[{"title":"Pájaro amarillo","url":"https://edgarivanhinojosa.xyz/photos/p%C3%A1jaro-en-el-usit/","content":"","summary":"","date":"2026-07-08","section":"photos"},{"title":"Test Series Entry","url":"https://edgarivanhinojosa.xyz/notes/test-series/","content":"","summary":"","date":"2026-07-08","section":"notes"},{"title":"The Reflection Identity of the Gamma Function via Complex Integration","url":"https://edgarivanhinojosa.xyz/blogs/gamma-identidad/","content":"The reflection identity of the gamma function is:\n\\begin{equation} \\Gamma( z ) \\Gamma( 1-z ) = \\frac{\\pi}{\\sin( \\pi z ) } \\end{equation}\nWe can start from the definition of $\\beta( 1- \\alpha, \\alpha ) $\n\\begin{align} \\Gamma( 1- \\alpha ) \\Gamma( \\alpha ) \u0026amp;= \\beta ( 1- \\alpha, \\alpha ) = \\int^\\infty_0 \\frac{t ^{1- \\alpha-1}dt}{( 1+ t )^{1- \\alpha+ \\alpha} } \\\\ \u0026amp;= \\int^\\infty_0 \\frac{t ^{-\\alpha} dt}{1 + t} \\\\ \u0026amp;= \\int^\\infty_{-\\infty} \\frac{e ^{-\\alpha x} e^x dx}{1+ e^x} = \\int^\\infty_{-\\infty} \\frac{e ^{( 1-\\alpha )x }}{ 1 + e^{x}} = I \\end{align}\nwhere we used the substitution $t=e^x$, so $dt = e^xdx$ with limits $t=0 \\implies x=-\\infty$ and $t=\\infty\\implies x=\\infty $.\nNow let us integrate over the rectangular contour $C_R$ with vertices $-R, R, R + 2\\pi i , - R + 2\\pi i$:\n\\begin{align} \\oint_{C_R} \\frac{e ^{( 1-\\alpha ) z} dz}{1 + e^z}\t\u0026amp;= \\int^R_{-R} \\frac{\\exp( ( 1- \\alpha ) x ) dx}{1 + e^x} + \\int ^{2\\pi}_0 \\frac{\\exp( ( 1-\\alpha ) ( R+iy ) ) dx}{1 + \\exp( R+i y ) } +\\\\ \u0026amp;+ \\int^{-R}_{R} \\frac{\\exp( ( 1- \\alpha ) ( x+2\\pi i ) ) dx}{1 + \\exp( x + 2\\pi i ) } + \\int ^{0}_{2\\pi i} \\frac{\\exp( ( 1-\\alpha ) ( -R+iy ) ) dy}{1 + \\exp( -R+iy ) } \\end{align}\nLet us evaluate the limit $R\\to\\infty$ of the integrands along the sides parallel to the imaginary axis. For the first one we notice that the limit is an indeterminate form $\\infty/\\infty$, so we can apply L\u0026rsquo;Hôpital\u0026rsquo;s rule\n\\begin{align} \\lim_{R \\to \\infty} \\frac{\\exp( ( 1- \\alpha ) R ) \\exp( ( 1- \\alpha ) iy ) }{1 + \\exp( R ) \\exp( iy ) } \u0026amp;=\\lim_{R \\to \\infty} \\frac{( 1-\\alpha ) \\exp( ( 1-\\alpha ) R ) \\exp( ( 1-\\alpha ) iy ) }{\\exp( R ) \\exp( iy ) } \\\\ \u0026amp;=\\lim_{R \\to \\infty} ( 1-\\alpha ) \\exp( -\\alpha R ) \\exp(-\\alpha iy ) = 0 \\end{align}\nfor the second integrand it is only a matter of simplification\n\\begin{align} \\lim_{R \\to \\infty} \\frac{\\exp( ( \\alpha -1 ) R ) \\exp( ( 1- \\alpha ) iy ) }{1 + \\exp( -R ) \\exp( iy ) } \u0026amp;= \\frac{\\lim_{R \\to \\infty} \\exp( ( \\alpha -1) R ) \\exp( ( 1-\\alpha ) iy ) }{\\lim_{R \\to \\infty} 1 +\\exp( -R ) \\exp( iy ) } \\\\ \u0026amp;=\\lim_{R \\to \\infty} \\exp( \\alpha R ) \\exp(-\\alpha iy ) = 0 \\end{align}\nwhich also vanishes since we assume $\\alpha-1\u0026lt;0$. So when we take the limit in the contour integral we get\n\\begin{align} \\lim_{R \\to \\infty} \\oint \\frac{\\exp( ( 1- \\alpha ) z ) }{1 + \\exp( z ) } \u0026amp;= \\lim_{R \\to \\infty} \\left\\lbrace \\int^R_{-R} \\frac{\\exp( ( 1- \\alpha ) x ) dx}{1 + e^x} + \\int^{-R} _{R} \\frac{\\exp( ( 1- \\alpha ) ( x+2\\pi i ) ) dx}{1 + \\exp( x + 2\\pi i) } \\right\\rbrace \\\\ \u0026amp;= \\int^\\infty_{-\\infty} \\frac{\\exp( ( 1- \\alpha ) x ) dx}{1 + e^x} - \\int^\\infty_{\\infty} \\frac{\\exp( ( 1- \\alpha ) x) \\exp(( 1- \\alpha ) 2\\pi i ) dx}{1 + \\exp( x ) \\exp( 2\\pi i ) } \\\\ \u0026amp;= ( 1- \\exp( ( 1- \\alpha ) 2\\pi i ) ) \\int ^\\infty_{-\\infty} \\frac{\\exp( ( 1- \\alpha )x ) }{1 + \\exp( x ) } \\\\ \u0026amp;= [ 1 - \\exp( ( 1- \\alpha )2\\pi i ) ] I \\end{align}\nwith this equality we can evaluate the contour using the residue theorem\n\\begin{align} \\lim_{R \\to \\infty} \\oint \\frac{\\exp( ( 1- \\alpha ) z ) }{1 + \\exp( z ) } \u0026amp;= 2\\pi i\\sum_k \\text{Res}\\left( \\frac{\\exp( ( 1- \\alpha ) z ) }{1 + \\exp( z ) } , z_k \\right) \\end{align}\nIt is easy to see there is only one simple pole $z_0$ inside the contour, given by $1 + \\exp( z )= 0 \\implies z_0= i\\pi$. We evaluate its residue with L\u0026rsquo;Hôpital\u0026rsquo;s rule\n\\begin{align} \\text{Res}\\left( \\frac{\\exp( ( 1- \\alpha ) z ) }{1 + \\exp( z ) } , i\\pi \\right) \u0026amp;= \\lim_{z \\to i \\pi} ( z-i\\pi ) \\frac{\\exp( ( 1- \\alpha ) z ) }{1 + \\exp( z ) } \\\\ \u0026amp;= \\lim_{z \\to i\\pi} \\frac{\\exp( ( 1-\\alpha )z ) + ( z- i\\pi ) ( 1- \\alpha ) \\exp( ( 1- \\alpha ) z ) }{\\exp( z ) } \\\\ \u0026amp;= \\frac{e ^{i\\pi} e ^{-\\alpha \\pi i}}{e^{i\\pi}} = e ^{- \\alpha \\pi i}. \\end{align}\nWe can now solve for the integral of interest $I$\n\\begin{align} I \u0026amp;= \\frac{ 2\\pi i e ^{- \\alpha \\pi i}}{1- e ^{2\\pi i} e^{- \\alpha \\pi i}} = \\pi \\frac{2i }{e ^{ \\alpha \\pi i} - e ^{- \\alpha \\pi i}} \\\\ \u0026amp;= \\frac{\\pi}{\\sin( \\pi \\alpha ) } \\\\ \\therefore \\Gamma( \\alpha ) \\Gamma( 1 - \\alpha ) \u0026amp;= \\frac{\\pi}{\\sin( \\pi z ) } . \\end{align}\n","summary":"\u003cp\u003eThe reflection identity of the gamma function is:\u003c/p\u003e\n\u003cp\u003e\\begin{equation}\n\\Gamma( z )  \\Gamma( 1-z )  = \\frac{\\pi}{\\sin( \\pi z ) }\n\\end{equation}\u003c/p\u003e\n\u003cp\u003eWe can start from the definition of $\\beta( 1- \\alpha,  \\alpha ) $\u003c/p\u003e\n\u003cp\u003e\\begin{align}\n\\Gamma( 