# Theoretical Physics Reference guide based on: How to become a GOOD Theoretical Physicist by Gerard 't Hooft

## 1 Languages

1. The acquisition-learning hypothesis is a hypothesis that forms part of Stephen Krashen's

theory of second language acquisition. It states that there are two independent ways in which we develop our linguistic skills: acquisition and learning. According to Krashen acquisition is more important than learning. https://en.wikipedia.org/wiki/Input_hypothesis

1. Learn through audiobook:
2. Learning:

### 1.1English.

English is a prerequisite. All sciences nowadays are in English.

1. Learning:

1. Learning:

## 3 Classical Mechanics

### 3.5 Poisson’s brackets.

A function $$f(p, q, t)$$ of the phase space coordinates of the system and time as total time derivative

$\frac{d f}{d t}=\frac{\partial f}{\partial t}+\sum_{i}\left(\frac{\partial f}{\partial q_{i}} \dot{q}_{i}+\frac{\partial f}{\partial p_{i}} \dot{p}_{i}\right),$

often written as

$\frac{d f}{d t}=\frac{\partial f}{\partial t} + \{H, f\},$

where $$\{H, f\}$$ is the Poisson bracket. In canonical coordinates $$(q_i, p_i)$$ given (any) two functions $$f(p_i, q_i, t)$$ and $$g(p_i, q_i, t)$$ the Poisson bracket takes the form,

$\{f, g\}=\sum_{i=1}^{N}\left(\frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}}-\frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}}\right).$

The canonical coordinates satisfy the fundamental Poisson bracket relations

• $$\left\{q^{i}, q^{j}\right\}=0$$
• $$\left\{p_{i}, p_{j}\right\}=0$$
• $$\left\{q_{i}, p_{j}\right\}=\delta_{i j}$$

The Poisson bracket has the following properties:

1. $$\{f, g\} = -\{g, f\}$$,
2. $$\{f, c\} = 0$$,
3. $$\{f, q_i\} = -\frac{\partial f}{\partial p_i}$$,
4. $$\{f, p_i\} = \frac{\partial f}{\partial q_i}$$,
5. $$\{f_1+f_2, g\} = \{f_1, g\} + \{f_2, g\}$$,
6. $$\{f_1f_2, g\} = f_1\{f_2, g\} + \{f_1, g\}f_2$$,
7. $$\frac{\partial}{\partial t}\{f, g\} = \left\{\frac{\partial f}{\partial t}, g\right\} + \left\{f, \frac{\partial g}{\partial t}\right\}$$,

In case two, $$c$$ is a constant of the motion, also called an integral of the motion.

An important identity satisfied by the Poisson brackets is the Jacob identity

$\{f,\{g, h\}\} + \{g,\{h, f\}\} + \{h,\{f, g\}\} = 0.$

1. Hamilton's equations are

$\dot{q}=\frac{\partial H}{\partial p}\\ \dot{p}=-\frac{\partial H}{\partial q}$

with the Poisson bracket we have

\begin{aligned} \dot{q} &=\{q, H(q, p)\} \\ \dot{p} &=\{p, H(q, p)\} \end{aligned}

## 8 Computational Physic

### 8.1 Numerical Integration and Differentiation

#### 8.1.1 Derivative by:

1. Finite difference https://en.wikipedia.org/wiki/Finite_difference_method

The derivative of a function $$f(x)$$ can be approximated by the forward difference based on the size $$h$$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \approx \frac{f(x+h)-f(x)}{h} - \frac{h}{2}f^{\prime\prime}(\xi).$

First order of a Taylor expansion of $$f(x)$$ with truncation error $$O(h) = hf(\xi)^{\prime\prime}/2$$ linearly related to $$h$$. The approximation is poor unless $$h$$ is sufficient small. A final expression of this example and its order

$\frac{f(x+h)-f(x)}{h} = f^{\prime}(x) + O(h).$

The derivative can be also approximated by the backward difference, replacing $$h$$ by $$-h$$ in the Taylor expansion,

