2.9 Bootstrap in $D 3$

Historically, the first sigma model was discovered by Newton. Newton directly gave the equations of motion (EoM) on the worldline \& interpreted the meaning of the worldline EoM directly on target space. Later, Lagrange gave the worldline Lagrangian, but he also directly interpreted its meaning in the target space.

Galilean spacetime

See the below image and ^[1] and chapter 5 of ^[2]. Spacetime with Galilean invariance is a fibre bundle with the worldline as the base space. Nonrelativistic metric: In the limit $c\to\infty$, $ {\displaystyle ds^{2}=-c^{2}dt^{2}+dx^{2}+dy^{2}+dz^{2}}$ gives 2 metrics $\dfrac{ds^{2}}{c^2}=-dt^{2}=g_{tt}dt^{2}\Rightarrow g_{tt}=-1$ and the spatial metric $ {\displaystyle ds^{2}=dx^{2}+dy^{2}+dz^{2}}$. The spatial metric is not invariant under Galilean boosts, which is why Galilean spacetime is not a trivial fibre bundle (Cartesian product). But for the sake of intution, I will write as $x^i:\mathbb R\to \mathbb {R} ^{n}$ even though, technically speaking, the target space is a fibre bundle, and I should write ${\displaystyle \pi:{ST}\to \mathbb{T},}$ where $\mathbb{ST}$ is the spacetime and $\mathbb{T}$ is the worldline.

Wick rotation of the Worldline $\mathbb{T}$ gives an Euclidean line $\mathbb{R}$.

\begin{equation} {\displaystyle S= \int dt \left( {\frac {1}{2}}m\sum_i (\dot {x}^i)^{2}-V({x}^i)\right)} \end{equation} The above is the worldline action, and the below is my interpretation of what the quantities mean on the worldline

$x^i:\mathbb R\to \mathbb {R} ^{n}$ is a group of classical scalar fields living on the worldline. Each field outputs a single real number. They are not particles¹ living on the worldline because particles don’t have a value at each point on the worldline.
Target spacetime symmetry: In the target space, we have Galilean coordinate invariance. This on the worldline becomes field redefinitions, it is not a real flavour symmetry, and it’s a fake redundancy that doesn’t give any conservation laws. Note that under field redefinitions, the target space metric $g_{ij}$ becomes non-trivial. If the target space has real symmetries, which need \href{https://en.wikipedia.org/wiki/Killing_vector_field}{Killing vector fields}, then that gives real flavour symmetry. If we neglect the electromagnetic interaction between protons and neutrons and since their masses are almost the same, we have an approximate $SU(2)$ flavour symmetry, which gives the conservation of Isospin ($I_3$) from the Noether’s theorem. Similarly, energy and momentum are also charges of the flavour symmetry.
$m$: Mass on the target space has the interpretation as the measure of inertia. More precisely, the mass on the target space is the central charge of the Bargmann group (it is a central extension of the Galilei group). On the worldline, it is the central charge of a flavour symmetry instead of the spacetime symmetry. If we add another mass $m’$ then its coordinates again have another flavour symmetry whose central charge is $m’$.
Worldline mass: Target spacetime mass becomes the central charge of a flavour symmetry on the worldline. But if we have a potential $V({x}^i)=\frac{1}{2}k\sum_i ( {x}^i)^{2}$. Then, the Lagrangian will separate, and each scalar field will not interact with others. \begin{equation} {\displaystyle S= \int dt \sum_im\left( {\frac {1}{2}} (\dot {x}^i)^{2}-\frac{1}{2}\omega^2 ( {x}^i)^{2}\right)} \end{equation} This is literally the Klein–Gordon Lagrangian for each scalar field separately with mass $\omega=\sqrt{\frac{k}{m}}$. The scalar field solutions in higher dimensions will be of the form $e^{-ip\cdot x}$ where $p^2=-m^2$. In 1D, it just becomes $p^\mu p_\mu=g_{tt}E^2=-\omega^2\Rightarrow E=\pm\omega$ and the solutions are $e^{\pm i\omega t}$. We get many solutions in higher dimensions, so we need to sum them up using the Fourier transformation. Here, we just sum over the 2 to get sinusoidal solutions. If we have $V({x}^i)=\frac{1}{2}\sum_i k_i( {x}^i)^{2}$, then the target spacetime symmetry is broken, and similarly the flavour symmetry on the worldline is broken, and each field has different mass $\omega_i=\sqrt{\frac{k_i}{m}}$.
More complicated interactions: Free particles on the target space gave us massless free scalar fields. If there was a Harmonic potential interaction on the target space, then each free scalar field became massive with mass given by the harmonic mode frequencies $\omega_i$. But still, on the worldline, there was no interaction so far between scalar fields. All these scalar fields get coupled to each other for more complicated potentials like the target space radial inverse square law. We might simplify the problem by using a field redefinition corresponding to the target space spherical coordinate transformation. But Cartesian coordinates had a nice Klein–Gordon interpretation on the worldline. But the fields $\theta,\phi$ don’t have such nice interpretations as they are coupled to the field $r$. Note that if we Taylor expand the $\operatorname{s i n}^{2} \theta$ we get infinite interactions between the 3 worldline fields. \begin{equation}\label{worldlineinteractingmetric} L=\frac{1} {2} m [ \dot{r}^{2}+r^{2} ( \dot{\theta}^{2}+\operatorname{s i n}^{2} \theta\dot{\phi}^{2} ) ]-V ( r ). \end{equation} Similarly, in generic coordinates, the metric term in the Lagrangian $\frac{1}{2} m g_{ij} (\vec{x}) \dot{x}^i \dot{x}^j - V(\vec{x})$ can be Taylor expanded to get infinite interactions between the worldline classical scalar fields.

Vertex Operators in CM

Rutherford scattering

Subsections

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Particles on the worldline are trivial. Their position must always be the same \& their velocity is $0$. ↩