Quick Calculation 3.5

A First Course in String Theory

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Show that

\displaystyle{     \begin{aligned}           \text{vol}(B^d)   &= \frac{\pi^{\frac{d}{2}}}{\Gamma \left( 1 + \frac{d}{2} \right)}   \\     \end{aligned} }

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Geometric proof

The relations \displaystyle{V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)} and \displaystyle{A_{n+1}(R)=(2\pi R)V_{n}(R)} and thus the volumes of n-balls and areas of n-spheres can also be derived geometrically. As noted above, because a ball of radius \displaystyle{R} is obtained from a unit ball \displaystyle{B_{n}} by rescaling all directions in \displaystyle{R} times, \displaystyle V_{n}(R) is proportional to \displaystyle{R^{n}}, which implies \displaystyle{{\frac {dV_{n}(R)}{dR}}={\frac {n}{R}}V_{n}(R)}.

Also, \displaystyle{A_{n-1}(R)={\frac {dV_{n}(R)}{dR}}} because a ball is a union of concentric spheres and increasing radius by \displaystyle{\epsilon} corresponds to a shell of thickness \displaystyle{\epsilon}. Thus, \displaystyle{V_{n}(R)={\frac {R}{n}}A_{n-1}(R)}; equivalently, \displaystyle{V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)}.

— Wikipedia on Volume of an n-ball

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\displaystyle{  \begin{aligned}    V &= \int_0^R S dr \\    V(B^d) &= \int_0^R V(S^{d-1}) dR \\     &= \frac{2 \pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})} \int_0^R r^{d-1} dr \\     &= \frac{2 \pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})} \frac{1}{d} R^{d} \\     \end{aligned}}

— Me@2022-05-18 09:08:11 AM

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2022.05.19 Thursday (c) All rights reserved by ACHK

Quick Calculation 3.2

A First Course in String Theory

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Verify that the gauge transformation (3.10) are correctly summarized by (3.21).

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Eq. (3.21):

\displaystyle{ \begin{aligned}   A_\nu' &= A_\nu + \partial_\nu \epsilon \\   \end{aligned} }

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\displaystyle{ \begin{aligned}   \left( A_0', A_1', ... \right) &= \left( - \Phi + \frac{\partial \epsilon}{\partial x^0}, A^1 + \frac{\partial \epsilon}{\partial x^1}, ... \right)  \\   \left( -\Phi', {A^1}', ... \right) &= \left( - \Phi + \frac{1}{c} \frac{\partial \epsilon}{\partial t}, A^1 + \frac{\partial \epsilon}{\partial x^1}, ... \right)  \\   \end{aligned} }

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\displaystyle{ \begin{aligned}   \Phi' &= \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t}  \\     \left( {A^1}', {A^2}', {A^3}' \right) &= \left( {A^1}, {A^2}, {A^3} \right) + \left( \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial x^3} \right) \epsilon \\     \end{aligned} }

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Eq. (3.10):

\displaystyle{ \begin{aligned} \Phi' &= \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} \\ \vec A' &= \vec A + \nabla \epsilon \\ \end{aligned} }

— Me@2022-04-07 07:05:29 PM

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2022.04.07 Thursday (c) All rights reserved by ACHK

Quick Calculation 3.1

A First Course in String Theory

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Verify that \displaystyle{\vec E}, as given in (3.8), is invariant under the gauge transformation (3.10).

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Eq. (3.8):

\displaystyle{\vec E = - \frac{1}{c} \frac{\partial \vec A}{\partial t} - \nabla \Phi}

Eq. (3.10):

\displaystyle{  \begin{aligned}  \Phi' &= \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} \\  \vec A' &= \vec A + \nabla \epsilon \\  \end{aligned}  }

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\displaystyle{  \begin{aligned}  \vec E'     &= - \frac{1}{c} \frac{\partial \vec A'}{\partial t} - \nabla \Phi' \\      &= - \frac{1}{c} \frac{\partial \vec A}{\partial t}      - \frac{1}{c} \frac{\partial}{\partial t} \left( \nabla \epsilon \right)      - \nabla \Phi     + \frac{1}{c} \nabla \frac{\partial \epsilon}{\partial t}  \\            &= \vec E    \\    \end{aligned}  }

