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Safi_Final_Defense.tex
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Safi_Final_Defense.tex
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\documentclass[aspectratio=169,usenames,dvipsnames]{beamer}
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% the Lund University theme
\usetheme{LundUniversity}
% this tells LaTeX how the latex file is encoded
\usepackage[utf8]{inputenc}
\usepackage[british]{babel}
% align the text
\usepackage{ragged2e}
% colors
\usepackage{xcolor}
% to draw feynman diagrams
\usepackage{axodraw2}
% list in line
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% Dirac notation
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%For use the minipage environment (figures side-by-side)
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% add note to the table
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\usepackage[many]{tcolorbox}
\usepackage{empheq}
\usetikzlibrary{shadows}
%\begin{empheq}[box=\tcbhighmath]{align}
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highlight math style={
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% For hyper references that can be browsable
\usepackage{hyperref}
\hypersetup{
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\setbeamercovered{transparent}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title[Master Thesis Presentation]{\large{SIMPLIFYING QUANTUM GRAVITY CALCULATIONS}}
\titlecolor{LUGreen} % Choose between LUPink, LULBlue, LUIvory, LUGreen
\titleimage{\includegraphics[scale=.5]{images/astro.png}}
%\titleimage{\includegraphics[scale=.6]{images/lumainb.jpg}}
\author{Safi Rafie-Zinedine}
\subtitle{\qquad\, \lowercase{\uppercase{S}afi \uppercase{R}afie-\uppercase{Z}inedine \;\qquad --- \;\qquad \uppercase{S}upervisor: \uppercase{P}rof. \uppercase{J}ohan \uppercase{B}ijnens}}
\date{\today}
\institute{Lund University\\Department of Portals}
\begin{document}
\titleframe
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Overview} \bf
\begin{center}
\begin{minipage}{0.42\textwidth}
\begin{itemize}
\item[$\bullet$] Introduction \vspace{5.0pt} \\
\item[$\bullet$] Deriving Lagrangian \\ \vspace{5.0pt}
\item[$\bullet$] Manipulating Lagrangian \\ \vspace{5.0pt}
\item[$\bullet$] Loop Integral \\ \vspace{5.0pt}
\item[$\bullet$] Our Strategies \\ \vspace{5.0pt}
\item[$\bullet$] Our Results \\ \vspace{5.0pt}
\item[$\bullet$] Scattering at Tree Level \\ \vspace{5.0pt}
\item[$\bullet$] Scattering at One-loop Level \\ \vspace{5.0pt}
\item[$\bullet$] Conclusions \\ \vspace{40.0pt}
\end{itemize}
\end{minipage}
\vfill
\end{center}
\vspace*{100mm}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bf Introduction}
\begin{frame}{What are the main reasons behind this thesis?}
\begin{itemize}
\item[$\bullet$] The Feynman rules are very complicated although the resulting amplitudes are often simple.
\vspace{8mm}
\item[$\bullet$]
A better understanding of the math in this theory when fewer terms actually contribute.
\vspace{8mm}
\item[$\bullet$] Reducing the running time in FORM program.
\vspace{8mm}
\item[$\bullet$] Following the belief that nature should be described in a beautiful and simple mathematical way.
\vspace{8mm}
\end{itemize}
%did you get some deeper understanding??
\end{frame}
\begin{frame}{Spin-2 Graviton}
\begin{itemize}
\item[$\bullet$] Spin 0: \\
$\Rightarrow$ \quad a Newtonian gravitational potential which considers that the mass, fixed Yukawa coupling, is the only source for gravity.
\linebreak
\item[$\bullet$] Spin 1: \\
$\Rightarrow$ \quad an attractive and repulsive gravitational potential.
\linebreak
\item[$\bullet$] Spin 2: \\
$\Rightarrow$ \quad a gravitational potential which considers that the EMT is the source for gravity.
\linebreak
\item[$\bullet$] The higher spin is not consistent with QFT.
\end{itemize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bf Deriving Lagrangian}
\begin{frame}{Analogy With Yang-Mills Theory}
\setbeamercovered{invisible}
\begin{itemize}
\item[] Considering the real scalar field Lagrangian:\\[2mm]
\centering $\mathcal{L}_{\text{Matter}} = \frac{1}{2} \eta^{ab}
\partial_a\phi \partial_b\phi - \frac{1}{2} m^2 \phi^2 $\\[7mm] \justifying
\item[] Invariant under the global translational symmetry: \qquad $y^a = y^a + d^a $ \\[7mm]
\item[] Gauging the global symmetry \; $\Rightarrow$ \; The general coordinate transformations.\\[2mm]
\centering \fbox{ $y^a = y^a + d^a $ \; $\Rightarrow$ \; $x^{\mu} = x^{\mu} +
d^{\mu}(x)$} \\[5mm]
\item[] \hspace{19mm} \textbf{(A)} \hspace{79mm} \textbf{(B)} \\[1mm]
$ a,b,c,\cdots \;\rightarrow\;
\mu,\nu,\alpha,\cdots$ \hfill $ d^a \;\rightarrow\; d^a(x) $ \\[1mm]
\raggedright\scriptsize{Interval invariant: $ds^2 = \eta_{ab} dy^a dy^b =
g_{\mu\nu} dx^{\mu} dx^{\nu}$} \\[3mm]
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{Measure correction: $d^4y=\sqrt{-\det(g_{\mu\nu})}\, d^4x=\sqrt{-g}\, d^4x$}\\[50mm]
\end{itemize}
\vspace{50mm}
\end{frame}
\begin{frame}{Analogy With Yang-Mills Theory}
\begin{itemize}
\item[] The Lagrangian for matter becomes: \scriptsize{(invariant under GCT)} \normalsize \\[1mm]
\centering \fbox{$\mathcal{L}_{\text{Matter}} = \frac{1}{2} g^{\mu\nu}
\partial_{\mu}\phi \partial_{\nu}\phi - \frac{1}{2} m^2 \phi^2$} \\[6mm]
\raggedright\item[] The commutator of covariant derivatives: \scriptsize{($\Rightarrow$
Analogy with the field strengh tensor)} \normalsize \\[1mm]
\centering $[D_{\mu},D_{\nu}] V^{\beta} =R_{\mu\nu\alpha}^{\;\;\;\;\;\;\beta}
V^{\alpha} $ \qquad $\Rightarrow$ \qquad ($R_{\mu\nu\alpha\beta},
R_{\nu\alpha},R$) \\[6mm]
\raggedright\item[] The Lagrangian for gravity: \scriptsize{(invariant under
GCT)} \normalsize \\[1mm]
\centering \fbox{$\mathcal{L}_{\text{Gravity}} = - \frac{2}{\kappa^2} R $} \\[6mm]
\raggedright\item[] Equation of motion: \scriptsize{(Einstein’s equation)} \normalsize \\[1mm]
\centering $R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = \frac{\kappa^2}{4}
\mathcal{T}_{\mu\nu}$\\[1mm]
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\raggedright\tiny{Where $\frac{\kappa^2}{4} = 8 \pi G$ \hfill $
\frac{2}{\sqrt{-g}} \frac{\delta \mathcal{S}_{\text{Matter}}}{\delta g^{\mu\nu}} = \mathcal{T}_{\mu\nu} $ } \\[8mm]
\end{itemize}
\end{frame}
\begin{frame}{Effective Field Theory (EFT)} \small
\setbeamercovered{invisible}
\begin{itemize}
\item[$\bullet$] EFT is to study the physics in particular ranges
of energy while neglecting the physics at higher energy.\\[1mm]
\centering $\mathcal{L}_{\textsf{eff}} = \mathcal{L}_0 + \mathcal{L}_1 + \mathcal{L}_2
+ \mathcal{L}_3 + \cdots$ \\[6mm] \justifying
\item[$\bullet$] Weinberg’s power counting theorem: $\mathcal{D} = 2 +
\sum_{n} V_n (n-2) + 2 L$ \\[6mm]
\item[$\bullet$] The most general effective Lagrangian for gravity in energy expansion:
\begin{empheq}[box=\fbox]{align*}
\mathcal{L}_{\text{eff}} & = \mathcal{L}_0 + \mathcal{L}_2 + \mathcal{L}_4 + \cdots \\
& = - \Lambda - \frac{2}{\kappa^2} R + c_1 R^2 + c_2 R_{\mu\nu} R^{\mu\nu} + c_3 R_{\mu\nu\alpha\beta} R^{\mu\nu\alpha\beta} + \cdots
\end{empheq} \footnotesize
$\circ$\; \(\Lambda \) is Cosmological constant: this term is \( \mathcal{O}(E^0) \)\\
$\circ$\; \(\kappa \) is Newtonian strength of gravitational interactions:
this term is \( \mathcal{O}(E^2) \) $\Rightarrow$ (tree level)\\
$\circ$\; \(c_i \) are higher order corrections: these terms are \(
\mathcal{O}(E^4) \) $\Rightarrow$ (one-loop level) \\ \vspace{1mm}
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\raggedright\tiny{Where $R_{\mu\nu\alpha\beta},
R_{\nu\alpha},R\sim(\partial\Gamma,\Gamma\Gamma)\sim(\partial g\partial
g,\partial\partial g)$ \hfill $n$ the \# derivatives in vertex, $L$ the \# loops } \\[50mm]
\end{itemize}
%can be renormalized at one loop.
