# Notes on string theory #3: Nambu-Goto action

1. Introduction

I haven’t been keeping up with this as much as I would like, mainly because I have been busy. But I am committed to continuing to reupload many of my notes on Polchinski’s textbooks. It is fun for me to go through it all again in my spare time, and I’ve noticed that since the time of first working through the textbooks there is more I can add to many topics.

It is worth remembering that, in the last note, we reviewed the classical worldline and polynomial action for the relativistic point particle. We also discussed reparameterisation invariance and calculated the equations of motion. In this note, the focus is to construct the first-principle Nambu-Goto action for the relativistic string as given in equations (1.2.9a-1.2.9b) in Polchinski’s textbook.

Often in popular literature and discourse I read descriptions of the string that almost shroud it in mystery. How could the fundamental constituents of matter be described by a bunch of strings? Other times, caricatures of string theory can leave the impression that to view all elementary particles as vibrating strings is somewhat arbitrary. Why not some other type of objects? It is suggestive of a certain arbitrariness to the idea of modelling fundamental particles as strings; but the development of string theory is, in fact, well motivated. Ultimately, all that we’re doing is extending the concept of point particles that we all know and love, and this is first and foremost evidenced in the Nambu-Goto action. But, in terms of the bigger picture, what we see is that in studying the string and its dynamics an entire universe of implications emerge. It reminds me of a great line in David Tong’s lecture notes that is worth paraphrasing: we find that the requirements demanded by the tiny string are so stringent that we are led naturally to a description of how the entire universe moves. On many occasions it is, indeed, like “the tail is wagging the dog”.

As we’re following Polchinski’s textbook, which only covers the Nambu-Goto action in a few words, if the interested reader would like to spend more time studying this action I would recommend ‘String theory and M-theory’ by Katrin Becker, Melanie Becker, and John H. Schwarz, especially the exercises, or for an even more gentle introduction see Barton Zwiebach’s textbook ‘A first course in string theory’.

2. Area functional

To arrive at the Nambu-Goto action, let us first recall from the last note that a p-brane may be described as a p-dimensional object moving through D-dimensional flat spacetime with ${D \geq p}$. If a 0-dimensional point particle (0-brane) traces out a (0+1)-dimensional worldline, it follows that a 1-dimensional string (1-brane) sweeps out a (1+1)-dimensional surface that we call the string worldsheet. And just as we can parameterise the relativistic point particle’s (0+1)-dimensional worldline, we can parameterise the (1+1)-dimensional worldsheet traced by the string. Coming to grips with this idea is the first task.

The main idea is that the worldline of a particle is replaced by the worldsheet ${\Sigma}$, which is a surface embedded into D-dimensional Minkowksi spacetime. Given that the path of a point particle can be described by a single parameter, the proper time ${\tau}$, which multiplied by c, is the Lorentz invariant proper length of the worldline; for strings, we will define the Lorentz invariant proper area of the worldsheet in a completely analogous way. As we’ll see, the first-principle string action is indeed proportional to this proper area.

To start, we see that because the string worldsheet is a (1+1)-dimensional surface, its requires two parameters which we will denote as ${\xi^{1}}$ and ${\xi^{2}}$. We will also limit our present considerations to the case of an open string (we will talk about closed strings in a later note). In order to define the appropriate area functional, we want to sketch a grid on the spacial surface of the string worldsheet with lines of constant ${\xi^{1}}$ and ${\xi^{2}}$; then we want to embed this spatial surface in the background target space.

The target space is the world where the 2-dimensional surface lives. Ultimately, we want to distinguish between the area we parameterise and the actual physical string worldsheet. In order to accomplish this, we define a one-to-one map, which we may call the string map. The purpose of the string map is therefore to take us from the parameter space that we have constructed to the target space in which the physical surface propagates. Indeed, as we’ll see, the string action is in this precise sense defined as a functional of smooth maps.

To construct the string map, we first formalise the notion of area in parameter space, with this parameter space itself defined by the range of the parameters ${\xi^{1}}$ and ${\xi^{2}}$. One can, in principle, view the parameters we have selected as local coordinates on the surface. And so, as emphasised above, we can think of the worldsheet as a physical surface, which is in fact the image of the parameter space under the one-to-one string map written as ${\vec{x}(\xi^{1}, \xi^{2})}$. The parameterised surface can therefore be described by the coordinate functions

$\displaystyle \vec{x}(\xi^1 , \xi^2) = x^1 (\xi^1 , \xi^2), x^2 (\xi^1 , \xi^2), x^3 (\xi^1 , \xi^2). \ \ \ \ \ (1)$

The area to which we want to give mathematical description is more accurately an infinitesimal area element. Since we begin working in a parameter space, and since our very small square is mapped onto the surface in target space, when we map this very small area from the parameter space to the surface we achieve a parallelogram the sides of which may be denoted as ${d\vec{v}_1}$ and ${d\vec{v}_2}$. We can express this as follows:

$\displaystyle d \vec{v}_1 = \frac{\partial \vec{x}}{\partial \xi^1} d\xi^1$

$\displaystyle d \vec{v}_2 = \frac{\partial \vec{x}}{\partial \xi^2}d\xi^2, \ \ \ \ \ (2)$

in which we have defined the rate of variation of the coordinates with respect to the parameters ${\xi}$. If we multiply this rate by the length ${d\xi}$ of the horizontal side of the infinitesimal parallelogram, we get the vector ${d \vec{v}_1}$ that represents this side in the target space.

The main objective is to now compute the area ${dA}$ of this parallelogram.

IMAGE

Since we have already labelled the sides of the infinitesimal area in the parameter space, we simply need to invoke the formula for the area of a parallelogram:

$\displaystyle d^2 A = \mid d\vec{v}_1 \mid \mid d\vec{v}_2 \mid \mid \sin \theta \mid$

$\displaystyle = \mid d\vec{v}_1 \mid \mid d\vec{v}_2 \mid \sqrt{1 - \cos^2 \theta}$

$\displaystyle = \sqrt{\mid d\vec{v}_1 \mid^2 \mid d\vec{v}_2 \mid^2 - \mid d\vec{v}_1 \mid^2 \mid d\vec{v}_2 \mid^2 \cos^2 \theta}. \ \ \ \ \ (3)$

Here ${\theta}$ denotes the angle between the vectors ${dv_1}$ and ${dv_2}$. Written in terms of dot products in which ${(\vec{A} \times \vec{B}) \cdot (\vec{A} \times \vec{B}) = \mid A \mid^2 \mid B \mid^2 - (A \cdot B)^2}$ such that

$\displaystyle (d\vec{v}_1 \times d\vec{v}_2) \cdot (d\vec{v}_1 \times d\vec{v}_2) = (d\vec{v}_1)^2 (d\vec{v})^2 - (d\vec{v}_1 \cdot d\vec{v}_2)^2$

we have

$\displaystyle = \sqrt{(d\vec{v}_1 \cdot d\vec{v}_1) (d\vec{v}_2 \cdot d\vec{v}_2) - (d\vec{v}_1 \cdot d\vec{v}_2)^2}. \ \ \ \ \ (4)$

From this result, notice that we can now substitute for ${d\vec{v}_1}$ and ${d\vec{v}_2}$ using (2). Doing so gives

$\displaystyle dA = \sqrt{(\frac{\partial \vec{x}}{\partial \xi^1} \cdot \frac{\partial \vec{x}}{\partial \xi^1})(\frac{\partial \vec{x}}{\partial \xi^2} \cdot \frac{\partial \vec{x}}{\partial \xi^2}) - (\frac{\partial \vec{x}}{\partial \xi^1} \cdot \frac{\partial \vec{x}}{\partial \xi^2})^2} d\xi^1 d\xi^2. \ \ \ \ \ (5)$

We have now obtained a general expression for the area element of the parameterised spatial surface. Written as the full area functional we have

$\displaystyle A = \int d\xi^1 d\xi^2 \ \sqrt{(\frac{\partial \vec{x}}{\partial \xi^1} \cdot \frac{\partial \vec{x}}{\partial \xi^1})(\frac{\partial \vec{x}}{\partial \xi^2} \cdot \frac{\partial \vec{x}}{\partial \xi^2}) - (\frac{\partial \vec{x}}{\partial \xi^1} \cdot \frac{\partial \vec{x}}{\partial \xi^2})^2}, \ \ \ \ \ (6)$

where the integral extends over the ranges of the parameters ${\xi^{1}}$ and ${\xi^{2}}$. This functional is reparameterisation invariant, which can be quickly verified by reparameterising the surface with tilde parameters ${(\tilde{\xi}^1, \tilde{\xi}^2)}$ that then gives back (6) when ${\tilde{\xi}^1 = \tilde{\xi}^1 (\xi^1)}$ and ${\tilde{\xi}^2 = \tilde{\xi}^2 (\xi^2)}$.

The main issue is that the area functional (6) is not very nice, and reparameterisation invariance is not completely general. We want reparameterisation invariance to be manifest.

3. Induced Metric

Suppose we have some vector ${d\vec{x}}$ on the surface ${\Sigma}$ that we have so far pencilled into the target space. We know that we can describe this surface through the string mapping functions ${\vec{x}(\xi^{1}, \xi^{2})}$. What if we then consider ${d\vec{x}}$ tangent to the surface ${\Sigma}$? We could then let ${ds}$ denote the length of this tangent vector, and hence we could invoke some early idea of a metric on ${\Sigma}$.

Given the vector tangent to the surface, with ${ds}$ its length, we can write

$\displaystyle ds^2 = d\vec{x} \cdot d\vec{x}. \ \ \ \ \ (7)$

But what is ${d\vec{x}}$ in terms of the parameter space coordinates that we constructed? In other words, can we relate ${d\vec{x}}$ with ${\xi^{1}, \xi^{2}}$? This is precisely what our mapping accomplishes such that we can express ${d\vec{x}}$ in terms of partial derivatives and derivatives of ${\xi^{1}, \xi^{2}}$:

$\displaystyle d\vec{x} = \frac{\partial \vec{x}}{\partial \xi^1} d\xi^1 + \frac{\partial \vec{x}}{\partial \xi^2} d\xi^2 = \frac{\partial \vec{x}}{\partial \xi^i} d\xi^i, \ \ \ \ \ (8)$

with the summation convention assumed for the repeated indices over possible values 1 and 2. If we now return to (7) and plug ${d\vec{x}}$ back into our equation for ${ds^2}$ we see that we can now write

$\displaystyle ds^2 = \frac{\partial \vec{x}}{\partial \xi^i} d\xi^i \cdot \frac{\partial \vec{x}}{\partial \xi^j} d\xi^j. \ \ \ \ \ (9)$

But notice something interesting. If we set ${h_{ij}(\xi) = \frac{\partial \vec{x}}{\partial \xi^i} d\xi^i \cdot \frac{\partial \vec{x}}{\partial \xi^j}}$, this means we can write a more simplified equation of the form

$\displaystyle ds^{2} = h_{ij}(\xi) d\xi^i d\xi^j, \ \ \ \ \ (10)$

in which the quantity ${h_{ij}(\xi)}$ is called the induced metric. It is a metric on the target space surface precisely in the sense that, as ${\xi_i}$ play the role of coordinates on ${\Sigma}$, we see that (10) determines distances on this surface. It is said to be induced because it uses the metric on the ambient space in which ${\Sigma}$ lives to determine distances on ${\Sigma}$. More technically, we say that the induced metric is the pullback of the target space metric onto the worldsheet.

A question we can now ask is whether, upon constructing a metric on the target space surface, does this then lead us to an equivalent expression for (6)? Observe that, in matrix form, we have for the induced metric

$\displaystyle h_{ij} = \begin{pmatrix} \frac{\partial \vec{x}}{\partial \xi^1} \cdot \frac{\partial \vec{x}}{\partial \xi^1} & \frac{\partial \vec{x}}{\partial \xi^1} \cdot \frac{\partial \vec{x}}{\partial \xi^2} \\ \frac{\partial \vec{x}}{\partial \xi^2} \cdot \frac{\partial \vec{x}}{\partial \xi^1} & \frac{\partial \vec{x}}{\partial \xi^2} \cdot \frac{\partial \vec{x}}{\partial \xi^2} \\ \end{pmatrix}. \ \ (11)$

What is this telling us? Notice that if you compute the determinant of the matrix ${h_{ij}}$, you find the same quantity that resides under the square root in (6). This is a massive hint that the construction is on the right track. So, let’s substitute the appropriate matrix elements into our earlier expression for the infinitesimal area. This is what we find,

$\displaystyle dA = \sqrt{(h_{11})h_{22} - h_{12}^2} \ d\xi^1 d\xi^2$

$\displaystyle = \sqrt{\det h} \ d\xi^1 d\xi^2$

$\displaystyle \therefore dA = \sqrt{h} \ d\xi^1 d\xi^2, \ \ \ \ \ (12)$

where ${h \equiv \det h_{ij} (\xi)}$. This implies,

$\displaystyle A = \int d\xi^1 d\xi^2 \sqrt{h}. \ \ \ \ \ (13)$

This new way to express the area, ${A}$, is now given in terms of the determinant of the induced metric. And although we are not yet done constructing the Nambu-Goto action, we see quite clearly from (13) why Polchinski says that the action ${S_{NG}}$ in equations (1.2.9a-1.2.9b) in his textbook is proportional to the area of the worldsheet.

