Problem 2.3b3

Now we lower the indices, by expressing the upper-index coordinates (contravariant components) by lower-index coordinates (covariant components), in order to find the Lorentz transformation for the covariant components:

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\   \eta^{\rho \mu} (x')_\rho &= L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \sum_\rho \eta^{\rho \mu} (x')_\rho &= \sum_\sigma \sum_\nu L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \end{aligned}}

After raising the indices, we lower the indices again:

\displaystyle{ \begin{aligned}   \eta_{\alpha \mu} \eta^{\rho \mu} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \eta_{\alpha \mu} \eta^{\mu \rho} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \end{aligned}}

\displaystyle{ \begin{aligned}   \delta_{\alpha \rho} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  (x')_\alpha &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \end{aligned}}

Prove that \displaystyle{\eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} = \left[L^{-1}\right]^\sigma_{~\alpha}}.

By index renaming, \displaystyle{ \begin{aligned}   (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} x_\nu \\  \end{aligned}}, the question becomes

Prove that \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} = \left[L^{-1}\right]^\nu_{~\mu}}.

Denote \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} as \displaystyle{L^{~\nu}_{\mu}}. Then the question is simplified to

Prove that \displaystyle{ L^{~\nu}_{\mu} = \left[L^{-1}\right]^\nu_{~\mu}}.

\displaystyle{ \begin{aligned}   (x')^\mu &= L^\mu_{~\nu} x^\nu \\  (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \\  \end{aligned}}

\displaystyle{ \begin{aligned}  (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\  (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\  \\  \end{aligned}}

\displaystyle{ \begin{aligned}  \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right)  \\  \end{aligned}}

\displaystyle{ \begin{aligned}  \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \left( \sum_{\nu = 0}^4 L^\mu_{~\nu} x^\nu \right) \left( \sum_{\beta = 0}^4 {L^{~\beta}_{\mu}} x_\beta \right)  \\  \end{aligned}}

\displaystyle{ \begin{aligned}  &(x')^0 (x')_0 + (x')^1 (x')_1 + (x')^2 (x')_2 + (x')^3 (x')_3 \\  &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right)   \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right)  \\  \end{aligned}}

The right hand side has 64 terms.

Since the spacetime interval is Lorentz-invariant, \displaystyle{ (x')^\mu (x')_\mu = x^\mu x_\mu }. So the left hand side can be replaced by \displaystyle{ x^\mu x_\mu }.

\displaystyle{ \begin{aligned}  &x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 \\  &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right)   \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right)  \\  \end{aligned}}

Note that the 4 terms on the left side also appear on the right hand side.

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\  (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\  (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\    x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\    \end{aligned}}

\displaystyle{ \begin{aligned}  x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\    \end{aligned}}

Since this equation is true for any coordinates, it is an identity. By comparing coefficients, we have:

1. For any terms with \displaystyle{\nu \ne \beta}, such as \displaystyle{\nu = 0} and \displaystyle{\beta=1},

\displaystyle{ \begin{aligned}  \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} x^0 x_1 &\equiv 0 \\  \left( \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} \right) x^0 x_1 &\equiv 0\\    \end{aligned}}

So

\displaystyle{ \begin{aligned}  \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} &= 0 \\    \end{aligned}}

2. For any terms with \displaystyle{\nu = \beta}.

\displaystyle{ \begin{aligned}  x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = \nu} L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\    x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \left( L^\mu_{~0} L^{~0}_{\mu} x^0 x_0   +  L^\mu_{~1} L^{~1}_{\mu} x^1 x_1   +  L^\mu_{~2} L^{~2}_{\mu} x^2 x_2   +  L^\mu_{~3} L^{~3}_{\mu} x^3 x_3  \right) \\    \end{aligned}}

So

\displaystyle{ \begin{aligned}  \sum_{\mu = 0}^4 L^\mu_{~0} L^{~0}_{\mu} &= 1 \\  \sum_{\mu = 0}^4 L^\mu_{~1} L^{~1}_{\mu} &= 1 \\  \sum_{\mu = 0}^4 L^\mu_{~2} L^{~2}_{\mu} &= 1 \\  \sum_{\mu = 0}^4 L^\mu_{~3} L^{~3}_{\mu} &= 1 \\  \end{aligned}}

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Denoting \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} as \displaystyle{L^{~\nu}_{\mu}} is misleading, because that presupposes that \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} is directly related to the matrix \displaystyle{L}.