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bf Manipulating Lagrangian}
\begin{frame}{\centering \fbox{\tiny{First Freedom -- Choosing A Gauge}}\\ Why do we need a gauge condition?} \small
\setbeamercovered{invisible}
\begin{itemize}
\item[$\bullet$] Considering an infinitesimal coordinate transformation: \footnotesize
\Big (\unboldmath $x^{\mu} \; \rightarrow x'^{\mu} = x^{\mu} - \xi^{\mu}(x)$
\Big)
$$ h_{\mu\nu} \; \rightarrow h'_{\mu\nu} = h_{\mu\nu} + \partial_{\mu}
\xi_{\nu}(x) + \partial_{\nu} \xi_{\mu}(x) + h_{\mu\sigma} \partial_{\nu}
\xi^{\sigma}(x) + h_{\nu\sigma} \partial_{\mu} \xi^{\sigma}(x) +
\xi^{\sigma}(x) \partial_{\sigma} h_{\mu\nu} $$ \\
\begin{itemize}
\scriptsize\item[$\circ$] $\mathcal{L}$ invariant under GCT \, $\Rightarrow$ \, a redundancy of the description.
\item[$\circ$] Insert $\mathcal{L}_{\text{FG}},\mathcal{L}_{\text{GH}}$ \, $\Rightarrow$ \, break this gauge symmetry \, $\Rightarrow$ \, remove this redundancy.
\end{itemize} \vspace{2.5mm}
\small\item[$\bullet$]{In the path integral formalism:}\footnotesize
$$ \mathcal{Z} = \int D[h] \; \exp( i \, S(h) ) = \int D[h] \; \exp( i
\,\int d^4x \mathcal{L} (h) ) $$\\
\begin{itemize}
\scriptsize\item[$\circ$] The measure $\int [h] $\, $\Rightarrow$ \, over all configurations of \(h\).
\item[$\circ$] Insert $\mathcal{L}_{\text{FG}},\mathcal{L}_{\text{GH}}$ \,
$\Rightarrow$ \, over the correct
configurations of \(h\).
\end{itemize} \vspace{2.5mm}
\small\item[$\bullet$]{Degrees of freedom:}\\
\begin{itemize}
\scriptsize\item[$\circ$] $\mathcal{L}$ has more degrees of freedom than its gauge boson.
\item[$\circ$] Insert $\mathcal{L}_{\text{FG}},\mathcal{L}_{\text{GH}}$ \, $\Rightarrow$ \, get rid of the extra degrees of freedom.
\end{itemize}
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\raggedright\tiny{Where $ g_{\mu\nu} \; \rightarrow g'_{\mu\nu}(x') = g_{\alpha\beta}(x) (\frac{\partial x^{\alpha}}{\partial x'^{\mu}}) (\frac{\partial x^{\beta}}{\partial x'^{\nu}}) $ \hfill $g_{\mu\nu}= \eta_{\mu\nu} + \kappa \; h_{\mu \nu}$ } \\[50mm]
\end{itemize}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{First Freedom -- Choosing A Gauge}}\\ \small How
do we derive $\mathcal{L}_{\text{FG}},\mathcal{L}_{\text{GH}}$ using Faddeev-Popov Method?} \footnotesize
\setbeamercovered{invisible}
\begin{itemize}
\onslide<1-> \item[$\bullet$] Using the identities:\\
\fbox{$ 1 = \int D[\xi_{\nu}] \delta \Big ( \mathcal{C}_{\mu} (h) - F_{\mu}(x) \Big
) \; \Delta(h) $} \;\; \;\; \fbox{$ 1 = N(\epsilon) \int D[F] \exp \Big ( -
\frac{i}{2\epsilon} \int d^4 x F_{\mu}(x)F^{\mu}(x) \Big )$} \\ \vspace{1mm}
\scriptsize Where $\mathcal{C}_{\mu}(h) = F_{\mu}(x)$ is the gauge condition and $\Delta(h)$ is
Faddeev-Popov determinant.\\ \vspace{2mm} \footnotesize
\item<only@1>[$\bullet$] Inserting these identities into the generating functional: \footnotesize
$$ \mathcal{Z} = \int D[h] \; \exp( i \, S(h) ) = \int D[h] \; \exp( i
\,\int d^4x \mathcal{L} (h) ) $$ \\
\item<2->[$\bullet$] It yields: \footnotesize
$$ \mathcal{Z} = N(\epsilon) \; N'^{-1} \; \int D[F] \; D[h] \; D[\xi_{\nu}]
\; \delta \Big ( \mathcal{C}_{\mu} (h) - F_{\mu}(x) \Big ) \; \Delta(h) \; \exp \Big ( i
S - \frac{i}{2\epsilon} \int d^4 x F_{\mu}(x)F^{\mu}(x) \Big ) $$ \\ \vspace{2mm}
\item<2->[$\bullet$] Integrating over \(\xi_{\nu}, F(x) \): \footnotesize
$$ \mathcal{Z} = N^{-1} \; \int D[h] \; D[\bar{\chi}_{\mu}] \; D[\chi_{\nu}] \; \exp \Big ( i S -
\frac{i}{2\epsilon} \int d^4 x \; \mathcal{C}_{\mu}(h) \mathcal{C}^{\mu}(h) + i \int d^4 x \; \bar{\chi}_{\mu}
\; \frac{\partial \mathcal{C}_{\mu}(h) }{\partial \xi_{\nu}} \; \chi_{\nu} \Big ) $$
\\ \vspace{2mm}
\onslide<2-> \small \hspace{15mm} \textbf{(A)} \onslide<2-> \hspace{75mm} \textbf{(B)} \\
\fbox{$\mathcal{L}_{\text{FG}}(h) = \frac{1}{2\epsilon} \mathcal{C}_{\mu}(h) \mathcal{C}^{\mu}(h)$} \hfill \fbox{$\mathcal{L}_{\text{GH}}(\bar{\chi}_{\mu},\chi_{\mu},h) = \bar{\chi}_{\mu} \; \frac{\partial \mathcal{C}_{\mu}(h) }{\partial \xi_{\nu}}
\; \chi_{\nu} $} \\ \vspace{2mm}
\onslide<2-> \textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{The Faddeev-Popov determinant $ \Delta(h) = \det \Big (
\frac{\partial \mathcal{C}_{\mu}(h) }{\partial \xi_{\nu}} \Big ) = \int
D[\bar{\chi}_{\mu}] D[\chi_{\nu}] \exp \Big ( i \int d^4 x \; \bar{\chi}_{\mu}
\; \frac{\partial \mathcal{C}_{\mu}(h) }{\partial \xi_{\nu}} \; \chi_{\nu} \Big ) $} \vspace{30mm}
\end{itemize}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{First Freedom -- Choosing A Gauge}}\\ What
gauge condition did we use?} \small
\setbeamercovered{invisible}
\begin{itemize}
\item[$\bullet$] The de Donder (harmonic) gauge condition: \footnotesize
$$\mathcal{C}_{\mu}(h) = \partial_{\nu} h_{\mu}^{\;\;\nu} - \frac{1}{2} \partial_{\mu}
h_{\lambda}^{\;\;\lambda} \hspace{30cm} $$ \\[2mm]
\small \item[$\bullet$] The general parameterized gauge condition: \footnotesize
\begin{align*}
\mathcal{C}_{\mu}(h) = & \;\; \kappa \Big [
b_1 \partial^{\nu} h_{ \nu \mu}
