The LATEX rendering of an article by Barry Barett on E = mc2

So, after trying to grapple with the LATEX code, I have produced an amateurish rendering of an Article by Barry Barett  and though the result is less than satisfactory, I have learned loads on how to render text as per the LATEX code and formatting, how to use Sagetex  and embed the output from within the system into a TEX file and ofc, how to share a file on the blog as well; A good learning experience, for sure.

With this wonderful experience under my belt, I would definitely improve the quality of the PDF's that I will upload in the future;P

Below is the PDF version of the article which I wanted to convert into a LATEX file.

Link to the PDF version of the article


This is the result of all the effort that I put in, quite amateurish but a good start, I believe. I skipped a lot of stuff and focused on the most basic elements of the articles, which made sense to me at the time. So, here it is the partial rendering.


And this is the LATEX code which produced the file

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\documentclass[12pt, letterpaper]{article} \usepackage[utf8]{inputenc} %%\DeclareMathSizes{12}{30}{16}{12} \usepackage{amsmath} % for the "align" and "align*" environments \title{The Most Famous Equation in Physics} \author{Abhinav Sharma \thanks{ Thanks to sharelatex.com}} \date{February 2014} \begin{document} \begin{titlepage} \maketitle \end{titlepage} \textbf{\textit{PART A: Does the inertia of a body depend upon its energy content?}} Students of physics know the answer to the question Einstein asked in the title of his celebrated paper, "Does the inertia of an object body depend upon its energy content?" because in it Einstein derived the equation $E=mc^2$ . The inertia (or mass) of an object at rest is equal to $\frac{E}{c^2}$, where E is the energy content of the object. A remarkable fact about this paper and many of his other early papers is their relative simplicity. I believe I can briefly state the few mathematical and physical assumptions used in the paper. These are conservation of energy, the principle of relativity, the formulas for the kinetic energy of a particle and the relativistic energy of light and the binomial theorem. Given these assumptions, and a cleverly devised thought experiment to apply them to, can be derived as though it were a simple undergraduate word problem. \textbf{\textit{PART B: The Toolkit: Physical and Mathematical Assumptions}} \textbf{Physical Assumptions:} 1. \textit{Conservation of Energy}: Energy is neither created or destroyed. The total energy in an isolated system remains constant. In Einstein's thought experiment, the system consists of all the volume enclosing a box. $$ E_{initial} = E_{final}$$ A classic example of conservation of energy. the pendulum The bob of a pendulum has kinetic $ \frac{1}{2} m v^2$ and potential energy $ m g h$ ,where $m$ is the mass of the bob, $v$ is its velocity, $h$ is its height and $g$ is a number that is determined by the strength of gravity. According to conservation of energy $$ \frac{1}{2} m v^2 + m g h = constt$$ As the bob ascends and its velocity decreases, h increases to keep the equation constant. When the height is maximum, the velocity is zero and when the height is zero, the velocity is maximized. As the bob swings back and forth, kinetic and potential energy vary, but the total energy of the system remains constant. 2. \textit{The relativistic equation for the energy of light} The energy of a light ray in a frame moving with a constant velocity v can be determined from its energy in a stationary frame from {\Large {$$l^{'} = {l} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right)$$}} [assuming the moving frame is parallel to the x-axis of the stationary frame and is the angle of the light ray from the x-axis] If the light ray is moving parallel with the x-axis and in the same direction as the moving reference frame then $\phi = 0$ and $cos(0)=1$ and therefore {\Large {$$l^{'} = {l} \left(\frac{ 1 - \frac{v}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right)$$}} If the light ray were moving in the opposite direction, then (because $\cos (2*pi)= -1\)$ the negative sign (_) before the $ \frac{v}{c}$ term is replaced by a positive sign(+). Einstein used this fact to simplify the mathematics of his derivation. 3. \textit{relativity: the freedom to choose different reference frames } The principle of Special Relativity states that if a physical principle or equation holds in a stationary frame, it must also hold in a frame moving relative to it at constant velocity. We can evaluate a process from either reference frame. In the derivation of $E=m c^2$ below, the energy of a simple system is calculated in different frames of reference moving at constant velocity $v$ relative to one another and, in accordance with the principle of relativity, it is assumed that the principle of conservation of energy holds equally in both frames of reference. 