To avoid this bug, instead, we denote \displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}} as \displaystyle{M ^\nu_{~\mu}}. So

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\  (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\  (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\    x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\    \end{aligned}}


\displaystyle{ \begin{aligned}  \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4  L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\    \nu = \mu:&~~~~~~\sum_{\mu = 0}^4  L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\    \end{aligned}}

— Me@2020-09-12 09:33:00 PM

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2020.09.13 Sunday (c) All rights reserved by ACHK

Problem 2.3b2

Prove that a metric tensor is symmetric.

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Assume \displaystyle{\eta_{\alpha\beta} \neq \eta_{\beta\alpha}}. Because it’s irrelevant what letter we use for our indices,

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}.

Then

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}

So only the symmetric part of \displaystyle{\eta_{\alpha\beta}} would survive the sum. As such we may as well take \displaystyle{\eta_{\alpha\beta}} to be symmetric in its definition.

— edited Jun 15 ’15 at 22:48

— rob

— answered Jun 15 ’15 at 17:52

— FenderLesPaul

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— Why is the metric tensor symmetric?

— Physics StackExchange

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1.

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}

means that

\displaystyle{\sum_{\alpha, \beta} \eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\sum_{\alpha, \beta}\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}

So in

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}},

we cannot cancel out \displaystyle{dx^{\alpha}dx^{\beta}} on both sides. In other words, we do NOT assume that \displaystyle{\eta_{\alpha\beta} = \eta_{\beta\alpha}} in the first place.

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2.

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}

means that

\displaystyle{\sum_{\alpha, \beta}\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}\sum_{\alpha, \beta}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} \sum_{\alpha, \beta}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}

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3. “… only the symmetric part of \displaystyle{\eta_{\alpha\beta}} would survive the sum” means that only the sum \displaystyle{\left(\eta_{\alpha\beta} + \eta_{\beta\alpha}\right)} is physically meaningful.

— Me@2020-08-14 03:34:05 PM

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2020.08.14 Friday (c) All rights reserved by ACHK

Problem 2.3b1

A First Course in String Theory

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2.3 Lorentz transformations, derivatives, and quantum operators.

(b) Show that the objects \displaystyle{\frac{\partial}{\partial x^\mu}} transform under Lorentz transformations in the same way as the \displaystyle{a_\mu} considered in (a) do. Thus, partial derivatives with respect to conventional upper-index coordinates \displaystyle{x^\mu} behave as a four-vector with lower indices – as reflected by writing it as \displaystyle{\partial_\mu}.

~~~

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \frac{\partial}{\partial (x')^\mu}  &= \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} \\  &= \frac{\partial x^0}{\partial (x')^\mu} \frac{\partial}{\partial x^0}  + \frac{\partial x^1}{\partial (x')^\mu} \frac{\partial}{\partial x^1}  + \frac{\partial x^2}{\partial (x')^\mu} \frac{\partial}{\partial x^2}  + \frac{\partial x^3}{\partial (x')^\mu} \frac{\partial}{\partial x^3}  \\  \end{aligned}}

The Lorentz transformation:

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\  \end{aligned}}

Lowering the indices to create covariant vectors:

\displaystyle{ \begin{aligned}  x_\mu &= \eta_{\mu \nu} x^\nu \\  \end{aligned}}

In matrix form, covariant vectors are represented by row vectors:

\displaystyle{ \begin{aligned}  \left[ x_\mu \right] &= \left( [\eta_{\mu \nu}] [x^\nu] \right)^T \\  \end{aligned}}

Change the subject:

\displaystyle{ \begin{aligned}  \left[ x_\mu \right]^T &= [\eta_{\mu \nu}] [x^\nu] \\  [\eta_{\mu \nu}] [x^\nu] &= \left[ x_\mu \right]^T  \\  [x^\nu] &= [\eta_{\mu \nu}]^{-1} \left[ x_\mu \right]^T  \\  \end{aligned}}

With \displaystyle{ \begin{aligned}  \eta^{\mu \nu} &\stackrel{\text{\tiny def}}{=} \left[ \eta_{\mu \nu} \right]^{-1}   \\  \end{aligned}}, we have:

\displaystyle{ \begin{aligned}  \left[ x^\nu \right] &= \left[ \eta^{\mu \nu} \right] \left[ x_\mu \right]^T  \\  \end{aligned}}

\displaystyle{ \begin{aligned}  x^\nu &= x_\mu \eta^{\mu \nu}  \\  \end{aligned}}

Now we lower the indices in order to find the Lorentz transformation for the covariant components:

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\  \eta^{\rho \mu} x_\rho &= L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  x_\rho &= \eta_{\rho \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \end{aligned}}

— Me@2020-07-21 10:46:32 AM

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2020.07.22 Wednesday (c) All rights reserved by ACHK

Problem 2.3a

A First Course in String Theory

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2.2 Lorentz transformations, derivatives, and quantum operators.

(a) Give the Lorentz transformations for the components \displaystyle{a_{\mu}} of a vector under a boost along the \displaystyle{x^1} axis.