+b_2 \partial_{\mu} h_{ \nu}^{\;\;\nu}
\Big ]
\\ & +\kappa^2 \Big [
b_3 \partial_{\mu} h_{ \nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha}
+b_4 \partial_{\mu} h^{ \nu \alpha} h_{\nu \alpha}
+b_5 \partial^{\nu} h_{ \mu \nu} h_{\alpha}^{\;\;\alpha}
+b_6 \partial_{\nu} h_{ \mu \alpha} h^{\nu \alpha}
\\ & \qquad\quad +b_7 \partial_{\nu} h^{ \nu \alpha} h_{\mu \alpha}
+b_8 \partial^{\nu} h_{ \alpha}^{\;\;\alpha} h_{\mu \nu}
\Big ]
\\ & +\kappa^3 \Big [
b_9 \partial_{\mu} h_{ \nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+ b_{10} \partial_{\mu} h_{ \nu}^{\;\;\nu} h^{\alpha \beta} h_{\alpha \beta}
+ b_{11} \partial_{\mu} h^{ \nu \alpha} h_{\nu \alpha} h_{\beta}^{\;\;\beta}
+ b_{12} \partial_{\mu} h^{ \nu \alpha} h_{\alpha}^{\;\;\beta} h_{\beta \nu}
\\ & \qquad\quad + b_{13} \partial^{\nu} h_{ \mu \nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+ b_{14} \partial^{\nu} h_{ \mu \nu} h^{\alpha \beta} h_{\alpha \beta}
+ b_{15} \partial_{\nu} h_{ \mu \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+ b_{16} \partial^{\nu} h_{ \mu \alpha} h^{\alpha \beta} h_{\beta \nu}
\\ & \qquad\quad + b_{17} \partial_{\nu} h^{ \nu \alpha} h_{\mu \alpha} h_{\beta}^{\;\;\beta}
+ b_{18} \partial^{\nu} h^{ \alpha \beta} h_{\mu \alpha} h_{\nu \beta}
+ b_{19} \partial^{\nu} h_{ \nu \alpha} h_{\mu \beta} h^{\alpha \beta}
+ b_{20} \partial_{\alpha} h_{ \nu}^{\;\;\nu} h_{\mu \beta} h^{\alpha \beta}
\\ & \qquad\quad + b_{21} \partial^{\nu} h_{ \alpha}^{\;\;\alpha} h_{\mu \nu} h_{\beta}^{\;\;\beta}
+ b_{22} \partial^{\nu} h^{ \alpha \beta} h_{\mu \nu} h_{\alpha \beta}
\Big ] + \cdots
\end{align*}
\end{itemize}
\vspace{50mm}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Second Freedom -- Adding Total Derivative}}\\ Why is it allowed to add total derivative terms?} \small
\setbeamercovered{invisible}
\begin{itemize}
\item[$\bullet$] Principle of least action: \scriptsize
\begin{align*}
\tilde{\mathcal{L}} & = \mathcal{L} + \partial_{\mu} F^{\mu} (h) \quad \Rightarrow \\[3mm]
\delta \mathcal{\tilde{S}} = \delta \int d^4x \tilde{\mathcal{L}} = \delta \int d^4x \big( \mathcal{L} + \partial_{\mu} & F^{\mu} (h) \big ) = \delta \int d^4x \mathcal{L} + \delta \int d^4x \partial_{\mu} F^{\mu} (h) = 0
\end{align*}
\scriptsize{Where \(\delta \mathcal{S} =\delta \int d^4x \mathcal{L} = 0\), and the infinitesimal variation let the total derivative part to vanish at the boundary of the integration.} \\[4mm]
\small\item[$\bullet$] Integration by parts: \\[1mm] \scriptsize
\begin{align*}
& \int d^4x\, ({\color{blue}\phi\; \partial_{\mu}\partial^{\mu}\phi} ) = \phi
\; \partial^{\mu}\phi \Big\rvert_{\delta} - \int d^4 x\, \partial_{\mu}\phi\;
\partial^{\mu}\phi = - \int d^4 x\, \partial_{\mu}\phi\; \partial^{\mu}\phi \\[3mm]
& \int d^4 x\, ({\color{blue}\phi\; \partial_{\mu}\partial^{\mu}\phi} - \partial_{\mu} (\phi\;\partial^{\mu}\phi)) = \int d^4 x\, (\phi\;\partial_{\mu}\partial^{\mu}\phi - \partial_{\mu}\phi\; \partial^{\mu}\phi - \phi\;
\partial_{\mu}\partial^{\mu}\phi) = - \int d^4 x\, \partial_{\mu}\phi\; \partial^{\mu}\phi
\end{align*}
\vspace{1mm}
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{The transformation of fields will be a symmetry transformation if the
Lagrangian changes by a total derivative.\\ }
\tiny{The momentum conservation in a vertex.}
\end{itemize}
\vspace{30mm}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Second Freedom -- Adding Total
Derivative}}\\ Total Derivative Lagrangian For $h$} \scriptsize
\noindent\begin{flalign*}
\mathcal{L}_{\text{TD}}(h) = \frac{1}{\kappa^{2}} \partial^{\mu} \Bigg [ &
\kappa \Big [
a_1 \partial_{\mu} h_{ \nu}^{\;\;\nu}
+a_2 \partial^{\nu} h_{ \mu \nu}
\Big ]
+\kappa^2 \Big [
a_3 \partial_{\mu} h_{ \alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+a_4 \partial_{\mu} h^{ \alpha \nu} h_{\alpha \nu}
+a_5 \partial^{\alpha} h_{ \mu \alpha} h_{\nu}^{\;\;\nu}
\\ & +a_6 \partial_{\alpha} h_{ \mu \nu} h^{\alpha \nu}
+a_7 h_{\mu \nu} \partial_{\alpha} h^{ \alpha \nu}
+a_8 h_{\mu \alpha} \partial^{\alpha} h_{ \nu}^{\;\;\nu}
\Big ]
+\kappa^3 \Big [
a_9 \partial_{\mu} h_{ \nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
\\ & +a_{10} \partial_{\mu} h_{ \nu}^{\;\;\nu} h^{\alpha \beta} h_{\alpha \beta}
+a_{11} \partial_{\mu} h_{ \nu \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+a_{12} \partial_{\mu} h^{ \nu \alpha} h_{\nu}^{\;\; \beta} h_{\alpha \beta}
+a_{13} \partial^{\nu} h_{ \mu \nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
\\ & +a_{14} \partial^{\nu} h_{ \mu \nu} h^{\alpha \beta} h_{\alpha \beta}
+a_{15} \partial_{\nu} h_{ \mu \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+a_{16} \partial^{\nu} h_{ \mu \alpha} h_{\nu \beta} h^{\alpha \beta}
+a_{17} \partial_{\nu} h^{ \nu \alpha} h_{\mu \alpha} h_{\beta}^{\;\;\beta}
\\ & +a_{18} \partial_{\nu} h^{ \nu \alpha} h_{\mu \beta} h_{\alpha}^{\;\; \beta}
+a_{19} \partial^{\nu} h_{ \alpha}^{\;\;\alpha} h_{\mu \nu} h_{\beta}^{\;\;\beta}