4. Reparameterisation invariance

The wonderful thing about this last result (13) is that we can now show manifest reparameterisation invariance in a much simpler way, as it may now be described by way of how the induced metric transforms.

To do this, we invoke a different set of parameters and therefore also a different metric, and then we show that the original vector ${d\vec{x}}$ does not depend on our original parameterisation.

We begin with

$\displaystyle ds^2 = h_{ij}(\xi) d\xi^i d\xi^j = \tilde{h}_{ij}(\tilde{\xi}) d\tilde{\xi}_1 d\tilde{\xi}_2. \ \ \ \ \ (14)$

We then use the chain rule

$\displaystyle ds^{2} = \tilde{h}_{pq}(\tilde{\xi}) \frac{\partial \tilde{\xi}^p}{\partial \xi^i} \frac{\partial \tilde{\xi}^q}{\partial \xi^j} d\xi^i d\xi^j$

$\displaystyle h_{ij}(\xi) = \tilde{h}_{pq} (\tilde{\xi}) \frac{\partial \tilde{\xi}^p}{\partial \xi^i} \frac{\partial \tilde{\xi}^q}{\partial \xi^j}. \ \ \ \ \ (15)$

Next, recall that the change of variable theorem tells us how the integration measure transforms

$\displaystyle d\xi^{1} d \xi^2 = \mid \det \frac{d \xi^i}{d \tilde{\xi}^j} \mid d \tilde{\xi}^1 d \tilde{\xi}^2 = \mid \det M \mid d \tilde{\xi}^1 d \tilde{\xi}^2, \ \ \ \ \ (16)$

where ${M}$ is the matrix defined by ${M_{ij} = \partial \xi^1 / \partial \tilde{\xi}j}$ and similarly

$\displaystyle d\tilde{\xi}^{1} d \tilde{\xi}^2 = \mid \det \frac{d \tilde{\xi}^i}{d \xi^j} \mid d \xi^1 d \xi^2 = \mid \det \tilde{M} \mid d \xi^1 d \xi^2, \ \ \ \ \ (17)$

where ${\tilde{M}}$ is defined by ${\tilde{M}_{ij} = \partial \tilde{\xi}^i / \partial \xi^j}$. Using this and returning to (15) we can rewrite this equation for ${h}$ and ${\tilde{h}}$ such that

$\displaystyle h_{ij}(\xi) = \tilde{h}_{pq} \tilde{M}_{pi}\tilde{M}_{qj} = (\tilde{M}^T)_{ip}\tilde{h}_{pq} \tilde{M}_{qj}. \ \ \ \ \ (18)$

If we denote ${h \equiv \det h_{ij}}$, and if take the determinant of the right-hand side of (18) we find

$\displaystyle h = (\det \tilde{M}^T) \tilde{h} (\det \tilde{M}) = \tilde{g}(\det \tilde{M})^2. \ \ \ \ \ (19)$

Clearly, then, if we take the square root we obtain

$\displaystyle \sqrt{h} = \sqrt{\tilde{h}} \mid \det \tilde{M} \mid, \ \ \ \ \ (20)$

which is the transformation property for the square root of the determinant of the metric.

Finally, we conclude using (16) and (20) with the fact that ${\mid \det M \mid \mid \det \tilde{M} \mid = 1}$ we can show that (13) is reparameterisation invariant

$\displaystyle \int d\xi^1 d\xi^2 \sqrt{h} = \int d\tilde{\xi}^1 d\tilde{\xi}^2 \mid \det M \mid \sqrt{\tilde{h}} \mid \det \tilde{M} \mid = \int d\tilde{\xi}_1 d\tilde{\xi}_2 \sqrt{\tilde{h}}. \ \ \ \ \ (21)$

There is perhaps a much more elegant way to show this proof. But for now, one should focus on how (21) is just a standard metric transformation inasmuch that ${\int d\xi^1 d\xi^2 \sqrt{h}}$ transforms via a Jacobian determinant of ${\xi}$ with respect to ${\tilde{\xi}}$ as ${\int d\tilde{\xi}_1 d\tilde{\xi}_2 \sqrt{\tilde{h}}}$.

5. String propagating in spacetime

Let us now work toward constructing the Nambu-Goto action as it appears in equations (1.2.9a-1.2.9b). Up to this point we have taken the approach of mapping from a parameter space to a target space in which the surface ${\Sigma}$ lives. But we are interested in the case of surfaces in spacetime. These surfaces are obtained by representing in spacetime the history of the string as it propagates, in the same way the worldline of the point particle is described by representing its history.

Spacetime surfaces, such as string worldsheets, are not all that different from the spatial surfaces we considered in the previous sections. Instead of the coordinates ${\xi^{1}}$ and ${\xi^{2}}$, for a relativistic string we should parameterise the string worldsheet in such a way that we account for both the proper time and the string’s spatial extension. Another way to put this is that, if our interest is to consider surfaces in spacetime (the worldsheet traced by the string), we now use ${\tau}$ to denote the proper time and ${\sigma}$ to denote the spacial extension of the surface. Given usual spacetime coordinates, which we write following string theory conventions ${X^{\mu} = (X^0, X^1, ..., X^d)}$, the surface is then described by the mapping functions

$\displaystyle X^{\mu}(\tau, \sigma). \ \ \ \ \ (22)$

Hence, we come to the point emphasised at the outset of this note. The string worldsheet action formally defines the map ${\Sigma : (\tau, \sigma) \mapsto X^{\mu}(\tau, \sigma) \in \mathbb{R}^{1, d-1}}$. If it is still not clear, remember that what we’re working toward is a description of the string worldsheet ${\Sigma}$ as a curved surface embedded in spacetime. This embedding is given by the fields ${X^{\mu}(\tau, \sigma)}$, in which the parameters ${\tau}$ and ${\sigma}$ can be viewed (locally) as coordinates on the worldsheet. So the string map tells us that given some fixed point ${X^{\mu}(\tau, \sigma)}$ in the parameter space, we are performing a direct mapping to a fixed point in spacetime coordinates. Typically we drop the arguments ${(\tau, \sigma)}$ and leave them implicit, with the inverse of the map ${X^{\mu}}$ taking the worldsheet to the parameter space.

It is also worth noting that the functions ${X^{\mu}}$ describe how the string propagates and oscillates through spacetime, while the endpoints of the string are parameterised by ${\tau}$ such that ${\frac{\partial X^{\mu}}{\partial \tau} (\tau, \sigma) \neq 0}$. In our present case, we are considering an open string; but if ${\sigma}$ is periodic then we’d be working with a closed string embedded in the background spacetime.

Getting back to the task at hand: to find the area element we proceed in similar fashion as before, except now we must use relativistic notation. So for the area element we have ${d\tau}$ and ${d\sigma}$ describing the sides of an infinitesimal parallelogram in parameter space. In spacetime, this becomes a quadrilateral area element. We therefore set-up a direct analogue with our expression for ${dA}$ in (4) where we consider the vectors ${dv^{\mu}_{1}}$ and ${dv_{2}^{\mu}}$ spanning the quadrilateral,

$\displaystyle dv^{\mu}_{1} = \frac{\partial X^{\mu}}{\partial \tau} d\tau, \ \ dv^{\mu}_{2} = \frac{\partial X^{\mu}}{\partial \sigma} d\sigma. \ \ \ \ \ (23)$

Notice that we may substitute for ${dv^{\mu}_{1}}$ and ${dv_{2}^{\mu}}$ into (4),

$\displaystyle dA = d\tau d\sigma \sqrt{(\frac{\partial X^{\mu}}{\partial \tau} \frac{\partial X_{\mu}}{\partial \tau})(\frac{\partial X^{\nu}}{\partial \sigma} \frac{\partial X_{\nu}}{\partial \sigma}) - (\frac{\partial X^{\mu}}{\partial \tau} \frac{\partial X_{\mu}}{\partial \sigma})^2}. \ \ \ \ \ (24)$

We now invoke relativistic dot product notation so that we ensure that what we are working with is the proper area. The object under the square root turns out to be negative, but we can switch the sign without violation of any rules. The basic idea is that, for a surface with a timelike vector and a spacelike vector the square root is always positive such that Cauchy-Schwarz inequality flips. This means, ${(\dot{X}^2 \cdot X^{\prime})^2 - (\dot{X})^2 (X^{\prime})^2 > 0}$. We also want to integrate (24). So, putting everything together, we have

$\displaystyle A = \int_{\sum} d\tau d\sigma \sqrt{(\frac{\partial X}{\partial \tau} \cdot \frac{\partial X}{\partial \sigma})^2 - (\frac{\partial X}{\partial \tau} \cdot \frac{\partial X}{\partial \tau})(\frac{\partial X}{\partial \sigma} \cdot \frac{\partial X}{\partial \sigma})}. \ \ \ \ \ (25)$

We can still simplify our expression for the area using the more compact notation, ${\dot{X}^{\mu} \equiv \frac{\partial X^{\mu}}{\partial \tau}}$ and ${X^{\prime \mu} \equiv \frac{\partial X^{\mu}}{\partial \sigma}}$. This means we can write,

$\displaystyle A = \int_{\Sigma} d\tau d\sigma \sqrt{(\dot{X})^2 (X^{\prime})^2 - (\dot{X} \cdot X^{\prime})^2}. \ \ \ \ \ (26)$

Now comes the important part. From (26) there are a few ways we can approach the Nambu-Goto action. The most direct approach is to remember how, inasmuch that we are generalising the point particle action, we may anticipate the existence of some constant of proportionality. Indeed, it is completely reasonable to anticipate an action of the form general form ${S = -T \int dA}$. And this proves to be the case, because it follows that we may write the Nambu-Goto action for the string as

$\displaystyle S_{NG} = -\frac{T_0}{c} \int_{\tau_i}^{\tau_f} d\tau \int_{0}^{\sigma_1} d\sigma \sqrt{(\dot{X} \cdot X^{\prime})^2 - \dot{X}^2 \cdot X^{\prime^2}}, \ \ \ \ \ (27)$

where ${\frac{T_0}{c}}$ is a constant of proportionality to ensure units of action. To explain this, consider the following. Given that the string action is proportional to the proper area of the worldsheet, the area functional has units of length squared. We see this because ${X^{\mu}}$ has unites of length, i.e., ${[X] = L}$, and there are four under the square root. Each term in the square root also has two ${\sigma}$ derivatives and two ${\tau}$ derivatives, with their units cancelling against the derivatives. Since ${S_{NG}}$ must have the units of action ${[S] = \hbar = ML^2/T}$ with ${A}$ having units ${L^2}$, the total proper area must be multiplied by the value ${M/T}$. We know that the string will have a tension, ${T_0}$, which has units of force. We also know that force divided by velocity has the units ${M/T}$; so to ensure units of action the proper area is multiplied by ${T_0 / c}$.

6. Manifest Reparameterisation Invariance of the Nambu-Goto Action

We still shouldn’t be completely satisfied with this early form of the Nambu-Goto action (28). How do we know, for instance, that what we have ended up with is manifestly reparameterisation invariant? It is crucial that the ${S_{NG}}$ action be dependent only on the embedding in spacetime and not the choice of parameterisation.

To explore the action (28) in a deeper way, we first need to invoke the target space Minkowski metric, ${\eta_{\mu \nu}}$, and we should consider a differential line element of the form

$\displaystyle -ds^{2} = dX^{\mu} dX_{\mu} = - \eta_{\mu \nu} dX^{\mu} dX^{\nu}. \ \ \ \ \ (28)$

We may now expand the derivatives acting on ${X}$,

$\displaystyle -ds^{2} = - \eta_{\mu \nu} \frac{\partial X^{\mu}}{\partial \xi^{\alpha}} \frac{\partial X^{\nu}}{\partial \xi^{\beta}} \ d\xi^{\alpha} d\xi^{\beta}, \ \ \ \ \ (29)$

where ${\alpha}$ and ${\beta}$ run from 1 and 2. Similar as before for the spatial surface, we can define an induced metric. In this case, the induced metric on the string worldsheet is given as ${h_{\alpha \beta}}$. It is simply the pullback of the target space Minkowski metric, ${\eta_{\mu \nu}}$. This allows us to define the induced metric as,

$\displaystyle h_{\alpha \beta} = \eta_{\mu \nu} \frac{\partial X^{\mu}}{\partial \xi^{\alpha}} \frac{\partial X^{\nu}}{\partial \xi^{\beta}}. \ \ \ \ \ (30)$

This means we can write the more compact equation for the line element

$\displaystyle -ds^{2} = h_{\alpha \beta} d\xi^{\alpha}d\xi^{\beta}, \ \ \ \ \ (31)$

because, while the induced metric describes distances on the string worldsheet, it also includes the metric of the background spacetime in its definition. But, to ensure clarity of knowledge, let’s think about this induced metric a bit more. In matrix form, it is a ${2 \times 2}$ matrix with components

$\displaystyle h_{\tau \tau} = \eta_{\mu \nu} \frac{\partial X^{\mu}}{\partial \tau} \frac{\partial X^{\nu}}{\partial \tau} = \dot{X}^{2},$

$\displaystyle h_{\sigma \tau} = \eta_{\mu \nu} \frac{\partial X^{\mu}}{\partial \sigma} \frac{\partial X^{\nu}}{\partial \tau} = \dot{X} \cdot X^{\prime} = h_{\tau \sigma},$

$\displaystyle h_{\sigma \sigma} = \eta_{\mu \nu} \frac{\partial X^{\mu}}{\partial \sigma} \frac{\partial X^{\nu}}{\partial \sigma} = X^{\prime 2}. \ \ \ \ \ (32)$

And so, the induced metric may be written in matrix form as

$\displaystyle h_{\alpha \beta} = \begin{pmatrix} \dot{X}^2 & \dot{X} \cdot X^{\prime} \\ \dot{X} \cdot X^{\prime} & X^2 \prime \\ \end{pmatrix}. \ \ \ \ \ (33)$

Therefore, as we showed in the case of the spatial surface, we extend the logic of the previous examples and manifest reparameterisation invariance is seen to be featured with the help of the induced metric

$\displaystyle S_{NG} = -\frac{T_{0}}{c} \int_{\sum} d\tau d\sigma \sqrt{-h}, \ \ \ \ \ (34)$

where ${h = \det h_{\alpha \beta}}$.