~~~

\displaystyle{\begin{aligned} \begin{bmatrix} c t' \\ z' \\ x' \\ y' \end{bmatrix} &= \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c\,t \\ z \\ x \\ y \end{bmatrix} \\ \end{aligned}}

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ a_\mu &= a^\nu \eta_{\mu \nu} \\ (a')_\mu &= L_\mu^{~\nu} a_\nu \\ [(a')_\mu] &= [a_\nu] [L^\mu_{~\nu}]^{-1} \\ [L^\mu_{~\nu}]^{-1} &= \begin{bmatrix} \gamma & \beta \gamma & 0 & 0 \\ \beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \end{aligned}}

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Another method is to start with:

\displaystyle{ \begin{aligned} a_0 &= -a^0 \\ a_1 &= a^1 \\ a_2 &= a^2 \\ a_3 &= a^3 \\ \end{aligned}}

— Me@2020-07-05 05:40:44 PM

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2020.07.05 Sunday (c) All rights reserved by ACHK

Problem 2.2c

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates \displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )} and the associated light-cone coordinates \displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}. Write the following Lorentz transformations in terms of the light-cone coordinates.

(c) A boost with velocity parameter \displaystyle{\beta} in the \displaystyle{x^3} direction.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ z' \\ x' \\ y' \end{bmatrix} &= \begin{bmatrix}         \gamma & -\beta \gamma & 0 & 0 \\         -\beta \gamma & \gamma & 0 & 0 \\              0 & 0 & 1 & 0 \\              0 & 0 & 0 & 1 \\     \end{bmatrix} \begin{bmatrix} c\,t \\ z \\ x \\ y \end{bmatrix}  \\ \end{aligned} }

~~~

\displaystyle{ \begin{aligned}  \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix}  &= \begin{bmatrix}         \gamma & 0 & 0 &  -\beta \gamma \\         0 & 1 & 0 & 0 \\              0 & 0 & 1 & 0 \\          -\beta \gamma & 0 & 0 & \gamma \\     \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}     \\ \end{aligned} }

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

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

— Me@2020-04-19 11:52:09 PM

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2020.04.21 Tuesday (c) All rights reserved by ACHK

Problem 2.2b

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates \displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )} and the associated light-cone coordinates \displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}. Write the following Lorentz transformations in terms of the light-cone coordinates.

(b) A rotation with angle \displaystyle{\theta} in the \displaystyle{x^1, x^2} plane.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix}  &= \begin{bmatrix}       1 & 0 & 0 & 0 \\       0 & \cos \theta & -\sin \theta & 0 \\       0 & \sin \theta & \cos \theta & 0 \\       0 & 0 & 0 & 1 \\     \end{bmatrix}     \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\   \end{aligned} }

~~~

\displaystyle{ \begin{aligned}    \begin{bmatrix}        \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}} & 0 & 0 \\        \frac{1}{\sqrt{2}}  & -\frac{1}{\sqrt{2}} & 0 & 0 \\                         0  & 0 & 1 & 0 \\                         0  & 0 & 0 & 1 \\     \end{bmatrix}  \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix}  &=   \begin{bmatrix}       1 & 0 & 0 & 0 \\       0 & \cos \theta & -\sin \theta & 0 \\       0 & \sin \theta & \cos \theta & 0 \\       0 & 0 & 0 & 1 \\     \end{bmatrix}    \begin{bmatrix}        \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}} & 0 & 0 \\        \frac{1}{\sqrt{2}}  & -\frac{1}{\sqrt{2}} & 0 & 0 \\                         0  & 0 & 1 & 0 \\                         0  & 0 & 0 & 1 \\     \end{bmatrix}  \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\  \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix}  &= \frac{1}{2}  \begin{bmatrix} \cos\theta + 1 & 1 - \cos\theta & -\sqrt{2} \sin\theta & 0 \\  1 - \cos\theta & \cos\theta + 1 &  \sqrt{2} \sin\theta & 0 \\  \sqrt{2} \sin{\theta} & -\sqrt{2} \sin{\theta} & 2 \cos{\theta} & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix}  \end{aligned} }