$f^{\prime}(x) \approx \frac{f(x)-f(x-h)}{h}.$

Subtracting the backward difference from the forward difference and dividing everything by $$2h$$, we find the central difference

\begin{aligned} f^{\prime}(x) &=\frac{f(x+h)-f(x-h)}{2 h}-\left(\frac{f^{\prime \prime \prime}(x)}{3 !} h^{2}+\frac{f^{(5)}(x)}{5 !} h^{4}+\cdots\right) \\ & \approx \frac{f(x+h)-f(x-h)}{2 h}, \end{aligned} with the second order truncation error $$O(h)$$. Adding the forward difference to the backward difference and solving for $$f(x)^{\prime\prime}$$ we can get the approximation for the second order derivative

$\boxed{f(x)^{\prime\prime} \approx \frac{f(x-h)-2 f(x)+f(x+h)}{h^{2}}}$

2. Interpolation

The derivative of a given function $$f(t)$$ can be approximated by that of its polynomial interpolation based on a set of $$n+1$$ points $$y_i = f(x_i), (i=0,\cdots, n)$$, such as the Lagrange polynomial interpolation (https://en.wikipedia.org/wiki/Polynomial_interpolation)

$f(x) \approx L_{n}(x)=\sum_{i=0}^{n} y_{i} l_{i}(x).$

Taking the derivative of $$f(x)$$,

$f^{\prime}(x) \approx L_{n}^{\prime}(x)=\sum_{i=0}^{n} y_{i} l_{i}^{\prime}(x)=\sum_{i=0}^{n} y_{i} d_{i}(x),$ where

$d_{i}(x)=\frac{d}{d x} l_{i}(x)=\frac{d}{d x}\left(\prod_{j=0, j \neq i}^{n} \frac{x-x_{j}}{x_{i}-x_{j}}\right).$

The error is

\begin{aligned} E_{n}(x)=R_{n}^{\prime}(x) &=\frac{1}{(n+1) !} \frac{d}{d x}\left(\left(f^{(n+1)}(\xi(x)) l(x)\right)\right.\\ &=\frac{1}{(n+1) !}\left[l^{\prime}(x) f^{(n+1)}(\xi(x))+l(x) \frac{d}{d x} f^{(n+1)}(\xi(x))\right]. \end{aligned}

The error is unknown in general since $$\xi$$ is not known explicit. But we can have for each node $$x_i$$ where $$w(x_i)$$,

$E_{n}(x)=R_{n}^{\prime}(x)=l^{\prime}(x) \frac{f^{(n+1)}(\xi(x))}{(n+1) !}.$

1. Special cases:
• $$n=1$$, linear interpolation based on $$n+1 = 2$$ points:

$\begin{array}{l} d_{0}=l_{0}^{\prime}(x)=\frac{d}{d x}\left(\frac{x-x_{1}}{x_{0}-x_{1}}\right)=\frac{1}{x_{0}-x_{1}}, \\ d_{1}=l_{1}^{\prime}(x)=\frac{d}{d x}\left(\frac{x-x_{0}}{x_{1}-x_{0}}\right)=\frac{1}{x_{1}-x_{0}}, \end{array}$

and

$f^{\prime}(x) \approx L_{1}^{\prime}(x)=y_{0} d_{0}(x)+y_{1} d_{1}(x)=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}, \quad f^{\prime \prime}(x)=0.$

• $$n=2$$, quadratic interpolation based on $$n+1 = 3$$ points:

\begin{aligned} d_{0}(x) &=l_{0}^{\prime}(x)=\frac{d}{d x}\left(\frac{\left(x-x_{1}\right)\left(x-x_{2}\right)}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)}\right)=\frac{2 x-x_{1}-x_{2}}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)}, \\ d_{1}(x) &=l_{1}^{\prime}(x)=\frac{d}{d x}\left(\frac{(x-x_{0})(x-x_{2})}{\left(x_1-x_{0}\right)\left(x_1-x_{2}\right)}\right)=\frac{2 x-x_{0}-x_{2}}{\left(x_{1}-x_{0}\right)\left(x_{1}-x_{2}\right)}, \\ d_{2}(x) &=l_{2}^{\prime}(x)=\frac{d}{d x}\left(\frac{\left(x-x_{0}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)}\right)=\frac{2 x-x_{0}-x_{1}}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)}, \end{aligned}

