3.1 Lorentz covariance for motion in electromagnetic fields, 2

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Is \displaystyle{\frac{d p_\mu}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds}} gauge invariant?

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… the defining property of a Lorentz transformation, \Lambda^\mu_{\;\;\nu}:

\eta_{\mu\nu} \Lambda^{\mu}_{\;\;\alpha}  \Lambda^\nu_{\;\;\beta} = \eta_{\alpha\beta}

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… 4-vectors and (Lorentz)-tensors are transformed like this:

U'^\mu = \Lambda^\mu_{\;\;\nu}U^\nu

and

F'_{\mu\nu} = \Lambda_\mu^{\;\;\alpha} \Lambda_\nu^{\;\;\beta}F_{\alpha\beta}= \Lambda_\mu^{\;\;\alpha} F_{\alpha\beta} (\Lambda^{-1})^{\beta}_{\;\;\nu}

where we have used the conventional notation

\displaystyle{\Lambda_\nu^{\;\;\mu} = (\Lambda^{-1})^\mu_{\;\;\nu}}

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Let us then take your equation and apply \displaystyle{\Lambda_\sigma^{\;\;\mu}} on both sides (recall this Lorentz transformation does not depend on \displaystyle{\tau}), and try rewriting everything in terms of prime quantities:

\displaystyle{  \begin{aligned}  m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau}     &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\     &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\    m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha} ((\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta}) U_\alpha\\    &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big)\\    &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\    &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu\Big) U'_\beta\\    &= e F'_{\sigma\nu}\eta^{\nu\beta} U'_\beta \\    \\    m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e F'_{\sigma\nu} U'^\nu \\  \end{aligned}}

This “game” can always be done with contracted indices, …

— answered Jul 7, 2020 at 15:06

— ohneVal

— Lorentz invariance of the Lorentz force law

— Physics StackExchange

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How come \displaystyle{ \begin{aligned}     \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}?

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Note that

\displaystyle{  \begin{aligned}    \sum ... \sum \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu    &= \sum ... \sum \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu  \\    \end{aligned}}

does not mean that

\displaystyle{ \begin{aligned}     \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}

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Also

\displaystyle{ \begin{aligned}     \sum \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \sum \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}

does not mean that

\displaystyle{ \begin{aligned}     \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}

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\displaystyle{  \begin{aligned}    A  = \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu U_\beta'    &=     \sum_{\alpha, \beta} F_{\mu\alpha}\eta^{\alpha\alpha}\Lambda^{\;\;\beta}_{\nu=\alpha} U_\beta' \\    &=     \sum_{\beta}     \left(     - F_{\mu 0} \Lambda^{\;\;\beta}_{0}    + F_{\mu 1} \Lambda^{\;\;\beta}_{1}    + F_{\mu 2} \Lambda^{\;\;\beta}_{2}    + F_{\mu 3} \Lambda^{\;\;\beta}_{3}      \right)     U_\beta' \\        \\    \end{aligned}}

\displaystyle{ \begin{aligned}     B = \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu U_\beta'    &=     \sum_{\alpha, \beta}     F_{\mu\alpha}     \eta^{\beta \beta}\Lambda^{\;\;\alpha}_{\nu=\beta} U_\beta      \\     &=     \sum_{\alpha}     F_{\mu\alpha}     \left(    - \Lambda^{\;\;\alpha}_{0} U_0'    + \Lambda^{\;\;\alpha}_{1} U_1'     + \Lambda^{\;\;\alpha}_{2} U_2'     + \Lambda^{\;\;\alpha}_{3} U_3'     \right)     \\       \end{aligned}}

At the first glance, it seems to be unlikely that

\displaystyle{  \begin{aligned}    \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu U_\beta'    &= \sum_{\alpha, \beta, \nu}  F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu U_\beta'    \\    \end{aligned}},

because while in A, for any \beta, U_\beta'‘s have visible negative terms; in B, only U_0'‘s do.

Without additional mathematical properties among those physical quantities, the identity is impossible to prove.

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Just for future reference:

From the invariance of the spacetime interval it follows

\displaystyle{\eta =\Lambda ^{\mathrm {T} }\eta \Lambda}

— Wikipedia on Lorentz transformation

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\displaystyle{  \begin{aligned}    \eta_{\mu \nu} = \eta'_{\mu \nu} &= \Lambda^{\alpha}{}_{\mu} \Lambda^{\beta}{}_{\nu} \eta_{\alpha \beta} \\     &= (\Lambda^T)_{\mu}{}^{\alpha} \eta_{\alpha \beta} \Lambda^{\beta}{}_{\nu} \\ \\     \eta_{\alpha\beta} &= \Lambda^{\mu}_{\;\;\alpha} \Lambda^{\nu}_{\;\;\beta}\eta_{\mu\nu} \\    \eta^{\alpha\beta} &= \Lambda^{\alpha}_{\;\;\mu} \Lambda^{\beta}_{\;\;\nu}\eta^{\mu\nu} \\    \end{aligned}}

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\displaystyle{  \begin{aligned}    m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\     & = e \Lambda_\nu^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\    \end{aligned}}

Inserting the identity \sum_\beta (\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta} into the expression:

\displaystyle{  \begin{aligned}    m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big) \\      &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\    \end{aligned}}

This path does not work. Also, the formula \displaystyle{ \Lambda^\alpha_{\;\;\beta} U_\alpha = U'_\beta } is plain wrong!