The final observation is that, as Polchinski notes (p.11), for the string tension an alternative parameter is ${\alpha^{\prime}}$. This proportionality constant ${\alpha^{\prime}}$ has been used since the early days of string theory; one may recognise it as the Regge slope, which has to do with the relation between the angular momentum, ${J}$, of a rotating string and the square of the energy ${E}$. In that ${\alpha^{\prime}}$ has units of spacetime-length-squared, we therefore observe

$\displaystyle T = \frac{1}{2 \pi \alpha \prime}, \ \ \ \ \ (35)$

where we’ve set ${\hbar = c = 1}$. This is equation (1.2.10) in Polchinski. Hence, we may now rewrite ${S_{NG}}$ in its more conventional form as read in equations (1.2.9a-1.29b):

$\displaystyle S_{NG} = - \frac{1}{2 \pi \alpha^{\prime}} \int_{\Sigma} d\tau d\sigma \ (- \det h_{\alpha \beta})^{1/2}. \ \ \ \ \ (36)$

The answer in Exercise 1.1b reveals more explicitly how the string tension is related to the Regge slope. It just requires that we write things in terms of the transverse velocity. To keep these notes focused and organised, at the conclusion of each chapter we’ll go over the solutions to the exercises and so we’ll return to this question then.

6.1. Equations of motion

Before we get to the symmetries of the action (36), let’s quickly look at its equations of motion. To simplify matters, we can write the Lagrangian as ${\mathcal{L} = \sqrt{-h}}$ with the Euler-Lagrange equations reading as

$\displaystyle \partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial_\alpha X^\mu}\right)=\frac{\partial\mathcal{L}}{\partial_\alpha X^\mu}=0. \ \ \ \ \ (37)$

From the chain rule, it is therefore clear that we need to calculate

$\displaystyle \partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial_\alpha X^\mu}\right)=\partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial h_{\beta\gamma}}\frac{\partial h_{\beta\gamma}}{\partial_\alpha X^\mu}\right). \ \ \ \ \ (38)$

For the first term in brackets, we use the identity for the variation of the determinant ${\delta\sqrt{-g}=-\frac{1}{2}\sqrt{-g}g_{\alpha\beta}\delta g^{\alpha\beta}}$, which can be easily verified. Hence,

$\displaystyle \frac{\partial\mathcal{L}}{\partial h_{\beta\gamma}}=\frac{\partial\sqrt{-h}}{\partial h_{\beta\gamma}}=-\frac{1}{2}\sqrt{-h}\frac{h_{\rho\kappa}\delta h^{\rho\kappa}}{\delta h_{\beta\gamma}}=-\frac{1}{2}\sqrt{-h}h^{\beta\gamma}. \ \ \ \ \ (39)$

For the next term we find

$\displaystyle \frac{\partial h_{\beta\gamma}}{\partial(\partial_\alpha X^\mu)} =\eta^{\mu\nu}\delta^{\alpha}_\beta\partial_\gamma X_\nu +\eta^{\mu\nu}\delta^\alpha_\gamma\partial_\beta X_\nu =\delta^\alpha_\beta\partial_\gamma X^\mu +\delta^\alpha_\gamma\partial_\beta X^\mu. \ \ \ \ \ (40)$

Putting everything together

$\displaystyle \partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial h_{\beta\gamma}}\frac{\partial h_{\beta\gamma}}{\partial_\alpha X^\mu}\right)=\partial_\alpha(-\frac{1}{2}\sqrt{-h}h^{\beta\gamma}(\delta_{\alpha}^\beta\partial_\gamma X^\mu+\delta_\alpha^\gamma\partial_\beta X^\mu))=0 \ \ \ \ \ (41)$

$\displaystyle \frac{1}{2}\partial_\alpha(\sqrt{-h}h^{\alpha\gamma}\partial_\gamma X^\mu+\sqrt{-h}h^{\beta\alpha}\partial_\beta X^\mu)=0 \ \ \ \ \ (42)$

$\displaystyle \partial_\alpha(\sqrt{-h}h^{\alpha\beta}\partial_\beta X^\mu)=0. \ \ \ \ \ (43)$

As the metric ${h}$ contains the embedding ${X^{\mu}}$, these equations are highly non-linear. But this is not unexpected given the fact that the action (35) is non-linear. One way to interrept these equations is that, as a minimal surface area is being demanded by the stationary action, in Zwiebach’s textbook one is motivated to think analogously of the image of static soap film in some Lorentz frame. In this case, we think of the film as a spatial surface in which every point is a saddle point.

6.2. Symmetries

Finally, the last topic covered concerns the symmetries of the action (36).

Poincare group: As the Nambu-Goto action is completely and directly analogous to the action for a relativistic point particle, one might rightly anticipate that the action for a string is invariant under the isometry group of flat spacetime, which is the D-dimensional Poincare group

$\displaystyle X^{\prime \mu}(\tau, \sigma) = \Lambda^{\mu}_{\nu}X^{\nu}(\tau, \sigma) + a^{\mu}. \ \ \ \ \ (44)$

This symmetry group consists of consists of Lorentz transformations ${\Lambda^{\mu}_{\nu}}$ satisfying ${SO(D-1, 1)}$ algebra and ${a^{\mu}}$ transformations. This symmetry is manifest and can be read-off from (36) since the Lorentz indices are contracted in the correct way to obtain a Lorentz scalar. But to see it explicitly just note that ${X^{\mu}}$ are flat spacetime vectors. Under the transformation ${X^{\mu} \rightarrow X^{\prime \mu} = \Lambda^{\mu}_{\nu}X^{\nu}(\tau, \sigma) + a^{\mu}}$ we see that ${\partial_{\alpha}X^{\prime \mu} = \Lambda^{\mu}_{\nu}\partial_{\beta}X^{\nu}}$. Hence

$\displaystyle \eta_{\mu}\partial_{\alpha} X^{\prime \mu}\partial_{\beta}X^{\prime \nu} = = {\Lambda^{\mu}}_{\gamma} \eta_{\mu \nu} {\Lambda^{\nu}}_{\sigma}\partial_{\alpha} X^{\gamma} \partial_{\beta} X^{\sigma} = \eta_{\gamma \sigma} \partial_{\alpha} X^{\gamma} \partial_{\beta} X^{\sigma}, \ \ \ \ \ (45)$

where we used ${\Lambda^{\mu}_{\gamma} \eta_{\mu \nu} \Lambda^{\nu}_{\sigma} = \eta_{\gamma \sigma}}$.

From the perspective of the worldsheet theory, the Nambu-Goto action is a 2-dimensional field theory of scalar fields ${X^{\mu}(\tau, \sigma)}$, and Poincare invariance is in fact an internal symmetry.

Diffeomorphism invariance The Nambu-Goto action is also invariant under diffeomorphism transformations, or reparamterisation of the coordinates, which we’ve already observed by the very nature of how we construct (36) such that ${X^{\prime \mu} (\tau^{\prime}, \sigma^{\prime}) = X^{\mu}(\tau, \sigma)}$.

6.3. Concluding remarks

The main issue with the action (36) is the presence of the square root, which complicates matters when we attempt to quantise the theory or take the massless limit ${m \rightarrow 0}$. That is why, analogous to the case of the relativistic point particle, we’ll want to get rid of this square root and construct a classically equivalent action. This is known as the Polyakov action and, following the progression in Polchinski, it is the topic of the next note.

In the meantime, I want to point out that there is still much more to be learned about the Nambu-Goto action and its dynamics. There are some quite famous and important results, which are not covered in Polchinski’s textbook. It is notable, for instance, that from an analysis of the worldsheet momentum densities

$\displaystyle P^{\alpha}_{\mu} = \frac{\partial \mathcal{L}}{\partial \partial_{\alpha}X^{\mu}} \ \ \ \ \ (46)$

we can evaluate the components of the canonical momenta explicitly

$\displaystyle \Pi = P_{\mu}^{\sigma} = \frac{\partial L}{\partial X^{\prime \mu}} = \frac{\partial}{\partial X^{\prime \mu}} (-T \sqrt{(\dot{X} \cdot X^{\prime})^{2} - (\dot{X}^2)(X^{\prime})^2})$

$\displaystyle = -\frac{T}{2}[(\dot{X} \cdot X^{\prime})^{2} - (\dot{X}^2)(X^{\prime})^2]^{-1/2} [2(\dot{X} \cdot X^{\prime})\dot{X}_{\mu} - 2 \dot{X}^{2}X_{\mu}^{\prime}]$

$\displaystyle = \frac{(\dot{X} \cdot X^{\prime})\dot{X}_{\mu} - \dot{X}^{2}X_{\mu}^{\prime}}{\sqrt{(\dot{X} \cdot X^{\prime})^{2} - (\dot{X}^2)(X^{\prime})^2}}, \ \ \ \ \ (47)$

and,

$\displaystyle P_{\mu}^{\tau} = \frac{\partial L}{\partial \dot{X}^{\mu}} = \frac{\partial}{\partial \dot{X}^{\mu}}(-T \sqrt{(\dot{X} \cdot X^{\prime})^{2} - (\dot{X}^2)(X^{\prime})^2})$

$\displaystyle = -\frac{T}{2}[(\dot{X} \cdot X^{\prime})^{2} - (\dot{X}^2)(X^{\prime})^2]^{-1/2} [2(\dot{X} \cdot X^{\prime})X_{\mu}^{\prime} - 2 X^{\prime} \dot{X}_{\mu}^{2}]$

$\displaystyle = \frac{(\dot{X} \cdot X^{\prime})X_{\mu}^{\prime} - X^{\prime 2} \dot{X}_{\mu}}{\sqrt{(\dot{X} \cdot X^{\prime})^{2} - (\dot{X}^2)(X^{\prime})^2}}. \ \ \ \ \ (48)$

From this analysis, we can obtain an equation that we can interpret as the generalised momentum flow of the particle worldline. This helps give a bit more insight and intuition into the direct analogue we’ve established between point particle theory and the theory of strings. Furthermore, by imposing the appropriate boundary conditions we can show for the equations of motion that

$\displaystyle \partial_{\alpha}P^{\alpha}_{\mu} = 0, \ \ \ \ \ (49)$

which, given the equations for the worldsheet momentum, we find the 2-dimensional wave equation given the choice of coordinates ${\dot{X} \cdot X^{\prime} = 0}$, ${\dot{X}^2 = -1}$, and ${X^{\prime} = 1}$. In the same analysis, we can find very important conditions such as the Virasaro constraints that govern the dynamics of the string.

There is also much more that can be studied: boundary conditions and motion of the string endpoints, which provides a first introduction to D-branes; tension and energy of the stretched string; transverse velocity; among other interesting topics. All of this of course comes up also in our study of the Polyakov action; but for the interested reader, Zwiebach’s textbook referenced at the outset covers all of these topics in detail in the context of the Nambu-Goto action.

# Notes on string theory #2: The relativistic point particle (pp. 9-11)

1. Introduction

In Chapter 1 of Polchinski’s textbook, we start with a discussion on the relativistic point particle (pp. 9-11).

String theory proposes that elementary particles are not pointlike, but rather 1-dimensional extended objects (i.e., strings). In fact, string theory (both the bosonic string in Volume 1 of Polchinski and the superstring that comprises much of Volume 2) can be seen as a special generalisation of point particle theory. But the deeper and more modern view is not one that necessarily begins with point particles and then strings, instead the story begins with branes. In that a number of features of string theory are shared by the point particle – as we’ll see in a later note, the point particle can be obtained in the limit the string collapses to a point – the bigger picture is that both of these objects can be considered as special cases of a p-brane.

We refer to p-branes as p-dimensional dynamical objects that have mass and can have other familiar attributes such as charge. As a p-brane moves through spacetime, it sweeps out a latex (p+1)-dimensional volume called its worldvolume. In this notation, a 0-brane corresponds to the case where p = 0. It simply describes a point particle that, as we’ll discuss in this note, traces out a worldline as it propagates through spacetime. A string (whether fundamental or solitonic) corresponds to the p = 1 case, and this turns out to be a very special case of p-branes (for many reasons we’ll learn in following notes). Without getting too bogged down in technical details that extend well beyond the current level of discussion, it is also possible to consider higher-dimensional branes. Important is the case for p = 2, which are 2-dimensional branes called membranes. In fact, the etymology for the word ‘brane’ can be viewed as derivative from `membrane’. As a physical object, a p-brane is actually a generalisation of a membrane such that we may assign arbitrary spatial dimensions. So, for the case ${p \geq 2}$, these are p-branes that appear in string theory as solitons in the corresponding low energy effective actions of various string theories (in addition to 0-branes and 1-branes).

