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東京工業大学 理学院 物理学系 2018年8月実施 午前 [3]

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Author​

Miyake

Description​

Kai​

(1)​

∂∂xln⁡∣r⃗∣=12∂∂xln⁡(x2+y2)=xx2+y2=xr2∂∂yln⁡∣r⃗∣=12∂∂yln⁡(x2+y2)=yx2+y2=yr2 \begin{aligned} \frac{\partial}{\partial x} \ln | \vec{r} | &= \frac{1}{2} \frac{\partial}{\partial x} \ln \left( x^2+y^2 \right) = \frac{x}{x^2+y^2} = \frac{x}{r^2} \\ \frac{\partial}{\partial y} \ln | \vec{r} | &= \frac{1}{2} \frac{\partial}{\partial y} \ln \left( x^2+y^2 \right) = \frac{y}{x^2+y^2} = \frac{y}{r^2} \end{aligned}

なので、

∇⃗ln⁡∣r⃗∣=(xr2,yr2)=r⃗r2 \begin{aligned} \vec{\nabla} \ln | \vec{r} | = \left( \frac{x}{r^2}, \frac{y}{r^2} \right) = \frac{\vec{r}}{r^2} \end{aligned}

(2)​

x=rcos⁡θ,y=rsin⁡θ,r=x2+y2,tan⁡θ=y/xx = r \cos \theta, y = r \sin \theta, r = \sqrt{x^2+y^2}, \tan \theta = y/x より、

∂r∂x=xr=cos⁡θ∂r∂y=yr=sin⁡θ∂θ∂x=−yx2cos⁡2θ=−sin⁡θr∂θ∂y=1xcos⁡2θ=cos⁡θr\begin{aligned} \frac{\partial r}{\partial x} &= \frac{x}{r} = \cos \theta \\ \frac{\partial r}{\partial y} &= \frac{y}{r} = \sin \theta \\ \frac{\partial \theta}{\partial x} &= - \frac{y}{x^2} \cos^2 \theta = - \frac{\sin \theta}{r} \\ \frac{\partial \theta}{\partial y} &= \frac{1}{x} \cos^2 \theta = \frac{\cos \theta}{r} \end{aligned}

であるから、

∂∂x=∂r∂x∂∂r+∂θ∂x∂∂θ=cos⁡θ∂∂r−sin⁡θr∂∂θ∂∂y=∂r∂y∂∂r+∂θ∂y∂∂θ=sin⁡θ∂∂r+cos⁡θr∂∂θ\begin{aligned} \frac{\partial}{\partial x} &= \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} = \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \\ \frac{\partial}{\partial y} &= \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta} = \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \end{aligned}