+a_{20} \partial^{\nu} h^{ \alpha \beta} h_{\mu \nu} h_{\alpha \beta}
+a_{21} \partial_{\nu} h_{ \alpha}^{\;\;\alpha} h_{\mu \beta} h^{\nu \beta}
\\ & +a_{22} \partial^{\nu} h^{ \alpha \beta} h_{\mu \alpha} h_{\nu \beta}
\Big ]
+\kappa^4 \Big [
a_{23} \partial_{\mu} h_{ \alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta} h_{\gamma}^{\;\;\gamma} h_{\delta}^{\;\;\delta}
+a_{24} \partial_{\mu} h_{ \alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta} h^{\gamma \delta} h_{\gamma \delta}
\\ & +a_{25} \partial_{\mu} h_{ \alpha}^{\;\;\alpha} h^{\beta \gamma} h_{\gamma}^{\;\; \delta} h_{\delta \beta}
+a_{26} \partial^{\alpha} h_{ \mu \alpha} h_{\beta}^{\;\;\beta} h_{\gamma}^{\;\;\gamma} h_{\delta}^{\;\;\delta}
+a_{27} \partial^{\alpha} h_{ \mu \alpha} h_{\beta}^{\;\;\beta} h^{\gamma \delta} h_{\gamma \delta}
\\ & +a_{28} \partial^{\alpha} h_{ \mu \alpha} h^{\beta \gamma} h_{\gamma}^{\;\; \delta} h_{\delta \beta}
+a_{29} \partial_{\mu} h_{ \alpha \beta} h^{\alpha \beta} h_{\gamma}^{\;\;\gamma} h_{\delta}^{\;\;\delta}
+a_{30} \partial_{\mu} h_{ \alpha \beta} h^{\alpha \beta} h_{\gamma \delta} h^{\gamma \delta}
\\ & +a_{31} \partial_{\alpha} h_{ \mu \beta} h^{\alpha \beta} h_{\gamma}^{\;\;\gamma} h_{\delta}^{\;\;\delta}
+a_{32} \partial_{\alpha} h_{ \mu \beta} h^{\alpha \beta} h^{\gamma \delta} h_{\gamma \delta}
+a_{33} \partial^{\beta} h_{ \alpha}^{\;\;\alpha} h_{\mu \beta} h_{\gamma}^{\;\;\gamma} h_{\delta}^{\;\;\delta}
\\[2mm] & + \cdots
\\ & +a_{48} \partial_{\alpha} h_{ \mu \beta} h^{\alpha \gamma} h^{\beta \delta} h_{\gamma \delta}
+a_{49} \partial_{\mu} h^{ \alpha \beta} h_{\alpha}^{\;\; \gamma} h_{\beta \gamma} h_{\delta}^{\;\;\delta}
+a_{50} \partial_{\mu} h^{ \alpha \beta} h_{\alpha}^{\;\; \gamma} h_{\beta }^{\;\;\delta} h_{\gamma \delta} \Big ]
\Bigg ] + \cdots &&
\end{flalign*}\vspace{30mm}
\vspace*{30mm}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Second Freedom -- Adding Total
Derivative}}\\ Total Derivative Lagrangian For $\phi,h$} \scriptsize
\begin{flalign*}
\mathcal{L}_{\text{TD}}(\phi,h) = \partial^{\mu} \Bigg [
& d_1 \phi \partial_{\mu} \phi
+\kappa \Big [
d_2 \phi^2\partial_{\mu} h_{ \nu}^{\;\;\nu}
+d_3 \phi^2 \partial^{\nu} h_{ \nu \mu}
+d_4 \phi \partial_{\mu} \phi h_{\nu}^{\;\;\nu}
+d_5 \phi \partial^{\nu} \phi h_{\nu \mu}
\Big ]
\\ & +\kappa^2 \Big [
d_6 \phi^2\partial_{\mu} h_{ \nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha}
+d_7 \phi^2\partial_{\mu} h^{ \nu \alpha} h_{\nu \alpha}
+d_8 \phi^2\partial^{\nu} h_{ \mu \nu} h_{\alpha}^{\;\;\alpha}
+d_9 \phi^2\partial_{\nu} h_{ \mu \alpha} h^{\nu \alpha}
\\ & +d_{10} \phi^2\partial_{\nu} h^{ \nu \alpha} h_{\mu \alpha}
+d_{11} \phi^2\partial^{\nu} h_{ \alpha}^{\;\;\alpha} h_{\mu \nu}
+d_{12} \phi \partial_{\mu} \phi h_{\nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha}
+d_{13} \phi \partial_{\mu} \phi h^{\nu \alpha} h_{\nu \alpha}
\\ & +d_{14} \phi \partial^{\nu} \phi h_{\mu \nu} h_{\alpha}^{\;\;\alpha}
+d_{15} \phi \partial_{\nu} \phi h_{\mu \alpha} h^{\nu \alpha}
\Big ]
+\kappa^3 \Big [
d_{16} \phi^2\partial_{\mu} h^{ \nu \alpha} h_{\nu \alpha} h_{\beta}^{\;\;\beta}
\\ & +d_{17} \phi^2\partial^{\nu} h_{ \mu \nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+d_{18} \phi^2\partial_{\nu} h_{ \mu \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+d_{19} \phi^2\partial_{\nu} h^{ \nu \alpha} h_{\mu \alpha} h_{\beta}^{\;\;\beta}
\\ & +d_{20} \phi^2\partial^{\nu} h_{ \alpha}^{\;\;\alpha} h_{\mu \nu} h_{\beta}^{\;\;\beta}
+d_{21} \phi \partial_{\mu} \phi h_{\nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+d_{22} \phi \partial_{\mu} \phi h^{\nu \alpha} h_{\nu \alpha} h_{\beta}^{\;\;\beta}
\\ & +d_{23} \phi \partial^{\nu} \phi h_{\mu \nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+d_{24} \phi \partial_{\nu} \phi h_{\mu \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+d_{25} \phi^2\partial_{\mu} h_{ \nu}^{\;\;\nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
\Big ]
\Bigg ] + \cdots &&
\end{flalign*}
\vspace*{30mm}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Second Freedom -- Adding Total
Derivative}}\\ Total Derivative Lagrangian For $\chi, \bar{\chi},h$} \scriptsize
\begin{flalign*}
\mathcal{L}_{\text{TD}} (\chi, \bar{\chi},h) = \partial^{\mu} \Bigg [
& h_1 \bar{\chi}^{\nu} \partial_{\mu} \chi_{ \nu}
+\kappa \Big [
h_2 h^{\nu \alpha} \bar{\chi}_{\nu} \partial_{\mu} \chi_{ \alpha}
+h_3 \bar{\chi}_{\nu} \partial_{\alpha} \chi_{ \mu} h^{\nu \alpha}
+h_4 \bar{\chi}^{\nu} \partial_{\mu} \chi_{ \nu} h_{\alpha}^{\;\;\alpha}
\\ & +h_5 \bar{\chi}^{\nu} \partial_{\nu} \chi_{ \mu} h_{\alpha}^{\;\;\alpha}
+h_6 \bar{\chi}^{\nu} \partial^{\alpha} \chi_{ \alpha} h_{\mu \nu}
+h_7 \bar{\chi}^{\nu} \partial^{\alpha} \chi_{ \nu} h_{\mu \alpha}
+h_8 \bar{\chi}_{\nu} \partial^{\nu} \chi^{ \alpha} h_{\mu \alpha}
\\ & +h_9 \bar{\chi}_{\mu} \partial^{\nu} \chi_{ \nu} h_{\alpha}^{\;\;\alpha}
+h_{10} \bar{\chi}_{\mu} \partial^{\nu} \chi^{ \alpha} h_{\nu \alpha}
+h_{11} \bar{\chi}^{\nu} \partial^{\alpha} \chi_{ \nu} h_{\mu \alpha}
