13.1 Commutation relations for oscillators

Posted on April 18, 2025 by rpflee

A First Course in String Theory

(a) Use the lower-sign version of equation (13.28) and the appropriate mode expansion to verify explicitly that the unbarred commutation relations of (13.29) emerge.

~~~

Eq. (13.28):

$[(\dot X^I - X^{I'}) (\tau, \sigma), (\dot X^J - X^{J'})(\tau, \sigma')] = - 4 \pi \alpha' i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma')$

Eq. (13.29):

$\left[ \alpha_m^I, \alpha_n^J \right] = m \delta_{m+n, 0} \eta^{IJ}$

Eq. (13.24):

$\displaystyle{X^\mu (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \neq 0} \frac{e^{-i n \tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}$

$\displaystyle{\dot{X}^\mu(\tau, \sigma) = \sqrt{2 \alpha'} \alpha_0^\mu + \sqrt{\frac{\alpha'}{2}} \sum_{n \neq 0} e^{-i n \tau} \left( \alpha_n^\mu e^{i n \sigma} + \bar{\alpha}_n^\mu e^{-i n \sigma} \right)}$

$\displaystyle{X^{\mu'}(\tau, \sigma) = \sqrt{\frac{\alpha'}{2}} \sum_{n \neq 0} e^{-i n \tau} \left( \bar{\alpha}_n^\mu e^{-i n \sigma} - \alpha_n^\mu e^{i n \sigma} \right)}$

$\displaystyle{\begin{aligned} &\int_0^{2 \pi} f(\sigma) d \sigma\frac{d}{d \sigma} \delta(\sigma - \sigma') \\ &= \int_0^{2 \pi} f(\sigma) d \delta(\sigma - \sigma') \\ &= f(2 \pi) \delta(2 \pi - \sigma') - f(0) \delta(0 - \sigma') - \int_0^{2 \pi} \delta(\sigma - \sigma') \frac{df(\sigma)}{d \sigma} d \sigma \\ \end{aligned}}$

Eq. (13.26):

$\displaystyle{\begin{aligned} \left[(\dot X^I - X^{I'}) (\tau, \sigma), (\dot X^J - X^{J'})(\tau, \sigma')\right] &= - 4 \pi \alpha' i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \left[ \sqrt{2 \alpha'} \sum_{m \in \mathbb{Z}} \alpha_m^I e^{-i m (\tau - \sigma)}, \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \alpha_n^J e^{-i n (\tau - \sigma')} \right] &= - 4 \pi \alpha' i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \sum_{m \in \mathbb{Z}} \sum_{n \in \mathbb{Z}} e^{-i m (\tau - \sigma)} e^{-i n (\tau - \sigma')} \left[ \alpha_m^I , \alpha_n^J \right] &= - 2 \pi i \eta^{IJ} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma e^{iq\sigma} \sum_{m \in \mathbb{Z}} \sum_{n \in \mathbb{Z}} e^{-i (m+n) \tau} e^{i m \sigma} e^{i n \sigma'} \left[ \alpha_m^I , \alpha_n^J \right] &= - 2 \pi i \eta^{IJ} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma e^{iq\sigma} \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \sum_{n \in \mathbb{Z}} e^{-i (-q+n) \tau} e^{i n \sigma'} \left[ \alpha_{-q}^I , \alpha_n^J \right] &= - i \eta^{IJ} \left[ \delta(2 \pi - \sigma') - \delta(0 - \sigma') - iq \int_0^{2 \pi} \delta (\sigma - \sigma') e^{i q \sigma} d\sigma \right] \\ \sum_{n \in \mathbb{Z}} e^{-i (-q+n) \tau} e^{i n \sigma'} \left[ \alpha_{-q}^I , \alpha_n^J \right]&= - i \eta^{IJ} \left[ \delta(2 \pi - \sigma') - \delta(0 - \sigma') - iq e^{i q \sigma'} \right] \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma' e^{ip\sigma'} \sum_{n \in \mathbb{Z}} e^{-i (-q+n) \tau} e^{i n \sigma'} \left[ \alpha_{-q}^I , \alpha_n^J \right]&= - i \eta^{IJ} \frac{1}{2 \pi} \int_0^{2 \pi} d \sigma' e^{ip\sigma'} \left[ \delta(2 \pi - \sigma') - \delta(0 - \sigma') - iq e^{i q \sigma'} \right] \\ \end{aligned}}$

$\displaystyle{\begin{aligned} e^{i (p+q) \tau} \left[ \alpha_{-q}^I , \alpha_{-p}^J \right]&= - \eta^{IJ} \frac{1}{2 \pi} q \int_0^{2 \pi} d \sigma' e^{ip\sigma'} e^{i q \sigma'} \\ e^{i (p+q) \tau} \left[ \alpha_{-q}^I , \alpha_{-p}^J \right]&= - \eta^{IJ} q \delta_{p+q,0} \\ \\ \\ e^{i (m+n) \tau} \left[ \alpha_{-n}^I , \alpha_{-m}^J \right]&= - n \delta_{m+n,0} \eta^{IJ} \\ \end{aligned}}$

— Me@2025-04-17 12:26:46 PM

Quick Calculation 13.5

Posted on February 4, 2025 by rpflee

A First Course in String Theory

Show that $[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] = \frac{q}{2} \bar{\alpha}_{- \frac{q}{2}}$ and explain why $\bar{N}^\perp$ is properly called a number operator.