In Type IIA and Type IIB string theories, which again is a subject of Volume 2, we see that there is entire family of p-brane solutions. From the viewpoint of perturbative string theory, which is the primary focus of Volume 1, solitons as p-branes are strictly non-perturbative objects. (There are also other classes of branes, such as Dp-branes that we’ll come across soon when studying the open string. The more complete picture of D/M-brane physics, including brane dynamics, is anticipated to be captured by M-theory. This is a higher dimensional theory that governs branes and, with good reason, is suspected to represent the non-perturbative completion of string theory).

In some sense, one can think of there being two equivalent ways to approach the idea of p-branes: a top-down higher dimensional view, or from the bottom-up as physical objects that generalise the notion of a point particle to higher dimensions. But given an introductory view of p-branes, perhaps it becomes slightly more intuitive why in approaching the concept of a string in string theory we may start (as Polchinski does) with a review of point particle theory. Indeed, it may at first seem odd to model the fundamental constituents of matter as strings. Indeed, it could seem completely arbitrary and therefore natural to ask, why not something else? But what is often missed, especially in popular and non-technical physics literature, is the natural generalising logic that leads us to study strings in particular. These are remarkable objects with remarkable properties, and what Polchinski does so well in Volume 1 is allow this generalising logic to come out naturally in the study of the simplest string theory: bosonic string theory.

In this note, we will construct the relativistic point particle action as given in p.10 (eqn. 1.2.2) and then work through the proceeding discussion in pages 10-11. The quantisation of the point particle is mentioned several pages later in the textbook, so we’ll address that topic then. In what follows, I originally also wanted to include notes on the superparticle and its superspace formulation (i.e., the inclusion of fermions to the point particle theory of bosons), as well as introduce other advanced topics; but I reasoned it is best to try to keep as close to the textbook as possible. The only exception to this rule is that, at the end of this note, we’ll finish by quickly looking at the p-brane action.

2. Relativistic point particle

Explanation of the action for a relativistic point particle as given in Polchinski (eqn. 1.2.2) is best achieved through its first-principle construction. So let us consider the basics of constructing the theory for a relativistic free point particle.

2.1. Minkowski space

As one may recall from studying Einstein’s theory of relativity, spacetime may be modelled by D-dimensional Minkowski space ${\mathbb{M}^D}$. In the abstract, the basic idea is to consider two (distinct) sets E and ${\vec{E}}$, where E is a set of points (with no given structure) and ${\vec{E}}$ is a vector space (of free vectors) acting on the set E. We view the elements of ${\vec{E}}$ as forces acting on points in E, which we in turn think of as physical particles. Applying a force (free vector) ${X \in \vec{E}}$ to a point ${P \in E}$ results in a translation. In other words, the action of a force X is to move every point P to the point ${P + X \in E}$ by translation that corresponds to X viewed as a vector.

In physics, the set E is viewed as the D-dimensional affine space ${\mathbb{M}^D}$, and then ${\vec{E}}$ is the associated D-dimensional vector space ${\mathbb{R}^{1,D-1}}$ defined over the field of real numbers. The choice to model spacetime as an affine space is quite natural, given that an affine space has no preferred or distinguished origin and, of course, the spacetime of special relativity possesses no preferred origin.

As the vectors ${X \in \mathbb{R}^{1,D-1}}$ do not naturally correspond to points ${P \in \mathbb{M}}$, but rather as displacements relating a point P to another point Q, we write ${X = \vec{PQ}}$. The points can be defined to be in one-to-one correspondence with a position vector such that ${\vec{X}_P = \vec{OP}}$, with displacements then defined by the difference ${\vec{PQ} = \vec{OQ} - \vec{OP}}$. The associated vector space possesses a zero vector ${\vec{0} \in \mathbb{R}^{1,D-1}}$, which represents the neutral element of vector addition. We can also use the vector space ${\mathbb{R}^{1,D-1}}$ to introduce linear coordinates on ${\mathbb{M}^{D}}$ by making an arbitrary choice of origin as the point ${O \in \mathbb{M}^D}$.

The elements or points ${P,Q,..., \in \mathbb{M}^D}$ are events, and they combine a moment of time with a specified position. With the arbitrary choice of origin made, we can refer to these points in Minkowski space in terms of their position vectors such that the components ${X^{\mu} = (X^0, X^i) = (t, \vec{X})}$, with ${\mu = 0,..., D-1, i = 1,...,D-1}$ of vectors ${X \in \mathbb{R}^{1,D-1}}$ correspond to linear coordinates on ${\mathbb{M}^D}$. The coordinates ${X^{0}}$ is related to the time t, which is measured by an inertial or free falling observer by ${X^0 =ct}$, with the c the fundamental velocity. The ${X^i}$ coordinates, which are combined into a (D-1)-component vector, parameterise space (from the perspective of the inertial observer).

It is notable that a vector ${X}$ has contravariant coordinates ${X^{\mu}}$ and covariant coordinates ${X_{\mu}}$ which are related by raising and lowering indices such that ${X_{\mu} = \eta_{\mu \nu}X^{\nu}}$ and ${X^{\mu} = \eta^{\mu \nu}x_{\nu}}$.

We still need to equip a Lorentzian scalar product. In the spacetime of special relativity, the vector space ${\mathbb{R}}$ is furnished with the scalar product (relativistic distance between events)

$\displaystyle \eta_{\mu \nu} = X^{\mu}X_{\mu} = -t^2 + \vec{X}^2 \begin{cases} <0 \ \text{for timelike disrance} \\ =0 \ \text{for lightlike distance} \\ >0 \ \text{for spacelike distance} \end{cases} \ \ (1)$

with matrix

$\displaystyle \eta = (\eta_{\mu \nu}) = \begin{pmatrix} - 1 & 0 \\ 0 & 1_{D-1} \end{pmatrix}, \ \ (2)$

where we have chosen the mostly plus convention. To make sense of (1), since the Minkowski metric (2) is defined by an indefinite scalar product, the distance-squared between events can be positive, zero or negative. This carries information about the causal structure of spacetime. If ${X = \vec{PQ}}$ is the displacement between two events, then these events are called time-like, light-like or space-like relative to each other, depending on X. The zeroth component of X then carries information about the time of the event P as related to Q relative to a given Lorentz frame: P is after Q (${X^0 > Q}$), or simultaneous with Q (${X^0 = 0}$), or earlier than Q (${X^0 < 0}$).

2.2. Lorentz invariance and the Poincaré group

Let’s talk more about Lorentz invariance and the Poincaré group. As inertial observers are required to use linear coordinates which are orthonormal with respect to the scalar product (1), these orthonormal coordinates are distinguished by the above standard form of the metric. It is of course possible to use other curvilinear coordinate systems, such as spherical or cylindrical coordinates. Given the standard form of the metric (2), the most general class of transformations which preserve its form are the Poincaré group, which represents the group of Minkowski spacetime isometries.

The Poincaré group is a 10-dimensional Lie group. It consists of 4 translations along with the Lorentz group of 3 rotations and 3 boosts. As a general review, let’s start with the Lorentz group. This is the set of linear transformations of spacetime that leave the Lorentz interval unchanged.

From the definitions in the previous section, the line element takes the form

$\displaystyle ds^2 = \eta_{\mu \nu}dX^{\mu}dX^{\nu} = - dt^2 + d\vec{X}^2. \ \ (3)$

For spacetime coordinates defined in the previous section, the Lorentz group is then defined to be the group of transformations ${X^{\mu} \rightarrow X^{\prime \mu}}$ leaving the relativistic interval invariant. Assuming linearity (we will not prove linearity here, with many proofs easily accessible), define a Lorentz transformation as any real linear transformation ${\Lambda}$ such that

$\displaystyle X^{\mu} \rightarrow X^{\prime \mu} = \Lambda^{\mu}_{\nu}X^{\nu} \ \ (4)$

with

$\displaystyle \eta_{\mu \nu} dX^{\prime \mu} dX^{\prime \nu} = \eta_{\mu \nu} dX^{ \mu} dX^{\nu}, \ \ (5)$

ensuring from (1) that

$\displaystyle X^{\prime 2} = X^{2}, \ \ (6)$

which, for arbitrary X, requires

$\displaystyle \eta_{\mu \nu} = \eta_{\alpha \beta} \Lambda^{\alpha}_{\mu} \Lambda^{\beta}_{\nu}. \ \ (7)$

Note that ${\Lambda = (\Lambda^{\mu}_{\nu})}$ is an invertible ${D \times D}$ matrix. In matrix notation (7) can be expressed as

$\displaystyle \Lambda^T \eta \Lambda = \eta. \ \ (8)$

Matrices satisfying (8) contain rotations together with Lorentz boosts, which relate inertial frames travelling a constant velocity relative to each other. The Lorentz transformations form a six-dimensional Lie group, which is the Lorentz group O(1,D-1).

For elements ${\Lambda \in O(1, D-1)}$ taking the determinant of (8) gives

$\displaystyle (\det \Lambda)^2 = 1 \implies \det \Lambda = \pm 1. \ \ (9)$

By considering the ${\Lambda^0_0}$ component we also find

$\displaystyle (\Lambda^0_0)^2 = 1 + \Sigma_i (\Lambda^0_i)^2 \geq 1 \Rightarrow \Lambda^0_0 \geq 1 \ \text{or} \ \Lambda^0_0 \leq -1. \ \ (10)$

So, the Lorentz group has four components according to the signs of ${\det \Lambda}$ and ${\Lambda^0_0}$. The matrices with ${\det \Lambda = 1}$ form a subgroup SO(1,D-1) with two connected components as given on the right-hand side of (10). The component containing the unit matrix ${1 \in O(1,D-1)}$ is connected and as ${SO_0(1,D-1)}$.

We may also briefly consider translations of the form

$\displaystyle X^{\mu} \rightarrow X^{\prime \mu} = X^{\mu} + a^{\mu}, \ \ (11)$

where ${a = (a^{\mu}) \in \mathbb{R}^{1, D-1}}$. Translations form a group that can be parametrised by the components of the translation vector ${a^{\mu}}$.

As mentioned, the Poincaré group is then the complete spacetime symmetry group that combines translations with Lorentz transformations. For a Lorentz transformation ${\Lambda}$ and a translation ${a}$ the combined transformation ${(\Lambda, a)}$ gives

$\displaystyle X^{\mu} \rightarrow X^{\prime \mu} = \Lambda^{\mu}_{\nu} X^{\nu} + a^{\mu}. \ \ (12)$

These combined transformations form a group since

$\displaystyle (\Lambda_2, a_2)(\Lambda_1, a_1) = (\Lambda_2 \Lambda_2, \Lambda_2 a_1 + a_2), \ (\Lambda, a)^{-1} = (\Lambda^{-1}, -\Lambda^{-1}a). \ \ (13)$

Since Lorentz transformations and translations do not commute, the Poincaré group is not a direct product. More precisely, the Poincaré group is the semi-direct product of the Lorentz and translation group, ${IO(1,D-1) = O(1,D-1) \propto \mathbb{R}^D}$.

2.3. Action principle

We now look to construct an action for the relativistic point particle (initially following the discussion in [Zwie09] as motivation).

The classical motion of a point particle as it propagates through spacetime is described by a geodesic on the spacetime. As Polchinski first notes, we can of course describe the motion of this particle by giving its position in terms of functions of time ${X(t) = (X^{\mu}(t)) = (t, \vec{X}(t))}$. For now, we may also consider some arbitrary origin and endpoint ${(ct_f, \vec{X}_{f})}$ for the particle’s path or what is also called its worldline. We also know from the principle of least action that there are many possible paths between these points.

It should be true that for any worldline all Lorentz observers compute the same value for the action. Let ${\mathcal{P}}$ denote one such worldline. Then we may use the proper time as an Lorentz invariant quantity to describe this path. Moreover, from special relativity one may recall that the proper time is a Lorentz invariant measure of time. If different Lorentz observers will record different values for the time interval between the two events along ${\mathcal{P}}$, then we instead imagine that attached to the particle is a clock. The proper time is therefore the time elapsed between the two events on that clock, according to which all Lorentz observers must agree on the amount of elapsed time. This is the basic idea, and it means we want an action of the worldline ${\mathcal{P}}$ that is proportional to the proper time.

To achieve this, we first recall the invariant interval for the motion of a particle

$\displaystyle - ds^2 = -c^2 dt^2 + (dX^1)^2 + (dX^2)^2 + (dX^3)^2, \ \ (14)$

in which, from special relativity, the proper time

$\displaystyle -ds^2 = -c^2 dt_f \rightarrow ds = c dt_f \ \ (15)$

tells us that for timelike intervals ds/c is the proper time interval. It follows that the integral of (ds/c) over the worldline ${\mathcal{P}}$ gives the proper time elapsed on ${\mathcal{P}}$. But, if the proper time gives units of time, we still needs units of energy or units of mass times velocity-squared to ensure we have the full units of action (recall that for any dynamical system the action has units of energy times time, with the Lagrangian possessing units of energy). We also need to ensure that we preserve Lorentz invariance in the process of building our theory. One obvious choice is m for the rest mass of the particle, with c for the fundamental velocity in relativity. Then we have an overall multiplicative factor ${mc^2}$ that represents the the rest energy of the particle. As a result, the action takes the tentative form ${mc^2 (ds/c) = mc ds}$. This should make some sense in that ${ds}$ is just a Lorentz scalar, and we have the factor of relativity we expect. We also include a minus sign to ensure the follow integrand is real for timelike geodesics.