であり、

∂2∂x2=(cos⁡θ∂∂r−sin⁡θr∂∂θ)(cos⁡θ∂∂r−sin⁡θr∂∂θ)=cos⁡2θ∂2∂r2+sin⁡θcos⁡θr2∂∂θ−sin⁡θcos⁡θr∂2∂r∂θ+sin⁡2θr∂∂r−sin⁡θcos⁡θr∂2∂θ∂r    +sin⁡θcos⁡θr2∂∂θ+sin⁡2θr2∂2∂θ2=cos⁡2θ∂2∂r2+2sin⁡θcos⁡θr2∂∂θ−2sin⁡θcos⁡θr∂2∂r∂θ+sin⁡2θr∂∂r+sin⁡2θr2∂2∂θ2∂2∂y2=(sin⁡θ∂∂r+cos⁡θr∂∂θ)(sin⁡θ∂∂r+cos⁡θr∂∂θ)=sin⁡2θ∂2∂r2−sin⁡θcos⁡θr2∂∂θ+sin⁡θcos⁡θr∂2∂r∂θ+cos⁡2θr∂∂r+sin⁡θcos⁡θr∂2∂θ∂r    −sin⁡θcos⁡θr2∂∂θ+cos⁡2θr2∂2∂θ2=sin⁡2θ∂2∂r2−2sin⁡θcos⁡θr2∂∂θ+2sin⁡θcos⁡θr∂2∂r∂θ+cos⁡2θr∂∂r+cos⁡2θr2∂2∂θ2\begin{aligned} \frac{\partial^2}{\partial x^2} &= \left( \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \right) \left( \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \right) \\ &= \cos^2 \theta \frac{\partial^2}{\partial r^2} + \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r} \frac{\partial}{\partial r} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial \theta \partial r} \\ &\ \ \ \ + \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \\ &= \cos^2 \theta \frac{\partial^2}{\partial r^2} +2 \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} -2 \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r} \frac{\partial}{\partial r} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \\ \frac{\partial^2}{\partial y^2} &= \left( \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \right) \left( \sin \theta \frac{\partial}{\partial r} + \frac{\cos \theta}{r} \frac{\partial}{\partial \theta} \right) \\ &= \sin^2 \theta \frac{\partial^2}{\partial r^2} - \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r} \frac{\partial}{\partial r} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial \theta \partial r} \\ &\ \ \ \ - \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \\ &= \sin^2 \theta \frac{\partial^2}{\partial r^2} -2 \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta} +2 \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r} \frac{\partial}{\partial r} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2} \end{aligned}

したがって、

Δ=∂2∂x2+∂2∂y2=∂2∂r2+1r∂∂r+1r2∂2∂θ2=1r∂∂rr∂∂r+1r2∂2∂θ2\begin{aligned} \Delta &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \\ &= \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \\ &= \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \end{aligned}

を得る。

(3)​

∣r⃗∣≠0| \vec{r} | \neq 0 のとき、(2) で得た式をつかって、

Δln⁡∣r⃗∣=1r∂∂rr∂∂rln⁡r=1r∂∂rr1r=0 \begin{aligned} \Delta \ln | \vec{r} | = \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} \ln r = \frac{1}{r} \frac{\partial}{\partial r} r \frac{1}{r} = 0 \end{aligned}

を得る。

(4)​

領域 ∣r⃗∣≤a| \vec{r} | \leq a を SS , 閉曲線 ∣r⃗∣=a| \vec{r} | = a を CC で表して、 次のように計算できる：

I=∬SΔln⁡∣r⃗∣dxdy=∬S∇⃗⋅r⃗r2 dxdy=∬S(∂∂xxr2+∂∂yyr2)dxdy=∮C(−yr2dx+xr2dy)=∫02π(−asin⁡θa2⋅(−asin⁡θ)+acos⁡θa2⋅(acos⁡θ))dθ=∫02πdθ=2π\begin{aligned} I &= \iint_S \Delta \ln | \vec{r} | dx dy \\ &= \iint_S \vec{\nabla} \cdot \frac{\vec{r}}{r^2} \ dx dy \\ &= \iint_S \left( \frac{\partial}{\partial x} \frac{x}{r^2} + \frac{\partial}{\partial y} \frac{y}{r^2} \right) dx dy \\ &= \oint_C \left( - \frac{y}{r^2} dx + \frac{x}{r^2} dy \right) \\ &= \int_0^{2 \pi} \left( - \frac{a \sin \theta}{a^2} \cdot (-a \sin \theta) + \frac{a \cos \theta}{a^2} \cdot (a \cos \theta) \right) d \theta \\ &= \int_0^{2 \pi} d \theta \\ &= 2 \pi \end{aligned}

(5)​

(3), (4) より、

Δln⁡∣r⃗∣=2πδ2(r⃗)\begin{aligned} \Delta \ln | \vec{r} | = 2 \pi \delta^2 ( \vec{r} ) \end{aligned}

がわかる。 ただし、 δ2(⋅)\delta^2 ( \cdot ) は、 2次元のディラックのデルタ関数である。

(6)​

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