— Me@2022-04-01 03:34:28 PM

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2022.04.01 Friday (c) All rights reserved by ACHK

2.10 Extra dimension and statistical mechanics

A First Course in String Theory

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Write a double sum that represents the statistical mechanics partition function \displaystyle{Z(a, R)} for the quantum mechanical system considered in Section 2.10. Note that \displaystyle{Z(a, R)} factors as \displaystyle{Z(a, R) = Z(a) \tilde{Z}(R)}.

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Eq. (2.118):

\displaystyle{  \begin{aligned}  - \frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{d^2 \psi(x)}{dx^2}   - \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{d^2 \phi(x)}{dx^2}   &= E \\   \end{aligned}}

\displaystyle{  \begin{aligned}  \psi_{k, l} (x,y) &= \psi_k (x) \phi_l (y) \\   \end{aligned}}

Eq. (2.119):

\displaystyle{  \begin{aligned}  \psi_k (x) &= c_k \sin \left( \frac{k \pi x}{a} \right) \\   \end{aligned}}

\displaystyle{  \begin{aligned}  \frac{d^2 \psi_k (x)}{dx^2} &= - \left( \frac{k \pi}{a} \right)^2 \psi_k (x) \\   \end{aligned}}

\displaystyle{  \begin{aligned}  - \frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{d^2 \psi(x)}{dx^2}   - \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{d^2 \phi(x)}{dx^2}   &= E \\     \frac{\hbar^2}{2m} \left( \frac{k \pi}{a} \right)^2   + \frac{\hbar^2}{2m} \left( \frac{l}{R} \right)^2   &= E \\     \end{aligned}}

\displaystyle{    \begin{aligned}    E   &=     \frac{\hbar^2}{2m} \left[ \left( \frac{k \pi}{a} \right)^2 + \left( \frac{l}{R} \right)^2 \right] \\    \end{aligned}}

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[guess]

Index \displaystyle{k = 1, 2, \dotsb} but not negative integers because, for example, k = 1 and k=-1 give physically identical states

\displaystyle{  \begin{aligned}  \psi_{-1} (x) &= c_{-1} \sin \left( \frac{- \pi x}{a} \right) \\   \end{aligned}} and \displaystyle{  \begin{aligned}  \psi_{1} (x) &= c_{1} \sin \left( \frac{\pi x}{a} \right) \\   \end{aligned}}.

The two wave functions give the same probability density distribution, if c_{-1} = c_{1}.

However, that is not the case for

\displaystyle{  \begin{aligned}  \phi_l(y) &= a_l \sin \left(\frac{ly}{R}\right) + b_l \cos \left(\frac{ly}{R} \right) \\   \end{aligned}}.

So l should have also negative integers as possible values: l = \dotsb, -2, -1, 0, 1, 2, \dotsb.

[guess]

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Eq. (2.120):

\displaystyle{  \begin{aligned}  \phi_l(y) &= a_l \sin \left(\frac{ly}{R}\right) + b_l \cos \left(\frac{ly}{R} \right) \\   \end{aligned}}

Eq. (2.121):

\displaystyle{  \begin{aligned}  \phi_l(y) &= \phi_l(y+2\pi R) \\   \end{aligned}}

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\displaystyle{  \begin{aligned}  Z &= \sum_{k} \sum_{l} e^{- \beta E_{k,l}} \\    &= \sum_{k} \sum_{l} \exp\left\{- \beta \left( \frac{\hbar^2}{2m} \right) \left[ \left(\frac{k \pi}{a} \right)^2 + \left(\frac{l}{R}\right)^2 \right]\right\} \\  &= Z(a) \tilde{Z}(R) \\     \end{aligned}}

