DAILY PHYSICS

 Holonomy. Given a manifold, M, and a point p on M. We append a vector,v  at p, and subject it to parallel transport around a loop. The number of times we revolve around said loop to match the original orientation of v at p is the holonomy. Take a loop over $I=[0,1]  i.e f(0) = f(1)$ We can write the group of parallel transport maps around a loop  $Hol_p(g)$. Naturally $Hol_p(g) \rightarrow GL(T_p M)$

-Recall a manifold -Construct a Kahler manifold: complex manifold with Reimann metric -Force Ricci flat -make compact -Two topologically distinct Calabi Yau can have the same stringy physics  TO FORCE ME TO SHOW THE MATH LEAVE A COMMENT OR CALL ME 818-310-5698.  Actually, maybe read this article by Brian Green first: https://arxiv.org/pdf/hep-th/9702155.pdf

 Chiral anomaly: Current not conserved!

 Consider: $X(\tau,\sigma + 2 \pi) = X(\tau,\sigma) + m(2\pi R)$ Set  $X(\tau,\sigma + 2\pi) = X(\tau, \sigma) + 2\pi \alpha$ Follow through as usual.  - Do Fourier mode expansion -Compute the momentum, and write the Hamiltonian -Use creation and annihilation operators to compute the spectrum At the level of the spectrum: "the closed string spectrum for a compactification with radius R is identical to the closed string spectrum for a compactification with radius $\bar{R} = \frac{\alpha}{R}$."

 No time to write trivial mathematical details. Might update this with details in the future.  Recipe: Consider first the Free boson first, apply a similar technique to Majorana, then get Dirac directly     - Recall:            - Two point function  equations:  $G^{2} = \langle \phi(x) \phi(y) \rangle$ $ -g \partial_x^2 G^{(2)} (x,y) = \delta( x-y)$ $G^{2} = \langle \phi(x) \phi(y) \rangle =  \frac{1}{4 pi g} ln^2( x-y)$ The energy-momentum tensor can be computed: $T = g(\partial_\mu \phi \partial_nu \phi - \frac{1}{2} \eta_{\mu \nu} \partial_\rho \phi \partial^\rho \phi )$ OPE is calculated as $T(z)T(w) = (2 \pi  g)^2:\partial \phi(z) \partial \phi(z)::\partial \phi(w) \partial \phi(w):$ Read the central charge from the general form! $T(z)T(w) = \frac{c/2}{ (z-w)^4} + \frac{2}{(z-w)^2} T(w) + \frac{\partial  T(w)}{z-w} +  . . $ Read the conformal dimension! $T(z) \Phi(w, \bar(w)) =  (\frac{h}{(z-w)^2} + \frac{\partial}{z -w}) \phi(w,\bar(w))$ For free fermion - Write Majorana action -

 -- Guide: -Couple field EM to exotic scalar field -Compute -Call this dark matter --Mathematical details are obvious

 Take graviton apply susy transform get spin 3/2 fermion. Write Rarita -Schwinger lagrangian density: $\mathcal{L} = \bar{\psi}^{\mu \nu \rho} \partial_\mu \psi_\rho$ Play the game: This is the gravitino

  ====> Thermal Dark Matter Equilibrium reactions of dark matter particles with                Forward-rate $=$ Backward-rate   -- Probability of interacting in time t is: Number-density$\times$ cross-section $\times$ velocity $\times$ time   -- reaction rate: Number-density$\times$ cross-section $\times$ velocity    -- So $\Gamma =n <\sigma v>$ distribution  function  = $f_A = \frac{1}{e^{(E - u)/T} \pm 1}$ --non-relativistic:: $n^{eq} = \frac{g}{2 \pi^3}\int f(p) d^3p$ --relativistic:: $n^{eq} = \frac{\xi(3)}{\pi^2} g T^4$  multiplied by 1 if bosons or 3/4 if fermions For non-relativistic $\Gamma_{inelastic}$ ~$T^{\frac{3}{2}}e^{\frac{m}{T}} <\sigma v>$ Forrelativistic $\Gamma_{elastic}$ ~$T^3 <\sigma v> $ $\Gamma_{inelastic} \neq \Gamma_{elastic}$ -- Can identify with Hubble rate to get "freeze-out " values!                           ----- Kinetic decoupling ----- One process dominates and continues after the other shuts off   --- Consider delta in tim

