then solve for $alpha$. The first step is to take the Ricci tensor, contract indices, and then simplify:

[ g^{mp} mathcal{R}_{mnpq} = mathcal{R}_{nq} ]

[ implies mathcal{R}_{nq} = alpha (2 g_{qn} – g_{qn}) = alpha g_{qn} ]

[ implies mathcal{R} = g^{nq} mathcal{R}_{nq} = 2alpha ]

[ therefore alpha = frac{mathcal{R}}{2} ]

We can now substitute this result back into $mathcal{R}_{nq} = alpha g_{qn}$,

[ mathcal{R}_{nq} = frac{mathcal{R}}{2} g_{qn} ]

With this, we have have to use a bit of logic. Let’s return all the way back to our earlier formula,

[ frac{delta S}{delta g_{mn}} = 0 rightarrow mathcal{R}_{m} – frac{1}{2} g_{mn} mathcal{R} + Lambda g_{mn} = T_{mn} ]

What we find, now, is that the Einstein tensor vanishes. Moreover, we just discovered that,

[ mathcal{R}_{nq} = frac{mathcal{R}}{2} g_{qn} = 0 ]

The Einstein tensor is equal to zero. Therefore, the equations of motion from earlier suggest,

[ Lambda g_{mn} = T_{mn} ]

The pressing question, then, is what do we make of the energy-momentum tensor. Well, we already know that it is. When $M=2$,

[ T_{mn} = eta_{mu nu} (partial_{m} X^{mu} partial_{n} X^{nu} – frac{1}{2} h_{mn} h^{pq} partial_{p} X^{mu} partial_{q} X^{mu}) ]

At this point, we can also note that, as we take the trace of the EM tensor,

[ T_{m}^{m} = h^{mn}T_{mn} = 0 ]

But if $T_{mn} = 0$ then,

[ Lambda g_{mn} = T_{mn} rightarrow Lambda = 0 ]

Thus, the cosmological constant vanishes.

Here is the important conclusion. If we travel back to the action once more, the lingering question is what happens to gravity such that we see the Einstein tensor equal zero. And the answer is rather pleasant, as, when we translate the old formula using updated notation for the present context,

[ S = int dtau dsigma sqrt{-h} mathcal{R} ]

This integrand is just a constant. The number it churns out, in 2-dimensions, describes the topology in the space. If we denote the space as $sum$, $dtau dsigma$ is therefore being integrated over in this space. The $sqrt{-h} mathcal{R}$ is just some medium as $tau rightarrow tau_{i}$. It turns out, the total object $int_{sum} dtau dsigma sqrt{-h} mathcal{R}$ produces the Euler characteristic of this particular 2-dim space. So, we finally arrive at an action for our metric and fields (where we set $tau = sigma^{0}, sigma=sigma^{1})$,

[ S_{p} [h_{mn}, X^{mu}] = int d^2 sigma sqrt{-h} (- frac{1}{2} h^{mn} partial_{m} X^{mu} partial_{n} X^{nu} eta_{mu nu}) ]

This is the Polyakov action (without the inclusion of the slope parameter).