\displaystyle{  \begin{aligned}  Z(a) &= \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k \pi}{a} \right)^2 \right] \\  \tilde Z (R) &= \sum_{l=-\infty}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{l}{R}\right)^2 \right] \\  &= \sum_{l=-\infty}^{-1} \left( \dotsb \right) + \sum_{l=0} \left( \dotsb \right) + \sum_{l=1}^\infty \left( \dotsb \right) \\  &= 1 + 2 Z(R \pi) \\  \end{aligned}}

— Me@2022-01-19 08:45:05 PM

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2022.01.26 Wednesday (c) All rights reserved by ACHK

2.10 A spacetime orbifold in two dimensions, 6

A First Course in String Theory

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(d) Consider the two curves \displaystyle{x^+ x^- = a^2} for some fixed \displaystyle{a}. The identification (2) makes each of these curves into a circle. Find the invariant circumference of this circle by integrating the appropriate root of \displaystyle{ds^2} between two neighboring identified points. Give your answers in terms of \displaystyle{a} and \displaystyle{\lambda}. Answer: \displaystyle{\sqrt{2} a \lambda}.

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Identification (2):

\displaystyle{(x^+, x^-) \sim \left( e^{-\lambda} x^+, e^{\lambda} x^- \right)}, where \displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}.

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\displaystyle{  \begin{aligned}  ds &= \sqrt{- ds^2} \\  \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds     &= \int_{x^+}^{e^{-\lambda} x^+} \sqrt{2 \frac{a^2}{(x^+)^2} (dx^+)^2}  \\     &= \sqrt{2} a \int_{x^+}^{e^{-\lambda} x^+} \sqrt{\frac{1}{(x^+)^2}} dx^+  \\   \end{aligned}}

Choose a segment on which \displaystyle{x^+ > 0} .

\displaystyle{  \begin{aligned}  \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds     &= \sqrt{2} a \int_{x^+}^{e^{-\lambda} x^+} \frac{1}{x^+} dx^+  \\     &= \sqrt{2} a \ln \left|\frac{e^{-\lambda} x^+}{x^+} \right| \\     &= - \sqrt{2} a \lambda \\   \end{aligned}}

The lower limit should be smaller than the upper limit.

\displaystyle{  \begin{aligned}  \left| \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds \right|    &= \sqrt{2} a \int_{e^{-\lambda} x^+}^{x^+} \frac{1}{x^+} dx^+  \\     &= \sqrt{2} a \lambda \\   \end{aligned}}

— Me@2022-01-05 02:40:45 PM

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2022.01.05 Wednesday (c) All rights reserved by ACHK

2.10 A spacetime orbifold in two dimensions, 2

A First Course in String Theory

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(a) Use the result of Problem 2.2, part (a), to recast (1) as

\displaystyle{(x^+, x^-) \sim \left( e^{-\lambda} x^+, e^{\lambda} x^- \right)}, where \displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}.

What is the range of \lambda? What is the orbifold fixed point? Assume now that \beta > 0, and thus \lambda > 0.

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Range of \displaystyle{\lambda}:

\displaystyle{ \begin{aligned}  0 &< \beta < \infty \\  1 &< \frac{1 + \beta}{1 - \beta} < \infty \\  0 &< \ln \frac{1 + \beta}{1 - \beta} < \infty \\  0 &< \frac{1}{2} \ln \frac{1 + \beta}{1 - \beta} < \infty \\  0 &< \lambda < \infty \\  \end{aligned}}

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Fixed points:

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix} &= \begin{bmatrix} e^{- \lambda} x^+ \\ e^{\lambda} x^- \\ \end{bmatrix} \\ \end{aligned}}

\displaystyle{ \begin{aligned}  (x^+, x^-) &= (0, 0) \\  \end{aligned}}

— Me@2021-05-16 06:31:12 PM

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2021.05.17 Monday (c) All rights reserved by ACHK

2.10 A spacetime orbifold in two dimensions

A First Course in String Theory

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Consider a two-dimensional world with coordinates \displaystyle{x^0} and \displaystyle{x^1}.