and

\begin{aligned} & f^{\prime}(x) \approx L_{2}^{\prime}(x)=\sum_{i=1}^{2} d_{i} y_{i} \\ =& \frac{2 x-x_{1}-x_{2}}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)} y_{0}+\frac{2 x-x_{0}-x_{2}}{\left(x_{1}-x_{0}\right)\left(x_{1}-x_{2}\right)} y_{1}+\frac{2 x-x_{0}-x_{1}}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)} y_{2}. \end{aligned}

If the 3 points are equally spaced with $$x_1=x, x_0 = x - h, x_2 = x + h$$ one can have

$f^{\prime}(x) \approx \frac{x-(x+h)}{2 h^{2}} y_{0}+\frac{2 x-(x-h)-(x+h)}{-h^{2}} y_{1}+\frac{x-(x-h)}{2 h^{2}}=\frac{1}{2 h}\left(y_{2}-y_{0}\right).$

with error $$O(h^2)$$.

The 2nd order derivative:

\begin{aligned} & f^{\prime \prime}(x) \approx L_{2}^{\prime \prime}(x)=\frac{d}{d x} L_{2}^{\prime}(x) \\ =& \frac{d}{d x}\left[\frac{2 x-x_{1}-x_{2}}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)} y_{0}+\frac{2 x-x_{0}-x_{2}}{\left(x_{1}-x_{0}\right)\left(x_{1}-x_{2}\right)} y_{1}+\frac{2 x-x_{0}-x_{1}}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)} y_{2}\right] \\ =& 2\left[\frac{y_{0}}{\left(x_{0}-x_{1}\right)\left(x_{0}-x_{2}\right)}+\frac{y_{1}}{\left(x_{1}-x_{0}\right)\left(x_{1}-x_{2}\right)}+\frac{y_{2}}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right)}\right] \end{aligned}

If the 3 points are equally spaced we have

$f^{\prime}(x) \approx \frac{2 x-x_{1}-x_{2}}{2 h^{2}} y_{0}-\frac{2 x-x_{0}-x_{2}}{h^{2}} y_{1}+\frac{2 x-x_{0}-x_{1}}{2 h^{2}} y_{2},$

and

\begin{aligned} f^{\prime}\left(x_{0}\right)&=\frac{1}{2 h}\left(-3 y_{0}+4 y_{1}-y_{2}\right), \\ f^{\prime}\left(x_{1}\right)&=\frac{1}{2 h}\left(-y_{0}+y_{2}\right), \\ f^{\prime}\left(x_{2}\right)&=\frac{1}{2 h}\left(y_{0}-4 y_{1}+3 y_{2}\right). \end{aligned}

The second derivative is

$f^{\prime \prime}(x) \approx \frac{d}{d x}\left[\frac{2 x-x_{1}-x_{2}}{2 h^{2}} y_{0}-\frac{2 x-x_{0}-x_{2}}{h^{2}} y_{1}+\frac{2 x-x_{0}-x_{1}}{2 h^{2}} y_{2}\right]$ $\boxed{f^{\prime \prime}(x) = \frac{1}{h^{2}}\left(y_{0}-2 y_{1}+y_{2}\right)}.$

### 8.2 Programming languages

#### 8.2.1 Julia

1. Packages
1. Latexify.jl

This is a package for generating LaTeX maths from julia objects.

2. SugarBLAS

This package provides macros for BLAS functions representing polynomials

3. Flux

Flux is an elegant approach to machine learning.

4. DifferentialEquations.jl

This is a suite for numerically solving differential equations written in Julia and available for use in Julia, Python, and R.

5. Knet

Knet (pronounced "kay-net") is the Koç University deep learning framework implemented in Julia by Deniz Yuret and collaborators. It supports GPU operation and automatic differentiation using dynamic computational graphs for models defined in plain Julia.