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Inserting the identity \sum_\beta (\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta} into the expression before (actually without) lowering the index:

\displaystyle{  \begin{aligned}    m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau}     &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\    &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha} ((\Lambda^{-1})^{\alpha}_{\;\;\beta} \Lambda^\beta_{\;\;\nu})  U^\nu \\    m \frac{{\rm d}U_\sigma'}{{\rm d}\tau}      &= e (\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha} (\Lambda^{-1})^{\alpha}_{\;\;\beta}) (\Lambda^\beta_{\;\;\nu}  U^\nu) \\    &= e {F'}_{\sigma \beta} {U'}^\beta \\    \end{aligned}}

— Me@2022-08-23 12:03:54 PM

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2022.08.26 Friday (c) All rights reserved by ACHK

3.1 Lorentz covariance for motion in electromagnetic fields, 1

A First Course in String Theory

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The Lorentz force equation (3.5) can be written relativistically as

\displaystyle{\frac{d p_\mu}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds}},

where \displaystyle{p_{\mu}} is the four-momentum.

(a) Check explicitly that this equation reproduces (3.5) when \displaystyle{\mu} is a spatial index.

(b) What does (1) gives when \displaystyle{\mu = 0}? Does it make sense?

~~~

Eq. (3.5):

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

Eq. (2.20):

\displaystyle{ds \equiv \sqrt{ds^2}}    if    \displaystyle{ds^2 > 0}

Eq. (2.21):

\displaystyle{-ds^2 = \eta_{\mu \nu} dx^\mu dx^\nu}

The spacetime interval \displaystyle{ds^2} is Lorentz invariant. If \displaystyle{ds^2 > 0}, we have Eq. (2.27) and (2.28):

\displaystyle{\begin{aligned}     ds &= c dt_p \\     ds &= c dt \sqrt{1 - \beta^2}     \end{aligned}}

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\displaystyle{\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)

\displaystyle{    \begin{aligned}     \frac{d p_1}{dt}     &= \frac{q}{c} F_{1 \nu} \frac{d x^\nu}{dt} \\    &= \frac{q}{c} \left( F_{1 0} \frac{d x^0}{dt} + F_{1 1} \frac{d x^1}{dt} + F_{1 2} \frac{d x^2}{dt} + F_{1 3} \frac{d x^3}{dt} \right) \\    \frac{d p_x}{dt} &= \frac{q}{c} \left( E_x c \frac{d t}{dt} + (0) \frac{d x^1}{dt} + B_z \frac{d x^2}{dt} - B_y \frac{d x^3}{dt} \right) \\    &= q \left( E_x + \frac{1}{c} \left( \vec v \times \vec B \right)_x \right) \\    \end{aligned}}

\displaystyle{    \begin{aligned}     \frac{d p_0}{dt}     &= \frac{q}{c} \left( F_{0 0} \frac{d x^0}{dt} + F_{0 1} \frac{d x^1}{dt} + F_{0 2} \frac{d x^2}{dt} + F_{0 3} \frac{d x^3}{dt} \right) \\    &= \frac{q}{c} \left( 0 \frac{d x^0}{dt} - E_x \frac{d x^1}{dt} - E_y \frac{d x^2}{dt} - E_z \frac{d x^3}{dt} \right) \\    \frac{d }{dt} \left( \frac{-E}{c} \right)&= - \frac{q}{c} \left( \vec v \cdot \vec   E \right) \\    \frac{d E}{dt}  &= \vec v \cdot \vec F_E \\      \end{aligned}}

— Me@2022-08-04 04:17:59 PM

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2022.08.06 Saturday (c) All rights reserved by ACHK

Quick Calculation 3.10

A First Course in String Theory

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Equation (3.96) has content: if you move a particle along a closed loop in a static gravitational field, the net work that you do against the gravitational field is zero.

Prove the above statement.

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Eq. (3.96):

\displaystyle{\vec g = - \nabla V_g}

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By the gradient theorem (aka the fundamental theorem of calculus for line integrals):

\displaystyle{\int _{\gamma }\nabla \varphi (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} =\varphi \left(\mathbf {q} \right)-\varphi \left(\mathbf {p} \right)}

— Me@2022-07-03 04:27:19 PM

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2022.07.03 Sunday (c) All rights reserved by ACHK

Quick Calculation 3.8

A First Course in String Theory

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Show that this condition fixes uniquely \displaystyle{\alpha = \gamma = 1/2}, and \displaystyle{\beta = - 3/2}, thus reproducing the result in (3.90).

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Eq. (3.93):

\displaystyle{l_P = (G)^\alpha (c)^\beta (\hbar)^\gamma}

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\displaystyle{l_P = \left( \frac{l_p^3}{m_p t_P^2} \right)^\alpha \left( \frac{l_P}{t_P} \right)^\beta \left( \frac{m_P l_P^2}{t_P} \right)^\gamma}

