Energy methods connect a spring oscillator and an orbit, but their coordinates and physical assumptions differ.
Prerequisites: 26, 41.
- Construct a simple Hamiltonian
- Use canonical equations
- Apply gravitational and radiative scaling
AH · Hamiltonian mechanics and astrophysical scaling
GRE · GRE Subject Test · GRE 物理 · 知识点 9
9.1
哈密顿力学与天体物理标度律
9.2
Choose the system and model
The canonical momentum 正则动量 is p=∂L/∂qdot. The Legendre transform H=p qdot−L gives a Hamiltonian when velocities can be expressed using coordinates and momenta.
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| canonical momentum/kəˈnɒnɪkl məʊˈmentəm/ | 正则动量 | zhèng zé dòng liàng |
9.3
Use the governing relation
Hamilton’s equations are qdot=∂H/∂p and pdot=−∂H/∂q. For an ordinary oscillator H=p²/(2m)+kq²/2, these reproduce Newton’s equation.
9.4
Apply the conditions
For a circular gravitational orbit v²=GM/r and period²=4π²r³/(GM). The assumptions include a dominant central mass and a circular approximation; elliptical orbits use the semimajor axis in Kepler’s law. For a circular satellite of mass m, angular momentum magnitude is mrv=m sqrt(GMr), so identical satellites have L proportional to sqrt(r). A radius ratio 9 therefore gives angular-momentum ratio 3, while the period ratio is 27. This distinction follows from the same centripetal-force relation; do not use period scaling for angular momentum.
9.5
Check the conclusion
Luminosity 光度 and received flux obey F=L/(4πd²) for isotropic radiation. A blackbody has L=4πR²σT⁴; its spectral peak shifts inversely with temperature. Distinguish intrinsic luminosity from observed brightness.
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| luminosity/ˌluːmɪˈnɒsɪti/ | 光度 | guāng dù |
9.6
Worked method
For a regular Lagrangian $L=m\dot q^2/2-kq^2/2$, canonical momentum is
$$p=\partial L/\partial\dot q=m\dot q,\qquad \dot q=p/m.$$
Substitute the velocity into the Legendre transform:
$$H=p\dot q-L=p^2/(2m)+kq^2/2.$$
$$\dot q=\partial H/\partial p=p/m,\qquad\dot p=-\partial H/\partial q=-kq.$$
Combining these gives $\ddot q=-(k/m)q$. The inversion of momentum to velocity
is required; it cannot be assumed for every constrained or singular Lagrangian.
9.7
Check conditions and vocabulary
A Hamiltonian equals total mechanical energy only under the relevant system assumptions; do not infer this universally from its name.
canonical momentum: The derivative of the Lagrangian with respect to a generalised velocity.
luminosity: Total emitted power, distinct from flux at an observer.
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