$\displaystyle S = -mc \int_{\mathcal{P}} ds. \ \ (16)$

A good strategy now is to find an integral of our Lagrangian over time – say, ${t_i}$ and ${t_f}$ which are world-events that we’ll take to define our interval – because it will enable use to establish a more satisfactory expression that includes the values of time at the initial and final points of our particle’s path. If we fix a frame – which is to say if we choose the frame of a particular Lorentz observer – we may express the action (16) as the integral of the Lagrangian over time. To achieve this end, we must first return to our interval (14) and relate ${ds}$ to ${dt}$,

$\displaystyle -ds^2 = -c^2 dt^2 + (dX^1)^2 + (dX^2)^2 + (dX^3)^2$

$\displaystyle ds^2 = c^2 dt^2 - (dX^1)^2 - (dX^2)^2 - (dX^3)^2$

$\displaystyle ds^2 = [c^2 - \frac{(dX^1)^2}{dt} - \frac{(dX^2)^2}{dt} - \frac{(dX^3)^2}{dt}] dt^2$

$\displaystyle \implies ds^2 = (c^2 - v^2) dt^2$

$\displaystyle \therefore ds = \sqrt{c^2 - v^2} dt. \ \ (17)$

With this relation between ${ds}$ and ${dt}$, in the fixed frame the point particle action becomes

$\displaystyle S = -mc^{2} \int_{t_{i}}^{t_{f}} dt \sqrt{1 - \frac{v^{2}}{c^{2}}}, \ \ (18)$

with the Lagrangian taking the form

$\displaystyle L = -mc^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}}. \ \ (19)$

This Lagrangian gives us a hint that it is correct as its logic breaks down when the velocity exceeds the speed of light ${v > c}$. This confirms the definition of the proper time from special relativity (i.e., the velocity should not exceed the speed of light for the proper time to be a valid concept). In the small velocity limit ${v << c}$, on the other hand, when we expand the square root (just use binomial theorem to approximate) we see that it gives

$\displaystyle L \simeq -mc^2 (1 - \frac{1}{2}\frac{v^2}{c^2}) = - mc^2 + \frac{1}{2}m v^2. \ \ (20)$

returning similar structure for the kinetic part of the free non-relativistic particle, with (${-mc^2}$) just a constant.

2.4. Canonical momentum and Hamiltonian

We will discuss the canonical momentum of the point particle again in a future note on quantisation; but for the present form of the action it is worth highlighting that we can also see the Lagrangian (19) is correct by computing the momentum ${\vec{p}}$ and the Hamiltonian.

For the canonical momentum, we take the derivative of the Lagrangian with respect to the velocity

$\displaystyle \vec{p} = \frac{\partial L}{\partial \vec{v}} = -mc^{2}(-\frac{\vec{v}}{c^{2}})\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{m\vec{v}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}. \ \ (21)$

Now that we have an expression for the relativistic momentum of the particle, let us consider the Hamiltonian. The Hamiltonian may be written schematically as ${H = \vec{p} \cdot \vec{v} - L}$. All we need to do is make the appropriate substitutions,

$\displaystyle H = \frac{m\vec{v}^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}. \ \ (22)$

The Hamiltonian should make sense. Notice, if we instead write the result in terms of the particle’s momentum (rather than velocity) by inverting (22), we find an expression in terms of the relativistic energy ${\frac{E^{2}}{c^{2}} - \vec{p} \cdot \vec{p} = m^{2}c^{2}}$. This is a deep hint that we’re on the right track, as it suggests quite clearly that we’ve recovered basic relativistic physics for a point-like object.

3. Reparameterisation invariance

An important property of the action (16) is that it is invariant under whatever choice of parameterisation we might choose. This makes sense because the invariant length ds between two points on the particle’s worldline does not depend on any parameterisation. We’ve only insisted on integrating the line element, which, if you think about it, is really just a matter of adding up all of the infinitesimal segments along the worldline. But, typically, a particle moving in spacetime is described by a parameterised curve. As Polchinski notes, it is generally best to introduce some parameter and then describe the motion in spacetime by functions of that parameter.

Furthermore, how we parameterise the particle’s path will govern whether, for the classical motion, the path is one that extremises the invariant distance ds as a minimum or maximum. Our choice of ${\tau}$-parameterisation is such that the invariant length ds is given by

$\displaystyle ds^2 = -\eta_{\mu \nu}(X) dX^{\mu} dX^{\nu}, \ \ (23)$

then the choice of worldline parameter ${\tau}$ is considered to be increasing between some initial point ${X^{\mu} (\tau_i)}$ and some final point ${X^{\mu}(\tau_f)}$. So the classical paths are those which maximise the proper time. It also means that the trajectory of the particle worldline is now described by the coordinates ${X^{\mu} = X^{\mu}(\tau)}$. As a result, the space of the theory can now be updated such that ${X^{\mu}(\tau) \in \mathbb{R}^{1, D-1}}$ with ${\mu, \nu = 0,...,D-1}$.

In the use of ${\tau}$ parameterisation, an important idea is that time is in a sense being promoted to a dynamical degree of freedom without it actually being a dynamical degree of freedom. We are in many ways leveraging the power of gauge symmetry, with our choice of parameterisation enabling us to treat space and time coordinates on equal footing. The cost by trading a less symmetric description for a more symmetric one is that we pick up redundancies.

Given the previous preference of background spacetime geometry to be Minkowski, recall the metric

$\displaystyle \eta_{\mu \nu} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \ \ (24)$

such that for the integrand ds we now use

$\displaystyle -\eta_{\mu \nu}(X) dX^{\mu} dX^{\nu} = -\eta_{\mu \nu}(X) \frac{dX^{\mu}(\tau)}{d\tau} \frac{dX^{\nu}(\tau)}{d\tau} d\tau^2. \ \ (25)$

Therefore, the action (16) may be updated to take the form

$\displaystyle S_{pp} = -mc \int_{\tau_i}^{\tau_f} d\tau \ \sqrt{-\eta_{\mu \nu} \dot{X}^{\mu} \dot{X}^{\nu}} \ \ (26)$

with ${\dot{X}^{\mu} \equiv dX^{\mu}(\tau) / d\tau}$.

Setting ${c = 1}$, notice (26) is precisely the action (eqn. 1.2.2) in Polchinski. This is the simplest action for a relativistic point particle with manifest Poincaré invariance that does not depend on the choice of parameterisation.

How do we interpret this form of the action? In the exercise to obtain (26) we have essentially played the role of a fixed observer, who has calculated the action using some parameter ${\tau}$. The important question is whether the value of the action depends on this choice of parameter. Polchinski comments that, in fact, it is a completely arbitrary choice of parameterisation. This should make sense because, again, the invariant length ds on the particle worldline ${\mathcal{P}}$ should not depend on how the path is parameterised.

Proposition 1 The action (26) is reparameterisation invariant such that if we replace ${\tau}$ with the parameter ${\tau^{\prime} = f(\tau)}$, where f is monotonic, we obtain the same value for the action.

Proof: Consider the following reparameterisation of the particle’s worldline ${\tau \rightarrow \tau^{\prime} = f(\tau)}$. Then we have

$\displaystyle d\tau \rightarrow d\tau^{\prime} = \frac{\partial f}{\partial \tau}d\tau, \ \ (27)$

implying

$\displaystyle \frac{dX^{\mu}(\tau^{\prime})}{d\tau} = \frac{dX^{\mu}(\tau^{\prime})}{d\tau^{\prime}}\frac{d\tau^{\prime}}{d\tau} = \frac{dX^{\mu}(\tau^{\prime})}{d\tau^{\prime}} \frac{\partial f(\tau)}{\partial \tau}. \ \ (28)$

Plugging this into the action (26) we get

$\displaystyle S^{\prime} = -mc \int_{\tau_i}^{\tau_f} d\tau^{\prime} \ \sqrt{\frac{dX^{\mu}(\tau^{\prime})}{d\tau^{\prime}} \frac{dX_{\mu}(\tau^{\prime})}{d\tau^{\prime}}}$

$\displaystyle = -mc \int_{\tau_i}^{\tau_f} \frac{\partial f}{\partial \tau} \ d\tau \ \sqrt{\frac{dX^{\mu}}{d\tau} \frac{dX_{\mu}}{d\tau} (\frac{\partial f}{\partial tau})^{-2}}$

$\displaystyle = -mc \int_{\tau_i}^{\tau_f} (\frac{\partial f}{\partial \tau})(\frac{\partial f}{\partial \tau})^{-1} \ d\tau \ \sqrt{\frac{dX^{\mu}}{d\tau} \frac{dX_{\mu}}{d\tau}}$

$\displaystyle = -mc \int_{\tau_i}^{\tau_f} d\tau \ \sqrt{\frac{dX^{\mu}(\tau)}{d\tau} \frac{dX_{\mu}(\tau)}{d\tau}}. \ \ (29)$

$\Box$

This ends the proof. So we see the value of the action does not depend on the choice of parameter; indeed, the choice is arbitrary.

As alluded earlier in this section, reparameterisation invariance is a gauge symmetry. In some sense, this is not even an honest symmetry; because it means that we’ve introduced a redundancy in our description, as not all degrees of freedom ${X^{\mu}}$ are physically meaningful. We’ll discuss this more in the context of the string (an example of such a redundancy appears in the study of the momenta).

4. Equation of motion for ${S_{pp}}$

To obtain (eqn. 1.2.3), Polchinski varies the action (26) and then integrates by parts. For simplicity, let us temporarily maintain ${c = 1}$. Varying (26)

$\displaystyle \delta S_{pp} = -m \int d\tau \delta (\sqrt{-\dot{X}^{\mu}\dot{X}_{\mu}}) \ \ (30)$

$\displaystyle = -m \int d\tau \frac{1}{2}(-\dot{X}^{\mu}\dot{X}_{\mu})^{-1/2}(-\delta \dot{X}^{\mu}\dot{X}_{\mu}), \ \ (31)$

then from the last term we pick up a factor of 2 leaving

$\displaystyle = -m \int d\tau (-\dot{X}^{\mu}\dot{X}_{\mu})^{-1/2} + (-\dot{X}^{\mu}\delta \dot{X}_{\mu}). \ \ (32)$

Next, we make the substitution ${u^{\mu} = \dot{X}^{\mu}(-\dot{X}^{\nu}\dot{X}_{\nu})^{-1/2}}$ such that

$\displaystyle \delta S_{pp} = -m \int d\tau (-u_{\mu})\delta \dot{X}^{\mu}. \ \ (33)$

And now we integrate by parts, which shifts a derivative onto u using the fact we can commute the variation and the derivative ${\delta \dot{X}^{\mu} = \delta d / d\tau X^{\mu} = d/d\tau \delta X^{\mu}}$. We also drop the total derivative term that we obtain in the process

$\displaystyle \delta S_{pp} = -m \int d\tau \frac{d}{d\tau} (-u_{\mu}\delta X^{\mu}) - m \int d\tau \dot{u}_{\mu} \delta X^{\mu}, \ \ (34)$

which gives the correct result

$\displaystyle \delta S_{pp} = -m \int d\tau \dot{u}_{\mu}\delta X^{\mu}. \ \ (34)$

As Polchinski notes, the equation of motion ${\dot{u}^{\mu} = 0}$ describes the free motion of the particle.

With the particle mass m being the normalisation constant, we can also take the non-relativistic limit to find (exercise 1.1). Returning to (26), one way to do this is for ${\tau}$ to be the proper time, then, as before (reinstating c for the purpose of example)

$\displaystyle \dot{X}^{\mu}(\tau) = c \frac{dt}{d\tau} + \frac{d\vec{X}^{\mu}(\tau)}{d\tau} \ \ (35)$

so that we may define the quantity ${\gamma = (1 - v^2/c^2)^{-1/2}}$. Then, in the non-relativistic limit where ${v << c}$ we have ${dt/d\tau = \gamma = 1 + \mathcal{O}(v^2/c^2)}$. It follows

$\displaystyle \dot{X}^{\mu}\dot{X}_{\mu} = -c^2 + \mid \vec{v} \mid^2 + \mathcal{O}(v^2/c^2), \ \ (36)$

with ${\vec{v}}$ a spatial vector and we define the norm ${\mid \vec{v} \mid \equiv v}$. Now, equivalent as with the choice of static gauge, the action to order ${v/c}$ takes the form

$\displaystyle S_{pp} \approx -mc \int dt \sqrt{c^2 -\mid \vec{v} \mid^2}, \ \ (37)$

where we now taylor expand to give

$\displaystyle S_{pp} \approx -mc \int (1 - \frac{1}{2}\frac{\mid \vec{v} \mid^2}{c^2}) \ \ (38)$

Observe that we now have a time integral of a term with classical kinetic structure minus a potential-like term (actually a total time derivative) that is an artefact of the relative rest energy

$\displaystyle S_{pp} \approx \int dt \ (\frac{1}{2}m\mid \vec{v} \mid^2 - mc^2). \ \ (39)$

5. Deriving ${S_{pp}^{\prime}}$(eqn. 1.2.5)

The main problem with the action (18) and equivalently (26) is that, when we go to quantise this theory, the square root function in the integrand is non-linear. Analogously, we will find a similar issue upon constructing the first-principle string action, namely the Nambu-Goto action. Additionally, in our study of the bosonic string, we will be interested firstly in studying massless particles. But notice that according to the action (26) a massless particle would be zero.