1- \\alpha )  \\Gamma(  \\alpha ) \u0026amp;= \\beta (  1- \\alpha,  \\alpha ) = \\int^\\infty_0 \\frac{t ^{1- \\alpha-1}dt}{( 1+ t  )^{1- \\alpha+ \\alpha} } \\\\\n\u0026amp;= \\int^\\infty_0 \\frac{t ^{-\\alpha} dt}{1 + t} \\\\\n\u0026amp;= \\int^\\infty_{-\\infty} \\frac{e ^{-\\alpha x} e^x dx}{1+ e^x} = \\int^\\infty_{-\\infty} \\frac{e ^{( 1-\\alpha )x }}{ 1 + e^{x}} = I\n\\end{align}\u003c/p\u003e","date":"2026-02-15","section":"blogs"},{"title":"General Lorentz Transformation via Group Generators","url":"https://edgarivanhinojosa.xyz/blogs/lorentz-transform/","content":"In this post we derive the functional form of a generic Lorentz transformation from its group generators. At first glance I considered this trivial, as I was unaware of the role played by their non-commutativity — which made it an interesting exercise to present. We follow Chapter 11 of Jackson (Third Edition), which makes use of the mathematical properties of Minkowski spacetime. Clarifications missing from the book are supplied with original arguments.\nLet $ \\mathcal{M} $ be a Minkowski spacetime, thought of as physical spacetime, and let $ x: \\mathcal{M} \\to \\mathbb{R}^4 $ and $ x\u0026rsquo;: \\mathcal{M} \\to \\mathbb{R}^4 $ be charts mapping events of this spacetime to a Minkowski vector space. We seek a linear map $A$ that takes components $x_i$ to components $x^\\prime_i$\n\\begin{equation} Ax =x' \\end{equation}\nbeing surjective and injective, linear, and above all preserving the spacetime structure. This last condition means the spacetime interval is preserved, defined as follows: For $V$ a Minkowski vector space there exists an isomorphism $\\overline{g}$ between $V$ and its dual $V^\\ast$, $\\overline{g}: V \\to V^\\ast$, such that taking the canonical bases of both we have the relation\n$$ \\overline{g}(e_i) = e^i \\forall i \\neq 0 $$$$\\overline{g}(e_0) = - e^0$$Denote by $(u, v)$ the action of the covector $u \\in V^*$ on the vector $v \\in V$; the evaluation $(u,u)$ is the spacetime interval of the event $u$. Given this symmetric bilinear form there exists a quadratic form via a matrix $g$ such that\n$$ u^T g v = (\\overline{g}(u), v) $$Given the isomorphism $\\overline{g}$, the matrix $g$ must take the form\n$$g = \\begin{pmatrix} -1 \u0026 \u0026 \u0026 \\\\ \u0026 1 \u0026 \u0026 \\\\ \u0026 \u0026 1 \u0026 \\\\ \u0026 \u0026 \u0026 1 \\end{pmatrix} $$and Lorentz transformations given this quadratic form satisfy\n\\begin{equation} x^T g x = x\u0026rsquo;^T g x\u0026rsquo; \\implies (Ax)^T g (Ax) = xg x \\implies A^T g A = g \\label{lorentz} \\end{equation}\nFrom this we observe the first property, namely that\n$$det(A^T g A) det(A^T) det(g) det(A)= det(g)$$$$\\implies det(A^T) det(A) = det(A)^2= 1 \\implies det(A) = \\pm 1$$Since we consider only proper Lorentz transformations we restrict to $det(A) = 1$, which is simply a choice of orientation. From \\eqref{lorentz} the coefficients of $g$ satisfy the system of equations\n$$a_{ij} g_{jl} a_{lk} = g_{ik}$$The symmetry of $g$ reduces the number of independent equations from 16 to 10, which gives us a way to parameterize the components with 6 numbers. Three of them are angles for ordinary spatial rotations; the other three are for boosts (hyperbolic rotations), which are our primary interest. In this solution we use the generator $L$ for $A$\n$$A = e^L$$Let us investigate its properties. We know that the determinant of $A$ is unity, so we must have\n$$det(A) = e^{trL} =1 \\implies trL =0 $$Recall that the trace of a matrix is the sum of its eigenvalues, so if we suppose that $L$ is a pure boost its three eigenvalues $\\lambda_i$ satisfy\n\\begin{equation} \\lambda_1 + \\lambda_2 + \\lambda_3 =0. \\label{eigen} \\end{equation}\nLet us manipulate equation (1) knowing that $g^2=I$\n$$A^TgA A^{-1}g = g A ^{-1}g \\implies A^T = gA ^{-1}g$$taking the definition of the generator\n\\begin{align*} A^T \u0026amp;= e^{L^T} \\implies gA^Tg = e^{g L^Tg} \\\\ \\implies g^2 A^{-1} g^2 \u0026amp;= A^{-1}= e^{g L^T g}\\\\ \\implies e^{-L} \u0026amp;= e^{g L^T g} \\end{align*}\nThis gives us the following property\n\\begin{equation} -gL = gg L ^T g = L^T g = (gL)^T \\label{antisimetria} \\end{equation}\nmeaning that $gL$ is anti-symmetric, i.e. $(gL)_{ij} = -(gL)_{ji}$. Let us analyze this equation index by index. Expanding\n\\begin{align*} (gL)_{ij} = g_{il} L_{lj} i\u0026amp;= - g_{jl} L_{li} \\\\ \\implies g_{ii} L_{ij} \u0026amp;= - g_{jj} L_{ji} \\\\ \\text{si i=0, j=0: } g_{00} L_{00} \u0026amp;= -g_{00} L_{00} \\\\ \\implies -L_{00} \u0026amp;= -(-1) L_{00} \\implies L_{00} = 0\\\\ \\text{ si } i=0 j\\neq 0: \\ g_{00} L_{0j} \u0026amp;= - g_{jj} L_{j_0} \\implies -L_{0j} = - L_{j_0}\\\\ \\text{si \\ } i\\neq 0, j \\neq 0: \\ g_{ii} L_{ij} \u0026amp;= - g_{jj} L_{ji} \\implies L_{ij} = -L_{ji} .\\end{align*}\nTherefore $L$ takes the form\n\\[ L = \\begin{pmatrix} 0 \u0026 L_{01} \u0026 L_{02} \u0026 L{03} \\\\ L_{01} \u0026 0 \u0026 L_{12} \u0026 L_{13} \\\\ L_{02} \u0026 -L_{12} \u0026 0 \u0026 L_{23} \\\\ L_{03} \u0026 - L_{13} \u0026 -L_{23} \u0026 0 \\end{pmatrix} .