Lower bound on bosonic dark matter                    - Use the uncertainty principle                                             pack all particles into one cell of volume                            -$\delta x \delta p$~ $ 1$                                   -$\ 2R_{halo} \times m \times v$~ $ 1$                      $m_{boson} \geq 10^{-23}$  given some data Lower bound on Fermionic dark matter                          - From fermi exclusion spread out particles throughout the volume                                   $M_{halo} =m_{fermion} V \int f(p) d^3 p$                      $f(p) = 1$              -Add data          $m_{fermion} \geq 10eV$                                

- Mean free path of stellar collision      collision time  ~$\frac{mean-free-path }{velocity-of-stars}$~${10^{21}} years$ - Circular velocity    $\frac{v^2}{R} = \frac{GM(R)}{R^2}$     $v = \sqrt{\frac{GM(R)}{R}}$     $M(R) = \frac{v^2 R}{G}$ -rotation curves      v(R) -Contradicts observation  -- need new particles          -Getting observed flattening of the curve for dark matter halo                 $M$ ~ $R$                 $V$ ~$ R^3$                 $\rho$ ~ $1/R^2$       -- Getting some data from experiments      -$\rho ~0.3 \frac{GeV}{cm^3}$      -$R_{halo}$ ~ $100$ kilo-parsec                      - Halo mass is computable                      - Average velocity is computable  ~ 200km/s            

 -- Schrodinger functional from Schrodinger representation of Quantum field theory  Following Liu's notes  and arxiv paper combined $<\phi_2(\vec{x}, t_2)| \phi_1(\vec{x}, t_1)> = \int_{\phi(t_1,\vec{x})} ^{\phi(t_1,\vec{x})}  D \phi(\vec{x},t)e^{i S[\phi]}$ Vacuum functional $\Psi_0[\phi(\vec{x})] = <\phi(\vec{x})| 0> = \int_{\phi(t_E < 0)} ^{\phi(t_E =0, \vec{x})}  D \phi(\vec{x},t)e^{- S_E [\phi]}$ $H = \int dx [\frac{1}{2}(\frac{\partial \phi}{\partial t_E})^2 + \frac{1}{2}(\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2]$ $\Psi_0 [\phi(X)] = \langle \phi_R|e^{-i(-i \pi) H}| \phi_L \rangle$ $| 0\rangle  \propto \sum_n  e^{- \pi  E_n }| n \rangle \otimes | n \rangle$ $\rho_{Rindler}$ is then obvious

Following from  Hong Liu's Lectures on Holography and, Gustavo Cesar Valdivia Mera's https://arxiv.org/pdf/2001.09869.pdf Considering Rindler Space:    Starting with the accelerated observer:         $x^\mu = (x^0;x^1$         $u^\mu = (\frac{dx^0}{d\tau}; \frac{dx^1}{d \tau})$                    $= (\gamma;\gamma v)$               Instantaneous co-moving frame         $a^\mu = (0;g)$          $a^\mu a_nu = a^0a_0 + a^1 a_1 = g^2$  $t(\tau ) = \frac{1}{g} \sinh({g \tau})$  $x(\tau) = \frac{1}{g} \cosh ({g\tau})$ . . . . . ,  and so on from the arxiv article then in exponential terms     $x = \frac{ \frac{e^{g \tau}}{g} + \frac{e^{g \tau}}{g}}{2}$   $t = \frac{  \frac{e^{g \tau}}{g} -  \frac{e^{g \tau}}{g}}{2}$ $\bar{v} = t + x$ $\bar{u} = t - x$ $\bar{v} = \frac{e^{g \tau}}{g}$ $\bar{u} = - \frac{e^{g \tau}}{g}$ setting spatial coordinate in Rindler space to $\xi$ . . . . . $x = \frac{e^{g\xi}}{g}\cosh(g \tau)$ $t = \frac{e^{g\xi}}{g}\sinh(g \tau)$ $ds^2 = e^{2g \xi}(-d\tau^2 +

Thermal equilibrium state  $<\bar{X}>_T = \frac{1}{Z} Tr(\bar{X} e^{- \beta H})$                          =$Tr(\bar{X \rho_T})$ where $\rho_T = \rho_{\beta} = \frac{e^{- \beta H}}{Z}$   and         $Z = \sum_i e^{- \beta E_i} $ Double Language, Entanglement, and Umezawa -- Thermofield Double $\psi = \frac{1}{Z} \sum_n e^{- \frac{\beta}{2} E_n} |n\rangle_1 \otimes |n \rangle_2$ For  $\bar{X}_1$ $\langle \psi | \bar{X}_1 | \psi \rangle = \frac{1}{Z} \sum_n  e^{- \beta E_n} \langle n|\bar{X}| n \rangle  = <\bar{X}>$ $Tr_2(| \psi \rangle \langle \psi|) = \frac{1}{Z} \sum_n  e^{- \beta E_n}|n\rangle_1 {_1}\langle n|$                                                    $= \rho_\beta$