What we want to do is rewrite ${S_{PP}}$ in yet another equivalent form. To do this, we add an auxiliary field so that our new action takes the form

$\displaystyle S_{pp}^{\prime} = \frac{1}{2} \int d \tau (\eta^{-1} \dot{X}^{\mu} \dot{X}_{\mu} - \eta m^2), \ \ (40)$

where we define the tetrad ${\eta (\tau) = (- \gamma_{\tau \tau} (\tau))^{\frac{1}{2}}}$. The independent worldline metric ${\gamma_{\tau \tau}(\tau)}$ that we’ve introduce as an additional field is, in a sense, a generalised Lagrange multiplier. For simplicity we can denote this additional field ${e(\tau)}$ so that we get the action

$\displaystyle S_{pp}^{\prime} = \frac{1}{2} \int d\tau (e^{-1} \dot{X}^{2} - em^{2}), \ \ (41)$

where we have simplified the notation by setting ${\dot{X}^{2} = \eta_{\mu \nu}\dot{X}^{\mu}\dot{X}^{\nu}}$ and completely eliminated the square root. This is equivlant to what Polchinski writes in (eqn.1.2.5). The structure of (41) may look familiar, as it reads like a worldline theory coupled to 1-dimensional gravity (worth checking and playing with).

To see that ${S_{pp}^{\prime}}$ is classically equivalent (on-shell) to ${S_{pp}}$, we first consider its variation with respect to ${e(\tau)}$

$\displaystyle \delta S_{pp}^{\prime} = \frac{1}{2}\delta \int d\tau (e^{-1} \dot{X}^{2} - m^2 e)$

$\displaystyle = \frac{1}{2} \int d\tau (- \delta (\frac{1}{e})\dot{X}^{2} - \delta (m^{2} e))$

$\displaystyle = \frac{1}{2} \int d\tau (- \frac{1}{e^{2}}\dot{X}^{2} - m^{2}), \ \ (42)$

which results in the following field equations

$\displaystyle e^{2} = \frac{\dot{X}^{2}}{m^{2}}$

$\displaystyle \implies e = \sqrt{\frac{-\dot{X}^{2}}{m^{2}}} \ \ (43).$

This again aligns with Polchinski’s result (eqn. 1.2.7).

Proposition 2 If we substitute (43) back into (41), we recover the original ${S_{pp}}$ action (26).

Proof:

$\displaystyle S_{pp}^{\prime} = \frac{1}{2} \int d\tau [(-\frac{\dot{X}^2}{m^{2}})^{-1/2} \dot{X}^{2} - m^{2}(-\frac{\dot{X}^{2}}{m^{2}})^{1/2}]$

$\displaystyle = \frac{1}{2} \int d\tau [(-\frac{m^{2}}{\dot{X}^{2}})^{1/2} (\dot{X}^{2} - m^{2}(\frac{\dot{X}^{2}}{m^{2}})^{1/2})]$

$\displaystyle = \frac{1}{2} \int d\tau [(-\frac{m^{2}}{\dot{X}^{2}})^{1/2} (\dot{X}^{2} - m (- \dot{X}^{2})^{1/2})] \ \ (44)$

Recalling ${\dot{X}^{2} = \eta_{\mu \nu} \dot{X}^{\mu}\dot{X}^{\nu}}$, substitute for ${\dot{X}}$ in the square root on the right-hand side

$\displaystyle = \frac{1}{2} \int d\tau [(-\frac{m^{2}}{\dot{X}^{2}})^{1/2} \dot{X}^{2} - m (- \eta_{\mu \nu} \dot{X}^{\mu}\dot{X}^{\nu})^{1/2}. \ \ (45)$

For the first term we clean up with a bit of algebra. From complex variables recall ${i^{2} = -1}$.

$\displaystyle (-\frac{m^{2}}{\dot{X}^{2}})^{1/2} \dot{X}^{2} = (-1)(-1) -(\frac{m^{2}}{\dot{X}^{2}})^{1/2} \dot{X}^{2}$

$\displaystyle = -(-\frac{m^{2}}{\dot{X}^{2}})^{1/2} i^{2} \dot{X}^{2}$

$\displaystyle = -(-\frac{m^{2}}{\dot{X}^{2}} i^{4} \dot{X}^{2})^{1/2}$

$\displaystyle = -(-m^{2}i^{4}\dot{X}^{2})^{1/2} = -m (-i^{4}\dot{X}^{2})^{1/2}. \ \ (46)$

As ${i^{4} = 1}$, it follows ${-m(i^{4}\dot{X}^{2})^{1/2} = -m (-\dot{X}^{2})^{1/2}}$. Now, substitute for ${\dot{X}^{2}}$ and we find ${-m (-\eta_{\mu \nu}\dot{X}^{\mu}\dot{X}^{\nu})^{1/2}}$ giving

$\displaystyle S_{pp}^{\prime} = \frac{1}{2} \int d\tau [-m(- \eta_{\mu \nu}\dot{X}^{\mu}\dot{X}^{\nu})^{1/2} - m (- \eta_{\mu \nu} \dot{X}^{\mu}\dot{X}^{\nu})^{1/2}$

$\displaystyle = -m \int d\tau (- \eta_{\mu \nu}\dot{X}^{\mu}\dot{X}^{\nu})^{1/2} = S_{pp} \ \ (47).$

$\Box$

This ends the proof, demonstrating that ${S_{pp}}$and ${S_{pp}^{\prime}}$ are classically equivalent.

It is also possible to show that, like with ${S_{pp}}$, the action ${S_{pp}^{\prime}}$ is both Poincaré invariant and reparameterisation invariant.

6. Generalising to Dp-branes

As an aside, and to conclude this note, we can generalise the action for a point particle (0-brane) to an action for a p-brane. It follows that a p-brane in a ${D \geq p}$ dimensional background spacetime can be described in such a way that the action becomes,

$\displaystyle S_{pb}= -T_p \int d\mu_p \ \ (48).$

The term ${T_p}$is one that will become more familiar moving forward, especially when we begin to discuss the concept of string tension. However, in the above action it denotes the p-brane tension, which has units of mass/volume. The ${d\mu_p}$ term is the ${(p + 1)}$-dimensional volume measure,

$\displaystyle d\mu_p = \sqrt{- \det G_{ab}} \ d^{p+1} \sigma, \ \ (49)$

where ${G_{ab}}$ is the induced metric, which, in the ${p = 1}$ case, we will understand as the worldsheet metric. The induce metric is given by,

$\displaystyle G_{ab} (X) = \frac{\partial X^{\mu}}{\partial \sigma^{a}} \frac{\partial X^{\nu}}{\partial \sigma^{b}} h_{\mu \nu}(X) \ \ \ a, b \equiv 0, 1, ..., p \ \ (50)$p>

A few additional comments may follow. As ${\sigma^{0} \equiv \tau}$, spacelike coordinates in this theory run as ${\sigma^{1}, \sigma^{2}, ... \sigma^{p}}$ for the surface traced out by the p-brane. Under ${\tau}$ reparameterisation, the above action may also be shown to be invariant.

7. Summary

To summarise, one may recall how in classical (non-relativistic) theory [LINK] the evolution of a system is described by its field equations. One can generalise many of the concepts of the classical non-relativistic theory of a point particle to the case of the relativistic point particle. Indeed, one will likely be familiar with how in the non-relativistic case the path of the particle may be characterised as a path through space. This path is then parameterised by time. On the other hand, in the case of the relativistic point particle, we have briefly reviewed how the path may instead be characterised by a worldline through spacetime. This worldline is parameterised not by time, but by the proper time. And, in relativity, we learn in very succinct terms how freely falling relativistic particles move along geodesics.

It should be understood that the equations of motion for the relativistic point particle are given by the geodesics on the spacetime. This means that one must remain cognisant that whichever path the particle takes also has many possibilities, as noted in an earlier section. That is, there are many possible worldlines between some beginning point and end point. This useful fact will be explicated more thoroughly later on, where, in the case of the string, we will discuss the requirement to sum over all possible worldsheets. Other lessons related to the point particle will also be extended to the string, and will help guide how we construct the elementary string action.

References

[Moh08] T. Mohaupt, Liverpool lectures on string theory [lecture notes].

[Pol07] J. Polchinski, An introduction to the bosonic string. Cambridge, Cambridge University Press. (2007).

[Wray11] K. Wray, An introduction to string theory [lecture notes].

[Zwie09] B. Zwiebach, A first course in string theory. Cambridge, Cambridge University Press. (2009).

# Notes on string theory #1: The non-relativistic string

The study of waves has far reaching applications throughout physics (and across the sciences). Fundamental physics is no exception. Essential for understanding waves is understanding oscillation, and as one will recall from classical mechanics a concept of fundamental importance in this regard is the harmonic oscillator. A simple harmonic oscillator can then also be generalised to quantum mechanical systems. Of course, coupled oscillators is another important generalisation (i.e., oscillators interacting with each other). Think, for example, of atoms in a crystal and the eventual discovery of the Debye theory of heat capacity.

The importance of understanding oscillation is true for the study of many other phenomena in nature: for sound waves, water waves, and even gravitational waves. In sound waves air molecules oscillate in the longitudinal direction (i.e., the coordinate direction the sound is travelling). In gravitation waves, the sort of waves predicted by General Relativity and other theories of gravity, we may think of oscillations in the very fabric of spacetime, similar to how electromagnetic waves are oscillations of the electromagnetic field that propagate through spacetime. In string theory, we will similarly study waves and their oscillations (for instance, we will find a general solution for the fields describing the centre of mass motion and oscillations of the string as it propagates through spacetime), and from this analysis we will discover some incredibly exciting results.

But in this note we begin with a much simpler story: the study of transverse waves travelling along a classical, non-relativistic string. The value in a such a review is, in my opinion, to refresh the connection with classical concepts in what is a lengthy journey of generalisation toward higher concepts. Or, at least this is how I like to approach and motivate string theory. (The same emphasis will become apparent in the next note on the relativistic point particle and p-branes). It will also help lay some groundwork intuition before studying the first-principle action of the bosonic string: namely the Nambu-Goto action, which of course will be relativistic in nature. Much of what is reviewed below will be generalised in the relativistic setting. What is nice is how, upon constructing the Nambu-Goto action for the relativistic string (as studied in the first pages of Polchinski’s textbook), we will show that in the non-relativistic limit we can recover (up to a total derivative) the action for the ordinary, classical vibrating string discussed in this note.

As this is a lightning review of a small selection of topics from classical and wave mechanics, for further reading see Taylor [1] – or, in the context of an introduction to string theory, see Chapter 4 in Zwiebach [2]. David Morin’s lecture notes as well as these course notes by Matt Jarvis are particularly good, to name a few.

## A stretched string with transverse oscillations

Consider a string stretched with fixed endpoints. For simplicity, we may use the $(x, y)$ plane and pin the string endpoints at $(0, 0)$ and $(a, 0)$ such that the string is stretched along the x-axis. The direction along the string is longitudinal (i.e., the x-coordinate direction), while the direction orthogonal to the string is transverse (i.e., the y-coordinate direction). When describing a transverse oscillation, the x-coordinate of any point on the string will not vary in time. Rather, the transverse displacements at any point will be described by the y-coordinate.

Two pieces of information are required in order to describe the classical mechanics of the homogeneous string: tension and mass per unit length.

Tension has units of force. This means that we can first write the tension in terms of [Energy / Length]. But energy has units of mass times velocity squared, so we can write the following equation

$T_0 : [T_0] = [\text{Force}] = [\text{Energy / Length}] = \frac{M}{L}[v^2]. \ \ (1)$

Denote the Mass / Length as $\mu_0$ such that (1) may be updated as $T_0 \approx \mu_0 v^2$, where the natural velocity $v = \sqrt{T_0 / \mu_0}$. The tension $T_0$ and the mass per unit length $\mu_0$ are dynamical parameters. The velocity will eventually prove to be the velocity of transverse waves.

As a continuation of the above reasoning, consider the next simplifying assumptions. Assume that the string is infinitesimally thin and completely flexible. Since we have not yet considered boundary conditions, let us change our previous assumption and now consider that this stretched homogeneous string extends infinitely in both directions. Next, consider exciting the string such that, for two nearby points separated by a displacement dx in the longitudinal direction, there is small transverse displacement dy.

It is worth noting that, in the non-relativistic case, if the string is stretched an infinitesimal amount dx, its tension will remain approximately constant. Whatever change in energy will be equal to the work done $T_0 dx$. For the total mass, there will be no change. But in the relativistic case, any increase in energy will of course correspond to a larger rest mass. Furthermore, equation (1) suggests for a relativistic string $T_0$ and $\mu_0$ may be expressed by the relation $T_0 = \mu_0 c^2$, where c is the canonical velocity in relativity. When we eventually study the relativistic string in detail, we find this relation to indeed be correct.

For a small dy (which is to say, if the slope dy / dx is small), then we can approximate that all points in the string move only in the transverse direction. This means we consider there to be no longitudinal motion. To make sense of this statement, consider a point on the string in transverse displacement.