— Me@2020-03-22 10:16:09 PM

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2020.03.23 Monday (c) All rights reserved by ACHK

Problem 2.2a

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates \displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )} and the associated light-cone coordinates \displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}. Write the following Lorentz transformations in terms of the light-cone coordinates.

(a) A boost with velocity parameter \displaystyle{\beta} in the \displaystyle{x^1} direction.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix}  &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\   \end{aligned}}

~~~

\displaystyle{ \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}}

The matrix is its own inverse.

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

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

Apply the result to the original transformation:

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^0)' \\ (x^1)' \\ (x^2)' \\ (x^3)' \end{bmatrix}  &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix} \\   \end{aligned}}

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

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

— Me@2020-02-27 07:14:19 PM

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2020.02.27 Thursday (c) All rights reserved by ACHK

Problem 2.1b

A First Course in String Theory

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2.1 Exercises with units

(b) Explain the meaning of the unit K (degree kelvin) used for measuring temperatures, and explain its relation to the basic length, mass, and time units.

~~~

{\displaystyle {\frac {1}{T}}=\left({\frac {\partial S}{\partial U}}\right)_{V,N}},

where \displaystyle{U} is the internal energy.
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The units of \displaystyle{k_B T} and \displaystyle{E} are the same.

\displaystyle{[k_B T] = [E]}

In other words, the Boltzmann constant \displaystyle{k_B} translates the temperature unit \displaystyle{K} to the language of energy unit \displaystyle{J}.

However, although the temperature unit \displaystyle{K} and the energy unit \displaystyle{J} have the relation

\displaystyle{k_B K = J},

just \displaystyle{k_B T} would not give the correct value of energy \displaystyle{E}, not to mention that we have not yet specified of which the energy \displaystyle{E} is.

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For an ideal gas,

\displaystyle{pV=Nk_B T}

and the average translational kinetic energy is

{\displaystyle {\frac {1}{2}}m{\overline {v^{2}}}={\frac {3}{2}}k_BT}

for 3 degrees of freedom. In 3D space, if there are only translational motions, there are only 3 degrees of freedom.

In other words, just the value of {k_B T} itself gives no physical meaning. Instead, {\tfrac{1}{2}k_B T} can be interpreted as the average translational kinetic energy of the particles in an one dimensional space. Equivalently, \displaystyle{\tfrac{3}{2}k_BT} gives that in our three dimensional space.

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Another main difference is that although energy is an extensive property, temperature is an intensive property.

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We define the temperature unit Kelvin \displaystyle{K} by requiring the water triple point temperature,

\displaystyle{T_{tp} \equiv 273.16K}

Once this value is fixed, the Boltzmann constant \displaystyle{k_B} value can be estimated by using, for example, the ideal gas law

\displaystyle{pV = N k_B T},

because \displaystyle{k_B} always comes with \displaystyle{T}.

— Me@2020-02-16 11:14:24 AM

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2020.02.16 Sunday (c) All rights reserved by ACHK

Problem 2.1

A First Course in String Theory

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2.1 Exercises with units

Construct and evaluate a dimensionless number using the charge \displaystyle{e} of the electron (as defined in Gaussian system of units), \displaystyle{\hbar}, and \displaystyle{c}. (In Heaviside-Lorentz units, the Gaussian \displaystyle{e^2} is replaced by \displaystyle{\frac{e^2}{4 \pi}}.)

~~~

\displaystyle{ \begin{aligned} \alpha &= \frac{e^2}{(4 \pi \varepsilon_0)\hbar c} \\ \alpha &= \frac{1}{137.035\,999\,679(94)} \\ \end{aligned} }

— Me@2010

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2020.02.04 Tuesday (c) All rights reserved by ACHK

Quick Calculation 13.2

A First Course in String Theory

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Verify that

\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I + {X^I}' \right)},

\displaystyle{\left[ L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I - {X^I}' \right)},