A boost with velocity parameter \displaystyle{\beta} along the \displaystyle{x^1} axis is described by the first two equations in (2.36). We want to understand the two-dimensional space that emerges if we identify

\displaystyle{(x^0, x^1) \sim ({x'}^0, {x'}^1)}.

We are identifying spacetime points whose coordinates are related by a boost!

(a) Use the result of Problem 2.2, part (a), to recast (1) as

\displaystyle{(x^+, x^-) \sim \left( e^{-\lambda} x^+, e^{\lambda} x^- \right)}, where \displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}.

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\displaystyle{  \begin{aligned}  (x')^0 &= \gamma (x^0 - \beta x^1) \\  (x')^1 &= \gamma (- \beta x^0 + x^1) \\  \end{aligned}}

\displaystyle{  \begin{aligned}  \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix}   &= \begin{bmatrix}         \gamma (1-\beta) & 0 \\         0 & \gamma (1+\beta) \\      \end{bmatrix}   \begin{bmatrix} x^+ \\ x^- \end{bmatrix} \\  &=     \begin{bmatrix}         \frac{1}{\sqrt{1 - \beta^2}} (1-\beta) x^+ \\         \frac{1}{\sqrt{1 - \beta^2}} (1+\beta) x^- \\      \end{bmatrix} \\  &=     \begin{bmatrix}         e^{- \lambda} x^+ \\         e^{\lambda} x^- \\      \end{bmatrix} \\  \end{aligned}}

— Me@2021-05-04 10:48:28 PM

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2021.05.05 Wednesday (c) All rights reserved by ACHK

2.9 Lightlike compactification, d, 2

A First Course in String Theory

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Represent your answer to part (c) in a spacetime diagram. Show two points related by the identification (2) and the space and time axes for the Lorentz frame S’ in which the compactification is standard.

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[guess]

In case you would like to have a fundamental domain:

[guess]

— Me@2021-04-21 10:35:13 PM

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2021.04.24 Saturday (c) All rights reserved by ACHK

2.9 Lightlike compactification, d

A First Course in String Theory

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Represent your answer to part (c) in a spacetime diagram. Show two points related by the identification (2) and the space and time axes for the Lorentz frame S' in which the compactification is standard.

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Note 1:

The identifications \displaystyle{ x \sim x + 0 } and \displaystyle{ x \sim x + \infty } have the same meaning, which is that x has no identification at all. In other words, there is NO non-zero real number r such that

\displaystyle{ x \sim x + r }

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Note 2:

\displaystyle{ \begin{aligned} \begin{bmatrix} ct \\ x \\ \end{bmatrix} &\sim \begin{bmatrix} ct - 2 \pi R \\ x + 2 \pi \sqrt{R^2 + R_S^2} \\ \end{bmatrix} \end{aligned} }

This identification must be done to both the space and time coordinates. In other words, it cannot be done to only one of ct and x.

— Me@2021-04-13 12:02:13 PM

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2021.04.14 Wednesday (c) All rights reserved by ACHK

2.9 Lightlike compactification, b

A First Course in String Theory

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Consider coordinates \displaystyle{(ct', x')} related to \displaystyle{(ct, x)} by a boost with velocity parameter \displaystyle{\beta}. Express the identifications in terms of the primed coordinates.

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\displaystyle{  \begin{aligned}  \begin{bmatrix} c t' \\ x' \end{bmatrix}   &= \begin{bmatrix}   \gamma & -\beta \gamma \\   -\beta \gamma & \gamma \\   \end{bmatrix}   \begin{bmatrix} c\,t \\ x  \end{bmatrix} \\    \end{aligned}}

\displaystyle{  \begin{aligned}  \begin{bmatrix}   ct \\  x \\   \end{bmatrix}   &\sim   \begin{bmatrix}   ct - 2 \pi R \\   x + 2 \pi R \\   \end{bmatrix}   \end{aligned}}