~~~

$\bar N^\perp = \sum_{p=1}^{\infty} \bar \alpha^i_{-p} \bar \alpha^i_{~p} + \sum_{k \in \mathbb{Z}^+_\text{odd}} \bar \alpha_{- \frac{k}{2}} \bar \alpha_{\frac{k}{2}}$

Eq. (13.109):

$\left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right] = \frac{m}{2} \delta_{m+n, 0}$

$[AB,C]=A[B,C]+[A,C]B$

$\begin{aligned} &[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \sum_{p=1}^{\infty} [ \bar \alpha^i_{-p} \bar \alpha^i_{~p}, \bar{\alpha}_{- \frac{q}{2}} ] + \sum_{k \in \mathbb{Z}^+_\text{odd}} [ \bar \alpha_{- \frac{k}{2}} \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \end{aligned}$

If $q=0$ ,

$[\bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] = 0$

If $q>0$ ,

$\begin{aligned} &[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \sum_{k \in \mathbb{Z}^+_\text{odd}}\left( \bar \alpha_{- \frac{k}{2}} [ \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] + [ \bar \alpha_{- \frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \bar \alpha_{\frac{k}{2}} \right) \\ &= \sum_{k \in \mathbb{Z}^+_\text{odd}}\left( \bar \alpha_{- \frac{k}{2}} [ \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \right) + 0 \\ &= \bar \alpha_{- \frac{q}{2}} [ \bar \alpha_{\frac{q}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \bar \alpha_{- \frac{q}{2}} \frac{q}{2} \\ \end{aligned}$

If $q<0$ ,

$\begin{aligned} &[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] \\ &= \sum_{k \in \mathbb{Z}^+_\text{odd}}\left( \bar \alpha_{- \frac{k}{2}} [ \bar \alpha_{\frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] + [ \bar \alpha_{- \frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \bar \alpha_{\frac{k}{2}} \right) \\ &= 0 + \sum_{k \in \mathbb{Z}^+_\text{odd}} + [ \bar \alpha_{- \frac{k}{2}}, \bar{\alpha}_{- \frac{q}{2}} ] \bar \alpha_{\frac{k}{2}} \\ &= \frac{q}{2} \bar \alpha_{\frac{-q}{2}} \\ \end{aligned}$

$\bar{N}^\perp$ should not be called a number operator. Instead, $[ \bar{N}^\perp, \cdot ]$ should be.

It's through the commutation relation $[ \bar{N}^\perp, \bar{\alpha}_{- \frac{q}{2}} ] = \frac{q}{2} \bar{\alpha}_{- \frac{q}{2}}$ that we see $\bar{N}^\perp$ functioning as a number operator, not by $\bar{N}^\perp$ acting directly on $\bar{\alpha}_{- \frac{q}{2}}$ operators.

— Me@2025-02-04 11:56:57 AM

Problem 14.5d1.2.3

Posted on January 1, 2025 by rpflee

A First Course in String Theory

The generating function is an infinite product:

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

To evaluate the infinite product, you can use wxMaxima. However, it does not provide $\LaTeX$ rendering of answers yet. Instead, you can call Maxima‘s code in SageMath, if you use Jupyter Notebook to access SageMath.

reset()

%display latex

maxima('taylor((1/x)*product((1 + x^(r - 1/2))^32 / (1 - x^r)^8, r, 1, oo), x, 0, 6)')

$\displaystyle {{1}\over{x}}+{{32}\over{\sqrt{x}}}+504+5248\,\sqrt{x}+40996\,x+ 258624\,x^{{{3}\over{2}}}+1384320\,x^2+6512384\,x^{{{5}\over{2}}}+ 27623826\,x^3+107640288\,x^{{{7}\over{2}}}+390667136\,x^4+1334500992 \,x^{{{9}\over{2}}}+4325559288\,x^5+13390178752\,x^{{{11}\over{2}}}+ 39794729472\,x^6+\cdots$

— Me@2024-12-02 06:33:46 AM

SageMath for Ubuntu 24.04

Posted on December 24, 2024 by rpflee

0. In this tutorial, you will need to go to the official website of NixOS. Make sure that its website is the real, official one. Any instructions from an imposter website can get your machine infected with malware.

1. Assuming your computer’s OS is Ubuntu 24.04 or above, go to the NixOS official website. Follow the instructions to install the Nix package manager (not the NixOS operating system) onto your OS. Choose the “single-user installation” method.

2. Run this command to install SageMath:

nix-env -iA nixpkgs.sage

3. Run this to open:

sage --notebook=jupyterlab

— Me@2024-11-27 12:28:48 AM

Quick Calculation 13.4

Posted on October 28, 2024 by rpflee

A First Course in String Theory

Prove (13.109).

$\left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right] = \frac{m}{2} \delta_{m+n, 0}$

~~~

Eq. (13.107):

$\begin{aligned} \sum_{m',n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m'}{2}(\tau+\sigma)} e^{-i\frac{n'}{2}(\tau+\sigma')} \left[ \bar \alpha_{\frac{m'}{2}}, \bar \alpha_{\frac{n'}{2}} \right] &= 2 \pi i \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \end{aligned}$

$\begin{aligned} \frac{1}{2\pi} \int_0^{2 \pi} d \sigma e^{i \frac{m}{2} \sigma}\sum_{m',n'\in \mathbb{Z}_\text{odd}} \cdots &= \int_0^{2 \pi} d \sigma e^{i \frac{m}{2} \sigma} i \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \end{aligned}$

$\begin{aligned} \frac{1}{2\pi} \int_0^{2 \pi} d \sigma \sum_{m',n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m'}{2}\tau} e^{i\frac{(m-m')}{2}\sigma} \cdots &= \int_0^{2 \pi} d \sigma e^{i \frac{m}{2} \sigma} i \frac{d}{d \sigma} \delta(\sigma - \sigma') \\ \sum_{m',n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m'}{2}\tau} \delta_{m,m'} \cdots &= i \int_0^{2 \pi} e^{i \frac{m}{2} \sigma} d \delta(\sigma - \sigma') \\ \end{aligned}$

$\begin{aligned} \sum_{n'\in \mathbb{Z}_\text{odd}} e^{-i\frac{m}{2}\tau} \cdots &= i \left( \delta(2\pi - \sigma') - \delta(- \sigma') - \int_0^{2 \pi} \delta(\sigma - \sigma') d e^{i \frac{m}{2} \sigma} \right) \\ \end{aligned}$

$\begin{aligned} e^{-i\frac{m}{2}\tau} e^{-i\frac{n}{2}\tau} \left[ \cdots \right] &= i \frac{1}{2\pi} \left( - e^{i\frac{n}{2} 2 \pi} - e^{i\frac{n}{2} 0} - \int_0^{2 \pi} e^{i\frac{n}{2}\sigma} d e^{i \frac{m}{2} \sigma} \right) \\ e^{-i\frac{m+n}{2}\tau} \left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right] &= i \frac{1}{2\pi} \left( - i\frac{m}{2} \int_0^{2 \pi} e^{i\frac{m+n}{2}\sigma} d \sigma \right) \\ e^{-i\frac{m+n}{2}\tau} \left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right] &= i \frac{1}{2\pi} \left( - i\frac{m}{2} 2\pi \delta_{m+n,0} \right) \\ e^{-i\frac{m+n}{2}\tau} \left[ \bar \alpha_{\frac{m}{2}}, \bar \alpha_{\frac{n}{2}} \right] &= \frac{m}{2} \delta_{m+n,0} \\ \end{aligned}$

— Me@2024-10-28 01:34:14 PM

Quick Calculation 13.3

Posted on October 3, 2024 by rpflee

A First Course in String Theory

Justify $UHU^{-1} = H$ directly from the oscillator expansion of $H = L_0^\perp + \bar L_0^\perp -2$ .