\\]We know that among the transformations contained in the general form of $L$ are ordinary spatial rotations satisfying $A A^T = I \\implies L^T = -L$, so we discard the anti-symmetric parts of this general form. The generators of the hyperbolic rotation group in Minkowski space are therefore\n\\[ K_1 = \\begin{pmatrix} 0 \u0026 1 \u0026 0 \u0026 0 \\\\ 1\u0026 0 \u0026 \u0026 \\\\ 0 \u0026 \u0026 0 \u0026 \\\\ 0 \u0026 \u0026 \u0026 \\end{pmatrix} , \\ \\ \\ K_2 = \\begin{pmatrix} 0 \u0026 0 \u0026 1 \u0026 0 \\\\ 0\u0026 \u0026 \u0026 \\\\ 1 \u0026 \u0026 \u0026 \\\\ 0 \u0026 \u0026 \u0026 \\end{pmatrix} , \\ \\ \\ K_2 = \\begin{pmatrix} 0 \u0026 0 \u0026 0 \u0026 1 \\\\ 0\u0026 \u0026 \u0026 \\\\ 0 \u0026 \u0026 \u0026 \\\\ 1 \u0026 \u0026 \u0026 \\end{pmatrix} \\ \\ \\ .\\]We denote $\\epsilon \\cdot K$ as a linear combination of the matrices, i.e. $\\epsilon = \\epsilon_1 K_1 + \\epsilon_2 K_2 + \\epsilon_3 K_3$. If $\\epsilon$ is a unit vector, then the equality $(\\epsilon \\cdot K)^3 = \\epsilon \\cdot K$ holds. This follows from the eigenvalues of a cubically idempotent matrix. Let $F$ be a cubically idempotent matrix; assuming it is diagonalizable\n\\begin{align*} Fv \u0026amp;= \\lambda v \\ F^3 v \u0026amp;= \\lambda^3 v \\implies \\lambda^3 = \\lambda\\ \\lambda(\\lambda^2 -1) \u0026amp;= 0 \\end{align*}\nwe conclude that a cubically idempotent matrix has eigenvalues that are $+1$, $-1$, or $0$.\n$$det (gL^T) = det(-gL)$$$$\\implies det(L) = - det(L) \\implies det(L) =0$$ therefore $L$ has at least one zero eigenvalue $\\lambda_1=0$. Looking at the even more particular case of $\\epsilon \\cdot K$, it is straightforward to verify the eigenvector\n\\[ \\begin{pmatrix} 0 \u0026 \\epsilon_1 \u0026 \\epsilon_2 \u0026 \\epsilon_3 \\\\ \\epsilon_1 \u0026 \u0026 \u0026 \\\\ \\epsilon_2 \u0026 \u0026 \u0026 \\\\ \\epsilon_3 \u0026 \u0026 \u0026 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ \\epsilon_1 \\\\ \\epsilon_2 \\\\ \\epsilon_3 \\end{pmatrix} = \\begin{pmatrix} \\epsilon_1^2 + \\epsilon_2^2 + \\epsilon_3^2 \\\\ \\epsilon_1 \\\\ \\epsilon_2 \\\\ \\epsilon_3 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ \\epsilon_1 \\\\ \\epsilon_2 \\\\ \\epsilon_3 \\end{pmatrix} \\]$$A = \\exp(-\\hat{\\beta}\\cdot K \\tanh^{-1}\\left( \\beta \\right) )$$ where $\\tanh ^{-1} \\beta$ is the hyperbolic rotation angle. Expanding in series\n\\begin{align*} \\exp(-\\tanh^{-1}\\left( \\beta \\right) \\hat{\\beta} \\cdot K ) \u0026amp;= \\sum_{n=0}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K \\tanh ^{-1} (\\beta)\\right)^n }{n!} \\\\ \u0026amp;= \\sum_{n=0}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K \\tanh ^{-1} (\\beta)\\right)^{2n+1}\t}{(2n +1)!} + \\sum_{n=0}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K \\tanh ^{-1} (\\beta)\\right)^{2n}\t}{(2n )!} \\\\ \u0026amp;= \\sum_{n=0}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K \\right)^{2n+1} \\left( \\tanh ^{-1} (\\beta)\\right)^{2n+1}\t}{(2n +1)!} + \\sum_{n=1}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K\\right)^{2n} \\left(\\tanh ^{-1} (\\beta)\\right)^{2n}\t}{(2n )!} + \\\\ \u0026amp; + (-\\hat{\\beta}\\cdot K \\tanh ^{-1} \\beta)^0 + (\\hat{\\beta} \\cdot K)^2 - (\\hat{\\beta} \\cdot K)^2 \\\\ \u0026amp;= \\sum_{n=0}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K \\right) \\left( \\tanh ^{-1} (\\beta)\\right)^{2n+1}\t}{(2n +1)!} + \\sum_{n=0}^{\\infty} \\frac{\\left( - \\hat{\\beta} \\cdot K\\right)^{2} \\left(\\tanh ^{-1} (\\beta)\\right)^{2n}\t}{(2n )!} + I - (\\hat{\\beta} \\cdot K)^2\\\\ \u0026amp;= \\left( - \\hat{\\beta} \\cdot K \\right)\\sum_{n=0}^{\\infty} \\frac{ \\left( \\tanh ^{-1} (\\beta)\\right)^{2n+1}\t}{(2n +1)!} + \\left( \\hat{\\beta} \\cdot K\\right)^{2}\\sum_{n=0}^{\\infty} \\frac{ \\left(\\tanh ^{-1} (\\beta)\\right)^{2n}\t}{(2n )!} + I - (\\hat{\\beta} \\cdot K)^2\\\\ \u0026amp;= \\left( - \\hat{\\beta} \\cdot K \\right)\\sinh(\\tanh ^{-1}\\beta) + \\left( \\hat{\\beta} \\cdot K\\right)^{2} \\cosh(\\tanh ^{-1} \\beta) + I - (\\hat{\\beta} \\cdot K)^2\\\\ .\\end{align*}\nLet us evaluate the terms we have here. We first resort to the logarithmic and exponential definitions of the hyperbolic trigonometric functions. The hyperbolic sine and cosine of $\\tanh ^{-1} (\\beta) = \\frac{1}{2} \\ln\\left( \\frac{1+ \\beta}{1-\\beta} \\right) $ are\n\\begin{align*} \\sinh \\left( \\tanh ^{-1} \\beta \\right) \u0026amp;= \\frac{1}{2} \\left\\lbrace \\exp \\left( \\ln \\sqrt{\\frac{1+ \\beta}{1-\\beta}} \\right) - \\exp \\left( \\ln \\sqrt{\\frac{1- \\beta}{1+\\beta}} \\right) \\right\\rbrace \\\\ \u0026amp;= \\frac{1}{2} \\left( \\frac{\\sqrt{(1+ \\beta)^2} - \\sqrt{(1-\\beta)^2} }{\\sqrt{1-\\beta^2} } \\right) = \\frac{1}{2} \\left(\\frac{1+\\beta -1 +\\beta}{\\sqrt{1-\\beta ^2} } \\right) \\\\ \u0026amp;= \\gamma \\beta\\\\ \\cosh \\left( \\tanh ^{-1} \\beta \\right) \u0026amp;= \\frac{1}{2} \\left\\lbrace \\exp \\left( \\ln \\sqrt{\\frac{1+ \\beta}{1-\\beta}} \\right) + \\exp \\left( \\ln \\sqrt{\\frac{1- \\beta}{1+\\beta}} \\right) \\right\\rbrace \\\\ \u0026amp;= \\frac{1}{2} \\left( \\frac{\\sqrt{(1+ \\beta)^2} - \\sqrt{(1-\\beta)^2} }{\\sqrt{1-\\beta^2} } \\right) = \\frac{1}{2} \\left(\\frac{1+\\beta +1 -\\beta}{\\sqrt{1-\\beta ^2} } \\right) \\\\ \u0026amp;= \\gamma .