## 12 Nuclear Physics

### 12.1 Isotopes. Radio-activity. Fission and Fusion.

#### 12.1.1 Geiger-Nuttall law

The Geinger-Nuttall relates the decay constant of a radioactive isotope with the energy of the alpha particles.

$\log _{10} \lambda=-a_{1} \frac{Z}{\sqrt{E}}+a_{2}$

where $$\lambda$$ is the decay constant, $$Z$$ the atomic number, $$E$$ the total kinetic energy and $$a_1$$ and $$a_2$$ are constants.

H. Geiger and J.M. Nuttall (1911) "The ranges of the α particles from various radioactive substances and a relation between range and period of transformation," Philosophical Magazine, Series 6, vol. 22, no. 130, pages 613-621. See also: H. Geiger and J.M. Nuttall (1912) "The ranges of α particles from uranium," Philosophical Magazine, Series 6, vol. 23, no. 135, pages 439-445.

G. Gamow (1928) "Zur Quantentheorie des Atomkernes" (On the quantum theory of the atomic nucleus), Zeitschrift für Physik, vol. 51, pages 204-212.

## 15 Special Relativity

These annotations are based on http://jacobi.luc.edu/index.html and http://jacobi.luc.edu/Useful.html i.e., the website of Prof. Robert A. McNees

Spacetime coordinates, 4-vectors, metric tensor $$g_{\mu\nu}$$, etc:

Spacetime is any mathematical model which fuses the three dimensions of space and the one dimension of time into a single four-dimensional continuum, within of special relativity it relies on the Minkowski geometry.

### 15.1 4-vectors.

In relativistic expressions four-vectors are written as

$A^{\mu} = (A_0, \vec{A}) = (A_0, A^1, A^2, A^3).$

Four-vectors with superscript are called contravariant vectors (regarded as a column vector). Covariant vectors, denoted by subscript are defined as

$A_{\mu} = g_{\mu\nu}A^{\nu},$

with the metric tensor $$g_{\mu\nu}$$ that reads in Cartesian coordinates in Minkowski space.

$\eta_{\mu \nu}=\eta^{\mu \nu}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right); \qquad \eta^{\mu}{ }_{\nu}=\delta^{\mu}{ }_{\nu}=\left(\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right).$

Raising and lowering indices:

$A^{\mu}=\eta^{\mu \nu} A_{\nu}; \qquad A_{\mu}=\eta_{\mu \nu} A^{\nu};$

Einstein summation convention:

$A^{\mu}B_{\mu} = A^0B_0 + A^{1}B_{1} + A^2B_2 + A^3B_3 = A^0B^0 - \vec{A}\cdot\vec{B}.$

The inverse of the metric tensor is given by

$g^{\mu\nu} = g_{\mu\nu}.$

A scalar product of two four-vectors is then defined by

$A \cdot B = A_{\mu}B^{\mu} = A^{\mu}B_{\mu} = A^0B^0 - \vec{A}\cdot\vec{B}.$

The invariant square of the four-vector is thus

$A^2 = A_{\mu}A^{\mu} = A_0^2 - \vec{A}^2.$

#### 15.1.1 Spacetime coordinates:

$x^{\mu} = (x^0, \vec{x}) = (ct, \vec{x}); \qquad x_{\mu} = (ct, -\vec{x}); \qquad \vec{x} = (x^1, x^2, x^3) = (x, y, z);$

#### 15.1.2 Four-momentum:

$p^{\mu} = (E/c, \vec{p}); \qquad p_{\mu} = (E/c, -\vec{p}); \qquad p^{\mu} = (p^1, p^2, p^3) = (p_x, p_y, p_z);$

#### 15.1.3 Derivatives:

$\partial_{\mu} \equiv \frac{\partial}{\partial x^{\mu}} = \left(\frac{1}{c} \frac{\partial}{\partial t}, \vec\nabla\right); \qquad \partial^{\mu} \equiv \frac{\partial}{\partial x_{\mu}} = \left(\frac{1}{c} \frac{\partial}{\partial t}, -\vec\nabla\right);$