~~~

Eq. (13.37):

$\displaystyle{ \begin{aligned} \bar L_n^\perp &= \frac{1}{2} \sum_{p \in \mathbf{Z}} \bar \alpha_p^I \bar \alpha_{n-p}^I\text{,} ~~~L_n^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \alpha_p^I \alpha_{n-p}^I\text{.} \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} \bar L_0^\perp &= \frac{1}{2} \sum_{p \in \mathbf{Z}} \bar \alpha_p^I \bar \alpha_{-p}^I\text{,} ~~~L_0^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \alpha_p^I \alpha_{-p}^I\text{.} \\ \end{aligned} }$

Eq. (13.93):

$\displaystyle{ \begin{aligned} U x_0 U^{-1} &= -x_0 \text{,}~~~ U p U^{-1} = -p \text{,} \\ U \alpha_n U^{-1} &= -\alpha_n \text{,}~~~ U \bar \alpha_n U^{-1} = - \bar \alpha_n \text{,} \end{aligned} }$

where $n \ne 0$ .

Eq. (13.22):

$\displaystyle{ \begin{aligned} \alpha_0^\mu &= \sqrt{\frac{\alpha'}{2}} p^\mu \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} U \alpha_0^\mu U^{-1} &= - \alpha_0^\mu \\ \end{aligned} }$

Eq. (13.17)

$\displaystyle{ \begin{aligned} \bar \alpha_0^\mu &= \alpha_0^\mu \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} U \bar \alpha_0^\mu U^{-1} &= - \bar \alpha_0^\mu \\ \end{aligned} }$

— Me@2024-10-03 06:42:21 PM

Quick Calculation 13.3.0

Posted on September 8, 2024 by rpflee

A First Course in String Theory

Physical Meaning of Virasoro Operators

The Virasoro operators are fundamental in the context of two-dimensional conformal field theory (CFT) and string theory. They arise from the study of the symmetries of two-dimensional surfaces, particularly in how these surfaces can be mapped conformally (i.e., preserving angles) onto one another.

Conformal Symmetry

The Virasoro operators are associated with the Virasoro algebra, which is an infinite-dimensional Lie algebra that extends the algebra of diffeomorphisms on a circle. This algebra captures the symmetries of two-dimensional conformal transformations. In physical terms, these transformations are crucial for understanding how physical theories behave under changes of coordinates on the worldsheet of strings or in two-dimensional quantum field theories.

Role in String Theory

In string theory, the Virasoro operators are derived from the quantization of the string’s motion. They correspond to the modes of oscillation of the string and are denoted as $L_n$ . The operator $L_0$ is particularly significant because it acts as the Hamiltonian for the system, determining the energy levels of the string states. The commutation relations among these operators encode important physical information, such as the constraints on physical states (the Virasoro constraints) that must be satisfied for a consistent theory.

Physical States and Constraints

The Virasoro constraints arise from the requirement that physical states must be invariant under the action of the Virasoro operators. This leads to the notion of physical states in the theory, which are those that satisfy the condition $L_n | \text{physical state} \rangle =0$ for $n>0$ . This condition ensures that the states are not affected by the higher modes of the Virasoro algebra, which correspond to unphysical degrees of freedom.

Applications in CFT

In the context of conformal field theory, the Virasoro operators play a crucial role in the conformal bootstrap approach, where they help in constructing correlation functions and understanding the structure of the theory. The Virasoro conformal blocks, which are special functions derived from these operators, are essential for calculating observables in CFTs.

Summary

In summary, the Virasoro operators are central to the mathematical formulation of two-dimensional conformal field theories and string theory. They encapsulate the symmetries of these theories, dictate the physical state conditions, and facilitate the computation of physical observables. Their significance extends beyond mathematics into the core of theoretical physics, influencing our understanding of fundamental interactions and the structure of spacetime.

— AI

2024.09.08 Sunday ACHK

Benefits of Using Light-Cone Coordinates

Posted on August 23, 2024 by rpflee

Light-cone coordinates offer several advantages in the study of spacetime, particularly in the context of special and general relativity. Here are some key benefits:

1. Simplification of Calculations: Light-cone coordinates can simplify the mathematical expressions involved in calculations. For instance, the metric in light-cone coordinates often takes a simpler form, which can make solving equations, such as Einstein’s field equations, more straightforward.

2. Inclusion of Causal Structure: One of the significant benefits of light-cone coordinates is that they inherently incorporate the causal structure of spacetime. In this coordinate system, two of the coordinates are null vectors, which means they correspond to the paths that light can take. This feature helps in visualizing and understanding the relationships between different events in spacetime.

3. Separation of Physical and Gauge Degrees of Freedom: In gauge theories, light-cone coordinates can help separate physical degrees of freedom from gauge degrees of freedom more cleanly. This separation is particularly useful in string theory and quantum field theory, where it aids in quantization processes.