+h_{12} \bar{\chi}^{\nu} \partial^{\alpha} \chi_{ \alpha} h_{\nu \mu}
\\ & +h_{13} \bar{\chi}_{\nu} \partial_{\mu} \chi_{ \alpha} h^{\nu \alpha}
+h_{14} \bar{\chi}_{\nu} \partial_{\mu} \chi^{\nu} h_{\alpha}^{\;\;\alpha}
+h_{15} \bar{\chi}_{\mu} \partial_{\nu} \chi_{ \alpha} h^{\nu \alpha}
\Big ]
\\ & +\kappa^2 \Big [
h_{20} \bar{\chi}_{\mu} \partial^{\nu} \chi_{ \nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+h_{21} \bar{\chi}_{\mu} \partial_{\nu} \chi_{ \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+h_{22} \bar{\chi}_{\mu} \partial^{\nu} \chi_{ \alpha} h_{\nu \beta} h^{\alpha \beta}
\\ & +h_{23} \bar{\chi}_{\nu} \partial_{\mu} \chi^{ \nu} h_{\alpha}^{\;\;\alpha} h_{\beta}^{\;\;\beta}
+h_{24} \bar{\chi}_{\nu} \partial_{\mu} \chi^{\nu} h^{\alpha \beta} h_{\alpha \beta}
+h_{25} \bar{\chi}_{\nu} \partial_{\mu} \chi_{ \alpha} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
\\ & +h_{26} \bar{\chi}^{\nu} \partial_{\mu} \chi_{ \alpha} h_{\nu \beta} h^{\alpha \beta}
+h_{27} \bar{\chi}^{\nu} \partial^{\alpha} \chi_{ \nu} h_{\mu \alpha} h_{\beta}^{\;\;\beta}
+h_{28} \bar{\chi}^{\nu} \partial_{\alpha} \chi_{ \nu} h_{\mu \beta} h^{\alpha \beta}
\\ & +h_{29} \bar{\chi}_{\nu} \partial_{\alpha} \chi_{ \mu} h^{\nu \alpha} h_{\beta}^{\;\;\beta}
+h_{30} \bar{\chi}^{\nu} \partial_{\alpha} \chi_{ \mu} h_{\nu \beta} h^{\alpha \beta}
+h_{31} \bar{\chi}^{\nu} \partial^{\alpha} \chi_{ \alpha} h_{\nu \mu} h_{\beta}^{\;\;\beta}
\\ & +h_{32} \bar{\chi}_{\nu} \partial^{\alpha} \chi_{ \alpha} h^{\nu \beta} h_{\mu \beta}
+h_{33} \bar{\chi}^{\nu} \partial_{\alpha} \chi_{ \beta} h_{\nu \mu} h^{\alpha \beta}
+h_{34} \bar{\chi}^{\nu} \partial^{\alpha} \chi^{ \beta} h_{\nu \alpha} h_{\mu \beta}
\\ & +h_{35} \bar{\chi}_{\nu} \partial^{\alpha} \chi_{ \beta} h^{\nu\beta} h_{\mu \alpha}
+h_{36} \bar{\chi}_{\mu} \partial^{\nu} \chi_{ \nu} h^{\alpha \beta} h_{\alpha \beta}
\Big ]
\Bigg ] + \cdots &&
\end{flalign*}
\vspace*{30mm}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Third Freedom -- Field Redefinition}}\\ The Equivalence Theorem} \small
\setbeamercovered{invisible}
\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
The S-matrix in quantum field theory remains unchanged under reparameterization of the field operators.
\end{tcolorbox}
For scalar field $\phi$, the generating functional is given by:
\begin{align*}
\mathcal{Z} = \int D[\phi] \; \exp \Big ( i\; \int d^4x \; \mathcal{L}(\phi,\partial_{\mu} \phi) \Big )
\end{align*}
If we redefine the scalar field:
\begin{align*}
\phi = \tilde{\phi} \qquad\qquad \;\; \text{Where,}\; \tilde{\phi} = a_1 \phi + a_2 \phi^2 + \cdots
\end{align*}
We get:
\begin{align*}
\mathcal{Z} = \int D[\tilde{\phi}] \; \exp \Big ( i\; \int d^4x \; \mathcal{L} (\tilde{\phi},\partial_{\mu} \tilde{\phi}) \Big )
\end{align*}
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{This redefinition is allowed as long as the Jacobian of the integral is essentially one.}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Third Freedom -- Field Redefinition}}\\ How
can field redefinition simplify Lagrangian?} \small
\setbeamercovered{invisible}
\noindent\begin{align*}
h_{\mu \nu} \; \rightarrow \; \tilde{h}_{\mu \nu} = h_{\mu \nu} +\kappa \Big [
a_{1} h_{\mu \gamma} h_{\nu}^{\;\; \gamma} +a_{2} h_{\mu \nu} h_{\gamma}^{\;\; \gamma} \Big ] + \cdots
\end{align*}\vspace{3mm}
\small A term of the triple graviton vertex:
\begin{align*}
{\color{blue} h_{\mu\nu}}\, \partial^{\mu} h^{\nu\alpha} \partial_{\alpha}h_{\beta}^{\;\;\beta} \qquad \,\rightarrow\,\qquad {\color{blue} h_{\mu\nu}} \, \partial^{\mu} h^{\nu\alpha} \partial_{\alpha}h_{\beta}^{\;\;\beta} +& {\color{blue} a_{1}\, \kappa\, h_{\mu \gamma} h_{\nu}^{\;\; \gamma}} \, \partial^{\mu} h^{\nu\alpha} \partial_{\alpha}h_{\beta}^{\;\;\beta} \\
+& {\color{blue} a_{2}\, \kappa\, h_{\mu \nu} h_{\gamma}^{\;\; \gamma}} \, \partial^{\mu} h^{\nu\alpha} \partial_{\alpha}h_{\beta}^{\;\;\beta} + \cdots
\end{align*}\\[2mm]
\small Schematically:\\
\begin{figure}
~~~
\begin{axopicture}(50,50)
\SetWidth{1.}
\DoublePhoton(25,25)(25,50){2}{5}{2}
\DoublePhoton(25,25)(0,0){2}{5}{2}
\DoublePhoton(25,25)(50,0){2}{5}{2}
\SetWidth{0.8}
\Vertex(25,25){3}
\end{axopicture}\hspace{2mm}
~~~
\begin{axopicture}(10,50)(-10,0)
\SetWidth{1.5}
\Line(-10,25)(10,25)
\SetWidth{1.}
\Line(5,30)(10,25)
\Line(5,20)(10,25)
\end{axopicture}\hspace{5mm}
~~~
\begin{axopicture}(50,50)
\SetWidth{1.}
\DoublePhoton(25,25)(25,50){2}{5}{2}
\DoublePhoton(25,25)(0,0){2}{5}{2}
\DoublePhoton(25,25)(50,0){2}{5}{2}
\SetWidth{0.8}
\Vertex(25,25){3}
\end{axopicture}\hspace{3mm}
~~~
\begin{axopicture}(10,50)
\SetWidth{1}
\Line(0,25)(10,25)
\Line(5,20)(5,30)
\end{axopicture}\hspace{3mm}
~~~
\begin{axopicture}(50,50)
\SetWidth{1.}
\DoublePhoton(25,25)(0,50){2}{5}{2}
\DoublePhoton(25,25)(50,50){2}{5}{2}
\DoublePhoton(25,25)(0,0){2}{5}{2}
\DoublePhoton(25,25)(50,0){2}{5}{2}
\SetWidth{0.8}
\Vertex(25,25){3}
\end{axopicture}\hspace{3mm}
~~~
\begin{axopicture}(10,50)