\\end{align*}\nTherefore the transformation becomes\n$$A = - \\hat{\\beta} \\cdot K \\gamma \\beta + \\left( \\hat{\\beta} \\cdot K \\right)^2 (\\gamma - 1) + I $$It remains to evaluate the matrix $\\hat{\\beta}\\cdot K$ and its square\n\\begin{align*} \\hat{\\beta} \\cdot K \u0026amp;= \\frac{1}{\\beta} \\left( \\beta_1 K_1 + \\beta_2 K_2 + \\beta_3 K_3 \\right) \\\\ \u0026amp;= \\frac{1}{\\beta} \\begin{pmatrix} 0 \u0026amp; \\beta_1 \u0026amp; \\beta_2 \u0026amp; \\beta_3 \\\\ \\beta_1 \u0026amp; \u0026amp; \u0026amp; \\\\ \\beta_2 \u0026amp; \u0026amp; \u0026amp; \\\\ \\beta_3 \u0026amp; \u0026amp; \u0026amp; \\end{pmatrix} \\\\ \\implies (\\hat{\\beta} \\cdot K)^2 \u0026amp;= \\frac{1}{\\beta^2} \\begin{pmatrix} \\beta_1^2 + \\beta_2^2 + \\beta_3^2 \u0026amp; 0 \u0026amp; 0\u0026amp; 0 \\\\ 0 \u0026amp; \\beta_1^2 \u0026amp; \\beta_1 \\beta_2 \u0026amp; \\beta_1\\beta_3 \\\\ 0 \u0026amp; \\beta_1\\beta_2 \u0026amp; \\beta_2^2 \u0026amp; \\beta_2 \\beta_3 \\\\ 0 \u0026amp; \\beta_1\\beta_3 \u0026amp; \\beta_2 \\beta_3 \u0026amp; \\beta_3^2\\end{pmatrix}\n.\\end{align*}\nSubstituting into the expression for $A$ we get\n\\begin{align} A \u0026amp;= - \\frac{\\gamma \\beta}{\\beta} \\begin{pmatrix} 0 \u0026amp; \\beta_1 \u0026amp; \\beta_2 \u0026amp; \\beta_3 \\\\ \\beta_1 \u0026amp; \u0026amp; \u0026amp; \\\\ \\beta_2 \u0026amp; \u0026amp; \u0026amp; \\\\ \\beta_3 \u0026amp; \u0026amp; \u0026amp; \\end{pmatrix} + /gfrac{\\gamma - 1}{\\beta^2} \\begin{pmatrix} \\beta_1^2 + \\beta_2^2 + \\beta_3^2 \u0026amp; 0 \u0026amp; 0\u0026amp; 0 \\\\ 0 \u0026amp; \\beta_1^2 \u0026amp; \\beta_1 \\beta_2 \u0026amp; \\beta_1\\beta_3 \\\\ 0 \u0026amp; \\beta_1\\beta_2 \u0026amp; \\beta_2^2 \u0026amp; \\beta_2 \\beta_3 \\\\ 0 \u0026amp; \\beta_1\\beta_3 \u0026amp; \\beta_2 \\beta_3 \u0026amp; \\beta_3^2\\end{pmatrix} +I\\\\ \u0026amp;=\n\\begin{pmatrix}\\gamma-1 \u0026amp; -\\beta_1 \\gamma \u0026amp; -\\beta_2 \\gamma\u0026amp; -\\beta_3 \\gamma \\\\ -\\beta_1 \\gamma \u0026amp; \\beta_1^2\\frac{\\gamma -1}{\\beta} \u0026amp; \\beta_1 \\beta_2 \\frac{\\gamma -1}{\\beta}\u0026amp; \\beta_1\\beta_3 \\frac{\\gamma -1}{\\beta}\\ -\\beta_2\\gamma \u0026amp; \\beta_1\\beta_2 \\frac{\\gamma -1}{\\beta}\u0026amp; \\beta_2^2 \\frac{\\gamma -1}{\\beta}\u0026amp; \\beta_2 \\beta_3 \\frac{\\gamma -1}{\\beta}\\\\ -\\beta_3 \\gamma \u0026amp; \\beta_1\\beta_3 \\frac{\\gamma -1}{\\beta}\u0026amp; \\beta_2 \\beta_3 \\frac{\\gamma -1}{\\beta} \u0026amp; \\beta_3^2\\frac{\\gamma -1}{\\beta} \\end{pmatrix} +I\\\\ \u0026amp;=\t\\begin{pmatrix} \\gamma -1 + 1 \u0026amp; -\\gamma \\beta_1 \u0026amp; -\\gamma_2 \\beta_2\u0026amp; -\\gamma \\beta_3 \\\\ \\gamma \\beta_1 \u0026amp; 1+ (\\gamma -1)\\frac{\\beta_1^2}{\\beta} \u0026amp;(\\gamma -1) \\beta_1 \\beta_2 \u0026amp; (\\gamma -1) \\frac{\\beta_1\\beta_3 }{\\beta} \\\\ -\\gamma \\beta_2 \u0026amp; \\frac{\\gamma -1}{\\beta} \u0026amp; 1+(\\gamma - 1) \\frac{\\beta_2^2}{\\beta} \u0026amp; \\frac{\\gamma -1 }{}\\beta\\\\ -\\gamma \\beta_3 \u0026amp; (\\gamma -1) \\frac{\\beta_1\\beta_3}{\\beta} \u0026amp; (\\gamma -1) \\frac{\\beta_2 \\beta_3}{\\beta} \u0026amp; 1 + (\\gamma -1) \\frac{\\beta_3^2}{\\beta}\\end{pmatrix}\n.\\end{align}\nTherefore\n\\[ A=\t\\begin{pmatrix} \\gamma \u0026 -\\gamma \\beta_1 \u0026 -\\gamma_2 \\beta_2\u0026 -\\gamma \\beta_3 \\\\ \\gamma \\beta_1 \u0026 1+ (\\gamma -1)\\frac{\\beta_1^2}{\\beta} \u0026(\\gamma -1) \\beta_1 \\beta_2 \u0026 (\\gamma -1) \\frac{\\beta_1\\beta_3 }{\\beta} \\\\ -\\gamma \\beta_2 \u0026 \\frac{\\gamma -1}{\\beta} \u0026 1+(\\gamma - 1) \\frac{\\beta_2^2}{\\beta} \u0026 \\frac{\\gamma -1 }{}\\beta\\\\ -\\gamma \\beta_3 \u0026 (\\gamma -1) \\frac{\\beta_1\\beta_3}{\\beta} \u0026 (\\gamma -1) \\frac{\\beta_2 \\beta_3}{\\beta} \u0026 1 + (\\gamma -1) \\frac{\\beta_3^2}{\\beta}\\end{pmatrix} .\\]","summary":"\u003cp\u003eIn this post we derive the functional form of a generic Lorentz transformation from its group generators.\nAt first glance I considered this trivial, as I was unaware of the role played by their non-commutativity — which made it an interesting exercise to present.\nWe follow Chapter 11 of Jackson (Third Edition), which makes use of the mathematical properties of Minkowski spacetime.\nClarifications missing from the book are supplied with original arguments.\u003c/p\u003e","date":"2026-02-12","section":"blogs"},{"title":"La forma de una transformada de Lorentz general por generadores","url":"https://edgarivanhinojosa.xyz/lorentz-transform/","content":"En el presente documento daremos solución a la tarea siguiendo el capítulo 11 del Jackson Tercera Edición, el cuál hace uso de las propiedades matemáticas del espacio tiempo de Minkowski. Se realizarán precisiones donde falten en el libro con apoyo de argumentos originales.\n$$Ax =x'$$$$ \\overline{g}(e_i) = e^i \\forall i \\neq 0 $$$$\\overline{g}(e_0) = - e^0$$\\[ u^T g v = (\\overline{g}(u), v) .\\]$$g = \\begin{pmatrix} -1 \u0026 \u0026 \u0026 \\\\ \u0026 1 \u0026 \u0026 \\\\ \u0026 \u0026 1 \u0026 \\\\ \u0026 \u0026 \u0026 1 \\end{pmatrix} $$$$det(A^T g A) det(A^T) det(g) det(A)= det(g)$$$$\\implies det(A^T) det(A) = det(A)^2= 1 \\implies det(A) = \\pm 1$$$$a_{ij} g_{jl} a_{lk} = g_{ik}$$$$A = e^L$$$$det(A) = e^{trL} =1 \\implies trL =0 $$$$A^TgA A^{-1}g = g A ^{-1}g \\implies A^T = gA ^{-1}g$$\\[ A^T= e^{L^T} \\implies gA^Tg = e^{g L^Tg} \\implies g^2 A^{-1} g^2 = A^{-1}= e^{g L^T g} \\implies e^{-L} = e^{g L^T g} .