~~~

Equation (13.24):

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

Equation (13.39):

\displaystyle{{\dot{X}}^- + {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \bar \alpha_n^- e^{-in (\tau + \sigma)}}

\displaystyle{{\dot{X}}^- - {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}}  \alpha_n^- e^{-in (\tau - \sigma)}}

Equation (13.51):

\displaystyle{\left[\bar L_m^\perp, \bar \alpha_n^J \right]  = - n \bar{\alpha}_{m+n}^J}

\displaystyle{\left[L_m^\perp, \alpha_n^J \right] = - n \alpha_{m+n}^J}

Equation (13.52):

\displaystyle{\left[L_m^\perp, \bar \alpha_n^J \right] = 0}

\displaystyle{\left[\bar L_m^\perp, \alpha_n^J \right] = 0}

Equation (13.53):

\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m}

\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}

.

\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right]},

\displaystyle{= \left[ \bar L_0^\perp, x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^I e^{i n \sigma} + \bar \alpha_n^I e^{-in \sigma}) \right]}

\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} e^{-in \sigma} (-n \bar \alpha_n^I)}

\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 - i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} e^{-in(\tau + \sigma)} ( \bar \alpha_n^I)}

\displaystyle{= - \frac{i}{2} \sqrt{2\alpha'} \sum_{n \in \mathbb Z} \bar \alpha_n^I e^{-in(\tau + \sigma)}}

— Me@2020-01-06 11:30:38 PM

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2020.01.06 Monday (c) All rights reserved by ACHK

Quick Calculation 13.1

A First Course in String Theory

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Verify that

\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m},

\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}.

~~~

Equation (13.37):

\displaystyle{\bar L_m^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \bar \alpha_p^J \bar \alpha_{n-p}^J},\displaystyle{~~~L_m^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \alpha_p^J \alpha_{n-p}^J}.

.

\displaystyle{ \begin{aligned} \left[ \bar L_m^\perp, x_0^I \right] &= \frac{1}{2} \sum_{p \in \mathbb{Z}} \left[ \bar \alpha^J_p \bar \alpha^J_{m-p}, x_0^I \right] \\  &= \frac{1}{2} \sum_{p \in \mathbb{Z}} \bar \alpha^I_p \left[ \bar \alpha^J_{m-p}, x_0^I \right] + \frac{1}{2} \sum_{p \in \mathbb{Z}} \left[ \bar \alpha^J_p, x_0^I \right] \bar \alpha^I_{m-p} \\  &= \frac{1}{2} \bar \alpha^J_m \left[ \bar \alpha^J_{0}, x_0^I \right] + \frac{1}{2} \left[ \bar \alpha^J_0, x_0^I \right] \bar \alpha^J_{m} \\  \end{aligned} }

.

By Equation (13.33):

\displaystyle{ \begin{aligned}  \left[ \bar L_m^\perp, x_0^I \right]  &= - \frac{1}{2} \bar \alpha^J_m \left[ i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \right] - \frac{1}{2} \left[ i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \right] \bar \alpha^J_{m} \\  &= - i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \bar \alpha^I_m \\  \end{aligned} }

.

Since I and J are transverse coordinate indices, neither of them can be zero.

\displaystyle{ \begin{aligned}  \left[ \bar L_m^\perp, x_0^I \right]  &=  - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m \\  \end{aligned} }

— Me@2019-12-25 10:56:15 AM

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2019.12.25 Wednesday (c) All rights reserved by ACHK

Problem 13.5b

A First Course in String Theory

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13.6 Unoriented closed strings

This problem is the closed string version of Problem 12.12. The closed string \displaystyle{X^{\mu} (\tau, \sigma)} with \displaystyle{\sigma \in [0, 2 \pi]} and fixed \displaystyle{\tau} is a parameterized closed curve in spacetime. The orientation of a string is the direction of the increasing \displaystyle{\sigma} on this curve.

Introduce an orientation reversing twist operator \displaystyle{\Omega} such that

\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)

Moreover, declare that

\displaystyle{\Omega x_0^- \Omega^{-1} = x_0^-}

\displaystyle{\Omega p^+ \Omega^{-1} = p^+}

(b) Used the closed string oscillator expansion (13.24) to calculate

\displaystyle{\Omega x_0^I \Omega^{-1}}

\displaystyle{\Omega \alpha_0^I \Omega^{-1}}

\displaystyle{\Omega \alpha_n^I \Omega^{-1}}

\displaystyle{\Omega \bar \alpha_n^I \Omega^{-1}}

~~~

Equation (13.24):

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

.