Notice that the length of the hypotenuse equals

$\sqrt{dx^2 + dy^2} = dx \sqrt{1 + (\frac{dy}{dx})^2}$

$\approx dx(1 + \frac{1}{2}(\frac{dy}{dx})^2) = dx + dy\frac{1}{2}(\frac{dy}{dx}). \ \ (2)$

The length element that defines the farthest a given point can move longitudinally is generally different (indeed, smaller) than the length of the long side in the dx direction by dy(dy/dx)/2, which is only (dy/dx)/2 times as large as the transverse displacement dy. With the assumption that dy/dx is small, which is to say $\mid dy/dx \mid << 1$, the longitudinal motion can therefore be neglected. In fact, all points along this segment of string move only in the transverse direction. Hence, each point is considered to be labelled with a unique value of x. Again, given the string will stretch slightly, we can safely assume that the amount of mass in any given horizontal span stays essentially constant.

## Equations of motion

We want to calculate the equations of motion. The strategy is to invoke Newton’s law and write down the transverse $F = ma$ equation for the little piece of string in the span from x to x + dx (ignoring gravity for simplicity). To do this, we must first consider the forces acting on the string.

Let $T_1$ and $T_2$ be the tensions in the string at the end points. We define the angle $\theta_1$ at x and $\theta_2$ at x + dx. As shown in Figure 3, this infinitesimal segment of string is subject to opposing tension forces at its two ends. These tension forces act along the local tangent line to the string. As the working assumption is that the string displacement remains sufficiently small such that the tension does not vary in magnitude along the string, we may suppose the local tangent line to the string subtends angles $\theta_1$ and $\theta_2$ with the x-axis. Note that these angles are written as infinitesimal quantities because the string displacement is assumed to be infinitesimally small, which implies that the string is everywhere almost parallel with the x-axis (i.e., the string displacement is greatly exaggerated in Figure 3 for purpose of illustration). Precisely because the slope dy/dx is small, it is essentially equal to the $\theta$ angles.

The slope of the string is different at the points x and x + dx. A consequence of this difference in slope (however small) is that the tension changes direction, so there is a net force $F_{net}$.

Consider the net longitudinal or x-component of the tension force. We don’t need calculus here, just a simple reading of the force diagram shows

$F_{x} = T_2 \cos (\theta_1 + \theta_2) - T_1 \cos \theta_1. \ \ (3)$

Again, assuming the change in the angle is small at x and x + dx, then $F_x \approx 0$. Another way to look at this is to use the small angle approximation $\cos \theta \approx 1 - \theta^2 / 2$ that tells us that the longitudinal components of the tensions are equal to the tensions themselves (up to small corrections of order $\theta^2 \approx (dy / dx)^2$). And because there is no longitudinal motion (or because it is so small we can neglect it), the acceleration component in the longitudinal direction is zero and, as a consequence of this reasoning, the longitudinal forces must cancel. Therefore, we find $T_1 = T_2$.

Analysis of the transverse or y-components is obviously very different. The transverse components differ by a quantity that is first order in dy/dx. This difference cannot simply be neglected, because it causes the transverse acceleration.

For the y-component of the tension force, first recognise that the transverse force at x + dx is $T sin \ \theta_2$, which is approximately equal to T times the slope. So the upward force can be written as $Ty^{\prime}(x + dx)$. The downward force at x is then simply $-Ty^{\prime}(x)$. After a bit of calculus the net transverse force is shown to be

$F_{y} = T(y^{\prime}(x + dx) - y^{\prime}(x)) = T \frac{y^{\prime}(x + dx) - y^{\prime}(x)}{dx} \equiv T dx \frac{d^2 y(x)}{dx^2}, \ \ (4)$

where the definition of the derivative is used to obtain the equivalence on the right-hand side. Indeed, the difference in the first derivatives yields the second derivative.

The mass dx of this piece of string, originally stretched from x to x + dx, is given by the mass density $\mu_0$ times dx. By Newton’s law, the net vertical force equals mass times vertical acceleration. So we can simply write

$F_y = ma \implies T \frac{d^2y(x)}{dx^2} dx = (\mu_0 dx)\frac{d^2y(x)}{dt^2}. \ \ (5)$

Cancel dx on both sides and rearrange terms to give

$\frac{d^2y(x)}{dx^2} - \frac{\mu_0}{T}\frac{d^2 y(x)}{dt^2} = 0. \ \ (6)$

Since y is a function of x and t, we can explicitly include this dependence and write y as y(x, t). Then the standard derivatives become partial derivatives and we arrive at the expected wave equation.

$\frac{\partial^2 y(x,t)}{\partial x^2} - \frac{\mu_0}{T}\frac{\partial^2 y(x,t)}{\partial t^2}. \ \ (7)$

Now, compare (7) with the standard wave equation below

$\frac{\partial^2 y}{\partial x^2} - \frac{1}{c^{2}} \frac{\partial^2 y}{\partial t^2} = 0, \ \ (8)$

with c the parameter for the velocity of the waves. In the case for transverse waves on the classical stretched string, we find the velocity of the waves is

$c = \sqrt{T_0 / \mu_0}. \ \ (9)$

Physically, we see that the lighter the string or the higher the tension, the faster the wave moves. This makes a lot of sense.

## General solution

The wave equation provides a general equation for the propagation of waves, linking the displacements of a wave in the y-coordinate direction with the time and also the displacement along the perpendicular x-axis. We therefore require solutions that link together x and t dependences.

One approach to deriving such a solution follows d’Alembert. In this approach, the displacement in y is defined as a function of two new variables, u and v, such that

$u = x -ct, \ \ v = x + ct. \ \ (10)$

To relate these solutions to the wave equation, we first differentiate both u and v with respect to x and t. Using the chain rule,

$\frac{\partial y}{\partial x} = \frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial y}{\partial u} + \frac{\partial y}{\partial v},$

$\frac{\partial y}{\partial t} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial t} = -c \frac{\partial y}{\partial u} + c \frac{\partial y}{\partial v}. \ \ (11)$

For the second derivatives we obtain,

$\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x})$

$= 2 (\frac{\partial^2 u}{\partial x \partial u} + \frac{\partial^2 v}{\partial x \partial v}) \frac{\partial y}{\partial x},$

and

$\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial y}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial t})$

$= 2(\frac{\partial^2 u}{\partial t \partial u} + \frac{\partial^2 v}{\partial t \partial v}) \frac{\partial y}{\partial t}. \ \ (12)$

Using the equation for the respective first derivative in (11) we find,

$\frac{\partial^2 y}{\partial x^2} = (\frac{\partial u}{\partial x}\frac{\partial}{\partial u} + \frac{\partial v}{\partial x}\frac{\partial}{\partial v})(\frac{\partial y}{\partial u} + \frac{\partial y}{\partial v}), \ \ (13)$

and

$\frac{\partial^2 y}{\partial t^2} = (\frac{\partial u}{\partial t}\frac{\partial}{\partial u} + \frac{\partial v}{\partial t}\frac{\partial}{\partial v})(-c (\frac{\partial y}{\partial u} - \frac{\partial y}{\partial v})). \ \ (14)$

Rearranging terms and substituting the fact from (10)

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 1, \ \ -\frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} = c, \ \ (15)$

we find

$\frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial u^2} + 2 \frac{\partial^2 y}{\partial u \partial v} + \frac{\partial^2 y}{\partial v^2}$

and

$\frac{\partial^2 y}{\partial t^2} = c^2 (\frac{\partial^2 y}{\partial u^2} - 2 \frac{\partial^2 y}{\partial u \partial v} + \frac{\partial^2 y}{\partial v^2}). \ \ (16)$

Substituting this result into the wave equation (7) we obtain

$\frac{\partial^2 y}{\partial u \partial v} = 0. \ \ (17)$

Observe that y is separable into functions of u and v such that

$y(u,v) = g(u) + h(v). \ \ (18)$

Therefore, the general solution to the wave equation takes the form

$y(x,t) = g(x - ct) + h(x + ct), \ \ (19)$

where g and h are single variable arbitrary functions, representing the string’s initial shape and velocity. The values of these functions are to be determined by the initial conditions of the system.

## Boundary conditions and initial conditions

In order to find unique solutions to the partial differential equation (7) and (8), which involves space and time derivatives, we are required to apply both initial conditions and boundary conditions. The initial conditions we apply constrain the solution at some time, while the boundary conditions constrain the solution at the boundary of the system.

There are primarily two types of boundary conditions common when discussing the dynamics of a string: Dirichlet and Neumann boundary conditions. This is also true in string theory, and we will see in the context of the relativistic string that application of boundary conditions imply the existence of some interesting new objects. We will therefore return to this topic in a future note. But for now, there is an intuitive way to think about what happens at the endpoints of our stretched classical string.

Dirichlet boundary conditions specify the positions of the string endpoints. For example, imagine screwing each endpoint of a string into a wall or a wooden block. The endpoint of the string is fastened such that it cannot move up and down the wall in what we previously described as the y-coordinate direction. Therefore, when imposing Dirichlet boundary conditions we have

$y(t, x = 0) = y(t,x=a) = 0. \ \ (20)$

On the other hand, Neumann boundary conditions are akin to attaching a massless loop to each end of the string, with the loops allowed to slide along two frictionless poles. In this case, the Neumann boundary conditions mean that the endpoints are free to move along the y-coordinate direction. We specify the values of $\partial y / \partial x$ at the endpoints.

$\frac{\partial y}{\partial x}(t,x = 0) = \frac{\partial y}{\partial x}(t, x = a) = 0. \ \ (21)$

## Finite string with fixed endpoints

There are a number of particular solutions to the wave equation that we may derive for various boundary configurations and initial conditions. For instance, we can have a fixed end at $x = 0$ and a free end at $x = a$, or two free ends, or both ends fixed, and so on. First, we consider the case of an infinite string with one fixed endpoint. Then we’ll consider the case of a finite string with both endpoints fixed.

Consider a leftward-moving single sinusoidal wave that is incident on a wall located at $x = 0$. The most general form of a leftward-moving sinusoidal wave is given by

$y (x,t) = A \cos (kx - \omega t + \phi), \ \ (22)$

where $\omega / k = c = \sqrt{T / \rho}$, $\phi$ is arbitrary and depends on the location of the wave at $t = 0$. We model the wall as equivalent to a system of infinite impedance with reflection coefficient $r = -1$. This can, again, be reviewed in most classical mechanics texts. This means that at the wall we obtain a reflected wave with amplitude of the same magnitude as the incident wave but with the opposite sign and travelling in the opposite direction $y_r$ such that

$y_r (x,t) = -A \cos (kx + \omega t + \phi). \ \ (23)$

If we were to observe this system, we would see a summation of these two waves (using familiar trigonometric functions)

$y(x,t) = A \cos (kx - \omega t + \phi) - A\cos (kx + \omega t + \phi), \ \ (24)$

which, using familiar trig identities, we obtain an expression in terms of the $\sin$ function

$y(x,t) = -2A \sin (\omega t + \phi) \sin kx. \ \ (25)$

Alternatively, we can write

$y(x,t) = 2A \sin (\frac{2\pi x}{\lambda})\sin(\frac{2\pi t}{T} + \phi). \ \ (26)$

Of course, rather than assigning $r =-1$ for a wall, we could instead derive the change in sign given that at the wall where $x = 0$ it follows $y = 0$. Then for all t and working from the general solution of the wave equation we find

$y(x,t) = A_1 \sin(kx - \omega t) + A_2 \cos(kx - \omega t) + A_3 \sin (kx + \omega t) + A_4 \cos (kx + \omega t)$

$\implies y(x,t) = B_1 \cos kn \cos \omega t + B_2 \sin kx \sin \omega t + B_3 \sin kx \cos \omega t + B_4 \cos kn \sin \omega t \ \ (27)$

at $y(0,t) = 0$. Therefore, we should only have $\sin kx$ terms

$y(x,t) = B_2 \sin kn \sin \omega t + B_3 \sin kx \cos \omega t$

$= (B_2 \sin \omega t + B_3 \cos \omega t) \sin kx$

$= B \sin (\omega t + \phi) \sin kx. \ \ (28)$

If the coefficients $B_2 = B_3$ then $B = 2B_2 = 2B_3$.

For a system in which the string is fixed at both ends, i.e. at x = 0 and x = a, the boundary conditions we impose require both $y(0, t) = 0$ and $y(a, t) = 0$. Therefore, we see from the previous example that the only way to have $y(a, t) = 0$ for all t is to ensure that $\sin ka = 0$. This means that ka must be an integer number of $\pi$ such that

$k_n = \frac{n\pi}{a}, \ \ (29)$

where $n \in \mathbb{Z}$ describes which string mode is excited.

In the present configuration, each endpoint represents a node. This implies that we can only have wavelengths which are related to the length of the string by n such that

$\lambda_n = \frac{2\pi}{k_n} = \frac{2a}{n}. \ \ (30)$

Therefore, we can now write a solution of the form

$y(x,t) = -2A \sin (\omega t + \phi)\sin (\frac{n\pi}{a}) = -2A \sin (\omega t + \phi) \sin (\frac{n\pi}{a}). \ \ (31)$

Observe that the allowed wavelengths on the string are all integer divisors of twice the length of the string. For the mode $n = 0$, the string can be seen to be at rest and in equilibrium.