4. Facilitation of Lorentz Transformations: Light-cone coordinates allow for a more intuitive understanding of Lorentz transformations. The transformation properties become clearer, as the coordinates are designed to reflect the structure of spacetime in a way that aligns with the behavior of light.

5. Application in Cosmology: In cosmological contexts, light-cone coordinates can simplify the analysis of light propagation and the relationship between initial conditions and observable phenomena. This is particularly useful in perturbation theory, where light-cone coordinates can streamline calculations related to cosmological observations.

Overall, light-cone coordinates provide a powerful framework for analyzing problems in relativity and quantum field theory, making them a valuable tool for physicists.

— AI

2024.08.23 Friday ACHK

3.6 Analytic continuation for gamma function, 6

Posted on July 30, 2024 by rpflee

A First Course in String Theory

Residues

The behavior for non-positive $\displaystyle{z}$ is more intricate. Euler’s integral does not converge for $\displaystyle{\Re (z)\leq 0}$ , but the function it defines in the positive complex half-plane has a unique analytic continuation to the negative half-plane. One way to find that analytic continuation is to use Euler’s integral for positive arguments and extend the domain to negative numbers by repeated application of the recurrence formula,

$\displaystyle{\Gamma (z)={\frac {\Gamma (z+n+1)}{z(z+1)\cdots (z+n)}}}$ ,

choosing $\displaystyle{n}$ such that $\displaystyle{z+n}$ is positive. The product in the denominator is zero when $\displaystyle{z}$ equals any of the integers $\displaystyle{0,-1,-2,\ldots}$ . Thus, the gamma function must be undefined at those points to avoid division by zero; it is a meromorphic function with simple poles at the non-positive integers.

— Wikipedia on Gamma function

2024.07.30 Tuesday ACHK

3.6 Analytic continuation for gamma function, 5

Posted on February 21, 2024 by rpflee

A First Course in String Theory

…

Explain why the above right-hand side is well defined for $\displaystyle{ \Re (z) > - N - 1 }$ .

…

$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

Prove that $\displaystyle{\int_{1}^{\infty}t^{z-1}e^{-t}dt}$ converges.

…

Prove that $\displaystyle{\int_{0}^{1}t^{z-1}e^{-t}dt}$ converges.

…

Since both $\displaystyle{e^{-t}}$ and $\displaystyle{\sum_{n=0}^N \frac{(-t)^n}{n!}}$ are finite, $\displaystyle{ \begin{aligned} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \end{aligned} \\ }$ is also finite.

$\displaystyle{ \begin{aligned} &\int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \\ &= \sum_{n=N+1}^\infty \int_{0}^{1} t^{z-1} \frac{(-t)^n}{n!} dt \\ \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} &= \sum_{n=N+1}^\infty \frac{(-1)^n}{n!} \frac{1}{z+n} \left[ 1 - 0^{z+n} \right] \\ \end{aligned} \\ }$

This term converges as long as $\displaystyle{ \Re (z) > - N - 1 }$ .

So this condition is not only necessary but also sufficient for the convergence of the analytic continuation of Gamma function,

$\displaystyle{ \begin{aligned} \Gamma_c (z) &= \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} \\ &+ \int_{0}^{1} t^{z-1} \left( e^{-t} - \sum_{n=0}^N \frac{(-t)^n}{n!} dt \right) + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ \end{aligned} \\ }$

— Me@2023-09-07 07:38:17 PM

3.6 Analytic continuation for gamma function, 4

Posted on January 10, 2024 by rpflee

A First Course in String Theory

…

Explain why the above right-hand side is well defined for $\displaystyle{ \Re (z) > - N - 1 }$ .

…

$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

Prove that $\displaystyle{\int_{1}^{\infty}t^{z-1}e^{-t}dt}$ converges.

…

Prove that $\displaystyle{\int_{0}^{1}t^{z-1}e^{-t}dt}$ converges.

…

Under this condition,

$\displaystyle{ \begin{aligned} &\int_{0}^{1}t^{z-1}e^{-t}dt \end{aligned} }$
$\displaystyle{ = \int_{0}^{1} t^{z-1} \left( \sum_{n=0}^N \frac{(-t)^n}{n!} +\sum_{n=N+1}^\infty \frac{(-t)^n}{n!} \right) dt, }$
$\displaystyle{ = \int_{0}^{1} t^{z-1} \sum_{n=0}^N \frac{(-t)^n}{n!} dt + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt}$

$\displaystyle{ = \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt}$

So for $\displaystyle{\Re (z) > 0}$ ,

$\displaystyle{ \begin{aligned} &\Gamma (z) \\ &= \int_{0}^{\infty}t^{z-1}e^{-t}dt \\ &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ &= \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ \end{aligned} \\ }$

If we define this as the analytic continuation of Gamma function:

$\displaystyle{ \begin{aligned} \Gamma_c (z) \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} = \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\ \end{aligned} \\ }$ ,

its convergence depends on the convergence of the middle term:

$\displaystyle{ \begin{aligned} \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \\ \end{aligned} \\ }$

By the argument above, a necessary condition for this term to converge is

$\displaystyle{ \Re (z) + n_{\text{min}} > 0 }$

$\displaystyle{ \Re (z) + N + 1 > 0 }$

$\displaystyle{ \Re (z) > - N - 1 }$

However, besides the integral convergence, we have to consider also the summation convergence.

To check whether the order of integration and summation can be interchanged, we rewrite the term as

$\displaystyle{ \begin{aligned} &\int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt \\ &= \int_{0}^{1} t^{z-1} \left( e^{-t} - \sum_{n=0}^N \frac{(-t)^n}{n!} \right) dt \\ \end{aligned} \\ }$

…

— Me@2023-09-07 07:38:17 PM

3.6 Analytic continuation for gamma function, 3

Posted on December 8, 2023 by rpflee

A First Course in String Theory

…

Explain why the above right-hand side is well defined for $\displaystyle{ \Re (z) > - N - 1 }$ .

…

$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

Prove that $\displaystyle{\int_{1}^{\infty}t^{z-1}e^{-t}dt}$ converges.

…

Prove that $\displaystyle{\int_{0}^{1}t^{z-1}e^{-t}dt}$ converges.