\SetWidth{1}
\Line(0,25)(10,25)
\Line(5,20)(5,30)
\end{axopicture}
~~~
\begin{axopicture}(50,50)
\SetWidth{0.8}
\Vertex(25,25){1}
\Vertex(15,25){1}
\Vertex(35,25){1}
\end{axopicture}
\end{figure}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Third Freedom -- Field Redefinition}}\\
Field Redefinition For $h$ and $\phi$} \small
\noindent\begin{flalign*}
h_{\mu \nu} = h_{\mu \nu}
& +\kappa \Big [
c_{1} h_{\mu \alpha} h_{\nu}^{\;\; \alpha}
+c_{2} h_{\mu \nu} h_{\alpha}^{\;\; \alpha}
\Big ]
\\ & +\kappa^2 \Big [
c_{3} h_{\mu \nu} h_{\alpha}^{\;\; \alpha} h_{\beta}^{\;\; \beta}
+c_{4} h_{\mu \nu} h^{\alpha \beta} h_{\alpha \beta}
+c_{5} h_{\mu \alpha} h_{\nu}^{\;\; \alpha} h_{\beta}^{\;\; \beta}
+c_{6} h_{\mu \alpha} h_{\nu \beta} h^{\alpha \beta}
\Big ]
\\ & +\kappa^3 \Big [
c_{7 } h_{\mu \nu} h_{\alpha}^{\;\; \alpha} h_{\beta}^{\;\; \beta} h_{\gamma}^{\;\; \gamma}
+c_{8 } h_{\mu \nu} h_{\alpha}^{\;\; \alpha} h^{\beta \gamma} h_{\beta \gamma}
+c_{9} h_{\mu \nu} h^{\alpha \beta} h_{\beta}^{\;\; \gamma} h_{\gamma \alpha}
\\ & \qquad\quad +c_{10} h_{\mu \alpha} h_{\nu}^{\;\; \alpha} h_{\beta}^{\;\; \beta} h_{\gamma}^{\;\; \gamma}
+c_{11} h_{\mu \alpha} h_{\nu}^{\;\; \alpha} h_{\beta \gamma} h^{\beta \gamma}
+c_{12} h_{\mu \alpha} h_{\nu \beta} h^{\alpha \beta} h_{\gamma}^{\;\; \gamma}
\\ & \qquad\quad +c_{13} h_{\mu \alpha} h_{\nu}^{\;\; \beta} h^{\alpha \gamma} h_{\beta \gamma}
\Big ] + \cdots &&
\end{flalign*}
\begin{flalign*}
\phi = \phi
& +\kappa \Big [
e_1 h_{\alpha}^{\;\; \alpha} \phi
\Big ]
\\ & +\kappa^2 \Big [
e_2 h_{\alpha}^{\;\; \alpha} h_{\beta}^{\;\; \beta} \phi
+e_3 h_{\alpha \beta} h^{\alpha \beta} \phi
\Big ]
\\ & +\kappa^3 \Big [
e_4 h_{\alpha }^{\;\;\alpha} h_{\beta}^{\;\; \beta} h_{\gamma}^{\;\; \gamma} \phi
+e_5 h_{\alpha \beta} h^{\alpha \beta} h_{\gamma}^{\;\; \gamma} \phi
+e_6 h^{\alpha \beta} h_{\beta}^{\;\; \gamma} h_{\gamma \alpha} \phi
\Big ] + \cdots &&
\end{flalign*} \vspace{2mm}
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{For the lowest order vertices.}
\vspace{30mm}
\end{frame}
\begin{frame}{\centering \fbox{\tiny{Third Freedom -- Field Redefinition}}\\ Field Redefinition For
$\chi$ and $\bar{\chi}$} \small
\begin{flalign*}
\chi_{\mu} = \chi_{\mu}
& +\kappa \Big [
g_{1} h_{\alpha }^{\;\;\alpha} \chi_{\mu}
+g_{2} h_{\alpha \mu} \chi^{\alpha}
\Big ]
\\ & +\kappa^2 \Big [
g_{3} h_{\alpha}^{\;\; \alpha} h_{\beta}^{\;\; \beta} \chi_{\mu}
+g_{4} h^{\alpha \beta} h_{\alpha \beta} \chi_{\mu}
+g_{5} h_{\alpha }^{\;\;\alpha} h_{\beta \mu} \chi_{\beta}
+g_{6} h^{\alpha \beta} h_{\alpha \mu} \chi_{\beta}
\Big ] + \cdots &&
\end{flalign*}
\vspace{4mm}
\begin{flalign*}
\bar{\chi}_{\mu} = \bar{\chi}_{\mu}
& +\kappa \Big [
f_{1} h_{\alpha}^{\;\; \alpha} \bar{\chi}_{\mu}
+f_{2} h_{\alpha \mu} \bar{\chi}^{\alpha}
\Big ]
\\ & +\kappa^2 \Big [
f_{3 }h_{\alpha}^{\;\; \alpha} h_{\beta}^{\;\; \beta} \bar{\chi}_{\mu}
+f_{4} h^{\alpha \beta} h_{\alpha \beta} \bar{\chi}_{\mu}
+f_{5} h_{\alpha}^{\;\; \alpha} h_{\beta \mu} \bar{\chi}^{\beta}
+f_{6} h^{\alpha \beta} h_{\alpha \mu} \bar{\chi}_{\beta}
\Big ] + \cdots &&
\end{flalign*}
\vspace*{14.2mm}
\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{For the lowest order vertices.}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bf Loop Integral}
\begin{frame}[t]{Dimensional Regularization} \small
\setbeamercovered{invisible}
%\only<1>{\begin{columns}[onlytextwidth]
% \column{.30\textwidth}
%{\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
% \centering Dimension $x$
%\end{tcolorbox}
%\centering \footnotesize{The loop integral diverges}}
% \column{.40\textwidth}
% \Large\centering $\xRightarrow{\qquad d=x-2\epsilon \qquad}$ \\
% \Large\centering $\xLeftarrow[\text{\small{analytic continuation}}]{}$
% \column{.30\textwidth}
%{\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
% \centering Dimension $d$
%\end{tcolorbox}
%\centering \footnotesize{The loop integral converges}}
%\end{columns}}
\onslide<1>\begin{columns}[onlytextwidth]
\column{.30\textwidth}
\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
\centering Dimension $4$
\end{tcolorbox}
\centering \footnotesize{The loop integral diverges}
\column{.40\textwidth}
\Large\centering $\xRightarrow{\qquad d=4-2\epsilon \qquad}$\\
\Large\centering $\xLeftarrow[\text{\small{analytic continuation}}]{}$
\column{.30\textwidth}
\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
\centering Dimension $d$
\end{tcolorbox}
\centering \footnotesize{The loop integral converges}
\end{columns}
\begin{itemize}
\item[$\bullet$] Some considerations: \footnotesize
\begin{align*}
g^{\mu \nu}_4 \qquad &\rightarrow \qquad g^{\mu \nu}_d \qquad\qquad\qquad\qquad \Rightarrow \qquad g^{\mu
\nu} g_{\mu \nu} = \delta^\mu_\mu = d = 4-2\epsilon \\
\int \frac{d^4 p}{(2\pi)^4} \qquad &\rightarrow \qquad \int \frac{(\mu)^{2\epsilon} \; d^d p}{(2\pi)^d}
\end{align*}
\scriptsize Where $\mu$ is a regulator parameter of dimensional regularization with dimension $[\mu]=M^{\epsilon}$.
