CGS制单位转化

今天我发现网上有人问了关于SRIM那本书的问题,原来参数看起来不一样的原因是SRIM书上的单位是厘米-克-秒单位制(centimetre-gram-second system,CGS),而不是现在比较常用的国际单位制(Système International d'Unités(French),SI),那么就让我记一下有关的笔记吧。

Quantity Quantity symbol CGS unit name Unit symbol Unit definition In coherent SI units
length, position \(L, x\) centimetre cm 1/100 of metre \(10^{−2}\) m
mass \(m\) gram g 1/1000 of kilogram \(10^{−3}\) kg
time \(t\) second s 1 second 1 s
velocity \(v\) centimetre per second cm/s cm/s \(10^{−2}\) m/s
acceleration \(a\) gal Gal cm/\(\text{s}^2\) \(10^{−2}\) m/\(\text{s}^2\)
force \(F\) dyne dyn g⋅cm/\(\text{s}^2\) \(10^{−5}\) N
energy \(E\) erg erg g⋅\(\text{cm}^2\)/\(\text{s}^2\) \(10^{−7}\) J
power \(P\) erg per second erg/s g⋅\(\text{cm}^2\)/\(\text{s}^{3}\) \(10^{−7}\) W
pressure \(p\) barye Ba g/(cm⋅\(\text{s}^2\)) \(10^{−1}\) Pa
dynamic viscosity \(μ\) poise P g/(cm⋅s) \(10^{−1}\) Pa⋅s
kinematic viscosity \(ν\) stokes St \(\text{cm}^2\)/s \(10^{−4}\) \(\text{m}^2\)/s
wavenumber \(k\) kayser (K) \(\text{cm}^{-1}\) \(\text{cm}^{-1}\) 100 \(\text{m}^{-1}\)
Quantity Symbol SI unit ESU unit Gaussian unit EMU unit
electric charge \(q\) 1 C ≘ (\(10^{-1}\)\(c\)) statC (Franklin) ≘ (\(10^{-1}\)\(c\)) statC (Franklin) ≘ (\(10^{-1}\)) abC
electric flux \(Φ_E\) 1 V⋅m ≘ (4π × \(10^{-1}\)\(c\)) statC (Franklin) ≘ (4π × \(10^{-1}\)\(c\)) statC (Franklin) ≘ (\(10^{-1}\)) abC
electric current \(I\) 1 A ≘ (\(10^{-1}\)\(c\)) statA (Fr⋅\(\text{s}^{−1}\)) ≘ (\(10^{-1}\)\(c\)) statA (Fr⋅\(\text{s}^{−1}\)) ≘ (\(10^{-1}\)) Bi
electric potential / voltage \(\Phi / V, U\) 1 V ≘ (\(10^8 c^{−1}\)) statV ≘ (\(10^8 c^{−1}\)) statV ≘ (\(10^8\)) abV
electric field \(E\) 1 V/m ≘ (\(10^6 c^{−1}\)) statV/cm ≘ (\(10^6 c^{−1}\)) statV/cm ≘ (\(10^6\)) abV/cm
electric displacement field \(D\) 1 C/\(\text{m}^2\) ≘ (\(10^{−5}c\)) statC/\(\text{cm}^2\) (Fr/\(\text{cm}^2\)) ≘ (\(10^{−5}c\)) statC/\(\text{cm}^2\) (Fr/\(\text{cm}^2\)) ≘ (\(10^{−5}\)) abC/\(\text{cm}^2\)
electric dipole moment \(p\) 1 C⋅m ≘ (10\(c\)) statC⋅cm ≘ (10\(c\)) statC⋅cm ≘ (10) abC⋅cm
magnetic dipole moment \(μ\) 1 A⋅\(\text{m}^2\) ≘ (\(10^3c\)) statC⋅\(\text{cm}^2\) ≘ (\(10^3\)) Bi⋅\(\text{cm}^2\) = (\(10^3\)) erg/G ≘ (\(10^3\)) Bi⋅\(\text{cm}^2\) = (\(10^3\)) erg/G
magnetic B field \(B\) 1 T ≘ (\(10^4 c^{−1}\)) statT ≘ (\(10^4\)) G ≘ (\(10^4\)) G
magnetic H field \(H\) 1 A/m ≘ (4π × \(10^{−3}c\)) statA/cm ≘ (4π × \(10^{−3}\)) Oe ≘ (4π × \(10^{−3}\)) Oe
magnetic flux \(Φ_m\) 1 Wb ≘ (\(10^8 c^{−1}\)) statWb ≘ (\(10^8\)) Mx ≘ (\(10^8\)) Mx
resistance \(R\) 1 Ω ≘ (\(10^9 c^{−2}\)) s/cm ≘ (\(10^9 c^{−2}\)) s/cm ≘ (\(10^9\)) abΩ
resistivity \(ρ\) 1 Ω⋅m ≘ (\(10^{11} c^{−2}\)) s ≘ (\(10^{11} c^{−2}\)) s ≘ (\(10^{11}\)) abΩ⋅cm
capacitance \(C\) 1 F ≘ (\(10^{−9} c^2\)) cm ≘ (\(10^{−9} c^2\)) cm ≘ (\(10^{−9}\)) abF
inductance \(L\) 1 H ≘ (\(10^9\) \(c^{−2}\)) \(\text{cm}^{−1}⋅\text{s}^2\) ≘ (\(10^9\) \(c^{−2}\)) \(\text{cm}^{−1}⋅\text{s}^2\) ≘ (\(10^9\)) abH

对于公式 \[ \begin{aligned} \varepsilon &=\frac{a E_{c}}{Z_{1} Z_{2} e^{2}} \\ E_{c} &=\frac{E_{0} M_{2}}{M_{1}+M_{2}} \\ a &=\frac{0.8853 a_{0}}{Z_{1}^{0.23}+Z_{2}^{0.23}} \end{aligned} \] 其中,玻尔半径\(a_0=0.529\)Å,\(E_0\)单位为keV,要得到: \[ \varepsilon=\frac{32.53 M_{2} E_{0}}{Z_{1} Z_{2}\left(M_{1}+M_{2}\right)\left(Z_{1}^{0.23}+Z_{2}^{0.23}\right)} \] 即在CGS单位制下有: \[ \frac{0.8853a_0}{e^2}=32.53 \] 该问题下,有人给出了: \[ 32.53 \approx 0.8853 \cdot 5.29 \cdot 10^{-9} \cdot 1000 \cdot 4.8 \cdot 10^{-10} \cdot \frac{1}{300} /\left(4.8 \cdot 10^{-10}\right)^{2} \] 唔,看起来和国际单位制的变换还是有些不一样的,让我们一步步地看看:

\[ \begin{aligned} &a_0=0.529\text{Å}=0.529*10^{-8}\text{cm}=5.29*10^{-9}\text{cm}\\ &e=4.803 204 27*10^{-10}\text{Fr} \end{aligned} \] 由于CSG(或者说CSG中,ESU)单位制中电压单位(statV)关于SI单位制的电压单位(V)有: \[ 1\text{V}=10^{8}c^{-1}\text{statV}\approx \frac{1}{300}\text{statV} \] 由于\(E_0\)给的是千电子伏特(keV)单位,所以要再添上比例系数: \[ 1000*\frac{1}{300}*e \] 将上面的全部代入式子即得结果。