$\displaystyle{ \begin{aligned} \Re (z) &> 0 \\ \Gamma (z) &= \int_{0}^{\infty}t^{z-1}e^{-t}dt \\ &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} &\int_{0}^{1}t^{z-1}e^{-t}dt \\ &= \int_{0}^{1} t^{z-1} \sum_{n=0}^\infty \frac{(-t)^n}{n!} dt \\ &= \int_{0}^{1} t^{z-1} \left( \sum_{n=0}^N \frac{(-t)^n}{n!} +\sum_{n=N+1}^\infty \frac{(-t)^n}{n!} \right) dt \\ \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} &= \int_{0}^{1} t^{z-1} \left( \sum_{n=0}^N \frac{(-t)^n}{n!} + e^{-t} - \sum_{n=0}^N \frac{(-t)^n}{n!} \right) dt \\ \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} &\int_{0}^{1} t^{z-1} \frac{(-t)^n}{n!} dt \\ &= \frac{(-1)^n}{n!} \int_{0}^{1} t^{z+n-1} dt \\ &= \frac{(-1)^n}{n!} \frac{1}{z+n} \left[ 1 - 0^{z+n} \right] \\ \end{aligned} \\ }$

Therefore, for the integral to converge, $\displaystyle{ \Re (z) + n > 0 }$ is required. So for the overall integral $\displaystyle{ \int_{0}^{1} t^{z-1} \sum_{n=0}^\infty \frac{(-t)^n}{n!} dt}$ to converge, a necessary condition is

$\displaystyle{ \Re (z) + n_{\text{min}} > 0 }$

Therefore,

$\displaystyle{ \Re (z) > 0 }$

Under this condition,

$\displaystyle{ = \sum_{n=0}^N \frac{(-1)^n}{n!} \frac{1}{z+n} + \int_{0}^{1} t^{z-1} \sum_{n=N+1}^\infty \frac{(-t)^n}{n!} dt}$

So for $\displaystyle{\Re (z) > 0}$ ,

…

— Me@2023-09-07 07:38:17 PM

3.6 Analytic continuation for gamma function, 2

Posted on October 8, 2023 by rpflee

A First Course in String Theory

…

$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

Prove that $\displaystyle{\int_{1}^{\infty}t^{z-1}e^{-t}dt}$ converges.

Since $\displaystyle{ \begin{aligned} e^t &= \sum_{s=0}^\infty \frac{t^s}{s!} \\ \end{aligned} \\ }$ ,

for any integer $r \ge 0$ and any real $t > 0$ ,

$\displaystyle{ \begin{aligned} e^t &\ge \frac{t^r}{r!} \\ \end{aligned}~~~}$

Then

$\displaystyle{ \begin{aligned} e^{-t} &\le \frac{1}{\frac{t^r}{r!}} \\ \end{aligned} \\ }$

Let $\displaystyle{ \begin{aligned} x &= \Re(z) \\ \end{aligned} \\ }$ :

$\displaystyle{ \begin{aligned} t^{x-1}e^{-t} &\le \frac{t^{x-1}}{\frac{t^r}{r!}} \\ t^{x-1}e^{-t} &\le r! t^{x - r - 1} \\ \end{aligned} \\ }$

By choosing an $r$ such that $r \ge 1$ and $r \ge \Re(z) + 1$ ,

for any $t \ge 1$ ,

$\displaystyle{ \begin{aligned} t^{x-1}e^{-t} &\le r! t^{x - (x+1) - 1} \\ t^{x-1}e^{-t} &\le r! t^{-2} \\ \end{aligned} \\ }$

Therefore,

$\displaystyle{ \begin{aligned} \int_1^\infty t^{x-1}e^{-t} dt &\le r! \int_1^\infty t^{-2} dt = - r! [ t^{-1} ]_1^\infty = r! \\ \end{aligned} \\ }$

In other words, for any complex $z$ ,

$\displaystyle{ \begin{aligned} \left| \int_1^\infty t^{z-1} e^{-t} dt \right| &\le r! \\ \end{aligned} \\ }$

for some $r$ such that $r \ge 1$ and $r \ge \Re(z) + 1$ .

$\displaystyle{ \begin{aligned} \left| \int_1^\infty t^{z-1} e^{-t} dt \right| &\le r! \\ \end{aligned} \\ }$

converges for any $z$ .

Prove that $\displaystyle{\int_{0}^{1}t^{z-1}e^{-t}dt}$ converges.

…

— Me@2023-09-07 07:38:17 PM

3.6 Analytic continuation for gamma function, 1

Posted on September 8, 2023 by rpflee

A First Course in String Theory

…

$\displaystyle{ \begin{aligned} \Gamma (z) = &\int_{0}^{1}dt~t^{z-1} \left( e^{-t} - \sum_{n=0}^N \frac{(-t)^n}{n!} \right) \\ &+ \sum_{n=0}^N \frac{(-t)^n}{n!} \frac{1}{z+n} + \int_1^\infty dt~e^{-t}t^{z-1} \\ \end{aligned} \\ }$

Explain why the above right-hand side is well defined for $\displaystyle{ \Re (z) > - N - 1 }$ .

…

$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{\infty}t^{z-1}e^{-t}dt \\ &= - \int_{0}^{\infty}t^{z-1}de^{-t} \\ &= - \left[ t^{z-1} e^{-t} \right]_{0}^{\infty} + (z-1) \int_{0}^{\infty} t^{z-2}e^{-t} dt \\ &= - \left[ \lim_{t \to \infty} t^{z-1} e^{-t} - 0^{z-1} e^{-0} \right] + (z-1) \int_{0}^{\infty} t^{z-2}e^{-t} dt \\ \end{aligned} \\ }$

Note that this can only prove that for convergence, it may be necessary to have $\displaystyle{ \Re (z) > 1 }$ , but not $\displaystyle{ \Re (z) > 0 }$ .

Also, the condition $\displaystyle{ \Re (z) > 1 }$ may be not necessary at all, because an infinity in one term may be cancelled out by another infinity in a latter term.