#### 15.1.4 Four-momentum operator

$\hat{P}^{\mu} = i\partial^{\mu}.$

#### 15.1.5 Lorentz-invariant d'Alembert operator

$\partial^{\mu}\partial_{\mu} = \frac{\partial^2}{\partial t^2} - \vec\nabla^2 = \Box.$

#### 15.1.6 Four-current

$j^{\mu} = (\rho, \vec{j}).$

#### 15.1.7 Electromagnetic potential

$A^{\mu} = (\phi, \vec{A}).$

### 15.2 The Lorentz transformation.

$A'_{\mu} = \Lambda_{\mu}{}^{\nu}A_{\nu}; \qquad A'^{\mu} = \Lambda^{\mu}{}_{\nu}A^{\nu}; \\ \eta^{\mu\nu} = \Lambda^{\mu}{}_{\alpha}\Lambda^{\nu}{}_{\beta}\,\eta^{\alpha\beta}; \qquad \eta_{\mu\nu} = \Lambda_{\mu}{}^{\alpha}\Lambda_{\nu}{}^{\beta}\,\eta_{\alpha\beta}.$

• Properties:
1. $\eta^{\mu \nu}=\Lambda_{\alpha}^{\mu} \Lambda_{\beta}^{\nu} \eta^{\alpha \beta}$
2. $\left|\operatorname{det} \Lambda_{\nu}^{\mu}\right|=1$
3. $\left(\Lambda_{1}\right)^{\mu} \alpha\left(\Lambda_{2}\right)_{\nu}^{\alpha}=\left(\Lambda_{12}\right)_{\nu}^{\mu}$
4. $\left|\Lambda_{0}^{0}\right| \geq 1$

## 18 General Relativity

### 18.1 4-tensors

#### 18.1.1 The metric tensor.

In local coordinates $$x^{\mu}$$ the metric can be written in the form

$g=g_{\mu \nu} d x^{\mu} \otimes d x^{\nu}.$

The factors $$dx^{\mu}$$ are one-form gradient of the scalar fields $$x^{\mu}$$. The coefficients $$g_{\mu\nu}$$ are a set of 16 real-valued functions. The metric is symmetric

$g_{\mu \nu} = g_{\nu \mu},$ giving 10 independents coefficients. Being $$dx^{\mu}$$ the components of an infinitesimal displacement four-vector, the metric determines the invariant square of an infinitesimal line element, interval, denoted as

$d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}.$

The interval $$ds^2$$ imparts information about the causal structure of the spacetime.

1. $$ds^2<0 \rightarrow$$ timelike
2. $$ds^2=0 \rightarrow$$ lightlike
3. $$ds^2>0 \rightarrow$$ spacelike

Under a change of coordinates $$x^{\mu} \rightarrow x^{\overline{\mu}}$$, the components transform as

$g_{\bar{\mu} \bar\nu}=\frac{\partial x^{\rho}}{\partial x^{\bar{\mu}}} \frac{\partial x^{\sigma}}{\partial x^{\bar{\nu}}} g_{\rho \sigma}=\Lambda_{\bar{\mu}}^{\rho} \Lambda_{\bar{\nu}}^{\sigma} g_{\rho \sigma}.$

1. Flat spacetime

The simplest of Lorentz manifold is flat spacetime which can be given as $$R^4$$ with coordinates,

$d s^{2}=-c^{2} d t^{2}+d x^{2}+d y^{2}+d z^{2}=\eta_{\mu \nu} d x^{\mu} d x^{\nu}$

#### 18.1.2 Space-time curvature.

The metric $$g$$ completely determines the curvature of spacetime. On any semi-Riemannian manifold there is a unique connection $$\nabla$$ that is compatible with the metric and torsion-free. The Levi-Civita connection. The Christoffel symbols of this connection are given in terms of partial derivatives of the metric in local coordinates