\\] Esto nos da la siguiente propiedad \\begin{equation}\ngL = gg L ^T g = L^T g = (gL)^T \\label{antisimetria} \\end{equation} quiere decir que $gL$ es anti simétrica i.e $(gL){ij} = - (gL){ji}$, analicemos indicialmente esta ecuación. Expandamos \\begin{align*} (gL){ij} = g{il} L_{lj} i\u0026amp;= - g_{jl} L_{li} \\ \\implies g_{ii} L_{ij} \u0026amp;= - g_{jj} L_{ji} \\ \\text{si i=0, j=0: } g_{00} L_{00} \u0026amp;= -g_{00} L_{00} \\ \\implies -L_{00} \u0026amp;= -(-1) L_{00} \\implies L_{00} = 0\\ \\text{ si } i=0 j\\neq 0: \\ g_{00} L_{0j} \u0026amp;= - g_{jj} L_{j_0} \\implies -L_{0j} = - L_{j_0}\\ \\text{si \\ } i\\neq 0, j \\neq 0: \\ g_{ii} L_{ij} \u0026amp;= - g_{jj} L_{ji} \\implies L_{ij} = -L_{ji} .\\end{align*} Por lo cuál $L$ es de la forma \\[ L = \\begin{pmatrix} 0 \u0026 L_{01} \u0026 L_{02} \u0026 L{03} \\\\ L_{01} \u0026 0 \u0026 L_{12} L_{13} \\\\ L_{02} \u0026 -L_{12} \u0026 0 \u0026 L_{23} \\\\ L_{03} \u0026 - L_{13} \u0026 -L_{23} \u0026 0 \\end{pmatrix} .\\] Sabemos que dentro de las posibles transformaciones que incluye la forma general de $L$ están las rotaciones espaciales ordinarias que satisfacen $A A^T = I \\implies L^T = -L$ por tanto ignoraremos las partes anti simétricas de esta forma general. Por lo tanto los generadores del grupo de rotaciones hiperbólicas en un espacio de Minkowski es el siguiente \\[ K_1 = \\begin{pmatrix} 0 \u0026 1 \u0026 0 \u0026 0 \\\\ 1\u0026 0 \u0026 \u0026 \\\\ 0 \u0026 \u0026 0 \u0026 \\\\ 0 \u0026 \u0026 \u0026 \\end{pmatrix} , \\ \\ \\ K_2 = \\begin{pmatrix} 0 \u0026 0 \u0026 1 \u0026 0 \\\\ 0\u0026 \u0026 \u0026 \\\\ 1 \u0026 \u0026 \u0026 \\\\ 0 \u0026 \u0026 \u0026 \\end{pmatrix} , \\ \\ \\ K_2 = \\begin{pmatrix} 0 \u0026 0 \u0026 0 \u0026 1 \\\\ 0\u0026 \u0026 \u0026 \\\\ 0 \u0026 \u0026 \u0026 \\\\ 1 \u0026 \u0026 \u0026 \\end{pmatrix} \\ \\ \\ .\\] Denotaremos a $\\epsilon \\cdot K$ como una combinación lineal de las matrices esto es $\\epsilon = \\epsilon_1 K_1 + \\epsilon_2 K_2 + \\epsilon_3 K_3$. Si $\\epsilon$ es un vector unitario, se cumple la siguiente igualdad $(\\epsilon \\cdot K)^3 = \\epsilon \\cdot K$. Esta propiedad se sostiene dado que los eigenvalores de una matriz cúbicamente idempotente son de la siguiente manera. Sea $F$ una matriz cúbicamente idempotente, supongamos que es diagonalizable \\begin{align*} Fv \u0026amp;= \\lambda v \\ F^3 v \u0026amp;= \\lambda^3 v \\implies \\lambda^3 = \\lambda\\ \\lambda(\\lambda^2 -1) \u0026amp;= 0 \\end{align*} en conclusión una matriz cúbica idempotente tiene sus eigen valores son unitarios positivo, negativo o nulo. $$det (gL^T) = det(-gL)$$$$\\implies det(L) = - det(L) \\implies det(L) =0$$\\[ \\begin{pmatrix} 0 \u0026 \\epsilon_1 \u0026 \\epsilon_2 \u0026 \\epsilon_3 \\\\ \\epsilon_1 \u0026 \u0026 \u0026 \\\\ \\epsilon_2 \u0026 \u0026 \u0026 \\\\ \\epsilon_3 \u0026 \u0026 \u0026 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ \\epsilon_1 \\\\ \\epsilon_2 \\\\ \\epsilon_3 \\end{pmatrix} = \\begin{pmatrix} \\epsilon_1^2 + \\epsilon_2^2 + \\epsilon_3^2 \\\\ \\epsilon_1 \\\\ \\epsilon_2 \\\\ \\epsilon_3 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ \\epsilon_1 \\\\ \\epsilon_2 \\\\ \\epsilon_3 \\end{pmatrix} \\]$$A = \\exp(-\\hat{\\beta}\\cdot K \\tanh^{-1}\\left( \\beta \\right) )$$$$A = - \\hat{\\beta} \\cdot K \\gamma \\beta + \\left( \\hat{\\beta} \\cdot K \\right)^2 (\\gamma - 1) + I $$\\[ A=\t\\begin{pmatrix} \\gamma \u0026 -\\gamma \\beta_1 \u0026 -\\gamma_2 \\beta_2\u0026 -\\gamma \\beta_3 \\\\ \\gamma \\beta_1 \u0026 1+ (\\gamma -1)\\frac{\\beta_1^2}{\\beta} \u0026(\\gamma -1) \\beta_1 \\beta_2 \u0026 (\\gamma -1) \\frac{\\beta_1\\beta_3 }{\\beta} \\\\ -\\gamma \\beta_2 \u0026 \\frac{\\gamma -1}{\\beta} \u0026 1+(\\gamma - 1) \\frac{\\beta_2^2}{\\beta} \u0026 \\frac{\\gamma -1 }{}\\beta\\\\ -\\gamma \\beta_3 \u0026 (\\gamma -1) \\frac{\\beta_1\\beta_3}{\\beta} \u0026 (\\gamma -1) \\frac{\\beta_2 \\beta_3}{\\beta} \u0026 1 + (\\gamma -1) \\frac{\\beta_3^2}{\\beta}\\end{pmatrix} .\\]","summary":"\u003cp\u003eEn el presente documento daremos solución a la tarea siguiendo el capítulo 11 del Jackson Tercera Edición, el cuál hace uso de las propiedades matemáticas del espacio tiempo de Minkowski.\nSe realizarán precisiones donde falten en el libro con apoyo de argumentos originales.\u003c/p\u003e\n$$Ax =x'$$$$\n\\overline{g}(e_i) = e^i \\forall i \\neq 0\n$$$$\\overline{g}(e_0) = - e^0$$\\[\nu^T g v = (\\overline{g}(u), v)\n.\\]$$g = \\begin{pmatrix} -1 \u0026 \u0026 \u0026 \\\\ \u0026 1 \u0026 \u0026 \\\\ \u0026 \u0026 1 \u0026 \\\\ \u0026 \u0026 \u0026 1 \\end{pmatrix} $$$$det(A^T g A) det(A^T) det(g) det(A)= det(g)$$$$\\implies det(A^T) det(A) = det(A)^2= 1 \\implies det(A) = \\pm 1$$$$a_{ij} g_{jl} a_{lk} = g_{ik}$$$$A = e^L$$$$det(A) = e^{trL} =1 \\implies trL =0 $$$$A^TgA A^{-1}g = g A ^{-1}g \\implies A^T = gA ^{-1}g$$\\[\n\t\tA^T= e^{L^T} \\implies gA^Tg = e^{g L^Tg} \\implies g^2 A^{-1} g^2 = A^{-1}= e^{g L^T g} \\implies e^{-L} = e^{g L^T g}\n.\\]\u003cp\u003e\nEsto nos da la siguiente propiedad\n\\begin{equation}\u003c/p\u003e","date":"2026-02-12","section":""},{"title":"Laser-track","url":"https://edgarivanhinojosa.xyz/projects/proto1/","content":"Graphical interface for laser path extraction and analysis for optical experimental arragements.\n","summary":"\u003cp\u003eGraphical interface for laser path extraction and analysis for optical experimental arragements.\u003c/p\u003e","date":"2026-01-10","section":"projects"},{"title":"Fractional Model of Gelatin Viscoelasticity","url":"https://edgarivanhinojosa.xyz/projects/gelatina-fraccional/","content":"We experimentally test a fractional calculus model with minimal instrumentation, using simple rheological methods.