\displaystyle{\begin{aligned}   X^{\mu} (\tau, \sigma) &= x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma}) \\   X^I (\tau, 2 \pi - \sigma)  &= x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} \left( \alpha_n^I e^{- in\sigma} + \bar \alpha_n^I e^{i n \sigma)} \right) \\   \end{aligned}}

.

\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)

.

By comparing \displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} with \displaystyle{X^I (\tau, 2 \pi - \sigma)}, we have:

\displaystyle{\begin{aligned}   \Omega x_0^I \Omega^{-1} &= x_0^I \\  \Omega \alpha_0^I \Omega^{-1} &= \alpha_0^I \\  \Omega \alpha_n^I \Omega^{-1} &= \bar \alpha_n^I \\  \Omega \bar \alpha_n^I \Omega^{-1} &= \alpha_n^I \\   \end{aligned}}

— Me@2019-11-24 04:33:23 PM

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2019.11.24 Sunday (c) All rights reserved by ACHK

Problem 13.6b

A First Course in String Theory | Topology, 2 | Manifold, 2

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13.6 Orientifold Op-planes

~~~

In the mathematical disciplines of topology, geometry, and geometric group theory, an orbifold (for “orbit-manifold”) is a generalization of a manifold. It is a topological space (called the underlying space) with an orbifold structure.

The underlying space locally looks like the quotient space of a Euclidean space under the linear action of a finite group.

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In string theory, the word “orbifold” has a slightly new meaning. For mathematicians, an orbifold is a generalization of the notion of manifold that allows the presence of the points whose neighborhood is diffeomorphic to a quotient of \displaystyle{\mathbf{R}^n} by a finite group, i.e. \displaystyle{\mathbf{R}^n/\Gamma}. In physics, the notion of an orbifold usually describes an object that can be globally written as an orbit space \displaystyle{M/G} where \displaystyle{M} is a manifold (or a theory), and \displaystyle{G} is a group of its isometries (or symmetries) — not necessarily all of them. In string theory, these symmetries do not have to have a geometric interpretation.

— Wikipedia on Orbifold

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In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an \displaystyle{n}-dimensional manifold has a neighborhood that is homeomorphic to the Euclidean space of dimension \displaystyle{n}.

— Wikipedia on Manifold

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In topology and related branches of mathematics, a topological space may be defined as a set of points, along with a set of neighbourhoods for each point, satisfying a set of axioms relating points and neighbourhoods. The definition of a topological space relies only upon set theory and is the most general notion of a mathematical space that allows for the definition of concepts such as continuity, connectedness, and convergence. Other spaces, such as manifolds and metric spaces, are specializations of topological spaces with extra structures or constraints.

— Wikipedia on Topological space

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2019.09.26 Thursday ACHK

Problem 13.6

A First Course in String Theory

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13.6 Orientifold Op-planes

(a) For an O23-plane the two normal directions \displaystyle{x^{24}, x^{25}} can be represented by a plane. A closed string at a fixed \tau appears as a parameterized closed curve \displaystyle{X^a(\tau, \sigma)} in this plane. Draw such an oriented closed string that lies fully in the first quadrant of the \displaystyle{(x^{24}, x^{25})} plane. Draw also the string \displaystyle{\tilde{X}^a(\tau, \sigma) = -X^a(\tau, 2\pi - \sigma)}.

~~~

d_2019_08_26__22_32_32_PM_a

This one is wrong.