\small\item[$\bullet$] Mathematically:\\
Transfer to Euclidean space, do the Wick rotation, apply Feynman parameters, shift the integration variable, perform the integral, go back to Minkowski space.
\end{itemize}\vspace{2mm}
\onslide<1->\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
\tiny{This scheme preserves the gauge and Lorentz invariances.}
\end{frame}
\begin{frame}[t]{\centering \fbox{\tiny{Loop Integrals -- Scalar Integrals}}\\ Classifying the loop Integrals}
\setbeamercovered{invisible}
\setlength{\abovedisplayskip}{0.5pt}
\setlength{\belowdisplayskip}{0pt}
\begin{columns}
\begin{column}{0.75\textwidth}
\centering\scriptsize\begin{align*}
& I^N ( p_1,...,p_{N-1},m_0,...,m_{N-1}) \sim \\
&\;\;\;\; \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-m_0^2+i\epsilon)((k+q_1)^2-m_1^2+i\epsilon)\, \cdots \, ((k+q_{N-1})^2-m_{N-1}^2+i\epsilon)}
\end{align*}
\vspace*{5mm}
\raggedright\tiny Where $q_i=p_1+...+p_i=\sum_{k=1}^i p_k$
\end{column}
\begin{column}{0.25\textwidth}
\begin{figure}
\centering
\scalebox{.42}{
\begin{axopicture}(140,120)
\SetWidth{1.5}
\Arc[arrow](70,70)(50,0,45)
\Arc[arrow](70,70)(50,45,90)
\Arc[arrow](70,70)(50,90,135)
\Arc[arrow](70,70)(50,135,270)
\Arc[arrow](70,70)(50,270,315)
\Arc[arrow](70,70)(50,315,360)
\Line[arrow](170,70)(120,70) %p1
\Text(112,55)[r]{$ m_0 $}
\Text(123,45)(40){$ k $}
\Text(160,75){$p_1$}
\Text(115,83)[r]{$ m_1 $}
\Text(132,100)(40){$ k+q_1 $}
\Line[arrow](70,-10)(70,20) %pN-1
\Text(65,0)[r]{$p_{N-1}$}
\Text(88,33)(35){$ m_{N-1} $}
\Line[arrow](70,160)(70,120) %p3
\Text(65,140)[r]{$p_3$}
\Text(65,106)[r]{$ m_3 $}
\Line[arrow](140,0)(105,35) %pN
\Text(115,5)[l]{$p_{N}$}
\Line[arrow](140,140)(105,105) %p2
\Text(135,130)[l]{$p_2$}
\Text(95,106)[r]{$ m_2 $}
\Text(97,135)(70){$ k+q_2 $}
\Line[arrow](0,140)(35,105)
\Vertex(0,80){2.5}
\Vertex(0,53){2.5}
\Vertex(12,30){2.5}
\Vertex(120,70){2}
\Vertex(70,120){2}
\Vertex(70,20){2}
\Vertex(105,105){2}
\Vertex(35,105){2}
\end{axopicture}}
\end{figure}
%\vspace*{1mm}
\end{column}
\end{columns}
\vspace{1mm}
\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
\scriptsize\begin{align*}
&A_0(m_0) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)} \\
&B_0(p_1,m_0,m_1) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)((k+q_1)^2-{m}^2_1+i\epsilon)} \\
& C_0(p_1,p_2,m_0,m_1,m_2) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)((k+q_1)^2-{m}^2_1+i\epsilon)((k+q_2)^2-{m}^2_2+i\epsilon)} \\
& D_0(p_1,p_2,p_3,m_0,m_1,m_2,m_3) = \\
& \qquad\;\,\, \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)((k+q_1)^2-{m}^2_1+i\epsilon)((k+q_2)^2-{m}^2_2+i\epsilon)((k+q_3)^2-{m}^2_3+i\epsilon)}
\end{align*}
\end{tcolorbox}
%\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
%\tiny{In our case $m_0,...,m_{N-1}=\{0,m\}$ \hfill calculated by the standard procedure}}
\vspace{100mm}
\end{frame}
\begin{frame}[t]{\centering \fbox{\tiny{Loop Integrals -- Tensor Integrals}}\\ Classifying the loop Integrals}
\setbeamercovered{invisible}
\setlength{\abovedisplayskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\begin{columns}
\begin{column}{0.75\textwidth}
\centering\scriptsize\begin{align*}
&I^N_{\mu_1, ..., \mu_M}(p_1,...,p_{N-1},m_0,...,m_{N-1}) \sim \\
&\;\;\;\; \int \frac{d^d k}{(2\pi)^d} \frac{k_{\mu_1} \,\cdots\, k_{\mu_M}}{(k^2-m_0^2+i\epsilon)((k+q_1)^2-m_1^2+i\epsilon)\,\cdots\, ((k+q_{N-1})^2-m_{N-1}^2+i\epsilon)}
\end{align*}
\vspace*{5mm}
\raggedright\tiny Where $q_i=p_1+...+p_i=\sum_{k=1}^i p_k$
\end{column}
\begin{column}{0.25\textwidth}
\begin{figure}
\centering
\scalebox{.42}{
\begin{axopicture}(140,120)
\SetWidth{1.5}
\Arc[arrow](70,70)(50,0,45)
\Arc[arrow](70,70)(50,45,90)
\Arc[arrow](70,70)(50,90,135)
\Arc[arrow](70,70)(50,135,270)
\Arc[arrow](70,70)(50,270,315)
\Arc[arrow](70,70)(50,315,360)
\Line[arrow](170,70)(120,70) %p1
\Text(112,55)[r]{$ m_0 $}
\Text(123,45)(40){$ k $}
\Text(160,75){$p_1$}
\Text(115,83)[r]{$ m_1 $}
\Text(132,100)(40){$ k+q_1 $}
\Line[arrow](70,-10)(70,20) %pN-1
\Text(65,0)[r]{$p_{N-1}$}
\Text(88,33)(35){$ m_{N-1} $}
\Line[arrow](70,160)(70,120) %p3
\Text(65,140)[r]{$p_3$}
\Text(65,106)[r]{$ m_3 $}
\Line[arrow](140,0)(105,35) %pN
\Text(115,5)[l]{$p_{N}$}
\Line[arrow](140,140)(105,105) %p2
\Text(135,130)[l]{$p_2$}
\Text(95,106)[r]{$ m_2 $}
\Text(97,135)(70){$ k+q_2 $}
\Line[arrow](0,140)(35,105)
\Vertex(0,80){2.5}
\Vertex(0,53){2.5}
