Notes on string theory: The nonrelativistic string

For pedagogical purposes, I think prefacing an engagement with Polchinski’s “String Theory” with a short study of nonrelativistic strings is reasonable. Although the first pages of the book immediately engage with the action for the relativistic particle, it is not uncommon in a lot of supplementary or individual texts to first spend some time building confidence. After all, the purpose is to work toward mastering the study of strings. There is a lot we can deduce from the nonrelativistic picture.

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To start, we know the action will take the form S = \int L dt, where L is the Lagrangian. One should also know that the Lagrangian will have both kinetic and potential energy. So we can build from this immediately.

    \[ L = T - V \]

    \[ \implies \int_{0}^{a} \frac{1}{2} \mu_0 dx (\frac{\partial y}{\partial t})^2 - \frac{1}{2} T_0 dx (\frac{\partial y}{\partial x})^2 \\ = \int_{0}^{a} \mathcal{L} (\dot{y}, y\prime) \]

Where \dot{y} = \frac{\partial y}{\partial t} and y\prime = \frac{\partial y}{\partial x}. One should also note that the integrand, which we’ve denoted as \mathcal{L}(\dot{y}, y\prime), is the Lagrangian density. \mu_0 is the mass per unit length of our string. T_0 is the tension.

For pedagogical purposes, I think it is also worth noting that to derive an expression for the potential, V, we need to assume small oscillation such that \mid \frac{dy}{dx} \mid << 1. In other words, in the standard way, we need to take \delta l = \sqrt{dx^2 + dy^2} - dx. Then, simplifying with some algebra, we can take the series expansion of that which is under the square root. For example,

    \[ \Delta l = \sqrt{(dx)^2 + (dy)^2} - dx = dx (\sqrt{1 + (\frac{dy}{dx})^2} - 1 \approx \frac{1}{2} (\frac{\partial y}{\partial x})^2 \]

Moving forward, in varying the action I like the approach that looks to substitute for the momentum density in the integrand. But first let’s explicitly write out the action for our nonrelativistic string. Notice that, because we’re considering the action, we also need to include a time component in addition to a spacial component. So you will see that we are now integrating from some initial time to some final time, and from an initial point to some final point. Thus,

    \[ S_{NR} = \int_{t_{i}}^{t_{f}} dt \int_{0}^{a} dx \mathcal{L}(\dot{y}, y\prime) \]

To the action we need to add variation in order to ultimately find our much needed EoM.

But before we carry on, it is perhaps worthwhile to first introduce some notation. This approach is owed to my notes from a Barton Zweibach lecture, and the reason for introducing this notation is that it will greatly simplify our calculations in the near future. This has to with the momentum density of our string that I alluded to a few moments ago. The momentum density will be denoted as \mathcal{P}^t and \mathcal{P}^x. We compute these terms as follows,

    \[ \mathcal{P}^t = \frac{\partial \mathcal{L}}{\partial \dot{y}} = \mu_0 \frac{\partial y}{\partial t} \]

    \[ \mathcal{P}^x = \frac{\partial \mathcal{L}}{\partial y\prime} = - T_0 \frac{\partial y}{\partial x} \]

Now we’re ready to proceed. Here, the function y(t, x) represents the path of our string. In varying the action, y (t, x) \rightarrow y(t, x) + \delta y (t, x). So,

    \[ \delta S = \int_{t_i}^{t_f} dt \int_{0}^{a} dx (\frac{\partial \mathcal{L}}{\partial \dot{y}} \delta \dot{y} + \frac{\partial \mathcal{L}}{\partial y \prime} \delta y\prime) \]

We now wish to substitute in for \mathcal{P}^t and \mathcal{P}^x. The result is as follows,

    \[ \delta S = \int_{t_i}^{t_f} dt \int_{0}^{a} dx (\mathcal{P}^t \frac{\partial}{\partial t} \delta y + \mathcal{P}^x \frac{\partial}{\partial x} \delta y) \]

    \[ = \int dt dx [\frac{\partial}{\partial t} (\mathcal{P}^t \delta y) + \frac{\partial}{\partial x} (\mathcal{P}^x \delta y) - \delta y (\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x})] \]

It’s now the time, as is common practice in the calculus of variations, to integrate by parts. To save space, I have avoided writing it all out. We get,

    \[ \delta S =\int_{0}^{a} dx (\mathcal{P}^t \delta y)]_{t_i}^{t_f} + \int_{t_i}^{t_f } dt (\mathcal{P}^x \delta y)]_{0}^{a} - \int_{t_i}^{t_f} dt \int_{0}^{a} dx \delta y (\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x}) \]

In our case, we don’t have to consider the first term regarding the initial and final conditions. That is to say, we can either just set the times to go off to infinity and put the term to the back of our mind or we can explicitly define specific initial and final conditions. But right now we’re not interested in this. The domain of interest here, for the sake of this example, is between 0 and a during the time interval from t_i to t_f. Unless interested in the behaviour of the endpoints via Hamilton-Jacobi – what the string looks like at t_i and t_f – in our case we’re just going to set the first term so it cancels.

Instead we’re going to focus explicitly on the second term. When we expand it out the result is below. As an aside, we want the variations of both sides of the minus sign in the expansion to equal zero independently of each other.

    \[ \int_{t_i}^{t_f } \mathcal{P}^x \delta y]_{0}^{a} = \int_{t_i}^{t_f } dt (\mathcal{P}^x(t, x=a) \delta y(t, x=a) - \mathcal{P}^t (t, x=0) \delta y (t, x=0)) \]

This requires invoking either Neumann or Dirichlet boundaries. This concept of boundary conditions will prove very important later on, as it will feature prominently when we begin studying the relativistic string (and advancing from there). More pointedly, there is a lot to be said about this matter, and perhaps it will be useful that I dedicate a separate post to it. But I should also point the interested reader to review Chapter 4 in Zwiebach (2004) – and, really, that goes for anything said here.

Meanwhile, the good news is that we’ve already found the EoM from an earlier result.

    \[ \frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x} = 0 \]

But what is \frac{\partial \mathcal{P}^t}{\partial t} and \frac{\partial \mathcal{P}^x}{\partial x}? Well, let me ask another question: what is \mathcal{P}^t and \mathcal{P}^x? We already computed these above. All we need to do is take their partial derivative with respect t. Thus,

    \[ \frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x} = 0 \\ \\ \implies \mu_0 \frac{\partial^2 y}{\partial t^2} - T_0 \frac{\partial^2 y}{\partial x^2} = 0 \\ \]

Rearranging, and noting that v^2 = \frac{T_0}{\mu_0}, we get

    \[ \frac{\partial^2 y}{\partial x^2} - \frac{1}{\frac{T_0}{\mu_0}}\frac{\partial^2 y}{\partial t^2} = 0 \]

What is nice about this result is that this is a wave equation.