$\displaystyle{ \begin{aligned} \Gamma (z) &= \int_{0}^{1}t^{z-1}e^{-t}dt + \int_{1}^{\infty}t^{z-1}e^{-t}dt \\\\ \end{aligned} \\ }$

Prove that $\displaystyle{ \int_{1}^{\infty}t^{z-1}e^{-t}dt }$ converges.

Since $\displaystyle{ \begin{aligned} e^t &= \sum_{s=0}^\infty \frac{t^s}{s!} \\ \end{aligned} \\ }$ , for any integer $r \ge 0$ and any real $t > 0$ :

$\displaystyle{ \begin{aligned} e^t &\ge \frac{t^r}{r!} \\ \end{aligned}~~~}$

…

— Me@2023-09-07 07:38:17 PM

3.5 Calculating the divergence in higher dimension

Posted on August 2, 2023 by rpflee

A First Course in String Theory

Let $\displaystyle{\vec f = f(r) \hat{\mathbf{r}}}$ be a vector function in $\displaystyle{\mathbb{R}^d}$ .

…

Derive a formula for $\displaystyle{\nabla \cdot \vec f}$ by applying the divergence theorem to a spherical shell of a radius $\displaystyle{r}$ and width $\displaystyle{dr}$ .

…

~~~

Volume and sphere area of an $\displaystyle{n}$ -ball:

$\displaystyle{ \begin{aligned} V_{n}(R)&={\frac {\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}+1\right)}}R^{n} \\ S_{n-1}(R)&={\frac {2\,\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}\right)}}R^{n-1} \end{aligned} \\ }$

Divergence theorem:

$\displaystyle{ \int_V \nabla \cdot \vec f dV = \int_S \vec f \cdot \hat n dS }$

Let $\displaystyle{V}$ be a spherical shell of radius $\displaystyle{r}$ and width $\displaystyle{dr}$ . Then

$\displaystyle{ \begin{aligned} \int_V \nabla \cdot \vec f dV &= \left. \nabla \cdot \vec f \right|_r \int_V dV \\ &= \left. \nabla \cdot \vec f \right|_r \left( S_{n-1}(r) dr \right) \\ \end{aligned} \\ }$

and

$\displaystyle{ \begin{aligned} \int_S \vec f \cdot \hat n dS &= f(r+dr) S_{n-1}(r+dr) - f(r) S_{n-1}(r) \\ \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} \left. \nabla \cdot \vec f \right|_r S_{n-1}(r) dr &= f(r+dr) S_{n-1}(r+dr) - f(r) S_{n-1}(r) \\ \end{aligned} \\ }$

$\displaystyle{ \begin{aligned} \left. \nabla \cdot \vec f \right|_r &= \frac{1}{S_{n-1}(r)} \frac{f(r+dr) S_{n-1}(r+dr) - f(r) S_{n-1}(r)}{dr} \\ &= \frac{1}{S_{n-1}(r)} \frac{d}{dr} \bigg( f(r) S_{n-1}(r) \bigg) \\ &= \frac{1}{r^{n-1}} \frac{d}{dr} \bigg( r^{n-1} f(r) \bigg) \\ \end{aligned} \\ }$

— Me@2023-08-02 09:28:32 AM

3.4 Electric fields and potentials of point charges, 2

Posted on May 17, 2023 by rpflee

A First Course in String Theory

(b) Show that with $d$ spatial dimensions, the potential $\Phi$ due to a point charge $q$ is given by

$\displaystyle{\Phi(r) = \frac{\Gamma \left( \frac{d}{2} - 1 \right)}{4 \pi^{d/2}} \frac{q}{r^{d-2}}}$

~~~

Eq. (3.74):

$\displaystyle{E(r) = \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{q}{r^{d-1}}}$

$\displaystyle{\begin{aligned} E(r) &= -\frac{d\Phi(r)}{dr} \\ \\ \int_r^{\infty} d\Phi &= - \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} q \int_r^{\infty} r^{-d+1} dr \\ \\ \Phi(\infty) - \Phi(r) &= \frac{\Gamma\left( \frac{d}{2} \right)}{4 \pi^{\frac{d}{2}}} \frac{q}{\frac{d}{2}-1} \left[\frac{1}{r^{d-2}} \right]_r^{\infty} \\ \\ \Gamma (z) &= \frac{\Gamma (z+1)}{z} \\ \\ \Phi(r) &= \frac{\Gamma\left( \frac{d}{2} - 1\right)}{4 \pi^{\frac{d}{2}}} \frac{q}{r^{d-2}} \\ \\ \end{aligned} }$

— Me@2023-05-17 09:11:07 AM

3.4 Electric fields and potentials of point charges

Posted on March 18, 2023 by rpflee

A First Course in String Theory

(a) Show that for time-independent fields, the Maxwell equation $\displaystyle{T_{0ij}=0}$ implies that $\displaystyle{\partial_i E_j - \partial_j E_i = 0}$ . Explain why this condition is satisfied by the ansatz $\displaystyle{\vec E = - \nabla \Phi}$ .

~~~

Eq. (3.23):

$\displaystyle{ \begin{aligned} T_{\mu \lambda \nu} &= \partial_\lambda F_{\mu \nu} + \partial_\mu F_{\nu \lambda} + \partial_\nu F_{\lambda \mu} \\ \end{aligned}}$

…

$\displaystyle{ \begin{aligned} &\vec E \\ &= - \nabla \Phi \\ &= - \left( \partial_x, \partial_y, \partial_z \right) \Phi \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &\partial_i E_j - \partial_j E_i \\ &= \partial_i \partial_j \Phi - \partial_j \partial_i \Phi \\ \end{aligned}}$

— Me@2023-03-18 11:08:24 AM

Problem 14.5d1.2.2

Posted on November 23, 2022 by rpflee

A First Course in String Theory

The generating function is an infinite product:

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

To evaluate the infinite product, you can use Mathematica (or its official free version Wolfram Engine) with the following commands:

TeXForm[
    HoldForm[
        (1/x)*Product[
                 (1+x^(r-1/2))^32/(1-x^r)^8,
                 {r, 1, Infinity}]]]

f[x_] := (1/x)*Product[
                 (1+x^(r-1/2))^32/(1-x^r)^8,
                 {r, 1, Infinity}]

Print[f[x]]

TeXForm[f[x]]

TeXForm[Series[f[x], {x,0,3}]]




                     1        32
    QPochhammer[-(-------), x]
                  Sqrt[x]
------------------------------------
        1    32                    8
(1 + -------)   x QPochhammer[x, x]
     Sqrt[x]

$\displaystyle{ \begin{aligned} &f_{L, NS+}(x) \\ \end{aligned}}$

$\displaystyle{ \approx \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}$

— Me@2022-11-23 04:40:28 PM

3.3 Electromagnetism in three dimensions, 2

Posted on November 10, 2022 by rpflee

A First Course in String Theory

(b) Repeat the analysis of three-dimensional electromagnetism starting with the Lorentz covariant formulation. Take $A^\mu = (\Phi, A^1, A^2)$ , examine $F_{\mu \nu}$ , the Maxwell equations (3.34), and the relativistic form of the force law derived in Problem 3.1.