\n","summary":"\u003cp\u003eWe experimentally test a fractional calculus model with minimal instrumentation, using simple rheological methods.\u003c/p\u003e","date":"2025-11-10","section":"projects"},{"title":"Measuring Hall Effect in Bismuth","url":"https://edgarivanhinojosa.xyz/projects/hall/","content":"Measurement of the Hall effect using thin bismuth films produced without specialized equipment.\n","summary":"\u003cp\u003eMeasurement of the Hall effect using thin bismuth films produced without specialized equipment.\u003c/p\u003e","date":"2025-04-10","section":"projects"},{"title":"Wavelength Measurement by Circular Diffraction","url":"https://edgarivanhinojosa.xyz/projects/fraunhoffer/","content":"An intensity analysis of photographs and their diffraction patterns was performed to estimate the wavelength of a laser.\n","summary":"\u003cp\u003eAn intensity analysis of photographs and their diffraction patterns was performed to estimate the wavelength of a laser.\u003c/p\u003e","date":"2025-03-10","section":"projects"},{"title":"Symplectic Geometry in Classical Mechanics","url":"https://edgarivanhinojosa.xyz/projects/geo-simplectica/","content":"We investigate the rich structural properties of phase space on which classical mechanics rests. Before doing so, we briefly review the mathematical results that will be used in developing more important findings. We physically motivate the proposal of the symplectic structure in classical mechanics. Finally, we present an example of solving a problem using these tools.\n","summary":"\u003cp\u003eWe investigate the rich structural properties of phase space on which classical mechanics rests. Before doing so, we briefly review the mathematical results that will be used in developing more important findings. We physically motivate the proposal of the symplectic structure in classical mechanics. Finally, we present an example of solving a problem using these tools.\u003c/p\u003e","date":"2024-10-10","section":"projects"},{"title":"Conducción de Calor y Gradiente de Temperatura en dos Dimensiones","url":"https://edgarivanhinojosa.xyz/projects/termo-bidim/","content":"An experimental setup was designed and executed to verify the two-dimensional temperature gradient described by Fourier\u0026rsquo;s law of heat conduction.\n","summary":"\u003cp\u003eAn experimental setup was designed and executed to verify the two-dimensional temperature gradient described by Fourier\u0026rsquo;s law of heat conduction.\u003c/p\u003e","date":"2023-11-24","section":"projects"},{"title":"Archives","url":"https://edgarivanhinojosa.xyz/archives/","content":"","summary":"","date":"0001-01-01","section":""},{"title":"Contact","url":"https://edgarivanhinojosa.xyz/contact/","content":" ","summary":"\u003cstyle\u003e\n \n.social-icons {\n  display: flex;\n  flex-wrap: wrap;\n  gap: 0.75rem;\n  align-items: center;\n}\n\n.social-icon {\n  display: inline-flex;\n  color: inherit;\n  opacity: 0.7;\n  transition: opacity 0.2s ease, color 0.2s ease;\n}\n\n.social-icon:hover {\n  opacity: 1;\n}\n\n.social-icon svg {\n  width: 1.5rem;\n  height: 1.5rem;\n}\n\u003c/style\u003e\u003cdiv class=\"social-icons\"\u003e\u003ca href=\"https://github.com/klingsor0\" class=\"social-icon\" title=\"GitHub\" rel=\"me noopener\" target=\"_blank\"\u003e\n      \u003csvg xmlns=\"http://www.w3.org/2000/svg\" viewBox=\"0 0 24 24\" fill=\"currentColor\"\u003e\u003cpath d=\"M12 .297c-6.63 0-12 5.373-12 12 0 5.303 3.438 9.8 8.205 11.385.6.113.82-.258.82-.577 0-.285-.01-1.04-.015-2.04-3.338.724-4.042-1.61-4.042-1.61C4.422 18.07 3.633 17.7 3.633 17.7c-1.087-.744.084-.729.084-.729 1.205.084 1.838 1.236 1.838 1.236 1.07 1.835 2.809 1.305 3.495.998.108-.776.417-1.305.76-1.605-2.665-.3-5.466-1.332-5.466-5.93 0-1.31.465-2.38 1.235-3.22-.135-.303-.54-1.523.105-3.176 0 0 1.005-.322 3.3 1.23.96-.267 1.98-.399 3-.405 1.02.006 2.04.138 3 .405 2.28-1.552 3.285-1.23 3.285-1.23.645 1.653.24 2.873.12 3.176.765.84 1.23 1.91 1.23 3.22 0 4.61-2.805 5.625-5.475 5.92.42.36.81 1.096.81 2.22 0 1.606-.015 2.896-.015 3.286 0 .315.21.69.825.57C20.565 22.092 24 17.592 24 12.297c0-6.627-5.373-12-12-12\"/\u003e\u003c/svg\u003e\n    \u003c/a\u003e\u003ca href=\"https://www.linkedin.com/in/edgar-ivan-hinojosa-salda%c3%b1a-95273b260\" class=\"social-icon\" title=\"LinkedIn\" rel=\"me noopener\" target=\"_blank\"\u003e\n      \u003csvg xmlns=\"http://www.w3.org/2000/svg\" viewBox=\"0 0 24 24\" fill=\"currentColor\"\u003e\u003cpath d=\"M20.447 20.452h-3.554v-5.569c0-1.328-.027-3.037-1.852-3.037-1.853 0-2.136 1.445-2.136 2.939v5.667H9.351V9h3.414v1.561h.046c.477-.9 1.637-1.85 3.37-1.85 3.601 0 4.267 2.37 4.267 5.455v6.286zM5.337 7.433a2.062 2.062 0 0 1-2.063-2.065 2.064 2.064 0 1 1 2.063 2.065zm1.782 13.019H3.555V9h3.564v11.452zM22.225 0H1.771C.792 0 0 .774 0 1.729v20.542C0 23.227.792 24 1.771 24h20.451C23.2 24 24 23.227 24 22.271V1.729C24 .774 23.2 0 22.222 0h.003z\"/\u003e\u003c/svg\u003e\n    \u003c/a\u003e\u003ca href=\"mailto:e_e.ivan.h_hs@outlook.com\" class=\"social-icon\" title=\"Email\" rel=\"me noopener\" target=\"_blank\"\u003e\n      \u003csvg xmlns=\"http://www.w3.org/2000/svg\" viewBox=\"0 0 24 24\" fill=\"currentColor\"\u003e\u003cpath d=\"M20 4H4c-1.1 0-1.99.9-1.99 2L2 18c0 1.1.9 2 2 2h16c1.1 0 2-.9 2-2V6c0-1.1-.9-2-2-2zm0 4-8 