From what has been derived above, one can proceed with an analysis of the angular frequency $\omega$ to find that the frequency of oscillations of the string are in fact all integer multiples of the fundamental frequency. But, most importantly here, it is noted that since the wave equation (7) is linear, the most general motion of a string with both endpoints fixed is simply a linear combination of the solution (25), where k can only $k = n\pi/L$ and $\omega / k = v$. Therefore, the general expression for y(x, t) may be written as a summation over all n,

$y(x,t) = \sum\limits^{\infty}_{n = 0} F_n \sin (\omega_n t + \phi_n) \sin(k_n x). \ \ (32)$

We have obtained a sum of all possible solutions with the coefficients $F_n$ given by the initial displacement.

## Constructing the Lagrangian

The remaining space will be focused on a review of how to construct the Lagrangian for the classical non-relativistic string, and then we will calculate its equations of motion within the Lagrangian formalism. It should be familiar that, in general, an action will take the form $S = \int \ L \ dt$, where L is the Lagrangian. The Lagrangian will have both kinetic (T) and potential energy (V) such that schematically we have something of the form $L = T - V$.

Returning to the picture of the string with constant mass density $\mu_0$, constant tension $T_0$, and with its endpoints located at $x = 0$ and $x = a$, the first step is to see that we should integrate the kinetic energy over the infinitesimal dx pieces along the string. In other words, the kinetic energy will be the sum of the kinetic energies of all the infinitesimal segments along the string. We also consider the rate at which each segment along the string is moving. Following this reasoning we obtain for the kinetic part of the Lagrangian

$T = \int_{0}^{a} \frac{1}{2}\mu_{0}(\frac{\partial y}{\partial t})^{2}. \ \ (33)$

The potential energy relates the work done when each segment along the string is stretched. Similar to the picture earlier, when a single infinitesimal segment of string is stretched from (x, y) to (x + dx, y + dy) this individual segment will change by some $\Delta \ L$. To derive an expression for the potential V, we assume small oscillation such that $\mid \frac{dy}{dx} \mid << 1$. That is to say, we take $\delta L = \sqrt{dx^2 + dy^2} - dx$.

As an expression of the work done by deformation, the potential energy V may be written as

$V = T(ds -dx), \ \ (34)$

where

$(ds)^2 = (dx)^2 + (dy)^2 = (dx)^2 (1 + (\frac{\partial y}{\partial x})^2). \ \ (35)$

Again, as found earlier, using the binomial series expansion (and because we are invoking the small angle approximation, we ignore the higher order terms in the expansion),

$ds \approx dx (1 + \frac{1}{2}(\frac{\partial y}{\partial x})^2 + ...). \ \ (36)$

We take this result and account for the fact that the work done relates to the stretching of each infinitesimal segment such that $T_{0} ds$.

$V = \int_{0}^{a} \frac{1}{2}T_{0}(\frac{\partial y}{\partial x})^{2} dx. \ \ (37)$

Bringing together our expressions for T and V gives

$L = \int_{0}^{a} [\frac{1}{2} \mu_0 (\frac{\partial y}{\partial t})^2 - \frac{1}{2} T_0 (\frac{\partial y}{\partial x})^2] dx \\ \equiv \int_{0}^{a} \mathcal{L} dx, \ \ (38)$

where $\mathcal{L}$ is the Lagrangian density

$\mathcal{L}(\frac{\partial y}{\partial t}, \frac{\partial y}{\partial x}) = \frac{1}{2}\mu_0 (\frac{\partial y}{\partial t})^2 - \frac{1}{2}T_0 (\frac{\partial y}{\partial x})^2. \ \ (39)$

We may simplify notation with $\partial y / \partial t = \dot{y}$ and $\partial y / \partial x = y^{\prime}.$ The action for the non-relativistic string therefore takes the form,

$S_{NR} = \int_{t_{i}}^{t_{f}} L(t) dt \int_{0}^{a} dx [\frac{1}{2} \mu_0 \dot{y}^2 - \frac{1}{2} T_0 y^{\prime 2}]. \ \ (40)$

Recall from a previous section that we need to include a time component in addition to a spacial component, because in the action the path is the function y(t, x). So we see, again, that we are integrating from some initial time to some final time, and from an initial point to some final point in a region of (t,x) space.

## Equations of motion

To find the equations of motion, we need to vary the action (37). But before that, let’s introduce some notation that will allow us to achieve a more general derivation. The simplification has to do with the momentum density of the string. The momentum density will be denoted as $\mathcal{P}^t$ and $\mathcal{P}^x$. We compute these terms as follows,

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

$\mathcal{P}^x = \frac{\partial \mathcal{L}}{\partial y\prime} = - T_0 \frac{\partial y}{\partial x}. \ \ (41)$

As the function y(t, x) represents the path of our string, in the variation $y (t, x) \rightarrow y(t, x) + \delta y (t, x)$ we may compute,

$\delta S_{NR} = \int_{t_i}^{t_f} dt \int_{0}^{a} dx (\frac{\partial \mathcal{L}}{\partial \dot{y}} \delta \dot{y} + \frac{\partial \mathcal{L}}{\partial y \prime} \delta y\prime). \ \ (42)$

Substituting for $\mathcal{P}^t$ and $\mathcal{P}^x$ in (40) we obtain,

$= \int_{t_i}^{t_f} dt \int_{0}^{a} dx [\mathcal{P}^t \delta \dot{y} + \mathcal{P}^x \delta y^{\prime}]$

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

$= \int dt dx [\frac{\partial}{\partial t} (\mathcal{P}^t \delta y) + \frac{\partial}{\partial x} (\mathcal{P}^x \delta y) - \delta y (\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x})]. \ \ (43)$

Following conventional procedure in the calculus of variations, we integrate by parts. Doing so gives

$\delta S =\int_{0}^{a} dx (\mathcal{P}^t \delta y)]_{t_i}^{t_f} + \int_{t_i}^{t_f } dt (\mathcal{P}^x \delta y)]_{0}^{a} - \int_{t_i}^{t_f} dt \int_{0}^{a} dx \ \delta y (\frac{\partial \mathcal{P}^t}{\partial t} + \frac{\partial \mathcal{P}^x}{\partial x}). \ \ (44)$

This final expression for the varied action contains three terms. Each term must vanish independently. Notice that the first term is determined by the string’s arrangement at $t_{i}$ and $t_{f}$. In the present case, our interest is not with these initial and final conditions (unless we were interested in the Hamilton-Jacobi), so we can just specify initial and final configurations which results in setting the variation to zero. We could even just set the times to infinity and forget about it entirely, the choice is ours.

Instead, our interest begins with the second term. Notice that it describes the motion of the string endpoints between 0 and a during the time interval from $t_i$ to $t_f$. We may expand this term as follows,

$\int_{t_i}^{t_f } [\mathcal{P}^x \delta y]_{0}^{a} = \int_{t_i}^{t_f } dt [\mathcal{P}^x(t, x=a) \delta y(t, x=a) - \mathcal{P}^t (t, x=0) \delta y (t, x=0)]. \ \ (45)$

What we require are boundary conditions for both terms.

Once again, putting the choice of mixed conditions to the side, we can either invoke Dirichlet or Neumann boundary conditions. As discussed in a previous section, there are physical implications dependent on either choice. Dirichlet boundary conditions means the string endpoints are fixed in time. Hence, consider some x coordinate at the endpoints, and with the choice of Dirichlet boundary conditions the variation $\delta y(t, x)$ must vanish for both terms. In this case, too, momenta along the string will not be conserved. Neumann boundary conditions, on the other hand, means that the endpoints are free to move, and thus under this choice $\delta y(t, x)$ would be unconstrained. If we were to do the calculations, we would see that momenta along the string is conserved.

Let us now focus on the remaining term in (42). This term is determined by the motion of the string for $x \in (0, a)$ and $t \in (t_{i}, t_{f})$. It gives us the equation of motion,

$\frac{\partial \mathcal{P}^{t}}{\partial t} + \frac{\partial \mathcal{P}^{x}}{\partial x} = 0. \ \ (46)$

Now let us ask: what is $\frac{\partial \mathcal{P}^t}{\partial t}$ and $\frac{\partial \mathcal{P}^x}{\partial x}$? Let’s 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 respectively.

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

$\implies \mu_0 \frac{\partial^2 y}{\partial t^2} - T_0 \frac{\partial^2 y}{\partial x^2} = 0. \ \ (47)$

Rearranging and remembering that $v^2 = \frac{T_0}{\mu_0}$ we find,

$\frac{\partial^2 y}{\partial x^2} - \frac{1}{\frac{T_0}{\mu_0}}\frac{\partial^2 y}{\partial t^2} = 0. \ \ (48)$

Once again, we have found the wave equation.

## Energy

In a final note, a vibrating string will of course carry energy. A natural question has to do with the quantity of energy. To answer this we may look at the relation between kinetic energy and potential energy as the string oscillates.

As before, we consider a small segment of string and study the linear density $\mu &fg=000000 &s=2$ between x and x + dx, displaced in the y-coordinate direction. Once again, assuming the displacement is small then we can calculate the kinetic energy density (i.e., the K.E. per unit length) and the potential energy density.

We know the kinetic energy (33) and the potential energy (37). We also know that the solutions to the wave equation take the form

$y(x,t) = f(x \pm ct). \ \ (49)$

Therefore,

$\frac{dK}{dx} = \frac{1}{2}\mu c^2 [f^{\prime}(x \pm ct)]^2, \ \frac{dV}{dx} = \frac{1}{2}T[f^{\prime}(x \pm ct)]^2. \ \ (50)$

Recall $c = \sqrt{T / \mu}$, and so we find the kinetic energy density to be equal to the potential energy density. Another way to observe this equality is by substituting a solution $y = A \sin (kx - \omega t)$ for the wave equation into these equations for the kinetic energy and potential energy density. We can then evaluate the energy over n wavelengths.

For the kinetic energy density,

$K = \frac{1}{2}\mu \int_{x}^{x + n\lambda} A^2 \omega^2 \cos^2 9kx - \omega t) dx$

$= \frac{1}{2} \mu A^2 \omega^2 \int_{x}^{x + n \lambda} \frac{1}{2}(1 + \cos[2(kx - \omega t)]) dx. \ \ (51)$

And for the potential energy density we find,

$V = \frac{1}{2}T \int_{x}^{x + n\lambda} A^2 k^2 \cos^2 (kx - \omega t) dx$

$= \frac{1}{2} T A^2 k^2 \int_{x}^{x + n\lambda} \frac{1}{2} (\cos [2(kx - \omega t)]) dx. \ \ (52)$

And so we see that

$K = \frac{1}{2} \mu A^2 \omega^2 \frac{n\lambda}{2}, \ V = \frac{1}{2} A^2 k^2 \frac{n\lambda}{2}. \ \ (53)$

But as $c = \sqrt{T / \mu} = \omega / k \rightarrow \mu \omega^2 = T k^2$, these expressions for K and V are once again equal. We can also deduce therefore that the total energy per unit length is $1/2 \mu A^2 \omega^2$. From this analysis it is possible to go on and study the energy flow per unit time and the power to generate the wave.

References

[1] Taylor, J., Classical mechanics. University Science Books, 2004.

[2] Zwiebach, B., A first course in string theory (2nd edition). Cambridge University Press, 2009.

# Reading Polchinski – My notes on string theory

When I was in the first year of my undergraduate degree, I started self-studying string theory. At that time, I set myself the task of working through Joe Polchinski’s two volume textbook, String Theory. It was honestly one of the best years of my life, despite the general uncertainty about my academic future at the time. I also began a project of sharing my own notes on this blog, as an extension of my enthusiasm.

Between the time as a first year undergraduate student and the commencement of my Masters the next year, I hadn’t much time to finish converting my notes to latex and uploading them to this blog. Things move fast, and quickly I was on to double field theory, developing my interests in M-theory, and other cool stuff with pressures of producing a thesis. Also, in that time, I had to change my web hosting service. As a result of transferring my blog, the latex broke in all of my old posts (including what string notes I had uploaded) which was a bit frustrating.

I have been intending to reupload what notes I had converted to latex and then to also continue my project for sometime. Now that I have settled into my PhD years, I feel able again to take up the task (of course, as and when the time becomes available). As part of my scholarship, I also feel obliged to participate in science communication. Sharing technical and explanatory notes in maths/physics is probably where I am most capable.

***

I consider my handwritten notes a product of rigorous examination and review, organised around the central motivation to rederive the whole of bosonic and then superstring theory from first principles, or, where appropriate, at least from as close to first principles as possible. They are a thorough companion to Polchinski’s textbook, providing a workthrough of its many pages and subtle details. They also offer a lot of comments from my own perspective, such as when making an interesting observation or offering more background work than presented in Polchinski’s textbook. I have also developed my own preferences, such as emphasising at the outset the notion of string theory as a generalisation point particle theory.

Due to the way I upload latex to my blog nowadays, what I currently plan to do is start from the very beginning and reupload everything that was already posted and then work toward uploading the rest. As the notes were originally written for myself, converting them to latex and organsing them in a presentable way takes quite a bit of time. I like the idea of uploading one batch of notes at a time as individual posts, such as in presenting one lecture note at a time (working page by page through Polchinski). Eventually I will also create a directory on my blog, where the reader may navigate by topic through the complete notes. (I will also file everything under the tag, ‘Notes on String Theory’). They may or may not prove useful for others, but, to be honest, I ultimately doing it because it is something I enjoy.

As for what is to come next: in the first entry we will start with a brief review of the non-relativistic string, before moving to a review of the relativistic free-point particle in a separate note. This will then take us to the first pages of Polchinski and the construction of the Nambu-Goto action.