\displaystyle{  \begin{aligned}  \begin{bmatrix} c t' \\ x' \end{bmatrix}   &\sim   \begin{bmatrix} c t' \\ x' \end{bmatrix}   +  \begin{bmatrix}   - \gamma - \beta \gamma \\   \gamma + \beta \gamma \\   \end{bmatrix} 2 \pi R \\    \end{aligned}}

— Me@2021-03-30 08:38:16 PM

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2021.03.31 Wednesday (c) All rights reserved by ACHK

Lightlike compatification

A First Course in String Theory

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2.9 Lightlike compatification

(a) Rewrite this identification using light-cone coordinates.

\begin{aligned}  \begin{bmatrix}   x \\   ct   \end{bmatrix}   &\sim   \begin{bmatrix}   x \\   ct   \end{bmatrix}   +   2 \pi  \begin{bmatrix}   R \\   -R   \end{bmatrix}   \end{aligned}

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\begin{aligned}  \begin{bmatrix} x^+ \\ x^- \end{bmatrix}   &= \frac{1}{\sqrt{2}}      \begin{bmatrix}         1 & 1 \\         1 & -1 \\      \end{bmatrix}      \begin{bmatrix} x^0 \\ x^1 \end{bmatrix} \\    \end{aligned}

\begin{aligned}   \frac{1}{\sqrt{2}}      \begin{bmatrix}         1 & 1 \\         1 & -1 \\      \end{bmatrix}      \begin{bmatrix}   x^0 \\  x^1 \\   \end{bmatrix}   &\sim    \frac{1}{\sqrt{2}}      \begin{bmatrix}         1 & 1 \\         1 & -1 \\      \end{bmatrix}   \begin{bmatrix}   x^0 \\   x^1 \\   \end{bmatrix}   +    \frac{1}{\sqrt{2}}      \begin{bmatrix}         1 & 1 \\         1 & -1 \\      \end{bmatrix}   \begin{bmatrix}   - 2 \pi R \\     2 \pi R \\   \end{bmatrix}   \\  \end{aligned}

\begin{aligned}  \begin{bmatrix}   x^+ \\  x^- \\   \end{bmatrix}   &\sim   \begin{bmatrix}   x^+ \\   x^- \\   \end{bmatrix}   +    \frac{1}{\sqrt{2}}      \begin{bmatrix}         0 \\         - 4 \pi R \\      \end{bmatrix} \\  \end{aligned}

— Me@2021-03-22 06:06:10 PM

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2021.03.23 Tuesday (c) All rights reserved by ACHK

Problem 2.8

A First Course in String Theory

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2.8 Spacetime diagrams and Lorentz transformations

Show that the \displaystyle{{x'}^0} and \displaystyle{{x'}^1} axes … appear in the original spacetime diagram as oblique axes.

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The Lorentz transform:

\displaystyle{  \begin{aligned}  (x')^0 &= \gamma (x^0 - \beta x^1) \\  (x')^1 &= \gamma (- \beta x^0 + x^1) \\  \end{aligned}}

The inverse Lorentz transform:

\displaystyle{  \begin{aligned}  x^0 &= \gamma ((x')^0 + \beta (x')^1) \\  x^1 &= \gamma (\beta (x')^0 + (x')^1) \\  \end{aligned}}

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The \displaystyle{(x')^0}-direction unit vector is \displaystyle{((x')^0, (x')^1) = (1, 0)}.

When \displaystyle{((x')^0, (x')^1) = (1,0)},

\displaystyle{(x^0, x^1) = (\gamma, \gamma \beta)}

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When \displaystyle{\beta > 0},

\displaystyle{  \begin{aligned}  \tan \phi &= \frac{\gamma \beta}{\gamma} \\  &= \beta \\  \end{aligned}}

where \displaystyle{\phi} is the angle between \displaystyle{x^0}-axis and \displaystyle{(x')^0}-axis.

— Me@2021-03-17 03:42:23 PM

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2021.03.17 Wednesday (c) All rights reserved by ACHK