ハミルトン力学および天体物理学的スケール
導入済み| English | 意味 |
|---|---|
| Hamiltonian/ˌhæmɪlˈtəʊnɪən/ | The function of coordinates and momenta that generates the equations of motion. |
| canonical momentum/kəˈnɒnɪkl məʊˈmentəm/ | 正準運動量 |
| luminosity/ˌluːmɪˈnɒsɪti/ | 光度 |
From Lagrangian to Hamiltonian 哈密顿量
- A spring oscillator and a planetary orbit can both be solved from energy-based formalisms — but the coordinates, momenta and scaling laws differ.
- GRE advanced mechanics tests Hamilton's equations and the scaling of orbits and stellar luminosities.
Canonical momentum 正则动量 and $H$
- $p = \dfrac{\partial L}{\partial \dot q}$; Legendre transform $H = p\dot q - L$, written in $q$ and $p$.
- Oscillator: $L = \tfrac12 m\dot q^2 - \tfrac12 kq^2$ gives $p = m\dot q$ and $H = \dfrac{p^2}{2m} + \tfrac12 kq^2$.
Hamilton's equations
- $\dot q = \dfrac{\partial H}{\partial p}$, $\dot p = -\dfrac{\partial H}{\partial q}$. Oscillator: $\dot q = p/m$, $\dot p = -kq$, so $\ddot q = -\tfrac{k}{m}q$.
- If $H$ lacks $q$, $p$ is conserved. $H = T + V$ when constraints are time-independent and $V$ does not depend on velocity; not in general.
For $H = \dfrac{p^2}{2m} + V(q)$, Hamilton's equations give $\dot p =$
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$-\partial H/\partial q$.
The Hamiltonian need not equal the total mechanical energy.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
True. With time-dependent constraints or velocity-dependent potentials, H can differ from T + V; it equals the energy only for natural systems with time-independent constraints.
Circular orbits
- $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ gives $v = \sqrt{GM/r}$, $T^2 = \dfrac{4\pi^2r^3}{GM}$, $L = m\sqrt{GMr}$.
An identical satellite at 9 times the radius has speed $\times\tfrac13$, angular momentum $\times 3$, period $\times 27$. Each quantity has its own power of $r$.
Larger circular orbit — up or down?
Use r-scaling for circular orbits about the same mass.
A satellite's orbital radius is multiplied by 4. By what factor does its period change?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Kepler's third law gives $T \propto r^{3/2}$, so $4^{3/2} = 8$.
Stars: luminosity 光度 and flux
- $L = 4\pi R^2\sigma T^4$ (intrinsic). Received flux $F = \dfrac{L}{4\pi d^2}$: twice as far, a quarter as bright.
- Wien: $\lambda_{\max}T \approx 2.9\times10^{-3}$ m·K; hotter stars peak bluer.
A star has twice the radius and twice the surface temperature of another. Luminosity ratio?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Luminosity is $L = 4\pi R^2\sigma T^4$, so doubling both gives $2^2 \times 2^4 = 64$.
A star at 5800 K. Peak wavelength in nm (nearest 10)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$2.9\times10^{-3}/5800$ m.
Recap
- $H = p\dot q - L$; $\dot q = H_p$, $\dot p = -H_q$.
- Circular orbits: $v \propto r^{-1/2}$, $L \propto r^{1/2}$, $T \propto r^{3/2}$.
- $L \propto R^2T^4$; $F \propto 1/d^2$; $\lambda_{\max} \propto 1/T$.
- Hamiltonian: The function of coordinates and momenta that generates the equations of motion.