Mekanika Hamiltonian dan penskalaan astrofisika
GRE Subject Test · GRE Physics Topik 9 4:58 Narasi bahasa Inggris · Subtitle bahasa Inggris + 中文 disematkan langsung
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Energy methods connect a spring oscillator and an orbit, but their coordinates and assumptions differ.
能量方法把弹簧振子和轨道运动联系起来，但两者的坐标和物理假设并不相同。
This topic builds simple Hamiltonians, uses Hamilton's canonical equations, and applies gravitational and radiative scaling laws.
本主题构建简单的哈密顿量，使用哈密顿正则方程，并应用引力和辐射的标度关系。
The canonical momentum p is the partial derivative of the Lagrangian with respect to q dot.
正则动量 p 等于拉格朗日量 L 对 q 点的偏导数。
The Legendre transform, H equals p q dot minus L, gives a Hamiltonian when velocities can be expressed using coordinates and momenta.
当速度可以用坐标和动量表示时，勒让德变换 H 等于 p 乘以 q 点减去 L，给出哈密顿量。
Hamilton's equations are q dot equals the partial derivative of H with respect to p, and p dot equals minus the partial derivative of H with respect to q.
哈密顿方程为：q点等于H对p的偏导数，p点等于负H对q的偏导数。
For an ordinary oscillator, H equals p squared over 2 m plus k q squared over 2, and these reproduce Newton's equation.
对于普通振子，H等于p平方除以2m，加上k乘以q平方除以2。 这些方程可导出牛顿方程。
For a circular gravitational orbit, v squared equals G M over r, and the period squared equals 4 pi squared r cubed over G M.
对于圆形引力轨道，v 的平方等于 GM 除以 r，周期的平方等于 4π 的平方乘以 r 的立方，再除以 GM。
The assumptions include a dominant central mass and a circular approximation; elliptical orbits use the semimajor axis in Kepler's law.
假设条件包括中心质量占主导以及圆轨道近似；椭圆轨道在开普勒定律中使用半长轴。
For a circular satellite of mass m, the angular momentum magnitude is m r v, which equals m times the square root of G M r, so identical satellites have L proportional to the square root of r.
对于质量为 m 的圆轨道卫星，角动量大小为 mrv，即 m 乘以 GMr 的平方根，因此相同卫星的 L 与 r 的平方根成正比。
A radius ratio of 9 therefore gives an angular-momentum ratio of 3, while the period ratio is 27.
半径之比为 9 时，角动量之比为 3，而周期之比为 27。
This distinction follows from the same centripetal-force relation; do not use period scaling for angular momentum.
这一区别源于同一个向心力关系；不要用周期的缩放来计算角动量。
Luminosity and received flux obey F equals L over 4 pi d squared for isotropic radiation.
对于各向同性辐射，光度与接收到的通量满足 F 等于 L 除以 4π 乘以 d 的平方。
A blackbody has L equal to 4 pi R squared sigma T to the fourth; its spectral peak shifts inversely with temperature.
黑体的光度 L 等于 4π 乘以 R 的平方、乘以 σ、再乘以 T 的四次方；其光谱峰值与温度成反比移动。
Distinguish intrinsic luminosity from observed brightness.
请区分本征光度与观测亮度。
For the oscillator Lagrangian, m q-dot squared over two minus k q squared over two, the canonical momentum is the partial of L with respect to q-dot: m q-dot.
对振子拉格朗日量 L = m q̇²/2 − k q²/2，正则动量等于 L 对 q̇ 的偏导数，即 m q̇。
So q-dot equals p over m.
所以 q̇ 等于 p 除以 m。
Substitute that velocity into the Legendre transform: H equals p q-dot minus L, which is p squared over two m plus k q squared over two, the total energy.
把这个速度代入勒让德变换：H 等于 p q̇ 减去 L，即 p²/2m 加 kq²/2，也就是总能量。
Hamilton's equations then give q-dot equals the partial of H with respect to p, which is p over m, and p-dot equals minus the partial of H with respect to q, which is minus k q.
哈密顿方程给出：q̇ 等于 H 对 p 的偏导数，即 p/m；ṗ 等于 H 对 q 的偏导数的相反数，即 −kq。
The state goes round an ellipse of constant energy.
系统状态沿一个能量恒定的椭圆运动。
Combining these gives q double dot equals negative k over m, times q.
将这两式合并可得 q 的二阶导数等于负的 k 除以 m，再乘以 q。
The inversion of momentum to velocity is required.
动量到速度的反演是必需的。
It cannot be assumed for every constrained or singular Lagrangian.
对于每一个受约束或奇异的拉格朗日量，都不能直接假设这一关系成立。
An isotropically radiating blackbody star has twice the radius and half the temperature of another.
一颗各向同性辐射的黑体恒星，其半径是另一颗的两倍，温度是另一颗的一半。
It is three times farther from the observer.
它距离观测者的距离是另一颗的三倍。
Find the luminosity ratio and received-flux ratio, ignoring absorption.
忽略吸收，求光度比和接收通量比。
Separately, derive Hamilton's equations for H equals p squared divided by two m plus a q to the fourth power, with a positive.
另外，推导哈密顿方程：H等于p平方除以2m加上a乘以q的四次方，其中a为正数。
L proportional to R squared T to the fourth gives L sub two divided by L sub one equals 2 squared times 1 over 2 to the fourth, which is 1 over 4.
L 正比于 R 平方乘以 T 的四次方，得出 L 下标 2 除以 L 下标 1 等于 2 的平方乘以 1/2 的四次方，即 1/4。
F proportional to L divided by d squared gives F sub two divided by F sub one equals 1 over 4 divided by 3 squared, which is 1 over 36.
F 正比于 L 除以 d 的平方，得出 F 下标 2 除以 F 下标 1 等于 1/4 除以 3 的平方，即 1/36。
Hamilton's equations give q dot equals p divided by m and p dot equals negative 4 a q cubed.
哈密顿方程给出 q 点等于 p 除以 m，以及 p 点等于负 4a 乘 q 的立方。
These describe an amplitude-dependent nonlinear oscillator. Its restoring force is cubic, not a Hooke-law force with constant nonzero small-oscillation stiffness at zero.
这些描述了一个振幅依赖的非线性振荡器；其恢复力是三次方的，而不是在零点具有恒定非零小振幅振动刚度的胡克定律力。
Award each ratio, each equation, and the nonlinear interpretation.
为每个比值、每个方程和非线性解释各给一分。
A Hamiltonian equals total mechanical energy only under the relevant system assumptions; do not infer this universally from its name.
哈密顿量仅在满足特定系统假设时才等于总机械能。 切勿仅凭其名称就普遍推断此结论成立。
canonical momentum: The derivative of the Lagrangian with respect to a generalised velocity.
正则动量：拉格朗日量对广义速度的导数。
luminosity: Total emitted power, distinct from flux at an observer.
光度：总辐射功率，不同于观测者处的通量。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆其含义，然后改变原始传递中的一个参数并检验极限情况。