4. \textit{kinetic energy} As mentioned above, the kinetic energy of a particle with mass $m$, is $$ \frac{1}{2} m v^2$$ Mathematical Assumptions: 1. Elementary Algebra The Binomial Series: $$ {\left( 1 + x\right)}^\alpha = \sum_{k = 0}^{+\infty} \binom{\alpha}{k} x^k$$ $$ & = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!} x^2 + \cdots $$ 2. Calculus (optional) The Taylor Series: $$ f(x) = f(0) + x f^{'}(0) + \cdots $$ These assumptions (and keen physical intuition) are all you need to derive $E=mc^2$. So, before I guide you through Einstein's derivation, try and derive it yourself. If I was a high school physics teacher I'd like to assign this word problem as extra credit. The mass-energy equivalence is described by the famous equation $$E=mc^2$$ discovered in 1905 by Albert Einstein. In natural units ($c$ = 1), the formula expresses the identity $$ E = m c^2 $$ \textit{\textbf{PART D: The Derivation. Solving the Word Problem }} The initial total energy of the box, relative to the stationary system (x,y,z), in Einstein's thought experiment is $E_0$ . This is the energy of the system, in the stationary frame, prior to the emission of the light rays. The initial total energy of the box, relative to the moving frame (t,u,v), is $H_0$ . This is the energy of the box, relative to the moving frame, prior to the emission of the light rays. If $E_1$ is the energy of the box, relative to the stationary frame, after the emission of the light rays and the total energy of both light rays is $L$ then, by conservation of energy $$ E_0 = E_1 + \frac{L}{2} + \frac{L}{2} $$ If $H_1$ is the energy of the box, relative to the moving frame, after the emission of the light rays and the energy of the light rays in the moving frame is given by {\Large {$$l^{'} = {l} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right)$$}} then, by conservation of energy and the relativistic equation for the energy of light $$ H_0 = H_1 + \frac{l}{2} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right) + \frac{l}{2} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right)$$ and therefore $$ H_0 = H_1 + {l} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right) $$ Einstein then calculates that $$ H_0 - E_0 - \( H_1 - E_1 \) = {l} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } - 1\right) $$ The first and second H-E terms on the left are measures of the change in energy of the box at the same instant in time due only to the relative motion of the two frames of reference. The first H-E term is the initial change in energy of the box due to relative motion of the two frames and the second term H-E term is the change in energy of the box after the emission of the light ray, but again, only due to the relative motion of the two frames. Because the H-E terms measure the change in energy of the box due to the relative motion of the two frames only, the additive constant C representing any other energy left over (such as the internal molecular energies of the box etc) is constant $$ H_0 - E_0 = K_0 + C $$ $$ H_1 - E_1 = K_1 + C $$ and therefore C cancels out leaving only the change in kinetic energy as $$ K_0 - K_1 = {l} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } - 1\right) $$ This can be approximated by the Binomial or Taylor Series "neglecting magnitudes of fourth order or higher." Applying the Binomial Theorem approximation above, $$ \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right) = 1 + \frac{1}{2} \frac{v^2}{c^2} $$ therefore $$ K_0 - K_1 = \frac{1}{2} \frac{v^2}{c^2} $$ By comparing this to the expression for Kinetic energy one can infer that the change in mass of the box due to the emission of light is equal to L/c^2. Recall that L is the total energy of the light rays emitted from the box, therefore $E=m c^2$. An Important Proviso: By using the binomial approximation in this derivation, it was assumed that v < < c. If v were close to c, then this approximation would be invalid. This is why mc^2 is the rest energy of an object. The slower the box in the derivation is moving, the more accurate the approximation becomes. If an object is moving close to the speed of light, then the E=mc^2 approximation must be replaced by E^2 = (mc^2)^2 + (pc)^2 where p is momentum. From this equation it directly follows that:—If a body gives off the energy L in the form of radiation, its mass diminishes by L/c². The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference, so that we are led to the more general conclusion that the mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 1020, the energy being measured in ergs, and the mass in grammes. \begin{align} $$ cos\left( \phi\right)$$ $$ 1 - \frac{v cos\left( \phi\right)}{c} $$ $$ \sqrt{\left( 1 - \frac{v^2}{c^2} \right)}$$ \end{align} $$l^{'} = {l} \frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } $$ {\Large {$$l^{'} = {l} \left(\frac{ 1 - \frac{v cos\left( \phi\right)}{c}}{\sqrt{\left( 1 - \frac{v^2}{c^2} \right)} } \right)$$}} %%\begin{align*} % the "starred" equation environments produce no %%equation numbers %%a &= b\\ % if no alignment is needed, use the gather* instead of %%the align* env. %% &= c %%\end{align*} %%$$ \sum_{n=-\infty}^{+\infty} f(x) $$ $$ \cdots = sin(\phi) $$ \end{document}