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d_2019_08_26__22_31_55_PM_b

— Me@2019-08-26 10:31:07 PM

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2019.08.26 Monday (c) All rights reserved by ACHK

Quick Calculation 15.1.2

A First Course in String Theory

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Recall that a group is a set which is closed under an associative multiplication; it contains an identity element, and each element has a multiplicative inverse. Verify that \displaystyle{U(1)} and \displaystyle{U(N)}, as described above, are groups.

~~~

Definition

A group is a set, G, together with an operation \displaystyle{\bullet} (called the group law of G) that combines any two elements a and b to form another element, denoted \displaystyle{a \bullet b} or \displaystyle{ab}. To qualify as a group, the set and operation, \displaystyle{(G, \bullet)}, must satisfy four requirements known as the group axioms:

Closure

For all a, b in G, the result of the operation, \displaystyle{a \bullet b}, is also in G.

Associativity

For all a, b and c in G, \displaystyle{(a \bullet b) \bullet c = a \bullet (b \bullet c)}.

Identity element

There exists an element e in G such that, for every element a in G, the equation \displaystyle{e \bullet a = a \bullet e = a} holds. Such an element is unique, and thus one speaks of the identity element.

Inverse element

For each a in G, there exists an element b in G, commonly denoted \displaystyle{a^{-1}} (or \displaystyle{-a}, if the operation is denoted “+”), such that \displaystyle{a \bullet b = b \bullet a = e}, where e is the identity element.

— Wikipedia on Group (mathematics)

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The axioms for a group are short and natural… Yet somehow hidden behind these axioms is the monster simple group, a huge and extraordinary mathematical object, which appears to rely on numerous bizarre coincidences to exist. The axioms for groups give no obvious hint that anything like this exists.

— Richard Borcherds in Mathematicians: An Outer View of the Inner World

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2019.07.28 Sunday ACHK

Quick Calculation 15.1

A First Course in String Theory

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Recall that a group is a set which is closed under an associative multiplication; it contains an identity element, and each element has a multiplicative inverse. Verify that \displaystyle{U(1)} and \displaystyle{U(N)}, as described above, are groups.

~~~

What is \displaystyle{U(1)}?

— Me@2019-05-24 11:25:41 PM

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The set of all \displaystyle{1 \times 1} unitary matrices clearly coincides with the circle group; the unitary condition is equivalent to the condition that its element have absolute value 1. Therefore, the circle group is canonically isomorphic to \displaystyle{U(1)}, the first unitary group.

— Wikipedia on Circle group

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In mathematics, a complex square matrix \displaystyle{U} is unitary if its conjugate transpose \displaystyle{U^*} is also its inverse—that is, if

\displaystyle{U^{*}U=UU^{*}=I,}

where \displaystyle{I} is the identity matrix.

In physics, especially in quantum mechanics, the Hermitian conjugate of a matrix is denoted by a dagger (\displaystyle{\dagger}) and the equation above becomes

\displaystyle{U^{\dagger }U=UU^{\dagger }=I.}

The real analogue of a unitary matrix is an orthogonal matrix. Unitary matrices have significant importance in quantum mechanics because they preserve norms, and thus, probability amplitudes.

— Wikipedia on Unitary matrix

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2019.05.25 Saturday ACHK

Problem 14.5d3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

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— This answer is my guess. —

~~~

.

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

.

\displaystyle{ \begin{aligned}   &f_{L, NS'+}(x) \\   &= a_{NS'+} (r) x^r \\  &= \frac{1}{2x} \left[ \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \right] \\   & \\  &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\   \end{aligned}}

.

\displaystyle{\begin{aligned}  &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ & \\ &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5} + ...   \end{aligned}}

.

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

\displaystyle{ \begin{aligned} f_{NS+}(x)  &= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ... \\ \end{aligned}}

\displaystyle{ \begin{aligned} f_{R-}(x)  &= 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ... \\ \end{aligned}}

.

So the total number of states in heterotic string theory at \displaystyle{ \begin{aligned} \alpha' M^2 = 8 \end{aligned}} is

\displaystyle{ \begin{aligned}   &\left(1384320 + 1310720 \right) \times \left(1152 + 1152\right) \\  \end{aligned}}.

\displaystyle{ \begin{aligned}   &= 6209372160 \\ \end{aligned}}.

~~~

— This answer is my guess. —

— Me@2019-01-26 04:49:37 PM

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2019.01.27 Sunday (c) All rights reserved by ACHK

Problem 14.5d2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

.