5-8-5V6l8 5 8-5v2z\"/\u003e\u003c/svg\u003e\n    \u003c/a\u003e\n\u003c/div\u003e","date":"0001-01-01","section":""},{"title":"CV","url":"https://edgarivanhinojosa.xyz/cv/","content":"Edgar Ivan Hinojosa Saldaña Physicist — San Nicolás de los Garza, México\nEducation B.S. in Physics Universidad Autónoma de Nuevo León — 2021–2025\nExperience Laboratory Technician and Instructor Department of Physics, FCFM — Jan. 2025 – Dec. 2025\nDesign and implementation of experimental setups using precision sensors adapted with microcontrollers, extending the lifespan of existing laboratory equipment. Calibration, validation, and maintenance of experimental instrumentation. Instruction of undergraduate mechanics laboratory courses and assistance in electrodynamics laboratory sessions. Research Assistant CICFIM, UANL — Jan. 2024 – Dec. 2025\nDesign and prototyping of open-source, low-cost scientific equipment while maintaining satisfactory experimental precision. Development and validation of geometry- and physics-based models to describe experimental phenomena, achieving satisfactory mean squared errors. Programming of automated data-processing pipelines, report generation tools, and interactive graphical user interfaces for experimental data analysis. Skills Programming Languages: Python, R, Bash, Fortran\nFrameworks \u0026amp; Tools: OpenCV, NumPy, Pandas, Mathematica, Microcontrollers, AutoCAD, Onshape\n","summary":"\u003ch2 id=\"edgar-ivan-hinojosa-saldaña\"\u003eEdgar Ivan Hinojosa Saldaña\u003c/h2\u003e\n\u003cp\u003e\u003cstrong\u003ePhysicist\u003c/strong\u003e — San Nicolás de los Garza, México\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"education\"\u003eEducation\u003c/h2\u003e\n\u003cp\u003e\u003cstrong\u003eB.S. in Physics\u003c/strong\u003e\nUniversidad Autónoma de Nuevo León — 2021–2025\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"experience\"\u003eExperience\u003c/h2\u003e\n\u003cp\u003e\u003cstrong\u003eLaboratory Technician and Instructor\u003c/strong\u003e\nDepartment of Physics, FCFM — Jan. 2025 – Dec. 2025\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003eDesign and implementation of experimental setups using precision sensors adapted with microcontrollers, extending the lifespan of existing laboratory equipment.\u003c/li\u003e\n\u003cli\u003eCalibration, validation, and maintenance of experimental instrumentation.\u003c/li\u003e\n\u003cli\u003eInstruction of undergraduate mechanics laboratory courses and assistance in electrodynamics laboratory sessions.\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp\u003e\u003cstrong\u003eResearch Assistant\u003c/strong\u003e\nCICFIM, UANL — Jan. 2024 – Dec. 2025\u003c/p\u003e","date":"0001-01-01","section":""},{"title":"Extended CV","url":"https://edgarivanhinojosa.xyz/extended-cv/","content":"Edgar Ivan Hinojosa Saldaña Physicist — San Nicolás de los Garza, México e_e.ivan.h_hs@outlook.com · +52 867 175 1288 · edgarivanhinojosa.xyz · LinkedIn\nSelf-driven professional with autonomous learning habits, focused on precision and continuous process improvement. Throughout my development I have acquired techniques and skills in mathematical modeling, statistical data analysis, software development, and instrumentation.\nEducation B.S. in Physics Facultad de Ciencias Físico Matemáticas, UANL — San Nicolás de los Garza Aug. 2021 – Dec. 2025\nRelevant courses: Programming Methodology, Computational Physics, Probability and Statistics, Scientific Modeling High School Instituto América de Estudios Superiores — Nuevo Laredo, Tamps. Aug. 2018 – Aug. 2021\nExperience Validation Testing Engineer (part-time) Creativeit Group — Mar. 2026 – Present\nDesign of a data acquisition system programmed in Python and PyVISA, using IEEE 488 SCPI protocols and PyQt graphical interface integration. Documentation and execution of validation tests; consulting standards and writing pre-certification reports and preventive maintenance assays. Laboratory Technician and Instructor Department of Physics, FCFM — Jan. 2025 – Dec. 2025\nDesign and instruction of laboratory practicals emphasizing quantitative and statistical analysis of results. Design and population of an SQLite database for equipment inventory management. Research \u0026amp; Development XXVI Scientific and Technological Summer Research — UANL, Aug. 2025 Design of a vortex machine for hydrodynamic systems.\nResearch Assistant in Mathematical Physics — CICFIM, 2023–2024 Physical-motivation-based deduction of symplectic structure in Classical Mechanics.\nThesis Writing — UANL, 2023 Documentation of calculations, bibliographic review, and interpretation of results.\nXXIV Scientific and Technological Summer Research — UANL, Aug. 2023 Documentation of deductive demonstrations in geometric optics.\nResearch Assistant in Materials Simulation — CICFIM, 2022–2023 Development of optimization algorithms on High Performance Computing systems.