~~~

$A^\mu = (\Phi, A^1, A^2)$

Eq. (3.20):

$F_{\mu \nu} = \begin{bmatrix} 0 & -E_x & -E_y & -E_z=0 \\ E_x & 0 & B_z & -B_y =0\\ E_y & -B_z & 0 & B_x = 0\\ E_z=0 & B_y=0 & -B_x=0 & 0\\ \end{bmatrix}$

Eq. (3.33):

$F^{\mu \nu} = \begin{bmatrix} 0 & E_x & E_y & E_z=0 \\ -E_x & 0 & B_z & -B_y =0\\ -E_y & -B_z & 0 & B_x = 0\\ -E_z=0 & B_y=0 & -B_x=0 & 0\\ \end{bmatrix}$

Eq. (3.34):

$\begin{aligned} \frac{\partial F^{\mu \nu}}{\partial x^\nu} &= \frac{1}{c} j^\mu \\ \end{aligned}$

$\begin{aligned} \frac{\partial F^{0 0}}{\partial x^0} + \frac{\partial F^{0 1}}{\partial x^1} + \frac{\partial F^{0 2}}{\partial x^2} + \frac{\partial F^{0 3}}{\partial x^3} &= \frac{1}{c} j^0 \\ \frac{\partial F^{1 0}}{\partial x^0} + \frac{\partial F^{1 1}}{\partial x^1} + \frac{\partial F^{1 2}}{\partial x^2} + \frac{\partial F^{1 3}}{\partial x^3} &= \frac{1}{c} j^1 \\ \frac{\partial F^{2 0}}{\partial x^0} + \frac{\partial F^{2 1}}{\partial x^1} + \frac{\partial F^{2 2}}{\partial x^2} + \frac{\partial F^{2 3}}{\partial x^3} &= \frac{1}{c} j^2 \\ \frac{\partial F^{3 0}}{\partial x^0} + \frac{\partial F^{3 1}}{\partial x^1} + \frac{\partial F^{3 2}}{\partial x^2} + \frac{\partial F^{3 3}}{\partial x^3} &= \frac{1}{c} j^3 \\ \end{aligned}$

$\begin{aligned} \frac{\partial F^{0 0}}{\partial x^0} + \frac{\partial F^{0 1}}{\partial x^1} + \frac{\partial F^{0 2}}{\partial x^2} &= \frac{1}{c} j^0 \\ \frac{\partial F^{1 0}}{\partial x^0} + \frac{\partial F^{1 1}}{\partial x^1} + \frac{\partial F^{1 2}}{\partial x^2} &= \frac{1}{c} j^1 \\ \frac{\partial F^{2 0}}{\partial x^0} + \frac{\partial F^{2 1}}{\partial x^1} + \frac{\partial F^{2 2}}{\partial x^2} &= \frac{1}{c} j^2 \\ \end{aligned}$

$\begin{aligned} \frac{\partial 0}{\partial x^0} + \frac{\partial E_x}{\partial x^1} + \frac{\partial E_y}{\partial x^2} &= \frac{1}{c} j^0 \\ \frac{\partial (-E_x)}{\partial x^0} + \frac{\partial 0}{\partial x^1} + \frac{\partial B_z}{\partial x^2} &= \frac{1}{c} j^1 \\ \frac{\partial (-E_y)}{\partial x^0} + \frac{\partial (-B_z)}{\partial x^1} + \frac{\partial 0}{\partial x^2} &= \frac{1}{c} j^2 \\ \end{aligned}$

$\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ - \frac{1}{c} \frac{\partial E_x}{\partial t} + \frac{\partial B_z}{\partial y} &= \frac{1}{c} j_x \\ - \frac{1}{c} \frac{\partial E_y}{\partial t} - \frac{\partial B_z}{\partial x} &= \frac{1}{c} j_y \\ \end{aligned}$

$\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ \frac{\partial B_z}{\partial y} &= \frac{1}{c} j_x + \frac{1}{c} \frac{\partial E_x}{\partial t} \\ - \frac{\partial B_z}{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} \frac{\partial E_y}{\partial t} \\ \end{aligned}$