— This answer is my guess. —

~~~

.

The left R’+ sector:

.

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

.

\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_{\alpha} \rangle_L \\  \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&|{\bar \alpha}_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\  \end{aligned}}

.

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

\displaystyle{N^\perp:}

\displaystyle{\begin{aligned}  \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...  \right)^8  \left(  1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ...  \right)^8  ...  \left( 1 + \lambda_{-1} x^{1} \right)^{32}  \left( 1 + \lambda_{-2} x^{2} \right)^{32}  ... \\  \end{aligned}}

.

However, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha\rangle_L} and \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_R}:

\displaystyle{\begin{aligned}  (2^{15} + 2^{15})   \left[ \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...  \right)^8  ...  \left( 1 + \lambda_{-1} x^{1} \right)^{32}   ... \right] \\   \end{aligned}}

\displaystyle{\begin{aligned}  2^{16} \prod_{r=1}^\infty  \frac{1}{(1 - x^r)^8}  (1 + x^{r})^{32} \\  \end{aligned}}

.

“Keep only states with \displaystyle{(-1)^{F_L} = +1}; this defines the left R’+ sector.”

\displaystyle{\begin{aligned}  \frac{2^{16}}{2} \prod_{r=1}^\infty  \frac{(1 + x^{r})^{32}}{(1 - x^r)^8}   \\  \end{aligned}}

.

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

\displaystyle{\begin{aligned}  &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\  & \\  &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5}+39739981824 \, x^{6} + ...   \end{aligned}}

.

~~~

— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

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2019.01.20 Sunday (c) All rights reserved by ACHK

Problem 14.5d1.1.2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function \displaystyle{f_L(x) = \sum_{r} a(r) x^r} for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where \displaystyle{a(r)} counts the number of states with \displaystyle{\alpha' M_L^2 = r}.

Use \displaystyle{f_L(x)} and an algebraic manipulator to find the total number of states in heterotic string theory at \displaystyle{\alpha' M_L^2 = 8}.

~~~

.

— This answer is my guess. —

~~~

p.322

\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}

\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}

\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}

.

The left NS’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

\displaystyle{N^\perp:}

\displaystyle{\begin{aligned}  \left(  1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ...   \right)^8  \left(  1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ...   \right)^8  ...   \left( 1 + \lambda_{-\frac{1}{2}} x^{\frac{1}{2}} \right)^{32}   \left( 1 + \lambda_{-\frac{3}{2}} x^{\frac{3}{2}} \right)^{32}  ... \\  \end{aligned}}

\displaystyle{\begin{aligned}  \prod_{r=1}^\infty   \frac{1}{(1 - x^r)^8}  (1 + x^{r-\frac{1}{2}})^{32} \\   \end{aligned}}

.

\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}}

.

“The left NS’ sector is built with oscillators \displaystyle{\bar \alpha_{-n}^I} and \displaystyle{\lambda_{-r}^A} acting on the vacuum \displaystyle{|NS' \rangle_L}, declared to have \displaystyle{(-1)^{F_L} = + 1}:”

\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}

So all the states with integer \displaystyle{N^{\perp}} have \displaystyle{(-1)^F = +1}.

.

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

\displaystyle{ = \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}} + ...}

.

Let

\displaystyle{ \begin{aligned}  &g (\sqrt{x}) \\  &= \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   &= 1+32 \, \sqrt{x}+504 \, x+5248 \, x^{\frac{3}{2}}+40996 \, x^{2}+258624 \, x^{\frac{5}{2}}+1384320 \, x^{3}+6512384 \, x^{\frac{7}{2}} + ... \\   \end{aligned}}

Then

\displaystyle{ \begin{aligned}  &g (-\sqrt{x}) \\  &= \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\   &= 1 - 32 \sqrt{x} + 504 x - 5248 \, x^{\frac{3}{2}} + 40996 \, x^{2} - 258624 \, x^{\frac{5}{2}}+1384320 \, x^{3} - 6512384 x^{\frac{7}{2}} + ... \\   \\ \end{aligned}}

.

\displaystyle{ \begin{aligned}  &f_{L, NS'+}(x) \\  &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\  &= \frac{1}{2x} \left[ g(\sqrt{x}) + g(-\sqrt{x}) \right] \\   &= \frac{1}{2x} \left[   \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8}   + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8}   \right] \\   \end{aligned}}

.

The left R’+ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

.

~~~

— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

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2019.01.14 Monday (c) All rights reserved by ACHK

Problem 14.5d1.2 | SageMath

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 SageMath with the following commands:

typeset_mode(True)

(1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..oo))

a = (1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..200))

F = a.taylor(x,0,6)

g = "+".join(map(latex, sorted([f for f in F.operands()], key=lambda exp:exp.degree(x))))

g

\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@2019-01-11 11:52:33 AM

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2019.01.11 Friday (c) All rights reserved by ACHK