\nScientific Modeling \u0026amp; Data Analysis Skills developed:\nMathematical model development and validation — Derived from first principles or statistical frameworks; experimentally contrasted for interpretation and successive correction. Literature review — Reading of documentation and bibliographic background, as well as writing of expository deliverables and interactive reports. Software development — Implementation of Python routines for cleaning, analysis, and visualization of experimental data. Pattern extraction algorithms with semi-automatic computer vision. Statistical techniques — Confidence intervals, hypothesis testing, normality tests, linear and logistic regression, cross-validation, bootstrapping, decision trees, random forests, DBSCAN, clustering, support vector machines, regularization methods (LASSO), principal component analysis, neural networks, among others. Projects:\nProfessor Approval Distribution — Application of normality tests (Kolmogorov–Smirnov and others) to characterize the distribution of professor ratings by users of misprofes.com. Numerical Methods Library in Modern Fortran — A Fortran and Python library using object-oriented Modern Fortran for organizing numerical and statistical methods. Laser Beam Extraction — Image processing for laser beam extraction via physically-motivated snake optimization method. Predictive Code in Python and Fortran — Computational implementation of a biologically plausible neural network with stochastic stimulus response, for learning probability distributions. Fractional Modeling of Gelatin Viscoelasticity — Experimental validation of a model via analysis of dynamic response experimental results. Predictive Maintenance — Decision trees and random forests for vibration characterization in power motors for successive failure prediction. Conferences \u0026amp; Colloquia National Physics Congress 2025 — Poster: Gravitational lens analogy via waveguide in a vorticity fluid system. Applied Optics and Energy Sustainability Seminar 2025 — Poster (2nd place): Friction Force Measurement with Time Series Analysis. FCFM Experimental Physics Fair, Spring 2025 — Poster: The Vortical Lens of a Fluid and its Similarity to the Gravitational Lens. AAPT Colloquium 2024 — Talk: Eddington Experiment: a Hydrodynamic Experimental Precedent. National Physics Congress 2024 — Poster: Symplectic Geometry in Classical Mechanics. Courses \u0026amp; Certifications Course Institution Date Advanced Integrated Technologies School 2025 — Optical technologies, plasmon theory CICESE, Monterrey 2025 Laser Safety Workshop CICESE, Monterrey 2025 Summer School in Statistical Computing 2024 — Machine learning, consulting, parallel computing CIMAT, Monterrey Jul. 2024 6th Thematic School in Analysis and Dynamical Systems — Spectral theory, chaotic dynamics CIMAT, Guanajuato Jun. 2024 Consulting Workshop — Applied statistical methods, executive reporting FCFM Jun. 2024 Statistical Learning (MOOC) — Linear and classification models, CNNs, bootstrapping Stanford Online 2023 Summer School in Modeling and Data Analysis, ed. V — FeynCalc for QFT symbolic computation UANL Jul. 2023 Summer School in Modeling and Data Analysis, ed. III — R, statistical methods, Bash FCFM Jul. 2023 AutoCAD Fundamentals UANL Oct. 2022 Summer School in Modeling and Data Analysis, ed. II — Bash, DFT, Molecular Dynamics UANL Jul. 2022 Python Programming: Basic and Intermediate UANL Jan. 2022 Distinctions Mexican Mathematics Olympiad (OMM) — State finalist, problem solving and mathematical reasoning, 2019–2021. Talent Program, UANL — Academic excellence scholarship, 2021. Extracurricular Activities Datathon 2024, TEC-MTY — Web scraping, sentiment analysis, clustering. Experiment Fair Judge 2025 — Participation as judge focusing on didactic experiment design and technological proposals. Skills Languages: Spanish, English\nProgramming: Python, R, Bash, Fortran, Mathematica\nTools \u0026amp; Frameworks: Jupyter, Streamlit, PyQt, OpenCV, NumPy, Scikit-Learn, PyTorch, SQLite\nSoftware: Office, AutoCAD, RStudio, Onshape, Vim, Linux environments\nInstrumentation: Oscilloscope, precision multimeter, electrometer, high-voltage testers, vector network analyzer, data acquisition systems, lasers, optical systems, high-voltage sources, soldering, digital circuits and electronics.\n","summary":"\u003ch2 id=\"edgar-ivan-hinojosa-saldaña\"\u003eEdgar Ivan Hinojosa Saldaña\u003c/h2\u003e\n\u003cp\u003e\u003cstrong\u003ePhysicist\u003c/strong\u003e — San Nicolás de los Garza, México\n\u003ca href=\"mailto:e_e.ivan.h_hs@outlook.com\"\u003ee_e.ivan.h_hs@outlook.com\u003c/a\u003e · +52 867 175 1288 · \u003ca href=\"https://edgarivanhinojosa.xyz\"\u003eedgarivanhinojosa.xyz\u003c/a\u003e · \u003ca href=\"https://www.linkedin.com/in/edgar-ivan-hinojosa-salda%C3%B1a-95273b260/\"\u003eLinkedIn\u003c/a\u003e\u003c/p\u003e\n\u003cp\u003eSelf-driven professional with autonomous learning habits, focused on precision and continuous process improvement. Throughout my development I have acquired techniques and skills in mathematical modeling, statistical data analysis, software development, and instrumentation.\u003c/p\u003e\n\u003c!-- Download button — uncomment when ready:\n[Download PDF](/cv/cv-02-26.pdf)\n--\u003e\n\u003chr\u003e\n\u003ch2 id=\"education\"\u003eEducation\u003c/h2\u003e\n\u003cp\u003e\u003cstrong\u003eB.S. in Physics\u003c/strong\u003e\nFacultad de Ciencias Físico Matemáticas, UANL — San Nicolás de los Garza\nAug. 2021 – Dec. 2025\u003c/p\u003e","date":"0001-01-01","section":""},{"title":"Search","url":"https://edgarivanhinojosa.xyz/search/","content":"","summary":"","date":"0001-01-01","section":""},{"title":"Sections","url":"https://edgarivanhinojosa.xyz/sections/","content":"","summary":"","date":"0001-01-01","section":""}]