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\displaystyle{\begin{aligned}   3 \alpha + \beta + 2\gamma &= 1 \\   -\alpha + \gamma &= 0 \\   - 2 \alpha - \beta - \gamma &= 0 \\   \end{aligned}}

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var('a b c')

solve([3*a+b+2*c==1, -a+c==0, -2*a-b-c==0], a, b, c)

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\displaystyle{\begin{aligned}   \alpha &= \frac{1}{2} \\ \\  \beta &= \frac{-3}{2} \\ \\  \gamma &= \frac{1}{2} \\ \\  \end{aligned}}

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Eq. (3.90):

\displaystyle{l_P = \sqrt{\frac{G \hbar}{c^3}}}

— Me@2022-06-23 10:46:22 AM

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2022.06.23 Thursday (c) All rights reserved by ACHK

Quick Calculation 3.7

A First Course in String Theory

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The force \displaystyle{\vec F} on a test charge \displaystyle{q} in an electric field \displaystyle{\vec E} is \displaystyle{\vec F = q \vec E}. What are the units of charge in various dimensions?

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Eq. (3.74):

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

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\displaystyle{     \begin{aligned}         \vec F &= q \vec E \\     &=  \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{qQ}{r^{d-1}} \\ \\ \\      [\vec F] &=  \left[ \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{qQ}{r^{d-1}} \right] \\     &=  \left[ \frac{1}{1} \frac{q^2}{r^{d-1}} \right] \\     &=  \left[ \frac{q^2}{r^{d-1}} \right] \\     \end{aligned}}

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\displaystyle{     \begin{aligned}             [r^{d-1} \vec F] &= \left[ q^2 \right] \\ \\        \left[ q \right] &= [\sqrt{r^{d-1} \vec F}] \\ \\    &= \sqrt{m^{d-1} N} \\ \\    \end{aligned}}

— Me@2022-06-08 11:09:27 AM

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Lorentz–Heaviside units (or Heaviside–Lorentz units) constitute a system of units (particularly electromagnetic units) within CGS, named for Hendrik Antoon Lorentz and Oliver Heaviside. They share with CGS-Gaussian units the property that the electric constant \displaystyle{\epsilon_0} and magnetic constant \displaystyle{\mu_0} do not appear, having been incorporated implicitly into the electromagnetic quantities by the way they are defined. Heaviside-Lorentz units may be regarded as normalizing \displaystyle{\epsilon_0 = 1} and \displaystyle{\mu_0 = 1}, while at the same time revising Maxwell’s equations to use the speed of light \displaystyle{c} instead.

Heaviside–Lorentz units, like SI units but unlike Gaussian units, are rationalized, meaning that there are no factors of \displaystyle{4 \pi} appearing explicitly in Maxwell’s equations. That these units are rationalized partly explains their appeal in quantum field theory: the Lagrangian underlying the theory does not have any factors of \displaystyle{4 \pi} in these units. Consequently, Heaviside-Lorentz units differ by factors of \displaystyle{\sqrt{4\pi}} in the definitions of the electric and magnetic fields and of electric charge. They are often used in relativistic calculations, and are used in particle physics. They are particularly convenient when performing calculations in spatial dimensions greater than three such as in string theory.

— Wikipedia on Lorentz–Heaviside units

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2022.06.08 Wednesday (c) All rights reserved by ACHK

Quick Calculation 3.6

A First Course in String Theory

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Verify that for \displaystyle{d=3} equation (3.74) coincides with (3.67).

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Eq. (3.67):

\displaystyle{E(r) = \frac{q}{4 \pi r^2}}

Eq. (3.74):

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

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When \displaystyle{d=3},

\displaystyle{\begin{aligned}    E(r)   &= \frac{\Gamma\left( \frac{3}{2} \right)}{2 \pi^{\frac{3}{2}}} \frac{q}{r^{2}} \\ \\    &= \frac{\sqrt{\pi}}{2} \frac{1}{2 \pi \pi^{\frac{1}{2}}} \frac{q}{r^{2}} \\ \\     &= \frac{1}{4 \pi} \frac{q}{r^{2}} \\ \\     \end{aligned}  }

— Me@2022-05-30 01:15:00 PM

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2022.05.30 Monday (c) All rights reserved by ACHK

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)}.

~~~

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}.

~~~

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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