$\Gamma^{\lambda}_{}{\mu \nu}=\frac{1}{2} g^{\lambda \rho}\left(\frac{\partial g_{\rho \mu}}{\partial x^{\nu}}+\frac{\partial g_{\rho \nu}}{\partial x^{\mu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\rho}}\right)=\frac{1}{2} g^{\lambda \rho}\left(g_{\rho \mu, \nu}+g_{\rho \nu, \mu}-g_{\mu \nu, \rho}\right),$

the curvature of spacetime is then given by the Riemann curvature tensor which is defined in terms of the Levi-Civita connection $$\nabla$$,

$R^{\rho}{}_{\sigma \mu \nu} = \partial_{\mu} \Gamma^{\rho}{}_{\nu \sigma}-\partial_{\nu} \Gamma^{\rho}{}_{\mu \sigma} + \Gamma^{\rho}{}_{\mu \lambda} \Gamma^{\lambda}{}_{\nu \sigma} - \Gamma^{\rho}{}_{\nu \lambda} \Gamma^{\lambda}{}_{\mu \sigma}.$

• Ricci tensor

$R_{\mu \nu}=\delta^{\sigma}{ }_{\rho} R^{\rho}{}_{\mu \sigma \nu}.$

• Scalar curvature

$R = g^{\mu\nu}R_{\mu\nu}.$

• Schouten tensor

$S_{\mu \nu}=\frac{1}{d-1}\left(R_{\mu \nu}-\frac{1}{2 d} g_{\mu \nu} R\right); \qquad \nabla^{\nu} S_{\mu \nu}=\nabla_{\mu} S^{\nu}{}_{\nu}.$

• Weyl tensor

$C^{\lambda}{}_{\mu \sigma \nu} = R^{\lambda}{}_{\mu \sigma \nu} + g^{\lambda}{}_{\nu} S_{\mu \sigma} - g^{\lambda}{}_{\sigma} S_{\mu \nu} + g_{\mu \sigma} S^{\lambda}{}_{\nu} - g_{\mu \nu} S^{\lambda}{}_{\sigma}.$

• Commutators of Covariant Derivatives

${\left[\nabla_{\mu}, \nabla_{\nu}\right] A_{\lambda} = R_{\lambda \sigma \mu \nu} A^{\sigma}}; \qquad {\left[\nabla_{\mu}, \nabla_{\nu}\right] A^{\lambda} = R^{\lambda}{}_{\sigma \mu \nu} A^{\sigma}}.$

• Bianchi Identity

$\nabla_{\kappa} R_{\lambda \mu \sigma \nu}-\nabla_{\lambda} R_{\kappa \mu \sigma \nu}+\nabla_{\mu} R_{\kappa \lambda \sigma \nu}=0; \qquad \nabla^{\nu} R_{\lambda \mu \sigma \nu}=\nabla_{\mu} R_{\lambda \sigma}-\nabla_{\lambda} R_{\mu\sigma}; \qquad \nabla^{\nu} R_{\mu \nu}=\frac{1}{2} \nabla_{\mu} R.$

• Bianchi identity for Weyl

$\nabla^{\nu} C_{\lambda \mu \sigma \nu}=(d-2) \left(\nabla_{\mu} S_{\lambda\sigma}-\nabla_{\lambda} S_{\mu \sigma}\right); \qquad \nabla^{\lambda} \nabla^{\sigma} C_{\lambda\mu \sigma \nu} = \frac{d-2}{d-1}\left[\nabla^{2} R_{\mu\nu}-\frac{1}{2 d} g_{\mu \nu} \nabla^{2} R-\frac{d-1}{2 d} \nabla_{\mu} \nabla_{\nu}R - \left(\frac{d+1}{d-1}\right) R_{\mu}{}^{\lambda} R_{\nu \lambda} + C_{\lambda \mu \sigma \nu} R^{\lambda \sigma}+\frac{(d+1)}{d(d-1)} R R_{\mu \nu}+\frac{1}{d-1} g_{\mu \nu}\left(R^{\lambda \sigma} R_{\lambda \sigma}-\frac{1}{d} R^{2}\right)\right]$

## 22 Supersymmetry/Supergravity

Created by: Ronaldo V. Lobato on 2020-11-11. Last Updated: 2021-02-14 Sun 13:49