P. (3.1):

$\displaystyle{ \begin{aligned} \frac{d p_\mu}{ds} &= \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds} \\ \frac{d p_\mu}{ds} \left( \frac{ds}{dt} \right) &= \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds} \left( \frac{ds}{dt} \right) \\ \frac{d p_\mu}{dt} &= \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{dt} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d p_0}{dt} &= \frac{q}{c} F_{0 0} \frac{d x^0}{dt} + \frac{q}{c} F_{0 1} \frac{d x^1}{dt} + \frac{q}{c} F_{0 2} \frac{d x^2}{dt} + \frac{q}{c} F_{0 3} \frac{d x^3}{dt} \\ \frac{d p_1}{dt} &= \frac{q}{c} F_{1 0} \frac{d x^0}{dt} + \frac{q}{c} F_{1 1} \frac{d x^1}{dt} + \frac{q}{c} F_{1 2} \frac{d x^2}{dt} + \frac{q}{c} F_{1 3} \frac{d x^3}{dt} \\ \frac{d p_2}{dt} &= \frac{q}{c} F_{2 0} \frac{d x^0}{dt} + \frac{q}{c} F_{2 1} \frac{d x^1}{dt} + \frac{q}{c} F_{2 2} \frac{d x^2}{dt} + \frac{q}{c} F_{2 3} \frac{d x^3}{dt} \\ \frac{d p_3}{dt} &= \frac{q}{c} F_{3 0} \frac{d x^0}{dt} + \frac{q}{c} F_{3 1} \frac{d x^1}{dt} + \frac{q}{c} F_{3 2} \frac{d x^2}{dt} + \frac{q}{c} F_{3 3} \frac{d x^3}{dt} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d p_0}{dt} &= \frac{q}{c} F_{0 0} \frac{d x^0}{dt} + \frac{q}{c} F_{0 1} \frac{d x^1}{dt} + \frac{q}{c} F_{0 2} \frac{d x^2}{dt} \\ \frac{d p_1}{dt} &= \frac{q}{c} F_{1 0} \frac{d x^0}{dt} + \frac{q}{c} F_{1 1} \frac{d x^1}{dt} + \frac{q}{c} F_{1 2} \frac{d x^2}{dt} \\ \frac{d p_2}{dt} &= \frac{q}{c} F_{2 0} \frac{d x^0}{dt} + \frac{q}{c} F_{2 1} \frac{d x^1}{dt} + \frac{q}{c} F_{2 2} \frac{d x^2}{dt} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d p_0}{dt} &= q (0) + \frac{q}{c} \left( - E_x \frac{d x}{dt} - E_y \frac{d y}{dt} \right) \\ \frac{d p_1}{dt} &= q E_x + \frac{q}{c} \left( (0) \frac{d x}{dt} + B_z \frac{d y}{dt} \right) \\ \frac{d p_2}{dt} &= q E_y + \frac{q}{c} \left( - B_z \frac{d x}{dt} + (0) \frac{d y}{dt} \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d E}{dt} &= \vec v \cdot \vec F_E \\ \frac{d p_x}{dt} &= q \left( E_x + \frac{v_y}{c} B_z \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{v_x}{c} B_z \right) \\ \end{aligned}}$

— Me@2022-11-08 03:46:01 PM

3.3 Electromagnetism in three dimensions

Posted on October 22, 2022 by rpflee

A First Course in String Theory

(a) Find the reduced Maxwell equations in three dimensions by starting with Maxwell’s equations and the force law in four dimensions, using the ansatz (3.11), and assuming that no field can depend on the $z$ direction.

~~~

$\displaystyle{\begin{aligned} \nabla \cdot \mathbf {E} &= \rho \\ \nabla \cdot \mathbf {B} &= 0 \\ \nabla \times \mathbf {E} &= - \frac{1}{c} {\frac {\partial \mathbf {B} }{\partial t}} \\ \nabla \times \mathbf {B} &= \frac{1}{c} \mathbf {J} + \frac{1}{c} {\frac {\partial \mathbf {E} }{\partial t}} \\ \end{aligned}}$

Eq. (3.11):

$\displaystyle{\begin{aligned} E_z &= 0 \\ B_x &= 0 \\ B_y &= 0 \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ \frac{\partial B_z}{\partial z} &= 0 \\ \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= - \frac{1}{c} {\frac {\partial B_z }{\partial t}} \\ \nabla \times \mathbf {B} &= \frac{1}{c} \mathbf {J} + \frac{1}{c} {\frac {\partial \mathbf {E} }{\partial t}} \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{\partial B_z }{\partial y} - \frac{\partial B_y }{\partial z} &= \frac{1}{c} j_x + \frac{1}{c} {\frac {\partial E_x }{\partial t}} \\ \frac{\partial B_x }{\partial z} - \frac{\partial B_z }{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} {\frac {\partial E_y }{\partial t}} \\ \frac{\partial B_y }{\partial x} - \frac{\partial B_x }{\partial y} &= \frac{1}{c} j_z + \frac{1}{c} {\frac {\partial E_z }{\partial t}} \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{\partial B_z }{\partial y} &= \frac{1}{c} j_x + \frac{1}{c} {\frac {\partial E_x }{\partial t}} \\ - \frac{\partial B_z }{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} {\frac {\partial E_y }{\partial t}} \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{d \vec p}{dt} &= q \left( \vec E + \frac{\vec v}{c} \times \vec B \right) \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{d p_x}{dt} &= q \left( E_x + \frac{1}{c} (v_y B_z - v_z B_y) \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{1}{c} (v_x B_z - v_z B_x) \right) \\ \frac{d p_z}{dt} &= q \left( E_z + \frac{1}{c} (v_x B_y - v_y B_x) \right) \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{d p_x}{dt} &= q \left( E_x + \frac{1}{c} v_y B_z \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{1}{c} v_x B_z \right) \\ \frac{d p_z}{dt} &= 0 \\ \end{aligned}}$

— Me@2022-10-22 04:17:10 PM

Physics Town

continues

Category Archives: Barton Zwiebach

13.1 Commutation relations for oscillators

Quick Calculation 13.5

Problem 14.5d1.2.3

SageMath for Ubuntu 24.04

Quick Calculation 13.4

Quick Calculation 13.3

Quick Calculation 13.3.0

Benefits of Using Light-Cone Coordinates

3.6 Analytic continuation for gamma function, 6

3.6 Analytic continuation for gamma function, 5

3.6 Analytic continuation for gamma function, 4

3.6 Analytic continuation for gamma function, 3

3.6 Analytic continuation for gamma function, 2

3.6 Analytic continuation for gamma function, 1

3.5 Calculating the divergence in higher dimension

3.4 Electric fields and potentials of point charges, 2

3.4 Electric fields and potentials of point charges

Problem 14.5d1.2.2

3.3 Electromagnetism in three dimensions, 2

3.3 Electromagnetism in three dimensions