\Vertex(12,30){2.5}
\Vertex(120,70){2}
\Vertex(70,120){2}
\Vertex(70,20){2}
\Vertex(105,105){2}
\Vertex(35,105){2}
\end{axopicture}}
\end{figure}
%\vspace*{1mm}
\end{column}
\end{columns}
\vspace{6.5mm}
\begin{tcolorbox}[enhanced,width=\textwidth,colframe=LUCopper,arc=4pt,boxrule=1pt,drop fuzzy shadow]
\scriptsize\begin{align*}
& A_{\mu}\qquad , \qquad A_{\mu\nu}\qquad , \qquad A_{\mu\nu\alpha}\qquad , \qquad A_{\mu\nu\alpha\beta} \\[3mm]
& B_{\mu}\qquad , \qquad B_{\mu\nu}\qquad , \qquad B_{\mu\nu\alpha}\qquad , \qquad B_{\mu\nu\alpha\beta}\qquad , \qquad B_{\mu\nu\alpha\beta\rho} \\[3mm]
& C_{\mu}\qquad , \qquad C_{\mu\nu}\qquad , \qquad C_{\mu\nu\alpha}\qquad , \qquad C_{\mu\nu\alpha\beta}\qquad , \qquad C_{\mu\nu\alpha\beta\rho}\qquad , \qquad C_{\mu\nu\alpha\beta\rho\sigma} \\[3mm]
& D_{\mu}\qquad , \qquad D_{\mu\nu}\qquad , \qquad D_{\mu\nu\alpha}\qquad , \qquad D_{\mu\nu\alpha\beta}
\end{align*}
\end{tcolorbox}
%\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
%\tiny{In our case $m_0,...,m_{N-1}=\{0,m\}$ \hfill calculated by the standard procedure}}
\vspace{100mm}
\end{frame}
%\begin{frame}[t]{\centering \fbox{\tiny{Loop Integrals -- Scalar Integrals}}\\ Our Scalar Integrals} \scriptsize
%\setbeamercovered{invisible}
%\begin{figure}
% \centering
% \scalebox{0.5}{
% \begin{axopicture}(100,60)(0,-10)
% \SetWidth{1.5}
% \Line(0,10)(100,10)
% \SetWidth{1.}
% \DoublePhoton(0,50)(100,50){2}{12}{2}
% \DoublePhoton(30,10)(30,50){2}{5}{2}
% \DoublePhoton(70,10)(70,50){2}{5}{2}
% \Vertex(30,10){3}
% \Vertex(70,10){3}
% \Vertex(30,50){3}
% \Vertex(70,50){3}
% \end{axopicture} \hspace{20mm}
% ~~~
% \begin{axopicture}(100,100)(0,-30)
% \SetWidth{1.5}
% \Line(0,10)(100,10)
% \SetWidth{1.}
% \DoublePhoton(50,40)(50,70){2}{5}{2}
% \DoublePhotonArc(50,10)(30,0,180){2}{10}{2}
% \DoublePhoton(50,10)(50,-20){2}{5}{2}
% \Vertex(20,10){3}
% \Vertex(80,10){3}
% \Vertex(50,40){3}
% \Vertex(50,10){3}
% \end{axopicture} \hspace{20mm}
% ~~~
% \begin{axopicture}(100,90)(0,-30)
% \SetWidth{1.5}
% \Line(0,10)(100,10)
% \SetWidth{1.}
% \DoublePhotonArc(50,10)(30,0,180){2}{10}{2}
% \DoublePhoton(40,-20)(40,10){2}{5}{2}
% \DoublePhoton(60,-20)(60,10){2}{5}{2}
% \Vertex(20,10){3}
% \Vertex(80,10){3}
% \Vertex(40,10){3}
% \Vertex(60,10){3}
% \end{axopicture}}
%\end{figure}
%\scriptsize\begin{align*}
% &A_0(m_0) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)} \\
% &B_0(p_1,m_0,m_1) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)((k+q_1)^2-{m}^2_1+i\epsilon)} \\
% & C_0(p_1,p_2,m_0,m_1,m_2) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)((k+q_1)^2-{m}^2_1+i\epsilon)((k+q_2)^2-{m}^2_2+i\epsilon)} \\
% & D_0(p_1,p_2,p_3,m_0,m_1,m_2,m_3) = \\
% & \qquad\quad\;\,\, \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2_0+i\epsilon)((k+q_1)^2-{m}^2_1+i\epsilon)((k+q_2)^2-{m}^2_2+i\epsilon)((k+q_3)^2-{m}^2_3+i\epsilon)}
%\end{align*}\\[1.9mm]
%\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
%\tiny{In our case $m_0,...,m_{N-1}=\{0,m\}$.}
%\\ \vspace{30mm}
%\end{frame}
%\begin{frame}[t]{\centering \fbox{\tiny{Loop Integrals -- Scalar Integrals}}\\
% Calculating scalar integrals}
%\setbeamercovered{invisible}
%\begin{figure}[H]
% \vspace{8mm}
% \centering
% \scalebox{0.9}{
% \begin{axopicture}(100,60)(0,-10)
% \SetWidth{1.5}
% \Arc(50,30)(20,0,360)
% \SetWidth{1.}
% \Line[arrow](0,20)(15,20)
% \DoublePhoton(0,30)(30,30){2}{5}{2}
% \DoublePhoton(70,30)(100,30){2}{5}{2}
% \Arc[arrow](50,30)(25,70,110)
% \Arc[arrow](50,30)(25,250,290)
% \Vertex(30,30){3}
% \Vertex(70,30){3}
% \Text(50,58)[b]{$k$}
% \Text(50,2)[t]{$k+p_1$}
% \Text(15,35)[rb]{$ h_{\mu\nu} (p_1) $}
% \Text(90,35)[lb]{$h_{\alpha\beta}(p_1)$}
% \end{axopicture}}
%\end{figure}
%
%\begin{empheq}[box=\fbox]{align*}
% B_0(p_1,m,m) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2-{m}^2+i\epsilon)((k+p_1)^2-{m}^2+ i \epsilon)}
%\end{empheq}
%\vspace{11mm}
%\textcolor{LUCopper}{\rule{\textwidth}{1pt}}
%\tiny{Where $q_1=p_1$ and $m_0=m_1=m$.}
%\end{frame}
%
%\begin{frame}[t]{\centering \fbox{\tiny{Loop Integrals -- Scalar Integrals}}\\
% Calculating scalar integrals} \scriptsize
%\setbeamercovered{invisible}
%\begin{itemize}
%\onslide<1->\item[$\bullet$] Transferring to Euclidean space $k_0 \rightarrow
%ik_4$ and doing Wick rotation:
%\only<1>{\begin{align*}
% B_0(p_1,m,m) =& \int \frac{i d^d k_E}{(2\pi)^d}