$\{Q,Q^{\dagger}\} = H$            ----- See later, but for central extension  :  $\{Q,Q^{\dagger}\} = H  + Z_{ij}$ $\{Q,Q\} = 0$ $\{Q^{\dagger},Q^{\dagger}\}= 0$ $\{Q_{\alpha}^I,Q_{\beta}^J\}= \sigma^0_{\alpha, \beta} H \delta^{IJ}$ Central Extension. $\{Q_{\alpha}^I,Q_{\beta}^J\}= \epsilon_{\alpha, \beta} H Z^{IJ}$ Central charges $Z^{IJ} = Z^{JI}$   $[Z^{IJ}, H] = 0$ Partial symmetry breaking      Given a few supercharges:             --Say two of them $Q_1$, $Q_2$             --It is possible to have                             -- $Q_2| 0 \rangle \neq 0$                             -- $Q_1| 0 \rangle = 0$          This is partial symmetry breaking R-Symmetry Given a global symmetry + A supersymmetry. Let $T_l$ be the generator of the global symmetry  $[T_l,  Q^I_\alpha] = 0$ $[T_l,  Q^I_\alpha] \neq 0$  R-symmetry

Getting to sigma models  Consider $\phi(x^i,t)$       Let $\phi(t)$ be  $0 + 1D$ QFT       Let $\mathcal{L} = \frac{1}{2} \frac{d \phi(t)}{dt} \frac{d \phi(t)}{dt}$        $\phi: \mathcal{R} \rightarrow \mathcal{R}$         $\phi$: base space $\rightarrow $ target space        Consider now $\mathcal{L} = \frac{1}{2} g_{ij} \frac{d \phi^i}{dt} \frac{d \phi^j}{dt}$                $\phi: \mathcal{R} \rightarrow \mathcal{M}$       $\mathcal{L} = \frac{1}{2} g_{ij} \frac{d \phi^i}{dt} \frac{d \phi^j}{dt}$          Setting $g_{ij}$ to $\delta_{ij}$          $\mathcal{L} = \frac{1}{2} \dot{\phi_1}^2 +  . . . . . \frac{1}{2} \dot{\phi_n}^2$        $\phi$ is a scalar             Let  Dirac spinors be $\psi^i$              $\mathcal{L_f} = g_{ij}\frac{1}{2} {\psi^i}_\alpha  (\gamma^{\mu}  \partial_\mu)_{\alpha \beta} {\psi^j} _{\beta}$      where $\alpha$ is the spinor index         $\mathcal{L_f} = \frac{i}{2} \delta_{ij}  \psi^i_\alpha \partial_t \psi^j_\alpha$         $\mathcal{L_f} = \frac{i

Bosonic Harmonic Oscillator: $H = a a^{\dagger} + \frac{1}{2}$ Where   $a = \frac{1}{\sqrt{2}}(p - i x)$ $a^{\dagger} = \frac{1}{\sqrt{2}}(p - i x)$ with commutation relations : $[a^{\dagger}, a] =1$ $[a^{\dagger}, a^{\dagger}] =0$ $[a, a] =0$ Fermionic Harmonic Oscillator: $H = b b^{\dagger} - \frac{1}{2}$ with anticommutation relations $\{b^{\dagger}, b^{\dagger}\} =0$ $\{b, b\}=0$ Consider:  $H = aa^{\dagger} + bb^{\dagger}$ with  Fock space:   $| \#bosons, \#fermions \rangle$ One seeks for example: $Q|1, 0 \rangle = |1, 1 \rangle$ $Q = a b^{\dagger}$ $Q^{\dagger} = b a^{\dagger}$ $Q$ is called a supercharge. Bosonic state: Even number of fermions Fermionic state: Odd number of fermions Consider $N = 2 $ supersymmetry $Q_1|boson 1 \rangle = |Fermion 1 \rangle$ $Q_2|boson 1 \rangle = |Fermion 2 \rangle$ $Q_1^{\dagger}|Fermion 1\rangle = |boson 1 \rangle$ $Q_2^{\dagger}|Fermion 1 \rangle = |boson 2\rangle$ Supersymmetry can be broken      - Spontaneously broken             - $[H, Q] =

I could not be more pleased to announce that this space of the internet will be dedicated to whatever physics topics I stumble upon in the day and get motivated enough to try to understand. I hope to engage in conversation with plenty of people over these topics. There are no rules here and there is no judgment of any form. Just type away any of your observations and comments on the content.